Calculus 9th Edition By Salas Hille Etgen Pdf
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Multivariable Calculus 8th Edition Stewart Solutions Manual Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-solutions-manual/ Multivariable Calculus 8th Edition Stewart Test Bank Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-test-bank/
10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1
1.
2.
3.
Curves Defined by Parametric Equations
−1
0
1
2
0
1
0
−3
−3
0
1
0
=1−
2
=2 −
,
2
−1 ≤ ≤ 2
,
−2
−1
0
1
2
−10
−2
0
2
10
6 = 3+ ,
3 =
−
−
2
0
2
−
−
2+1
0
2+1
−1 0 1 0 = + sin , = cos , − ≤ ≤
−2 2
−2
5 39 −2
−
=
5.
= 2 − 1, (a)
−1
0
−1
1
+2
=
1 2
−1
1 −1
1 72 −1
2 14 + , =
4.
AL
3 6 22 + 2, −2 ≤ ≤ 2
+1
1 37
+1
−1
1
1 37 − , −2 ≤ ≤ 2 +1
−4
−2
0
2
4
−9
−5
−1
3
7
−1
0
1
2
3
1 72
2 −2
+2
2 14 2
−2
5 39
=2 −1
(b)
=
864
6.
¤
1 2
+1 =
= 3 + 2,
(b)
1 2
1 2
+
⇒
+1 1 2
+1 =
1 2
= 1 4
1 4
+
+ 12 , so
+1
⇒
=
1 4
+
5 4
= 2 +3 −4
−2
0
2
4
−10
−4
2
8
14
−5
−1
3
7
11
⇒
=3 +2
=2 +3 =2
=
2
− 3,
(b)
−2
3 = −
1 3
2 3
+3 =
−1
1
3
6
−2
−2
6
−1
1
3
5
⇒
2 3
= −
4 3
−
1 3
2 3,
so
⇒
+3
=
2 3
+
5 3
≤3
−3
= +2
− 2, so
=
=
2
− 3 = ( − 2)2 − 3 =
=
2
− 4 + 1, −1 ≤
= 1 − cos ,
= sin ,
⇒
−3 ≤
= + 2,
(a)
8.
2 =
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(a)
7.
⇒
2
− 4 +4 − 3
⇒
≤5
0≤
≤2
(a) 0
(b)
3
2
2
0
1
0
−1
0
0
1
2
1
0
= 1 − cos
[or
= sin , 2
2
− 1 = − cos ]
+ ( − 1)2 = (sin )2 + (− cos )2
⇒
⇒ 2
+ ( − 1)2 = 1.
As varies from 0 to 2 , the circle with center (0 1) and radius 1 is traced out.
9.
=
√
,
= 1−
(a) 0
1
2
3
4
0
1
1 414
1 732
2
1
0
−1
−2
−3
AL
(b)
=
√
⇒
⇒
2
=
= 1− = 1−
2
. Since ≥ 0,
= 1−
So the curve is the right half of the parabola
2
≥ 0.
.
SECTION 10.1
10.
=
2
,
3
=
(a)
(b)
11. (a)
−2
−1
0
1
2
4
1
0
1
4
−8
−1
0
1
8
⇒
3
=
1
= sin
=
,
+
2
= cos
=
1 2
1 2
,− ≤
+ cos2
≤ 0 and 0 ≤
≤
2
3
=
1 2
2
≤ 1. For 0
≤
≥ 0.
≤ 0, we have ≤1
, we have 0
2
1
= cos2
≤
+ sin2
= 1
ALE (b)
. ⇒
4
2
+
1
= 1 ⇒
2
4
22
= 1, which is an equation of an ellipse with
-intercepts ± 12 and -intercepts ±2. For 0 ≤ ≥ 0 and 0 ≤
and 2
13. (a)
∈ R,
2
+
2)2
≥
∈ R,
(b)
2
1 2
.
.
= 1. For − ≤
= 2 sin , 0 ≤
cos ,
(2 )2 +
(1
2 3
=
≥ 0. The graph is a semicircle.
and 1
12. (a)
1
2
=
2
= sin2
−1 ≤
⇒
3
2 2
CURVES DEFINED BY PARAMETRIC EQUATIONS
= sin
≤ 2. For
2
≤
≤
2, we have ≥ − 21
, we have 0
≥ 0. So the graph is the top half of the ellipse.
= csc , 0
.
= csc =
1
=
1
.
(b)
¤
865
2
For 0
, we have 0
2
1 and
the portion of the hyperbola
14. (a)
866
¤
−2
= ( )−2 =
=1
−2
with
2
=1
1. Thus, the curve is
for
1.
0 since
=
⇒
= ln
(b)
√
=
, so
+1
= sinh ,
2
= ( )2 =
2
.
(b)
−
⇒
2
2
−
2
= cosh2
− sinh2 = 1.
(b)
= 1.
= sec , −
= sec2
−
≤ 0, we have
2
and 1
2
⇒
1 + tan2
0
(b)
2.
1+
=
≥ 0 and
2
⇒
=
≥ 1. For 0
2
− 1. For
=
2 to 0, the point (
approaches (0 1) along the parabola. As increases from 0 to point (
AL
2, we have
. Thus, the curve is the portion of the parabola
in the first quadrant. As increases from −
2
−1
) 2, the
) retreats from (0 1) along the parabola.
−5 = 5 + 2 cos −5 2
(b)
= +1
= cosh
= tan2 ,
18. (a)
2
= cosh ≥ 1, we have the upper branch of the hyperbola
Since 2
⇒
=
⇒ = 2 − 1. √ √ 2 − 2. The curve is the part of = − 1 = ( 2 − 1) − 1 = √ the hyperbola 2 − 2 = 2 with ≥ 2 and ≥ 0. =
17. (a)
19.
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
15. (a)
16. (a)
=
sin
2
+
,
= 3 + 2 sin −3 2
⇒
cos
=
2
−3 , sin
=
2
2
.
2
cos ( ) + sin ( ) = 1 ⇒
2
= 1. The motion of the particle takes place on a circle centered at (5 3) with a radius 2. As goes
−5 2
from 1 to 2, the particle starts at the point (3 3) and moves counterclockwise along the circle
2
+
−3 2
2
= 1 to
(7 3) [one-half of a circle]. −1 20.
= 2 + sin ,
= 1 + 3 cos
⇒
sin =
− 2, cos =
3
2
.
sin
2
2
+ cos
= 1
⇒
( − 2) +
−1 3
2
= 1.
The motion of the particle takes place on an ellipse centered at (2 1). As goes from
2 to 2 , the particle starts at the point
(3 1) and moves counterclockwise three-fourths of the way around the ellipse to (2 4).
SECTION 10.1 2
21.
= 5 sin ,
= 2 cos
⇒
sin =
5
, cos =
2
.
sin
CURVES DEFINED BY PARAMETRIC EQUATIONS 2
2
+ cos
= 1
⇒
867
2
+
5
¤
= 1. The motion of the
2
particle takes place on an ellipse centered at (0 0). As goes from − to 5 , the particle starts at the point (0 −2) and moves clockwise around the ellipse 3 times. 22.
= cos2 = 1 − sin2
= 1−
2
. The motion of the particle takes place on the parabola
= 1−
2
. As goes from −2 to
− , the particle starts at the point (0 1), moves to (1 0), and goes back to (0 1). As goes from − to 0, the particle moves to (−1 0) and goes back to (0 1). The particle repeats this motion as goes from 0 to 2 . 23. We must have 1 ≤
≤ 4 and 2 ≤
≤ 3. So the graph of the curve must be contained in the rectangle [1 4] by [2 3].
24. (a) From the first graph, we have 1 ≤
≤ 2. From the second graph, we have −1 ≤
≤ 1 The only choice that satisfies
either of those conditions is III. (b) From the first graph, the values of of
cycle through the values from −2 to 2 six times. Choice I satisfies these conditions.
(c) From the first graph, the values of 0≤
cycle through the values from −2 to 2 four times. From the second graph, the values
cycle through the values from −2 to 2 three times. From the second graph, we have
≤ 2. Choice IV satisfies these conditions.
(d) From the first graph, the values of
cycle through the values from −2 to 2 two times. From the second graph, the values of
do the same thing. Choice II satisfies these conditions. 25. When = −1, (
) = (1 1). As increases to 0,
As increases from 0 to 1,
increases from 0 to 1 and
−1. As increases beyond 1, decrease. For
−1,
and
and
both decrease to 0.
decreases from 0 to
continues to increase and
continues to
are both positive and decreasing. We could
achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points. 26. When = −1, (
while
) = (0 0). As increases to 0,
increases from 0 to 1,
first decreases to −1 and then increases to 0. As increases from 0 to 1,
decreases from 1 to 0, while
y 1
first increases to 1 and then decreases to 0. We
could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
0
t=_1, 1 (0, 0)
t=0 (1, 0) 1
x
_1
27. When = −1, (
) = (0 1). As increases to 0,
increases from 0 to 1 and
decreases from 1 to 0. As increases from 0 to 1, the curve is retraced in the opposite direction with
decreasing from 1 to 0 and
increasing from 0 to 1.
We could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
868
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
28. (a)
=
4
− +1 =(
+ 1) −
4
0 [think of the graphs of
=
4
+ 1 and
= ] and
=
2
≥ 0, so these equations
are matched with graph V. √
≥ 0.
2
(b)
=
=
(c)
= sin 2 has period 2
− 2 = ( − 2) is negative for 0 2=
2, so these equations are matched with graph I.
. Note that
( + 2 ) = sin[ + 2 + sin 2( + 2 )] = sin( + 2 + sin 2 ) = sin( + sin 2 ) = ( ), so cycles through the values −1 to 1 twice as
These equations match graph II since (d)
= cos 5 has period 2
5 and
= sin 2 has period , so
takes on the values −1 to 1. Note that when = 0, ( (e)
= + sin 4 ,
=
2
+ cos 3 .
, so the graph will look like the graph of (f )
=
29. Use
30. Use
sin 2 , 4+ 2
= and
1
= ,
1
=
cos 2 . 4+ 2
= − 2 sin
=
3
As → ∞,
= and
2
=
3
−4,
2
will take on the values −1 to 1, and then 1 to −1, before
2
become the dominant terms in the expressions for
and
, but with oscillations. These equations are matched with graph IV. both approach 0. These equations are matched with graph III.
with a -interval of [−
− 4 and
cycles through those values once.
) = (1 0). These equations are matched with graph VI
As becomes large, and 2
has period 2 .
].
= with a -interval of
[−3 3]. There are 9 points of intersection; (0 0) is fairly obvious. The point in quadrant I is approximately (2 2 2 2), and by symmetry, the point in quadrant III is approximately (−2 2 −2 2). The other six points are approximately (∓1 9 ±0 5), (∓1 7 ±1 7), and (∓0 5 ±1 9).
31. (a)
=
1
through
+(
2
2( 2
−
1) 2)
,
=
1
+(
2
−
when = 1. For 0
1)
,0≤ 1,
≤ 1. Clearly the curve passes through is strictly between 2
−
1
1
and
2
and
1( 1
1)
when = 0 and
is strictly between
1
and
2.
For
every value of , 1( 1
1)
and
and 2( 2
Finally, any point (
satisfy the relation
−
1
= 2
−
( −
1 ),
which is the equation of the line through
1
2 ).
) on that line satisfies
− 2 −
1 1
=
− 2 −
1 1
; if we call that common value , then the given
parametric equations yield the point (
); and any (
) on the line between
1( 1
in [0 1]. So the given parametric equations exactly specify the line segment from (b)
= −2 + [3 − (−2)] = −2 + 5 and
= 7 + (−1 − 7) = 7 − 8 for 0 ≤
SECTION 10.1
32. For the side of the triangle from
to
, use (
1)
1
= (1 1) and (
and
1( 1
1)
2( 2
to
2)
2( 2
yields a value of 2 ).
≤ 1.
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS 2)
2
1)
869
= (4 2).
Hence, the equations are = = Graphing
1
+( +(
to
− − 2 2
= 1 + (4 − 1) = 1 + 3 , ) = 1 + (2 − 1) = 1 + . 1 1)
= 1 + with 0 ≤
= 1 + 3 and
triangle from and
1
≤ 1 gives us the side of the
. Similarly, for the side
we use
= 4 − 3 and
= 2 + 3 , and for the side
we use
=1
= 1 +4 .
33. The circle
2
+ ( − 1)2 = 4 has center (0 1) and radius 2, so by Example 4 it can be represented by
= 1 + 2 sin , 0 ≤
≤ 2 . This representation gives us the circle with a counterclockwise orientation starting at (2 1).
(a) To get a clockwise orientation, we could change the equations to
= 1 − 2 sin , 0 ≤
= 2 cos ,
(b) To get three times around in the counterclockwise direction, we use the original equations the domain expanded to 0 ≤
≤2 .
= 2 cos ,
= 1 + 2 sin with
≤6 .
(c) To start at (0 3) using the original equations, we must have = 2 cos ,
= 2 cos ,
= 1 + 2 sin ,
2
≤
≤
2
1
= 0; that is, 2 cos = 0. Hence, =
2
. So we use
.
Alternatively, if we want to start at 0, we could change the equations of the curve. For example, we could use = −2 sin , 34. (a) Let
2
2
= 1 + 2 cos , 0 ≤
= sin2
and
= cos with 0 ≤ 2
2
+
2
2
2
2
≤
.
= cos2 to obtain
= sin and
≤ 2 as possible parametric equations for the ellipse
= 1.
(b) The equations are
= 3 sin and
= cos for ∈ {1 2 4 8}.
(c) As increases, the ellipse stretches vertically.
35. Big circle: It’s centered at (2 2) with a radius of 2, so by Example 4, parametric equations are
= 2 + 2 cos
= 2 + 2 sin
0≤
≤2
Small circles: They are centered at (1 3) and (3 3) with a radius of 0 1. By Example 4, parametric equations are
and
(left)
= 1 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
(right)
= 3 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
Semicircle: It’s the lower half of a circle centered at (2 2) with radius 1. By Example 4, parametric equations are = 2 + 1 cos
= 2 + 1 sin
≤
≤2
To get all four graphs on the same screen with a typical graphing calculator, we need to change the last -interval to[0 2 ] in order to match the others. We can do this by changing to 0 5 . This change gives us the upper half. There are several ways to get the lower half—one is to change the “+” to a “−” in the -assignment, giving us
= 2 + 1 cos(0 5 )
= 2 − 1 sin(0 5 )
0≤
≤2
870
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
36. If you are using a calculator or computer that can overlay graphs (using multiple -intervals), the following is appropriate.
Left side:
= 1 and
goes from 1 5 to 4, so use =
15≤
≤4
=
15≤
≤4
=15
1≤
≤ 10
=1 Right side:
= 10 and
goes from 1 5 to 4, so use = 10
Bottom:
goes from 1 to 10 and
= 1 5, so use =
Handle: It starts at (10 4) and ends at (13 7), so use = 10 +
0≤
= 4+
≤3
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
= 1 + 1 sin
6
≤
≤
13 6
6
≤
≤
13 6
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos
= 1 + 1 sin
If you are using a calculator or computer that cannot overlay graphs (using one -interval), the following is appropriate. We’ll start by picking the -interval [0 2 5] since it easily matches the -values for the two sides. We now need to find parametric equations for all graphs with 0 ≤ Left side:
= 1 and
≤ 2 5.
goes from 1 5 to 4, so use =1
Right side:
= 10 and
0≤
≤25
=15+
0≤
≤25
goes from 1 5 to 4, so use = 10
Bottom:
=15+
goes from 1 to 10 and
= 1 5, so use = 1 +36
0≤
=15
≤25
To get the x-assignment, think of creating a linear function such that when = 0, = 10. We can use the point-slope form of a line with ( −1 =
10 − 1 ( − 0) 25−0
⇒
1
1)
= (0 1) and (
2
= 1 and when = 2 5, 2)
= (2 5 10).
= 1 +36 .
Handle: It starts at (10 4) and ends at (13 7), so use = 10 + 1 2 (
1
1)
= (0 10) and (
2
2)
= (2 5 13) gives us
0≤
= 4 +1 2 − 10 =
≤ 25
13 − 10 ( − 0) 25−0
⇒
= 10 + 1 2 .
7− 4 (
1
1 ) = (0 4) and (
2
2 ) = (2 5 7) gives us − 4 =
25 −0
( − 0)
⇒
= 4 +12 .
SECTION 10.1
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
871
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
(
1)
1
= 0
and (
2)
2
+
15
gives us −
5
=
6
= 1 + 1 sin
6
2
5 6
=
+
15
−
13 6 5
6 2
5 6
5 6
( − 0)
0≤
≤ 25
⇒
=
−0
+ 6
. 15
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos 37. (a)
=
3
⇒
1 3
=
, so
We get the entire curve
=
+
15
2 3
= 1 + 1 sin
6
=
2
2 3
traversed in a left to
=
.
(b)
right direction.
(c)
= = If
−3 −2
=( =(
0, then
−
)3
−
)2 = (
and
−
[so
=
1 3 2
) =
1 3
=
2 3
0 and
+
= , so
=
−2
=
−2
= cos ,
= sec2
=
cos2
=
1
2.
1 6
≤ 25 , so
=
curve
=
2 3
. 0, then
and
0, the curve never quite
=1
2
traversed in a
Since sec ≥ 1, we only get the
4
=
4 6
=
2 3
≥ 0, we only get the right half of the
6
.
AL
. We get the entire curve
1
=
=
left-to-right direction.
(b)
6
⇒
6
reaches the origin. 38. (a)
0≤
5
Since
],
are both larger than 1. If
are between 0 and 1. Since
15
.
2
≥ 1. We get the first quadrant portion of
parts of the curve
=1
with
the curve when
0, that is, cos
portion of the curve when
(c)
=
,
=
−2
0, and we get the second quadrant
0, that is, cos
= ( )−2 =
−2
0.
. Since
only get the first quadrant portion of the curve
¤
872
are both positive, we 2
=1
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
39. The case
and
is illustrated.
2
has coordinates (
has coordinates (
+ cos( − )) = (
[since cos( − ) = cos cos
+ sin
) as in Example 7, (1 − cos ))
= − cos
sin
, so
has
− sin( − ) (1 − cos )) = ( ( − sin ) (1 − cos ))
coordinates (
[since sin( − ) = sin parametric equations
− cos sin
cos
= ( − sin ),
= sin
+ sin( − )
−
2,
. As in Example 7,
+ cos( − )) = (
diagram) has coordinates (
. Again we have the
= (1 − cos ).
40. The first two diagrams depict the case
(
−2
and
cos ). That i
−
has coordinates (
cos ), so a typical point
has coordinates (
), where
−
=
). Now
(in the second
of the trochoid has coordinates
sin
and
=
−
cos . When
= , these equations agree with those of the cycloid.
41. It is apparent that
= =
= cos and cos and
sin2
=
=(
2
+ cos2
has coordinates ( cos
42.
( sec
0). It follows tha
= (2 cot
43.
Then ∠
=
=
= sin . Thus, the parametric equations are
⇒
= = 1 =
2
. From the diagram,
we rearrange: sin cos2
2
+
=(
2
sin ). Since
2
is a right angle and ∠
2
) . Adding the two
. Thus, we have an ellipse.
has coordinates ( sec is
is a right triangle and
sin ). Thus, the parametric equations are
= 2 cot . Le
= , so
,∆
= 2 sin
has coordinates
= sec ,
= sin .
= (0 2 ). and
(2 sin ) sin ). Thus, the -coordinate of
= 2 sin2 .
44. (a) Let
AL
⇒
=
is perpendicular to
2 ), so the -coordinate of
= ((2 sin ) cos is
=
= sin . To eliminate
) and cos
equations: sin2
and
be the angle of inclination of segment
cos . Then
=
2
.
Le
= (2 0). Then by use of right triangle
we see that
= 2 cos . (b)
Now
=
=2
= 1 cos
− − cos
=2
1 − cos2 cos
=2
sin2 cos
= 2 sin
tan
So
has coordinates
= 2 sin
· cos = 2 sin2
tan
and
= 2 sin
tan
SECTION 10.1
45. (a)
· sin
= 2 sin2
tan .
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
873
There are 2 points of intersection: (−3 0) and approximately (−2 1 1 4).
(b) A collision point occurs when
=
1
2
and
1
=
2
for the same . So solve the equations:
3 sin = −3 + cos
(1)
2 cos = 1 + sin
(2)
From (2), sin = 2 cos − 1. Substituting into (1), we get 3(2 cos − 1) = −3 + cos cos = 0
⇒
=
occurs when =
2
3 2
or
2
. We check that =
satisfies (1) and (2) but =
2
2
⇒
5 cos = 0 ( )
does not. So the only collision point
, and this gives the point (−3 0). [We could check our work by graphing
functions of and, on another plot, 3 2
pairs of graphs intersect is =
1
and
2
⇒
1
and
2
together as
as functions of . If we do so, we see that the only value of for which both
.]
(c) The circle is centered at (3 1) instead of (−3 1). There are still 2 intersection points: (3 0) and (2 1 1 4), but there are ⇒
no collision points, since ( ) in part (b) becomes 5 cos = 6 = 30◦ and
46. (a) If
0
=
1 2 (9
8)
2
= 250 − 4 9 2 . √
250
49
≈ 51 s. Then
The formula for
= 250
2
with equality when =
= 0 when = 0 (when the gun is fired) and again when
3
49
≈ 22,092 m, so the bullet hits the ground about 22 km from the gun.
−
= −4 9
250
2
−
125
250
+
49
125 2
cos )
⇒
1252
= −4 9
49
s, so the maximum height attained is
125 2
−
125 2
+
49
125 2
≤
49
1252 49
≈ 3189 m.
49
As
OT 0
+
49
49
=(
√ 3 and
250
49
(c)
1.
is quadratic in . To find the maximum -value, we will complete the square:
= −4 9
(b)
6 5
= (500 cos 30◦ ) = 250
= 500 m s, then the equations become
= (500 sin 30◦ ) −
cos =
= 0
cos
(0◦
90◦ ) increases up to 45◦ , the projectile attains a
greater height and a greater range. As
increases past 45◦ , the
projectile attains a greater height, but its range decreases.
. 2
=(
0
sin ) −
1 2
2
⇒
=(
0
sin
) 0
cos
−
2
0
cos
= (tan ) −
2
2
0 2
cos2
,
which is the equation of a parabola (quadratic in ).
874
¤
47.
=
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 2
=
3
−
. We use a graphing device to produce the graphs for various values of with − ≤
the members of the family are symmetric about the -axis. For cusp at (0 0) and for
48.
=2
−4
3
=−
0 the graph crosses itself at
2
≤
. Note that all
0, the graph does not cross itself, but for = 0 it has a
= , so the loop grows larger as increases.
+ 3 4 . We use a graphing device to produce the graphs for various values of with − ≤
Note that all the members of the family are symmetric about the -axis. When of even degree, but when = 0 there is a corner at the origin, and when
= + cos
0, the graph crosses itself at the origin, and has
0. From the first figure, we see that
= + sin
curves roughly follow the line
= , and they start having loops when
is between 1 4 and 1 6. The loops increase in size as
increases.
While not required, the following is a solution to determine the exact values for which the curve has a loop, that is, we seek the values of ( + cos
for which there exist parameter values and
+ sin ) = ( + cos
such that
and
+ sin ). In the diagram at the left, and
denotes the point (
the point ( + cos
Since
=
angles,
=∠
=2 − + =
2
4
−
(1).
=
4
=∠ −
)
+ sin ) = ( + cos
= , the triangle and
.
0, the graph resembles that of a polynomial
two cusps below the -axis. The size of the “swallowtail” increases as increases.
49.
≤
the point (
),
+ sin ).
is isosceles. Therefore its base are equal. Since
the relation
=
= −
implies that
4
and
SECTION 10.1
Since cos
= distance(( =
− =
1 2
=
( − )
√ 2 cos
−
) ( √ 2
2( − )2 =
− =
√ 2 cos , that is,
(2). Now cos − 4 = sin 2 − − 4 = sin 4 − , √ − = 2 sin 34 − (20 ). Subtracting (20 ) from (1) and
4
so we can rewrite (2) as
−
dividing by 2, we obtain = 4
Since
√ 2
−
sin
2
implicitly assumed that 0 by + 2 merely increases √ 3 2 4 − (3), we get = sin
4
−
, or
3
− =
4
4
−
−
sin √ 2
4
it follows from (20 ) that sin
0 and
875
√ 2 ( − ), we see that
)) =
, so
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS
(3).
4
0. Thus from (3) we see that
4.
[We have
by the way we drew our diagram, but we lost no generality by doing so since replacing and
by 2 . The curve’s basic shape repeats every time we change by 2 .] Solving for
. Write
=
3
4
. Then −
=
√ 2 sin
, where
0. Now sin
for
0, so
√ 2.
in
→ 0+ , that is, as →
As
50. Consider the curves
− 4
= sin + sin
→
,
,
radius 2 centered at the origin. For
√ 2 .
= cos + cos
, where
is a positive integer. For
= 1, we get a circle of − 1 loops as
1, we get a curve lying on or inside that circle that traces out
ranges from 0 to 2 . 2
Note:
+
2
= (sin + sin
)2 + (cos + cos
= sin2 + 2 sin sin
+ sin2
= (sin2 + cos2 ) + (sin2 = 1 + 1 + 2 cos( − with equality for
)2 + cos2 + 2 cos cos
+ cos2
+ cos2
) + 2(cos cos
+ sin sin
)
) = 2 + 2 cos((1 − ) ) ≤ 4 = 22 ,
= 1. This shows that each curve lies on or inside the curve for
= 1, which is a circle of radius 2 centered
at the origin.
NOT = 1
= 2
= 3
51. Note that all the Lissajous figures are symmetric about the -axis. The parameters
- and -directions respectively. For
876
¤
= =
and simply stretch the graph in the
= 1 the graph is simply a circle with radius 1. For
= 2 the graph crosses
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
itself at the origin and there are loops above and below the -axis. In general, the figures have all of which are on the -axis, and a total of
= =1 52.
=5
= cos ,
= sin − sin
=2 .
If = 1, then
graphs are shown for = 2 3 4 and 5.
− 1 points of intersection,
closed loops.
=3
= 0, and the curve is simply the line segment from (−1 0) to (1 0). The
F R
It is easy to see that all the curves lie in the rectangle [−1 1] by [−2 2]. When is an integer, ( + 2 ) = ( ) and ( + 2 ) = ( ), so the curve is closed. When is a positive integer greater than 1, the curve intersects the x-axis + 1 times and has loops (one of which degenerates to a tangency at the origin when is an odd integer of the form 4 + 1). √ As increases, the curve’s loops become thinner, but stay in the region bounded by the semicircles = ± 1 + 1 − and the line segments from (−1 −1) to (−1 1) and from (1 −1) to (1 1). This is true because
2
= sin − sin
≤ sin + sin ≤
√ 1−
2
+ 1. This curve appears to fill the entire region when is very large, as
shown in the figure for = 1000.
LABORATORY PROJECT
RUNNING CIRCLES AROUND CIRCLES
¤
When is a fraction, we get a variety of shapes with multiple loops, but always within the same region. For some fractional values, such as = 2 359, the curve again appears to fill the region.
S LE
LABORATORY PROJECT Running Circles Around Circles 1. The cente
Arc
of the smaller circle has coordinates (( − )cos
on circ
has length
( − )sin ).
since it is equal in length to arc
(the smaller circle rolls without slipping against the larger.) Thus, ∠
and 2. With
=
and ∠
− , so
=
has coordinates
= ( − )cos + cos(∠
) = ( − )cos + cos
= ( − )sin − sin(∠
) = ( − )sin − sin
= 1 and
− −
a positive integer greater than 2, we obtain a hypocycloid of
cusps. Shown in the figure is the graph for
= 4. Let
= 4 and = 1. Using the
sum identities to expand cos 3 and sin 3 , we obtain = 3 cos + cos 3 = 3 cos + 4 cos3 and
.
− 3 cos
= 3 sin − sin 3 = 3 sin − 3 sin − 4 sin3
= 4 cos3 = 4 sin3 .
877
3. The graphs at the right are obtained with 1
1
1
1
2
3
4
10
= , , , and
with −2 ≤
= 1 and
≤ 2 . We
conclude that as the denominator increases, the graph gets smaller, but maintains the basic shape shown.
[continued]
¤
878
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Letting
So if
= 2 and
= 3, 5, and 7 with −2 ≤
is held constant and
obtain a hypocycloid of graphs have
4. If
varies, we get a graph with
cusps. As
= 32 , 54 , and
≤ 2 gives us the following:
cusps (assuming
=
+ 1, we
increases, we must expand the range of in order to get a closed curve. The following
11 10 .
S
= 1, the equations for the hypocycloid are = ( − 1) cos + cos (( − 1) )
which is a hypocycloid of
is in lowest form). When
= ( − 1) sin − sin (( − 1) )
cusps (from Problem 2). In general, if
1, we get a figure with cusps on the “outside ring” and if
1, the cusps are on the “inside ring”. In any case, as the values of get larger, we get a figure that looks more and more like a washer. If we were to graph the hypocycloid for all values of , every point on the washer would eventually be arbitrarily close to a point on the curve.
= 5. The cente
Arc and ∠
√ 2,
−10 ≤
= − 2,
≤ 10
of the smaller circle has coordinates (( + ) cos
has length =
−
(as in Problem 1), so that ∠ −
Thus, the coordinates of
=
since ∠
+
−
=
( + ) sin ).
,∠
=
−
,
= .
are
= ( + ) cos + cos
−
+
= ( + ) cos − cos
+ and
0≤
= ( + ) sin − sin
−
+
+ = ( + ) sin − sin
.
≤ 446
LABORATORY PROJECT
6. Let
= 1 and the equations become = ( + 1) cos − cos(( + 1) )
If
RUNNING CIRCLES AROUND CIRCLES
= 1, we have a cardioid. If
= ( + 1) sin − sin(( + 1) )
is a positive
integer greater than 1, we get the graph of an “ -leafed clover”, with cusps that are
units
from the origin. (Some of the pairs of figures are not to scale.) = 3, −2 ≤ If
=
with
≤2
= 10, −2 ≤
≤2
= 1, we obtain a figure that
does not increase in size and requires −
≤
≤
to be a closed curve traced
exactly once.
S LE
= 1 , −4 ≤
R
≤4
= 1 , −7 ≤
≤ 7
¤
879
4
Next, we keep
constant and let
7
vary. As
increases, so does the size of the figure. There is an -pointed star in the middle.
Now if
= 25 , −5 ≤
≤5
= 75 , −5 ≤
≤ 5
= 43 , −3 ≤
≤3
= 76 , −6 ≤
≤ 6
√ 2, 0 ≤
≤ 200
= − 2, 0 ≤
+ 1 we obtain figures similar to the
=
previous ones, but the size of the figure does not increase.
If
is irrational, we get washers that increase in
size as
increases.
=
¤
880
10.2 1.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Calculus with Parametric Curves ,
=
=
√ 1+
1+
⇒
=
=
3.
=
,
3
+ 1,
=
4
+ ;
(1 + )−1
2
= √
1
,
=
(1 + )(1) − (1) (1 + )2
2 1+
, and
(1 + )2
(1 + ) 1 3 2 = √ = (1 + ) . 2 2 1+
⇒
= 1 + cos ,
= −1.
=4
3
=
+
=
( + 1), and
= 3 2 , and
+ 1,
=
=
=
4
=
3
3
4.
1
=
2
1 (2 1 + )
= + sin
1 2
1 (1 + )2
2.
=
√
=
and
≤ 446
+1 2
1 + cos . ( + 1)
hen = −1, ( .W
) = (0 0)
= −3 3 = −1, so an equation of the tangent to the curve at the point corresponding to = −1 is
− 0 = −1( − 0), or √ = , = 2−2;
=− . = 4.
= 2 − 2,
1
= 2
√
, and
=
√ √ = (2 − 2)2 = 4( − 1) . When = 4,
(
) = (2 8) and
= 4(3)(2) = 24, so an equation of the tangent to the curve at the point corresponding to = 4 is
− 8 = 24( − 2), or 5.
= cos ,
= sin ;
When = , (
=
sin
= .
,
2
=
=−
is
−0 =
− (− )], or
;
= 0. 2
=
= cos + sin ,
0) and
) = (−
corresponding to = 6.
= 24 − 40.
=2
2
,
=
cos + sin . − sin + cos
=
2
+
( cos
.
) + (sin
)
=
( cos
+ sin
), and
2
= ( cos
+ sin
)
=
cos
. When = 0, (
+ sin
of the tangent to the curve at the point corresponding to = 0 is − 1 = 7. (a)
=
(−1) = , so an equation of the tangent to the curve at the point
=
2
= (− sin ) + cos , and
= 1 + ln ,
=
2
+ 2; (1 3).
=2
=
1
and
2
) = (0 1) and
( − 0), or
=
=
2
=2
2
=
, so an equation
+ 1.
= 2 2 . At (1 3),
1 = 1 + ln = 1 or (b)
ln = 0
⇒
= 1 and
−1
⇒
=
⇒
= 1 + ln
= 1+
0
√
= ,
ln =
2(1)−2
=
2
−1
, so
= 1+ =4
√
= 2
⇒
=
2
−1 2
+2 =(
) +2 =
2 −2
· 2 = 2, so an equation of the tangent is − 3 = 2( − 1), or ; (2 ).
=
2
2, ·
or
= 2, so an equation of the tangent is
− 3 = 2( − 1),
= 2 + 1.
At (1 3), 8. (a)
⇒
√
= 1
⇒
= 1 and
= 2
1 , and √
=
=
+ 2, and
0
=
2 −2
· 2.
= 2 + 1. 2
1 2
2
√
= 4 , so an equation of the tangent is
=4
−
3 2
2
. At (2 ),
= 4 ( − 2),
−7 .
Multivariable Calculus 8th Edition Stewart Solutions Manual Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-solutions-manual/ Multivariable Calculus 8th Edition Stewart Test Bank Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-test-bank/ °
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10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1
1.
2.
3.
Curves Defined by Parametric Equations
−1
0
1
2
0
1
0
−3
−3
0
1
0
=1−
2
=2 −
,
2
−1 ≤ ≤ 2
,
−2
−1
0
1
2
−10
−2
0
2
10
6 = 3+ ,
3 =
−
−
2
0
2
−
−
2+1
0
2+1
−1 0 1 0 = + sin , = cos , − ≤ ≤
−2 2
−2
5 39 −2
−
=
5.
= 2 − 1, (a)
−1
0
−1
1
+2
=
1 2
−1
1 −1
1 72 −1
2 14 + , =
4.
AL
3 6 22 + 2, −2 ≤ ≤ 2
+1
1 37
+1
−1
1
1 37 − , −2 ≤ ≤ 2 +1
−4
−2
0
2
4
−9
−5
−1
3
7
−1
0
1
2
3
1 72
2 −2
+2
2 14 2
−2
5 39
=2 −1
(b)
=
864
6.
¤
1 2
+1 =
= 3 + 2,
(b)
1 2
1 2
+
⇒
+1 1 2
+1 =
1 2
= 1 4
1 4
+
+ 12 , so
+1
⇒
=
1 4
+
5 4
= 2 +3 −4
−2
0
2
4
−10
−4
2
8
14
−5
−1
3
7
11
⇒
=3 +2
=2 +3 =2
=
2
− 3,
(b)
−2
3 = −
1 3
2 3
+3 =
−1
1
3
6
−2
−2
6
−1
1
3
5
⇒
2 3
= −
4 3
−
1 3
2 3,
so
⇒
+3
=
2 3
+
5 3
≤3
−3
= +2
− 2, so
=
=
2
− 3 = ( − 2)2 − 3 =
=
2
− 4 + 1, −1 ≤
= 1 − cos ,
= sin ,
⇒
−3 ≤
= + 2,
(a)
8.
2 =
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(a)
7.
⇒
2
− 4 +4 − 3
⇒
≤5
0≤
≤2
(a) 0
(b)
3
2
2
0
1
0
−1
0
0
1
2
1
0
= 1 − cos
[or
= sin , 2
2
− 1 = − cos ]
+ ( − 1)2 = (sin )2 + (− cos )2
⇒
⇒ 2
+ ( − 1)2 = 1.
As varies from 0 to 2 , the circle with center (0 1) and radius 1 is traced out.
9.
=
√
,
= 1−
(a) 0
1
2
3
4
0
1
1 414
1 732
2
1
0
−1
−2
−3
AL
(b)
=
√
⇒
⇒
2
=
= 1− = 1−
2
. Since ≥ 0,
= 1−
So the curve is the right half of the parabola
2
≥ 0.
.
SECTION 10.1
10.
=
2
,
3
=
(a)
(b)
11. (a)
−2
−1
0
1
2
4
1
0
1
4
−8
−1
0
1
8
⇒
3
=
1
= sin
=
,
+
2
= cos
=
1 2
1 2
,− ≤
+ cos2
≤ 0 and 0 ≤
≤
2
3
=
1 2
2
≤ 1. For 0
≤
≥ 0.
≤ 0, we have ≤1
, we have 0
2
1
= cos2
≤
+ sin2
= 1
ALE (b)
. ⇒
4
2
+
1
= 1 ⇒
2
4
22
= 1, which is an equation of an ellipse with
-intercepts ± 12 and -intercepts ±2. For 0 ≤ ≥ 0 and 0 ≤
and 2
13. (a)
∈ R,
2
+
2)2
≥
∈ R,
(b)
2
1 2
.
.
= 1. For − ≤
= 2 sin , 0 ≤
cos ,
(2 )2 +
(1
2 3
=
≥ 0. The graph is a semicircle.
and 1
12. (a)
1
2
=
2
= sin2
−1 ≤
⇒
3
2 2
CURVES DEFINED BY PARAMETRIC EQUATIONS
= sin
≤ 2. For
2
≤
≤
2, we have ≥ − 21
, we have 0
≥ 0. So the graph is the top half of the ellipse.
= csc , 0
.
= csc =
1
=
1
.
(b)
¤
865
2
For 0
, we have 0
2
1 and
the portion of the hyperbola
14. (a)
866
¤
−2
= ( )−2 =
=1
−2
with
2
=1
1. Thus, the curve is
for
1.
0 since
=
⇒
= ln
(b)
√
=
, so
+1
= sinh ,
2
= ( )2 =
2
.
(b)
−
⇒
2
2
−
2
= cosh2
− sinh2 = 1.
(b)
= 1.
= sec , −
= sec2
−
≤ 0, we have
2
and 1
2
⇒
1 + tan2
0
(b)
2.
1+
=
≥ 0 and
2
⇒
=
≥ 1. For 0
2
− 1. For
=
2 to 0, the point (
approaches (0 1) along the parabola. As increases from 0 to point (
AL
2, we have
. Thus, the curve is the portion of the parabola
in the first quadrant. As increases from −
2
−1
) 2, the
) retreats from (0 1) along the parabola.
−5 = 5 + 2 cos −5 2
(b)
= +1
= cosh
= tan2 ,
18. (a)
2
= cosh ≥ 1, we have the upper branch of the hyperbola
Since 2
⇒
=
⇒ = 2 − 1. √ √ 2 − 2. The curve is the part of = − 1 = ( 2 − 1) − 1 = √ the hyperbola 2 − 2 = 2 with ≥ 2 and ≥ 0. =
17. (a)
19.
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
15. (a)
16. (a)
=
sin
2
+
,
= 3 + 2 sin −3 2
⇒
cos
=
2
−3 , sin
=
2
2
.
2
cos ( ) + sin ( ) = 1 ⇒
2
= 1. The motion of the particle takes place on a circle centered at (5 3) with a radius 2. As goes
−5 2
from 1 to 2, the particle starts at the point (3 3) and moves counterclockwise along the circle
2
+
−3 2
2
= 1 to
(7 3) [one-half of a circle]. −1 20.
= 2 + sin ,
= 1 + 3 cos
⇒
sin =
− 2, cos =
3
2
.
sin
2
2
+ cos
= 1
⇒
( − 2) +
−1 3
2
= 1.
The motion of the particle takes place on an ellipse centered at (2 1). As goes from
2 to 2 , the particle starts at the point
(3 1) and moves counterclockwise three-fourths of the way around the ellipse to (2 4).
SECTION 10.1 2
21.
= 5 sin ,
= 2 cos
⇒
sin =
5
, cos =
2
.
sin
CURVES DEFINED BY PARAMETRIC EQUATIONS 2
2
+ cos
= 1
⇒
867
2
+
5
¤
= 1. The motion of the
2
particle takes place on an ellipse centered at (0 0). As goes from − to 5 , the particle starts at the point (0 −2) and moves clockwise around the ellipse 3 times. 22.
= cos2 = 1 − sin2
= 1−
2
. The motion of the particle takes place on the parabola
= 1−
2
. As goes from −2 to
− , the particle starts at the point (0 1), moves to (1 0), and goes back to (0 1). As goes from − to 0, the particle moves to (−1 0) and goes back to (0 1). The particle repeats this motion as goes from 0 to 2 . 23. We must have 1 ≤
≤ 4 and 2 ≤
≤ 3. So the graph of the curve must be contained in the rectangle [1 4] by [2 3].
24. (a) From the first graph, we have 1 ≤
≤ 2. From the second graph, we have −1 ≤
≤ 1 The only choice that satisfies
either of those conditions is III. (b) From the first graph, the values of of
cycle through the values from −2 to 2 six times. Choice I satisfies these conditions.
(c) From the first graph, the values of 0≤
cycle through the values from −2 to 2 four times. From the second graph, the values
cycle through the values from −2 to 2 three times. From the second graph, we have
≤ 2. Choice IV satisfies these conditions.
(d) From the first graph, the values of
cycle through the values from −2 to 2 two times. From the second graph, the values of
do the same thing. Choice II satisfies these conditions. 25. When = −1, (
) = (1 1). As increases to 0,
As increases from 0 to 1,
increases from 0 to 1 and
−1. As increases beyond 1, decrease. For
−1,
and
and
both decrease to 0.
decreases from 0 to
continues to increase and
continues to
are both positive and decreasing. We could
achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points. 26. When = −1, (
while
) = (0 0). As increases to 0,
increases from 0 to 1,
first decreases to −1 and then increases to 0. As increases from 0 to 1,
decreases from 1 to 0, while
y 1
first increases to 1 and then decreases to 0. We
could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
0
t=_1, 1 (0, 0)
t=0 (1, 0) 1
x
_1
27. When = −1, (
) = (0 1). As increases to 0,
increases from 0 to 1 and
decreases from 1 to 0. As increases from 0 to 1, the curve is retraced in the opposite direction with
decreasing from 1 to 0 and
increasing from 0 to 1.
We could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
868
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
28. (a)
=
4
− +1 =(
+ 1) −
4
0 [think of the graphs of
=
4
+ 1 and
= ] and
=
2
≥ 0, so these equations
are matched with graph V. √
≥ 0.
2
(b)
=
=
(c)
= sin 2 has period 2
− 2 = ( − 2) is negative for 0 2=
2, so these equations are matched with graph I.
. Note that
( + 2 ) = sin[ + 2 + sin 2( + 2 )] = sin( + 2 + sin 2 ) = sin( + sin 2 ) = ( ), so cycles through the values −1 to 1 twice as
These equations match graph II since (d)
= cos 5 has period 2
5 and
= sin 2 has period , so
takes on the values −1 to 1. Note that when = 0, ( (e)
= + sin 4 ,
=
2
+ cos 3 .
, so the graph will look like the graph of (f )
=
29. Use
30. Use
sin 2 , 4+ 2
= and
1
= ,
1
=
cos 2 . 4+ 2
= − 2 sin
=
3
As → ∞,
= and
2
=
3
−4,
2
will take on the values −1 to 1, and then 1 to −1, before
2
become the dominant terms in the expressions for
and
, but with oscillations. These equations are matched with graph IV. both approach 0. These equations are matched with graph III.
with a -interval of [−
− 4 and
cycles through those values once.
) = (1 0). These equations are matched with graph VI
As becomes large, and 2
has period 2 .
].
= with a -interval of
[−3 3]. There are 9 points of intersection; (0 0) is fairly obvious. The point in quadrant I is approximately (2 2 2 2), and by symmetry, the point in quadrant III is approximately (−2 2 −2 2). The other six points are approximately (∓1 9 ±0 5), (∓1 7 ±1 7), and (∓0 5 ±1 9).
31. (a)
=
1
through
+(
2
2( 2
−
1) 2)
,
=
1
+(
2
−
when = 1. For 0
1)
,0≤ 1,
≤ 1. Clearly the curve passes through is strictly between 2
−
1
1
and
2
and
1( 1
1)
when = 0 and
is strictly between
1
and
2.
For
every value of , 1( 1
1)
and
and 2( 2
Finally, any point (
satisfy the relation
−
1
= 2
−
( −
1 ),
which is the equation of the line through
1
2 ).
) on that line satisfies
− 2 −
1 1
=
− 2 −
1 1
; if we call that common value , then the given
parametric equations yield the point (
); and any (
) on the line between
1( 1
in [0 1]. So the given parametric equations exactly specify the line segment from (b)
= −2 + [3 − (−2)] = −2 + 5 and
= 7 + (−1 − 7) = 7 − 8 for 0 ≤
SECTION 10.1
32. For the side of the triangle from
to
, use (
1)
1
= (1 1) and (
and
1( 1
1)
2( 2
to
2)
2( 2
yields a value of 2 ).
≤ 1.
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS 2)
2
1)
869
= (4 2).
Hence, the equations are = = Graphing
1
+( +(
to
− − 2 2
= 1 + (4 − 1) = 1 + 3 , ) = 1 + (2 − 1) = 1 + . 1 1)
= 1 + with 0 ≤
= 1 + 3 and
triangle from and
1
≤ 1 gives us the side of the
. Similarly, for the side
we use
= 4 − 3 and
= 2 + 3 , and for the side
we use
=1
= 1 +4 .
33. The circle
2
+ ( − 1)2 = 4 has center (0 1) and radius 2, so by Example 4 it can be represented by
= 1 + 2 sin , 0 ≤
≤ 2 . This representation gives us the circle with a counterclockwise orientation starting at (2 1).
(a) To get a clockwise orientation, we could change the equations to
= 1 − 2 sin , 0 ≤
= 2 cos ,
(b) To get three times around in the counterclockwise direction, we use the original equations the domain expanded to 0 ≤
≤2 .
= 2 cos ,
= 1 + 2 sin with
≤6 .
(c) To start at (0 3) using the original equations, we must have = 2 cos ,
= 2 cos ,
= 1 + 2 sin ,
2
≤
≤
2
1
= 0; that is, 2 cos = 0. Hence, =
2
. So we use
.
Alternatively, if we want to start at 0, we could change the equations of the curve. For example, we could use = −2 sin , 34. (a) Let
2
2
= 1 + 2 cos , 0 ≤
= sin2
and
= cos with 0 ≤ 2
2
+
2
2
2
2
≤
.
= cos2 to obtain
= sin and
≤ 2 as possible parametric equations for the ellipse
= 1.
(b) The equations are
= 3 sin and
= cos for ∈ {1 2 4 8}.
(c) As increases, the ellipse stretches vertically.
35. Big circle: It’s centered at (2 2) with a radius of 2, so by Example 4, parametric equations are
= 2 + 2 cos
= 2 + 2 sin
0≤
≤2
Small circles: They are centered at (1 3) and (3 3) with a radius of 0 1. By Example 4, parametric equations are
and
(left)
= 1 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
(right)
= 3 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
Semicircle: It’s the lower half of a circle centered at (2 2) with radius 1. By Example 4, parametric equations are = 2 + 1 cos
= 2 + 1 sin
≤
≤2
To get all four graphs on the same screen with a typical graphing calculator, we need to change the last -interval to[0 2 ] in order to match the others. We can do this by changing to 0 5 . This change gives us the upper half. There are several ways to get the lower half—one is to change the “+” to a “−” in the -assignment, giving us
= 2 + 1 cos(0 5 )
= 2 − 1 sin(0 5 )
0≤
≤2
870
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
36. If you are using a calculator or computer that can overlay graphs (using multiple -intervals), the following is appropriate.
Left side:
= 1 and
goes from 1 5 to 4, so use =
15≤
≤4
=
15≤
≤4
=15
1≤
≤ 10
=1 Right side:
= 10 and
goes from 1 5 to 4, so use = 10
Bottom:
goes from 1 to 10 and
= 1 5, so use =
Handle: It starts at (10 4) and ends at (13 7), so use = 10 +
0≤
= 4+
≤3
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
= 1 + 1 sin
6
≤
≤
13 6
6
≤
≤
13 6
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos
= 1 + 1 sin
If you are using a calculator or computer that cannot overlay graphs (using one -interval), the following is appropriate. We’ll start by picking the -interval [0 2 5] since it easily matches the -values for the two sides. We now need to find parametric equations for all graphs with 0 ≤ Left side:
= 1 and
≤ 2 5.
goes from 1 5 to 4, so use =1
Right side:
= 10 and
0≤
≤25
=15+
0≤
≤25
goes from 1 5 to 4, so use = 10
Bottom:
=15+
goes from 1 to 10 and
= 1 5, so use = 1 +36
0≤
=15
≤25
To get the x-assignment, think of creating a linear function such that when = 0, = 10. We can use the point-slope form of a line with ( −1 =
10 − 1 ( − 0) 25−0
⇒
1
1)
= (0 1) and (
2
= 1 and when = 2 5, 2)
= (2 5 10).
= 1 +36 .
Handle: It starts at (10 4) and ends at (13 7), so use = 10 + 1 2 (
1
1)
= (0 10) and (
2
2)
= (2 5 13) gives us
0≤
= 4 +1 2 − 10 =
≤ 25
13 − 10 ( − 0) 25−0
⇒
= 10 + 1 2 .
7− 4 (
1
1 ) = (0 4) and (
2
2 ) = (2 5 7) gives us − 4 =
25 −0
( − 0)
⇒
= 4 +12 .
SECTION 10.1
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
871
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
(
1)
1
= 0
and (
2)
2
+
15
gives us −
5
=
6
= 1 + 1 sin
6
2
5 6
=
+
15
−
13 6 5
6 2
5 6
5 6
( − 0)
0≤
≤ 25
⇒
=
−0
+ 6
. 15
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos 37. (a)
=
3
⇒
1 3
=
, so
We get the entire curve
=
+
15
2 3
= 1 + 1 sin
6
=
2
2 3
traversed in a left to
=
.
(b)
right direction.
(c)
= = If
−3 −2
=( =(
0, then
−
)3
−
)2 = (
and
−
[so
=
1 3 2
) =
1 3
=
2 3
0 and
+
= , so
=
−2
=
−2
= cos ,
= sec2
=
cos2
=
1
2.
1 6
≤ 25 , so
=
curve
=
2 3
. 0, then
and
0, the curve never quite
=1
2
traversed in a
Since sec ≥ 1, we only get the
4
=
4 6
=
2 3
≥ 0, we only get the right half of the
6
.
AL
. We get the entire curve
1
=
=
left-to-right direction.
(b)
6
⇒
6
reaches the origin. 38. (a)
0≤
5
Since
],
are both larger than 1. If
are between 0 and 1. Since
15
.
2
≥ 1. We get the first quadrant portion of
parts of the curve
=1
with
the curve when
0, that is, cos
portion of the curve when
(c)
=
,
=
−2
0, and we get the second quadrant
0, that is, cos
= ( )−2 =
−2
0.
. Since
only get the first quadrant portion of the curve
¤
872
are both positive, we 2
=1
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
39. The case
and
is illustrated.
2
has coordinates (
has coordinates (
+ cos( − )) = (
[since cos( − ) = cos cos
+ sin
) as in Example 7, (1 − cos ))
= − cos
sin
, so
has
− sin( − ) (1 − cos )) = ( ( − sin ) (1 − cos ))
coordinates (
[since sin( − ) = sin parametric equations
− cos sin
cos
= ( − sin ),
= sin
+ sin( − )
−
2,
. As in Example 7,
+ cos( − )) = (
diagram) has coordinates (
. Again we have the
= (1 − cos ).
40. The first two diagrams depict the case
(
−2
and
cos ). That i
−
has coordinates (
cos ), so a typical point
has coordinates (
), where
−
=
). Now
(in the second
of the trochoid has coordinates
sin
and
=
−
cos . When
= , these equations agree with those of the cycloid.
41. It is apparent that
= =
= cos and cos and
sin2
=
=(
2
+ cos2
has coordinates ( cos
42.
( sec
0). It follows tha
= (2 cot
43.
Then ∠
=
=
= sin . Thus, the parametric equations are
⇒
= = 1 =
2
. From the diagram,
we rearrange: sin cos2
2
+
=(
2
sin ). Since
2
is a right angle and ∠
2
) . Adding the two
. Thus, we have an ellipse.
has coordinates ( sec is
is a right triangle and
sin ). Thus, the parametric equations are
= 2 cot . Le
= , so
,∆
= 2 sin
has coordinates
= sec ,
= sin .
= (0 2 ). and
(2 sin ) sin ). Thus, the -coordinate of
= 2 sin2 .
44. (a) Let
AL
⇒
=
is perpendicular to
2 ), so the -coordinate of
= ((2 sin ) cos is
=
= sin . To eliminate
) and cos
equations: sin2
and
be the angle of inclination of segment
cos . Then
=
2
.
Le
= (2 0). Then by use of right triangle
we see that
= 2 cos . (b)
Now
=
=2
= 1 cos
− − cos
=2
1 − cos2 cos
=2
sin2 cos
= 2 sin
tan
So
has coordinates
= 2 sin
· cos = 2 sin2
tan
and
= 2 sin
tan
SECTION 10.1
45. (a)
· sin
= 2 sin2
tan .
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
873
There are 2 points of intersection: (−3 0) and approximately (−2 1 1 4).
(b) A collision point occurs when
=
1
2
and
1
=
2
for the same . So solve the equations:
3 sin = −3 + cos
(1)
2 cos = 1 + sin
(2)
From (2), sin = 2 cos − 1. Substituting into (1), we get 3(2 cos − 1) = −3 + cos cos = 0
⇒
=
occurs when =
2
3 2
or
2
. We check that =
satisfies (1) and (2) but =
2
2
⇒
5 cos = 0 ( )
does not. So the only collision point
, and this gives the point (−3 0). [We could check our work by graphing
functions of and, on another plot, 3 2
pairs of graphs intersect is =
1
and
2
⇒
1
and
2
together as
as functions of . If we do so, we see that the only value of for which both
.]
(c) The circle is centered at (3 1) instead of (−3 1). There are still 2 intersection points: (3 0) and (2 1 1 4), but there are ⇒
no collision points, since ( ) in part (b) becomes 5 cos = 6 = 30◦ and
46. (a) If
0
=
1 2 (9
8)
2
= 250 − 4 9 2 . √
250
49
≈ 51 s. Then
The formula for
= 250
2
with equality when =
= 0 when = 0 (when the gun is fired) and again when
3
49
≈ 22,092 m, so the bullet hits the ground about 22 km from the gun.
−
= −4 9
250
2
−
125
250
+
49
125 2
cos )
⇒
1252
= −4 9
49
s, so the maximum height attained is
125 2
−
125 2
+
49
125 2
≤
49
1252 49
≈ 3189 m.
49
As
OT 0
+
49
49
=(
√ 3 and
250
49
(c)
1.
is quadratic in . To find the maximum -value, we will complete the square:
= −4 9
(b)
6 5
= (500 cos 30◦ ) = 250
= 500 m s, then the equations become
= (500 sin 30◦ ) −
cos =
= 0
cos
(0◦
90◦ ) increases up to 45◦ , the projectile attains a
greater height and a greater range. As
increases past 45◦ , the
projectile attains a greater height, but its range decreases.
. 2
=(
0
sin ) −
1 2
2
⇒
=(
0
sin
) 0
cos
−
2
0
cos
= (tan ) −
2
2
0 2
cos2
,
which is the equation of a parabola (quadratic in ).
874
¤
47.
=
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 2
=
3
−
. We use a graphing device to produce the graphs for various values of with − ≤
the members of the family are symmetric about the -axis. For cusp at (0 0) and for
48.
=2
−4
3
=−
0 the graph crosses itself at
2
≤
. Note that all
0, the graph does not cross itself, but for = 0 it has a
= , so the loop grows larger as increases.
+ 3 4 . We use a graphing device to produce the graphs for various values of with − ≤
Note that all the members of the family are symmetric about the -axis. When of even degree, but when = 0 there is a corner at the origin, and when
= + cos
0, the graph crosses itself at the origin, and has
0. From the first figure, we see that
= + sin
curves roughly follow the line
= , and they start having loops when
is between 1 4 and 1 6. The loops increase in size as
increases.
While not required, the following is a solution to determine the exact values for which the curve has a loop, that is, we seek the values of ( + cos
for which there exist parameter values and
+ sin ) = ( + cos
such that
and
+ sin ). In the diagram at the left, and
denotes the point (
the point ( + cos
Since
=
angles,
=∠
=2 − + =
2
4
−
(1).
=
4
=∠ −
)
+ sin ) = ( + cos
= , the triangle and
.
0, the graph resembles that of a polynomial
two cusps below the -axis. The size of the “swallowtail” increases as increases.
49.
≤
the point (
),
+ sin ).
is isosceles. Therefore its base are equal. Since
the relation
=
= −
implies that
4
and
SECTION 10.1
Since cos
= distance(( =
− =
1 2
=
( − )
√ 2 cos
−
) ( √ 2
2( − )2 =
− =
√ 2 cos , that is,
(2). Now cos − 4 = sin 2 − − 4 = sin 4 − , √ − = 2 sin 34 − (20 ). Subtracting (20 ) from (1) and
4
so we can rewrite (2) as
−
dividing by 2, we obtain = 4
Since
√ 2
−
sin
2
implicitly assumed that 0 by + 2 merely increases √ 3 2 4 − (3), we get = sin
4
−
, or
3
− =
4
4
−
−
sin √ 2
4
it follows from (20 ) that sin
0 and
875
√ 2 ( − ), we see that
)) =
, so
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS
(3).
4
0. Thus from (3) we see that
4.
[We have
by the way we drew our diagram, but we lost no generality by doing so since replacing and
by 2 . The curve’s basic shape repeats every time we change by 2 .] Solving for
. Write
=
3
4
. Then −
=
√ 2 sin
, where
0. Now sin
for
0, so
√ 2.
in
→ 0+ , that is, as →
As
50. Consider the curves
− 4
= sin + sin
→
,
,
radius 2 centered at the origin. For
√ 2 .
= cos + cos
, where
is a positive integer. For
= 1, we get a circle of − 1 loops as
1, we get a curve lying on or inside that circle that traces out
ranges from 0 to 2 . 2
Note:
+
2
= (sin + sin
)2 + (cos + cos
= sin2 + 2 sin sin
+ sin2
= (sin2 + cos2 ) + (sin2 = 1 + 1 + 2 cos( − with equality for
)2 + cos2 + 2 cos cos
+ cos2
+ cos2
) + 2(cos cos
+ sin sin
)
) = 2 + 2 cos((1 − ) ) ≤ 4 = 22 ,
= 1. This shows that each curve lies on or inside the curve for
= 1, which is a circle of radius 2 centered
at the origin.
NOT = 1
= 2
= 3
51. Note that all the Lissajous figures are symmetric about the -axis. The parameters
- and -directions respectively. For
876
¤
= =
and simply stretch the graph in the
= 1 the graph is simply a circle with radius 1. For
= 2 the graph crosses
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
itself at the origin and there are loops above and below the -axis. In general, the figures have all of which are on the -axis, and a total of
= =1 52.
=5
= cos ,
= sin − sin
=2 .
If = 1, then
graphs are shown for = 2 3 4 and 5.
− 1 points of intersection,
closed loops.
=3
= 0, and the curve is simply the line segment from (−1 0) to (1 0). The
F R
It is easy to see that all the curves lie in the rectangle [−1 1] by [−2 2]. When is an integer, ( + 2 ) = ( ) and ( + 2 ) = ( ), so the curve is closed. When is a positive integer greater than 1, the curve intersects the x-axis + 1 times and has loops (one of which degenerates to a tangency at the origin when is an odd integer of the form 4 + 1). √ As increases, the curve’s loops become thinner, but stay in the region bounded by the semicircles = ± 1 + 1 − and the line segments from (−1 −1) to (−1 1) and from (1 −1) to (1 1). This is true because
2
= sin − sin
≤ sin + sin ≤
√ 1−
2
+ 1. This curve appears to fill the entire region when is very large, as
shown in the figure for = 1000.
LABORATORY PROJECT
RUNNING CIRCLES AROUND CIRCLES
¤
When is a fraction, we get a variety of shapes with multiple loops, but always within the same region. For some fractional values, such as = 2 359, the curve again appears to fill the region.
S LE
LABORATORY PROJECT Running Circles Around Circles 1. The cente
Arc
of the smaller circle has coordinates (( − )cos
on circ
has length
( − )sin ).
since it is equal in length to arc
(the smaller circle rolls without slipping against the larger.) Thus, ∠
and 2. With
=
and ∠
− , so
=
has coordinates
= ( − )cos + cos(∠
) = ( − )cos + cos
= ( − )sin − sin(∠
) = ( − )sin − sin
= 1 and
− −
a positive integer greater than 2, we obtain a hypocycloid of
cusps. Shown in the figure is the graph for
= 4. Let
= 4 and = 1. Using the
sum identities to expand cos 3 and sin 3 , we obtain = 3 cos + cos 3 = 3 cos + 4 cos3 and
.
− 3 cos
= 3 sin − sin 3 = 3 sin − 3 sin − 4 sin3
= 4 cos3 = 4 sin3 .
877
3. The graphs at the right are obtained with 1
1
1
1
2
3
4
10
= , , , and
with −2 ≤
= 1 and
≤ 2 . We
conclude that as the denominator increases, the graph gets smaller, but maintains the basic shape shown.
[continued]
¤
878
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Letting
So if
= 2 and
= 3, 5, and 7 with −2 ≤
is held constant and
obtain a hypocycloid of graphs have
4. If
varies, we get a graph with
cusps. As
= 32 , 54 , and
≤ 2 gives us the following:
cusps (assuming
=
+ 1, we
increases, we must expand the range of in order to get a closed curve. The following
11 10 .
S
= 1, the equations for the hypocycloid are = ( − 1) cos + cos (( − 1) )
which is a hypocycloid of
is in lowest form). When
= ( − 1) sin − sin (( − 1) )
cusps (from Problem 2). In general, if
1, we get a figure with cusps on the “outside ring” and if
1, the cusps are on the “inside ring”. In any case, as the values of get larger, we get a figure that looks more and more like a washer. If we were to graph the hypocycloid for all values of , every point on the washer would eventually be arbitrarily close to a point on the curve.
= 5. The cente
Arc and ∠
√ 2,
−10 ≤
= − 2,
≤ 10
of the smaller circle has coordinates (( + ) cos
has length =
−
(as in Problem 1), so that ∠ −
Thus, the coordinates of
=
since ∠
+
−
=
( + ) sin ).
,∠
=
−
,
= .
are
= ( + ) cos + cos
−
+
= ( + ) cos − cos
+ and
0≤
= ( + ) sin − sin
−
+
+ = ( + ) sin − sin
.
≤ 446
LABORATORY PROJECT
6. Let
= 1 and the equations become = ( + 1) cos − cos(( + 1) )
If
RUNNING CIRCLES AROUND CIRCLES
= 1, we have a cardioid. If
= ( + 1) sin − sin(( + 1) )
is a positive
integer greater than 1, we get the graph of an “ -leafed clover”, with cusps that are
units
from the origin. (Some of the pairs of figures are not to scale.) = 3, −2 ≤ If
=
with
≤2
= 10, −2 ≤
≤2
= 1, we obtain a figure that
does not increase in size and requires −
≤
≤
to be a closed curve traced
exactly once.
S LE
= 1 , −4 ≤
R
≤4
= 1 , −7 ≤
≤ 7
¤
879
4
Next, we keep
constant and let
7
vary. As
increases, so does the size of the figure. There is an -pointed star in the middle.
Now if
= 25 , −5 ≤
≤5
= 75 , −5 ≤
≤ 5
= 43 , −3 ≤
≤3
= 76 , −6 ≤
≤ 6
√ 2, 0 ≤
≤ 200
= − 2, 0 ≤
+ 1 we obtain figures similar to the
=
previous ones, but the size of the figure does not increase.
If
is irrational, we get washers that increase in
size as
increases.
=
¤
880
10.2 1.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Calculus with Parametric Curves ,
=
=
√ 1+
1+
⇒
=
=
3.
=
,
3
+ 1,
=
4
+ ;
(1 + )−1
2
= √
1
,
=
(1 + )(1) − (1) (1 + )2
2 1+
, and
(1 + )2
(1 + ) 1 3 2 = √ = (1 + ) . 2 2 1+
⇒
= 1 + cos ,
= −1.
=4
3
=
+
=
( + 1), and
= 3 2 , and
+ 1,
=
=
=
4
=
3
3
4.
1
=
2
1 (2 1 + )
= + sin
1 2
1 (1 + )2
2.
=
√
=
and
≤ 446
+1 2
1 + cos . ( + 1)
hen = −1, ( .W
) = (0 0)
= −3 3 = −1, so an equation of the tangent to the curve at the point corresponding to = −1 is
− 0 = −1( − 0), or √ = , = 2−2;
=− . = 4.
= 2 − 2,
1
= 2
√
, and
=
√ √ = (2 − 2)2 = 4( − 1) . When = 4,
(
) = (2 8) and
= 4(3)(2) = 24, so an equation of the tangent to the curve at the point corresponding to = 4 is
− 8 = 24( − 2), or 5.
= cos ,
= sin ;
When = , (
=
sin
= .
,
2
=
=−
is
−0 =
− (− )], or
;
= 0. 2
=
= cos + sin ,
0) and
) = (−
corresponding to = 6.
= 24 − 40.
=2
2
,
=
cos + sin . − sin + cos
=
2
+
( cos
.
) + (sin
)
=
( cos
+ sin
), and
2
= ( cos
+ sin
)
=
cos
. When = 0, (
+ sin
of the tangent to the curve at the point corresponding to = 0 is − 1 = 7. (a)
=
(−1) = , so an equation of the tangent to the curve at the point
=
2
= (− sin ) + cos , and
= 1 + ln ,
=
2
+ 2; (1 3).
=2
=
1
and
2
) = (0 1) and
( − 0), or
=
=
2
=2
2
=
, so an equation
+ 1.
= 2 2 . At (1 3),
1 = 1 + ln = 1 or (b)
ln = 0
⇒
= 1 and
−1
⇒
=
⇒
= 1 + ln
= 1+
0
√
= ,
ln =
2(1)−2
=
2
−1
, so
= 1+ =4
√
= 2
⇒
=
2
−1 2
+2 =(
) +2 =
2 −2
· 2 = 2, so an equation of the tangent is − 3 = 2( − 1), or ; (2 ).
=
2
2, ·
or
= 2, so an equation of the tangent is
− 3 = 2( − 1),
= 2 + 1.
At (1 3), 8. (a)
⇒
√
= 1
⇒
= 1 and
= 2
1 , and √
=
=
+ 2, and
0
=
2 −2
· 2.
= 2 + 1. 2
1 2
2
√
= 4 , so an equation of the tangent is
=4
−
3 2
2
. At (2 ),
= 4 ( − 2),
−7 .
Multivariable Calculus 8th Edition Stewart Solutions Manual Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-solutions-manual/ Multivariable Calculus 8th Edition Stewart Test Bank Full clear download (no famatting errors) at: https://testbanklive.com/download/multivariable-calculus-8th-edition-stewart-test-bank/ °
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1.http://www.elsolucionario.blogspot.comLIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.2. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 200617:25 3. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006The first person to invent a car that runs on water may be sitting right in your classroom! Every one of your students has the potential to make a difference. And realizing that potential starts right here, in your course. When students succeed in your coursewhen they stay on-task and make the breakthrough that turns confusion into confidencethey are empowered to realize the possibilities for greatness that lie within each of them. We know your goal is to create an environment where students reach their full potential and experience the exhilaration of academic success that will last them a lifetime. WileyPLUS can help you reach that goal.WileyPLUS is an online suite of resourcesincluding the complete textthat will help your students: come to class better prepared for your lectures get immediate feedback and context-sensitive help on assignments and quizzes track their progress throughout the courseI just wanted to say how much this program helped me in studying I was able to actually see my mistakes and correct them. I really think that other students should have the chance to use WileyPLUS. Ashlee Krisko, Oakland Universitywww.wiley.com/college/wileyplus80%of students surveyed said it improved their understanding of the material.*i17:25 4. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006FOR INSTRUCTORS WileyPLUS is built around the activities you perform in your class each day. With WileyPLUS you can: Prepare & PresentCreate AssignmentsTrack Student ProgressCreate outstanding class presentations using a wealth of resources such as PowerPoint slides, image galleries, interactive simulations, and more. You can even add materials you have created yourself.Automate the assigning and grading of homework or quizzes by using the provided question banks, or by writing your own.Keep track of your students' progress and analyze individual and overall class results.Now Available with WebCT and Blackboard!It has been a great help, and I believe it has helped me to achieve a better grade. Michael Morris, Columbia Basin CollegeFOR STUDENTS You have the potential to make a difference! WileyPLUS is a powerful online system packed with features to help you make the most of your potential and get the best grade you can!With WileyPLUS you get: A complete online version of your text and other study resources. Problem-solving help, instant grading, and feedback on your homework and quizzes. The ability to track your progress and grades throughout the term.For more information on what WileyPLUS can do to help you and your students reach their potential, please visit www.wiley.com/college/wileyplus.76%of students surveyed said it made them better prepared for tests. **Based on a survey of 972 student users of WileyPLUSii17:25 5. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006iii17:25 6. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006iv17:25 7. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006TENTH EDITIONSALAS HILLE ETGENCALCULUS ONE AND SEVERAL VARIABLESJOHN WILEY & SONS, INC.v17:25 8. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006In fond remembrance of Einar HilleACQUISITIONS EDITOR PUBLISHER MARKETING MANAGER EDITORIAL ASSISTANT MARKETING ASSISTANT SENIOR PRODUCTION EDITOR DESIGNER SENIOR MEDIA EDITOR PRODUCTION MANAGEMENT FREELANCE DEVELOPMENTAL EDITOR COVER IMAGEMary Kittell Laurie Rosatone Amy Sell Danielle Amico Tara Martinho Sandra Dumas Hope Miller Stefanie Liebman Suzanne Ingrao Anne Scanlan-Rohrer Steven Puetzer/MasterleThis book was set in New Times Roman by Techbooks, Inc. and printed and bound by Courier(Westford). The cover was printed by Courier(Westford). This book is printed on acid-free paper.Copyright 2007 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201) 748-6008, website http://www.wiley.com/go/permissions. To order books or for customer service please, call 1-800-CALL WILEY (225-5945). Library of Congress Cataloging-in-Publication Data Salas, Saturnino L. Calculus10th ed/Saturnino Salas, Einar Hille, Garrett Etgen. ISBN-13 ISBN-10978-0471-69804-3 0-471-69804-0Printed in the United States of America 10 9 8 7 6 5 4 3 2 1vi17:25 9. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006PREFACEThis text is devoted to the study of single and multivariable calculus. While applications from the sciences, engineering, and economics are often used to motivate or illustrate mathematical ideas, the emphasis is on the three basic concepts of calculus: limit, derivative, and integral. This edition is the result of a collaborative effort with S.L. Salas, who scrutinized every sentence for possible improvement in precision and readability. His gift for writing and his uncompromising standards of mathematical accuracy and clarity illuminate the beauty of the subject while increasing its accessibility to students. It has been a pleasure for me to work with him.FEATURES OF THE TENTH EDITION Precision and Clarity The emphasis is on mathematical exposition; the topics are treated in a clear and understandable manner. Mathematical statements are careful and precise; the basic concepts and important points are not obscured by excess verbiage.Balance of Theory and Applications Problems drawn from the physical sciences are often used to introduce basic concepts in calculus. In turn, the concepts and methods of calculus are applied to a variety of problems in the sciences, engineering, business, and the social sciences through text examples and exercises. Because the presentation is exible, instructors can vary the balance of theory and applications according to the needs of their students.Accessibility This text is designed to be completely accessible to the beginning calculus student without sacricing appropriate mathematical rigor. The important theorems are explainedvii17:25 10. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13viii QC: PBU/OVYT1: PBU October 24, 2006PREFACEand proved, and the mathematical techniques are justied. These may be covered or omitted according to the theoretical level desired in the course.Visualization The importance of visualization cannot be over-emphasized in developing students understanding of mathematical concepts. For that reason, over 1200 illustrations accompany the text examples and exercise sets.Technology The technology component of the text has been strengthened by revising existing exercises and by developing new exercises. Well over half of the exercise sets have problems requiring either a graphing utility or a computer algebra system (CAS). Technology exercises are designed to illustrate or expand upon the material developed within the sections.Projects Projects with an emphasis on problem solving offer students the opportunity to investigate a variety of special topics that supplement the text material. The projects typically require an approach that involves both theory and applications, including the use of technology. Many of the projects are suitable for group-learning activities.Early Coverage of Differential Equations Differential equations are formally introduced in Chapter 7 in connection with applications to exponential growth and decay. First-order linear equations, separable equations, and second linear equations with constant coefcients, plus a variety of applications, are treated in a separate chapter immediately following the techniques of integration material in Chapter 8.CHANGES IN CONTENT AND ORGANIZATION In our effort to produce an even more effective text, we consulted with the users of the Ninth Edition and with other calculus instructors. Our primary goals in preparing the Tenth Edition were the following: 1. Improve the exposition. As noted above, every topic has been examined for possible improvement in the clarity and accuracy of its presentation. Essentially every section in the text underwent some revision; a number of sections and subsections were completely rewritten. 2. Improve the illustrative examples. Many of the existing examples have been modied to enhance students understanding of the material. New examples have been added to sections that were rewritten or substantially revised. 3. Revise the exercise sets. Every exercise set was examined for balance between drill problems, midlevel problems, and more challenging applications and conceptual problems. In many instances, the number of routine problems was reduced and new midlevel to challenging problems were added. Specic changes made to achieve these goals and meet the needs of todays students and instructors include:17:25 11. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006PREFACEComprehensive Chapter-End Review Exercise Sets The Skill Mastery Review Exercise Sets introduced in the Ninth Edition have been expanded into chapter-end exercise sets. Each chapter concludes with a comprehensive set of problems designed to test and to re-enforce students understanding of basic concepts and methods developed within the chapter. These review exercise sets average over 50 problems per set.Precalculus Review (Chapter 1) The content of this chapterinequalities, basic analytic geometry, the function concept and the elementary functionsis unchanged. However, much of the material has been rewritten and simplied.Limits (Chapter 2) The approach to limits is unchanged, but many of the explanations have been revised. The illustrative examples throughout the chapter have been modied, and new examples have been added.Differentiation and Applications (Chapters 3 and 4) There are some signicant changes in the organization of this material. Realizing that our treatments of linear motion, rates of change per unit time, and the Newton-Raphson method depended on an understanding of increasing/decreasing functions and the concavity of graphs, we moved these topics from Chapter 3 (the derivative) to Chapter 4 (applications of the derivative). Thus, Chapter 3 is now a shorter chapter which focuses solely on the derivative and the processes of differentiation, and Chapter 4 is expanded to encompass all of the standard applications of the derivativecurve-sketching, optimization, linear motion, rates of change, and approximation. As in all previous editions, Chapter 4 begins with the mean-value theorem as the theoretical basis for all the applications.Integration and Applications (Chapters 5 and 6) In a brief introductory section, area and distance are used to motivate the denite integral in Chapter 5. While the denition of the denite integral is based on upper and lower sums, the connection with Riemann sums is also given. Explanations, examples, and exercises throughout Chapters 5 and 6 have been modied, but the content and organization remain as in the Ninth Edition.The Transcendental Functions, Techniques of Integration (Chapters 7 and 8) The coverage of the inverse trigonometric functions (Chapter 7) has been reduced slightly. The treatment of powers of the trigonometric functions (Chapter 8) has been completely rewritten. The optional sections on rst-order linear differential equations and separable differential equations have been moved to Chapter 9, the new chapter on differential equations.Some Differential Equations (Chapter 9) This new chapter is a brief introduction to differential equations and their applications. In addition to the coverage of rst-order linear equations and separable equations notedix17:25 12. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13x QC: PBU/OVYT1: PBU October 24, 2006PREFACEabove, we have moved the section on second-order linear homogeneous equations with constant coefcients from the Ninth Editions Chapter 18 to this chapter.Sequences and Series (Chapters 11 and 12) Efforts were made to reduce the overall length of these chapters through rewriting and eliminating peripheral material. Eliminating extraneous problems reduced several exercise sets. Some notations and terminology have been modied to be consistent with common usage.Vectors and Vector Calculus (Chapters 13 and 14) The introduction to vectors in three-dimensional space has been completely rewritten and reduced from two sections to one. The parallel discussion of vectors in twoand three-dimensional space has been eliminatedthe primary focus is on threedimensional space. The treatments of the dot product, the cross product, lines and planes in Chapter 13, and vector calculus in Chapter 14 are unchanged.Functions of Several Variables, Gradients, Extreme Values (Chapters 15 and 16); Multiple Integrals, Line and Surface Integrals (Chapters 16 and 17) The basic content and organization of the material in these four chapters remain as in the ninth edition. Improvements have been made in the exposition, examples, illustrations, and exercises.Differential Equations (Chapter 19) This chapter continues the study of differential equations begun in Chapter 9. The sections on Bernoulli, homogeneous and exact equations have been rewritten, and elementary numerical methods are now covered in a separate section. The section on second-order linear nonhomogeneous equations picks up from the treatment of linear homogeneous equations in the new Chapter 9. The applications sectionvibrating mechanical systemsis unchanged.SUPPLEMENTS An Instructors Solutions Manual, ISBN 0470127309, includes solutions for all problems in the text. A Student Solutions Manual, ISBN 0470105534, includes solutions for selected problems in the text. A Companion Web site, www.wiley.com/college/salas, provides a wealth of resources for students and instructors, including: PowerPoint Slides for important ideas and graphics for study and note taking. Online Review Quizzes to enable students to test their knowledge of key concepts. For further review diagnostic feedback is provided that refers to pertinent sections of the text. Animations comprise a series of interactive Java applets that allow students to explore the geometric signicance of many major concepts of Calculus. Algebra and Trigonometry Refreshers is a self-paced, guided review of key algebra and trigonometry topics that are essential for mastering calculus.17:25 13. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006PREFACE Personal Response System Questions provide a convenient source of questions to use with a variety of personal response systems. Printed Test Bank contains static tests which can be printed for quick tests. Computerized Test Bank includes questions from the printed test bank with algorithmically generated problems.WILEYPLUS Expect More from Your Classroom Technology This text is supported by WileyPLUSa powerful and highly integrated suite of teaching and learning resources designed to bridge the gap between what happens in the classroom and what happens at home. WileyPLUS includes a complete online version of the text, algorithmically generated exercises, all of the text supplements, plus course and homework management tools, in one easy-to-use website.Organized Around the Everyday Activities You Perform in Class, WileyPLUS Helps You: Prepare and present: WileyPLUS lets you create class presentations quickly and easily using a wealth of Wiley-provided resources, including an online version of the textbook, PowerPoint slides, and more. You can adapt this content to meet the needs of your course. Create assignments: WileyPLUS enables you to automate the process of assigning and grading homework or quizzes. You can use algorithmically generated problems from the texts accompanying test bank, or write your own. Track student progress: An instructors grade book allows you to analyze individual and overall class results to determine students progress and level of understanding. Promote strong problem-solving skills: WileyPLUS can link homework problems to the relevant section of the online text, providing students with context-sensitive help. WileyPLUS also features mastery problems that promote conceptual understanding of key topics and video walkthroughs of example problems. Provide numerous practice opportunities: Algorithmically generated problems provide unlimited self-practice opportunities for students, as well as problems for homework and testing. Support varied learning styles: WileyPLUS includes the entire text in digital format, enhanced with varied problem types to support the array of different student learning styles in todays classroom. Administer your course: You can easily integrate WileyPLUS with another course management system, grade books, or other resources you are using in your class, enabling you to build your course your way.WileyPLUS Includes A Wealth of Instructor and Student Resources: Student Solutions Manual: Includes worked-out solutions for all odd-numbered problems and study tips. Instructors Solutions Manual: Presents worked out solutions to all problems. PowerPoint Lecture Notes: In each section of the book a corresponding set of lecture notes and worked out examples are presented as PowerPoint slides that are tied to the examples in the text.xi17:25 14. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13xii QC: PBU/OVYT1: PBU October 24, 2006PREFACEView an online demo at www.wiley.com/college/wileyplus or contact your local Wiley representative for more details.The Wiley Faculty NetworkWhere Faculty Connect The Wiley Faculty Network is a faculty-to-faculty network promoting the effective use of technology to enrich the teaching experience. The Wiley Faculty Network facilitates the exchange of best practices, connects teachers with technology, and helps to enhance instructional efciency and effectiveness. The network provides technology training and tutorials, including WileyPLUS training, online seminars, peer-to-peer exchanges of experiences and ideas, personalized consulting, and sharing of resources.Connect with a Colleague Wiley Faculty Network mentors are faculty like you, from educational institutions around the country, who are passionate about enhancing instructional efciency and effectiveness through best practices. You can engage a faculty mentor in an online conversation at www.wherefacultyconnect.com.Connect with the Wiley Faculty Network Web: www.wherefacultyconnect.com Phone: 1-866-FACULTYACKNOWLEDGMENTS The revision of a text of this magnitude and stature requires a lot of encouragement and help. I was fortunate to have an ample supply of both from many sources. The present book owes much to the people who contributed to the rst nine editions, most recently: Omar Adawi, Parkland College; Mihaly Bakonyi, Georgia State University; Edward B. Curtis, University of Washington; Boris A. Datskovsky, Temple University; Kathy Davis University of Texas-Austin; Dennis DeTurck, University of Pennsylvania; John R. Durbin, University of Texas-Austin; Ronald Gentle, Eastern Washington University; Robert W. Ghrist, Georgia Institute of Technology; Charles H. Giffen, University of VirginiaCharlottesville; Michael Kinyon, Indiana University-South Bend; Susan J. Lamon, Marquette University; Peter A. Lappan, Michigan State University; Nicholas Macri, Temple University; James Martino, Johns Hopkins University; James R. McKinney, California State Polytechnic University-Pomona; Jeff Morgan, Texas A & M University; Peter J. Mucha, Georgia Institute of Technology; Elvira Munoz-Garcia, University of California, Los Angeles; Ralph W. Oberste-Vorth, University of South Florida; Charles Odion, Houston Community College; Charles Peters, University of Houston; Clifford S. Queen, Lehigh University; J. Terry Wilson, San Jacinto College Central and Yang Wang, Georgia Institute of Technology. I am deeply indebted to all of them. The reviewers and contributors to the Tenth Edition supplied detailed criticisms and valuable suggestions. I offer my sincere appreciation to the following individuals: Omar Adawi,Parkland CollegeUlrich Albrecht,Auburn UniversityJoseph Borzellino,California Polytechnic State University, San Luis ObispoMichael R. Colvin,The University of St. Thomas17:25 15. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006PREFACEJames Dare, Nasser Dastrange, David Dorman, Martin E. Flashman, David Frank, Melanie Fulton, Isobel Gaensler, Frieda Ganter, Murli M. Gupta, Aida Kadic-Galeb, Mohammad Ghomi, Robert W. Ghrist, Semion Gutman, Rahim G. Karimpour, Robert Keller, Kevin P. Knudson, Ashok Kumar, Jeff Leader, Xin Li, Doron Lubinsky, Edward T. Migliore, Maya Mukherjee, Sanjay Mundkur, Michael M. Neumann, Charles Odion, Dan Ostrov, Shahrokh Parvini, Chuang Peng, Kanishka Perera, Denise Reid, Paul Seeburger, Constance Schober, Peter Schumer, Kimberly Shockey Henry Smith, James Thomas, James L. Wang, Ying Wang, Charles Waters, Mark Woodard, Yan Wu, Dekang Xu,Indiana-Purdue University at Fort Wayne Buena Vista University Middlebury College Humboldt State University/Occidental College University of Minnesota High Point University Georgia State University West Hills College, Lemoore George Washington University University of Tampa Georgia Institute of Technology University of Illinois, Urbana-Champaign University of Oklahoma Southern Illinois University- Edwardsville Loras College Mississippi State University Valdosta State University Rose-Hulman Institute of Technology University of Central Florida Georgia Institute of Technology Monterey Peninsula College/University of California, Santa Cruz Morehouse College Kennesaw State University Mississippi State University Houston Community College Santa Clara University San Diego Mesa College Morehouse College Florida Institute of Technology Valdosta State University Monroe Community College University of Central Florida Middlebury College River Parishes Community College Colorado State University University of Alabama Augusta State University Minnesota State University, Mankato Furman University Georgia Southern University University of Houstonxiii17:25 16. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13xiv QC: PBU/OVYT1: PBU October 24, 2006PREFACEI am especially grateful to Paul Lorczak and Neil Wigley, who carefully read the revised material. They provided many corrections and helpful comments. I would like to thank Bill Ardis for his advice and guidance in the creation of the new technology exercises in the text. I am deeply indebted to the editorial staff at John Wiley & Sons. Everyone involved in this project has been encouraging, helpful, and thoroughly professional at every stage. In particular, Laurie Rosatone, Publisher, Mary Kittell, Acquisitions Editor, Anne Scanlan-Rohrer, Freelance Developmental Editor and Danielle Amico, Editorial Assistant, provided organization and support when I needed it and prodding when prodding was required. Special thanks go to Sandra Dumas, Production Editor, and Suzanne Ingrao, Production Coordinator, who were patient and understanding as they guided the project through the production stages; Hope Miller, Designer, whose creativity produced the attractive interior design as well as the cover and Stefanie Liebman, Senior Media Editor, who has produced valuable media resources to support my text. Finally, I want to acknowledge the contributions of my wife, Charlotte; without her continued support, I could not have completed this work. Garret J. Etgen17:25 17. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006CONTENTSCHAPTER 1 1.1PRECALCULUS REVIEW1GirthWhat is Calculus? 11.3Coordinate Plane; Analytic Geometry 171.5Functions 241.6LeReview of Inequalities 111.4thng1.2 Review of Elementary Mathematics 3The Elementary Functions 321.7 Combinations of Functions 41 1.8A Note on Mathematical Proof; Mathematical Induction 47CHAPTER 2LIMITS AND CONTINUITY2.1The Limit Process (An Intuitive Introduction) 532.2Some Limit Theorems 732.4Continuity 82yDenition of Limit 642.3532.5The Pinching Theorem; Trigonometric Limits 912.6fL+ L LTwo Basic Theorems 97 Project 2.6 The Bisection Method for Finding the Roots of f (x) = 0 102ccc+xxv17:25 18. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13xvi QC: PBU/OVYT1: PBU October 24, 2006CONTENTSCHAPTER 3THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 1053.1The Derivative 1053.2Some Differentiation Formulas 1153.3The d/dx Notation; Derivatives of Higher Order 1243.4The Derivative As A Rate of Change 1303.5The Chain Rule 1333.6Differentiating The Trigonometric Functions 1423.7Implicit Differentiation; Rational Powers 147CHAPTER 4THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES4.1The Mean-Value Theorem 1544.2Increasing and Decreasing Functions 1604.3Local Extreme Values 1674.4Endpoint Extreme Values; Absolute Extreme Values 1744.5enttangSome Max-Min Problems 182 Project 4.5 Flight Paths of Birds 190lcbConcavity and Points of Inection 190 Vertical and Horizontal Asymptotes; Vertical Tangents and Cusps 1954.8(a, f (a))f (b))4.6 4.7(b,a154Some Curve Sketching 2014.9Velocity and Acceleration; Speed 209 Project 4.9A Angular Velocity; Uniform Circular Motion 217 Project 4.9B Energy of a Falling Body (Near the Surface of the Earth) 2174.10Related Rates of Change per Unit Time 2184.11Differentials 223 Project 4.11 Marginal Cost, Marginal Revenue, Marginal Prot 2284.12Newton-Raphson Approximations 229CHAPTER 5INTEGRATION5.1An Area Problem; a Speed-Distance Problem 2345.2The Denite Integral of a Continuous Function 2375.3The Function f (x) =xf (t) dt 246 a5.4The Fundamental Theorem of Integral Calculus 2545.5Some Area Problems 2605.6Indenite Integrals 2685.7Working Back from the Chain Rule; the u-Substitution 2745.8Additional Properties of the Denite Integral 2815.9Mean-Value Theorems for Integrals; Average Value of a Function 285Project 5.5 Integrability; Integrating Discontinuous Functions 26623417:25 19. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006CHAPTER 6SOME APPLICATIONS OF THE INTEGRAL6.1Volume by Parallel Cross Sections; Disks and Washers 2966.3Volume by the Shell Method 3066.4y292More on Area 2926.2xviiCONTENTSThe Centroid of a Region; Pappuss Theorem on Volumes 312 Project 6.4 Centroid of a Solid of Revolution 3196.5 *6.6The Notion of Work 319g(y)Fluid Force 327xCHAPTER 7THE TRANSCENDENTAL FUNCTIONS7.1The Logarithm Function, Part I 3427.3The Logarithm Function, Part II 3477.4The Exponential Function 3567.5Arbitrary Powers; Other Bases 3647.6Exponential Growth and Decay 3707.7yOne-to-One Functions; Inverses 3337.2333The Inverse Trigonometric Functions 378ffProject 7.4 Some Rational Bounds for the Number e 36411 1 fx1 11Project 7.7 Refraction 387 7.8 The Hyperbolic Sine and Cosine 388 *7.9The Other Hyperbolic Functions 392 fCHAPTER 8 8.1TECHNIQUES OF INTEGRATION398Integral Tables and Review 3988.2 Integration by Parts 402 fProject 8.2 Sine Waves y = sin nx and Cosine Waves y = cos nx 410 8.3 Powers and Products of Trigonometric Functions 411 8.4 Integrals Featuring a 2 x 2 , a 2 + x 2 , x 2 a 2 417 8.5 *8.6Rational Functions; Partial Fractions 422 Some Rationalizing Substitutions 430 x8.7 Numerical Integration 433yCHAPTER 9SOME DIFFERENTIAL EQUATIONS4433 29.1First-Order Linear Equations 4449.2Integral Curves; Separable Equations 4511 422Project 9.2 Orthogonal Trajectories 458 9.31The Equation y + ay + by = 0 4592 3Denotes optional section.4x17:25 20. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13xviii QC: PBU/OVYT1: PBU October 24, 2006CONTENTSCHAPTER 10THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS10.1Geometry of Parabola, Ellipse, Hyperbola 46910.2Polar Coordinates 47810.3Sketching Curves in Polar Coordinates 48410.4Area in Polar Coordinates 49210.5y469Curves Given Parametrically 496Project 10.3 Parabola, Ellipse, Hyperbola in Polar Coordinates 491 xProject 10.5 Parabolic Trajectories 503 10.6Tangents to Curves Given Parametrically 50310.7Arc Length and Speed 50910.8The Area of A Surface of Revolution; The Centroid of a Curve; Pappuss Theorem on Surface Area 517 Project 10.8 The Cycloid 525CHAPTER 11SEQUENCES; INDETERMINATE FORMS; IMPROPER INTEGRALS528y11.1The Least Upper Bound Axiom 52811.2Sequences of Real Numbers 53211.3Limit of a Sequence 538 Project 11.3 Sequences and the Newton-Raphson Method 54711.4Some Important Limits 55011.5The Indeterminate Form (0/0) 55411.6The Indeterminate Form (/); Other Indeterminate Forms 56011.7xImproper Integrals 565CHAPTER 12INFINITE SERIES12.1Sigma Notation 57512.2Innite Series 57712.3The Integral Test; Basic Comparison, Limit Comparison 58512.4The Root Test; the Ratio Test 59312.5Absolute Convergence and Conditional Convergence; Alternating Series 597312.6Taylor Polynomials in x; Taylor Series in x 602212.7Taylor Polynomials and Taylor Series in x a 61312.8Power Series 61612.9Differentiation and Integration of Power Series 623y 6P3(x) f (x) = e xP2(x)5 P1( x)4P2132 P31 P1P0(x)123xProject 12.9A The Binomial Series 633 Project 12.9B Estimating 63457517:25 21. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006CONTENTSCHAPTER 13VECTORS IN THREE-DIMENSIONAL SPACE13.1The Dot Product 65313.4The Cross Product 66313.5Lines 67113.6Planes 67913.7638Vectors in Three-Dimensional Space 64413.3xixRectangular Space Coordinates 63813.2Higher Dimensions 689P3PP1Project 13.3 Work 663P2Project 13.6 Some Geometry by Vector Methods 688CHAPTER 14VECTOR CALCULUS14.1Limit, Continuity, Vector Derivative 69414.2z692The Rules of Differentiation 70114.3 Curves 705 14.4 Arc Length 714 Project 14.4 More General Changes of Parameter 721 14.5Curvilinear Motion; Curvature 723 Project 14.5A Transition Curves 732 Project 14.5B The Frenet Formulas 73314.6 *14.7Vector Calculus in Mechanics 733yPlanetary Motion 741 xCHAPTER 15FUNCTIONS OF SEVERAL VARIABLES15.1A Brief Catalogue of the Quadric Surfaces; Projections 75115.3Graphs; Level Curves and Level Surfaces 75815.4Partial Derivatives 76715.5Open and Closed Sets 77415.6zElementary Examples 74815.2748Limits and Continuity; Equality of Mixed Partials 777Project 15.3 Level Curves and Surfaces 766y (0, 1, 0)Project 15.6 Partial Differential Equations 785 x17:25 22. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13xx QC: PBU/OVYT1: PBU October 24, 2006CONTENTSCHAPTER 16zGRADIENTS; EXTREME VALUES; DIFFERENTIALS16.1Differentiability and Gradient 78816.21788Gradients and Directional Derivatives 79616.3 The Mean-Value Theorem; the Chain Rule 805 16.4The Gradient as a Normal; Tangent Lines and Tangent Planes 818 Local Extreme Values 828 Absolute Extreme Values 83616.7y16.5 16.61Maxima and Minima with Side Conditions 841 Project 16.7 Maxima and Minima with Two Side Conditions 84916.8 Differentials 849 16.9 Reconstructing a Function from Its Gradient 855CHAPTER 17DOUBLE AND TRIPLE INTEGRALS17.1Double Integrals 86717.3The Evaluation of Double Integrals by Repeated Integrals 87817.4The Double Integral as the Limit of Riemann Sums; Polar Coordinates 88817.5Further Applications of the Double Integral 89517.6 yMultiple-Sigma Notation 86417.2zTriple Integrals 90217.7Reduction to Repeated Integrals 90717.8 Cylindrical Coordinates 916 17.9x86417.10The Triple Integral as the Limit of Riemann Sums; Spherical Coordinates 922 Jacobians; Changing Variables in Multiple Integration 930 Project 17.10 Generalized Polar Coordinates 935CHAPTER 18LINE INTEGRALS AND SURFACE INTEGRALS93818.1 18.2Work-Energy Formula; Conservation of Mechanical Energy 95118.4Another Notation for Line Integrals; Line Integrals with Respect to Arc Length 95418.5 T2The Fundamental Theorem for Line Integrals 94618.3T1Line Integrals 938Greens Theorem 95918.6Parametrized Surfaces; Surface Area 96918.7 Surface Integrals 980 18.8 The Vector Differential Operator 989 18.9 The Divergence Theorem 995 Project 18.9 Static Charges 1000 18.10Stokess Theorem 100117:25 23. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 27, 2006CONTENTSCHAPTER 19ADDITIONAL DIFFERENTIAL EQUATIONS19.1Exact Differential Equations; Integrating Factors 101319.3xxi1010Bernoulli Equations; Homogeneous Equations 101019.2Numerical Methods 1018x 1.5 1 0.5 yProject 19.3 Direction Fields 10220 0.519.4The Equation y + ay + by = (x) 10221 1.519.5Mechanical Vibrations 1030APPENDIX ASOME ADDITIONAL TOPICSA.1Rotation of Axes; Eliminating the x y-Term A-1A.2A-1Determinants A-3APPENDIX BSOME ADDITIONAL PROOFSB.1The Intermediate-Value Theorem A-8B.2Boundedness; Extreme-Value Theorem A-9B.3Inverses A-10B.4The Integrability of Continuous Functions A-11B.5The Integral as the Limit of Riemann Sums A-14ANSWERS TO ODD-NUMBERED EXERCISES A-15 Index I-1 Table of Integrals Inside CoversA-8y3210 x12314:50 24. P1: PBU/OVYP2: PBU/OVYJWDD023-FMJWDD023-Salas-v13QC: PBU/OVYT1: PBU October 24, 2006xxii17:25 25. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 2006CHAPTER1 PRECALCULUS REVIEWIn this chapter we gather together for reference and review those parts of elementary mathematics that are necessary for the study of calculus. We assume that you are familiar with most of this material and that you dont require detailed explanations. But rst a few words about the nature of calculus and a brief outline of the history of the subject. 1.1 WHAT IS CALCULUS? To a Roman in the days of the empire, a calculus was a pebble used in counting and gambling. Centuries later, calculare came to mean to calculate, to compute, to gure out. For our purposes, calculus is elementary mathematics (algebra, geometry, trigonometry) enhanced by the limit process. Calculus takes ideas from elementary mathematics and extends them to a more general situation. Some examples are on pages 2 and 3. On the left-hand side you will nd an idea from elementary mathematics; on the right, this same idea as extended by calculus. It is tting to say something about the history of calculus. The origins can be traced back to ancient Greece. The ancient Greeks raised many questions (often paradoxical) about tangents, motion, area, the innitely small, the innitely largequestions that today are claried and answered by calculus. Here and there the Greeks themselves provided answers (some very elegant), but mostly they provided only questions.Elementary MathematicsCalculusslope of a line y = mx + bslope of a curve y = f (x) (Table continues)115:42 26. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20062 CHAPTER 1 PRECALCULUS REVIEWtangent line to a circletangent line to a more general curvearea of a region bounded by line segmentsarea of a region bounded by curveslength of a line segmentlength of a curvevolume of a rectangular solidvolume of a solid with a curved boundarymotion along a straight line with constant velocitymotion along a curved path with varying velocitywork done by a constant forcework done by a varying force15:42 27. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.2 REVIEW OF ELEMENTARY MATHEMATICSmass of an object of constant densitymass of an object of varying densitycenter of a spherecenter of gravity of a more general solidAfter the Greeks, progress was slow. Communication was limited, and each scholar was obliged to start almost from scratch. Over the centuries, some ingenious solutions to calculus-type problems were devised, but no general techniques were put forth. Progress was impeded by the lack of a convenient notation. Algebra, founded in the ninth century by Arab scholars, was not fully systematized until the sixteenth century. Then, in the seventeenth century, Descartes established analytic geometry, and the stage was set. The actual invention of calculus is credited to Sir Isaac Newton (16421727), an Englishman, and to Gottfried Wilhelm Leibniz (16461716), a German. Newtons invention is one of the few good turns that the great plague did mankind. The plague forced the closing of Cambridge University in 1665, and young Isaac Newton of Trinity College returned to his home in Lincolnshire for eighteen months of meditation, out of which grew his method of uxions, his theory of gravitation, and his theory of light. The method of uxions is what concerns us here. A treatise with this title was written by Newton in 1672, but it remained unpublished until 1736, nine years after his death. The new method (calculus to us) was rst announced in 1687, but in vague general terms without symbolism, formulas, or applications. Newton himself seemed reluctant to publish anything tangible about his new method, and it is not surprising that its development on the Continent, in spite of a late start, soon overtook Newton and went beyond him. Leibniz started his work in 1673, eight years after Newton. In 1675 he initiated the basic modern notation: dx and . His rst publications appeared in 1684 and 1686. These made little stir in Germany, but the two brothers Bernoulli of Basel (Switzerland) took up the ideas and added profusely to them. From 1690 onward, calculus grew rapidly and reached roughly its present state in about a hundred years. Certain theoretical subtleties were not fully resolved until the twentieth century. 1.2 REVIEW OF ELEMENTARY MATHEMATICS In this section we review the terminology, notation, and formulas of elementary mathematics.Sets A set is a collection of distinct objects. The objects in a set are called the elements or members of the set. We will denote sets by capital letters A, B, C, . . . and use lowercase letters a, b, c, . . . to denote the elements.315:42 28. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20064 CHAPTER 1 PRECALCULUS REVIEW For a collection of objects to be a set it must be well-dened; that is, given any object x, it must be possible to determine with certainty whether or not x is an element of the set. Thus the collection of all even numbers, the collection of all lines parallel to a given line l, the solutions of the equation x 2 = 9 are all sets. The collection of all intelligent adults is not a set. Its not clear who should be included. Notions and Notationthe object x is in the set A the object x is not in the set A the set of all x which satisfy property P ({x : x 2 = 9} = {3, 3}) A is a subset of B, A is contained in B B contains A the union of A and B (A B = {x : x A or x B}) the intersection of A and B (A B = {x : x A and x B}) the empty setxA xA / {x : P} AB BA AB AB These are the only notions from set theory that you will need at this point.Real Numbers Classificationpositive integers integers rational numbers irrational numbers1, 2, 3, . . . 0, 1, 1, 2, 2, 3, 3, . . . p/q, with p, q integers, q = 0; for example, 5/2, 19/7, 4/1 = 4 real numbers not rational; that are for example 2, 3 7, Decimal RepresentationEach real number can be expressed as a decimal. To express a rational number p/q as a decimal, we divide the denominator q into the numerator p. The resulting decimal either terminates or repeats: 3 = 0.6, 527 = 1.35, 2043 = 5.375 8are terminating decimals; 15 2 = 0.6666 = 0.6, = 1.363636 = 1.36, 3 11 116 = 3.135135 = 3.135 37andare repeating decimals. (The bar over the sequence of digits indicates that the sequence repeats indenitely.) The converse is also true; namely, every terminating or repeating decimal represents a rational number. Also called natural numbers.15:42 29. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.2 REVIEW OF ELEMENTARY MATHEMATICSThe decimal expansion of an irrational number can neither terminate nor repeat. The expansions 2 = 1.414213562 and = 3.141592653 do not terminate and do not develop any repeating pattern. If we stop the decimal expansion of a given number at a certain decimal place, then the result is a rational number that approximates the given number. For instance, 1.414 = 1414/1000 is a rational number approximation to 2 and 3.14 = 314/100 is a rational number approximation to . More accurate approximations can be obtained by using more decimal places from the expansions. The Number Line (Coordinate Line, Real Line)On a horizontal line we choose a point O. We call this point the origin and assign to it coordinate 0. Now we choose a point U to the right of O and assign to it coordinate 1. See Figure 1.2.1. The distance between O and U determines a scale (a unit length). We go on as follows: the point a units to the right of O is assigned coordinate a; the point a units to the left of O is assigned coordinate a. In this manner we establish a one-to-one correspondence between the points of a line and the numbers of the real number system. Figure 1.2.2 shows some real numbers represented as points on the number line. Positive numbers appear to the right of 0, negative numbers to the left of 0. 27 412 2011 43 22 53Figure 1.2.2Order Properties(i) (ii) (iii) (iv) (v)Either a < b, b < a, or a = b. If a < b and b < c, then a < c. If a < b, then a + c < b + c for all real numbers c. If a < b and c > 0, then ac < bc. If a < b and c < 0, then ac > bc.(trichotomy)(Techniques for solving inequalities are reviewed in Section 1.3.) DensityBetween any two real numbers there are innitely many rational numbers and innitely many irrational numbers. In particular, there is no smallest positive real number. Absolute Value a =other characterizations geometric interpretationa, if a 0 a, if a < 0. a = max{a, a}; a = a 2 . a = distance between a and 0; a c = distance between a and c.01OUFigure 1.2.1515:42 30. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20066 CHAPTER 1 PRECALCULUS REVIEW properties(i) (ii) (iii) (iv) (v) (vi) a = 0 iff a = 0. a = a . ab = a b . a + b a + b . a b a b . a 2 = a 2 = a 2 .(the triangle inequality) (a variant of the triangle inequality)Techniques for solving inequalities that feature absolute value are reviewed in Section 1.3. IntervalsSuppose that a < b. The open interval (a, b) is the set of all numbers between a and b: (a, b) = {x : a < x < b}.abThe closed interval [a, b] is the open interval (a, b) together with the endpoints a and b: [a, b] = {x : a x b}.abThere are seven other types of intervals: (a, b] = {x : a < x b},ab[a, b) = {x : a x b},ab(a, ) = {x : a < x},a[a, ) = {x : a x},a(, b) = {x : x < b},b(, b] = {x : x b},b(, ) = the set of real numbers. Interval notation is easy to remember: we use a square bracket to include an endpoint and a parenthesis to exclude it. On a number line, inclusion is indicated by a solid dot, exclusion by an open dot. The symbols and , read innity and negative innity (or minus innity), do not represent real numbers. In the intervals listed above, the symbol is used to indicate that the interval extends indenitely in the positive direction; the symbol is used to indicate that the interval extends indenitely in the negative direction. Open and ClosedAny interval that contains no endpoints is called open: (a, b), (a, ), (, b), (, ) are open. Any interval that contains each of its endpoints (there may be one or two) is called closed: [a, b], [a, ), (, b] are closed. The intervals (a, b] and [a, b) are called half-open (half-closed): (a, b] is open on the left and closed on the right; [a, b) is closed on the left and open on the right. Points of an interval that are not endpoints are called interior points of the interval. By iff we mean if and only if. This expression is used so often in mathematics that its convenient to have an abbreviation for it. The absolute value of the sum of two numbers cannot exceed the sum of their absolute values. This is analogous to the fact that in a triangle the length of one side cannot exceed the sum of the lengths of the other two sides.15:42 31. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.2 REVIEW OF ELEMENTARY MATHEMATICS BoundednessA set S of real numbers is said to be: (i) Bounded above if there exists a real number M such that xMx S;for allsuch a number M is called an upper bound for S. (ii) Bounded below if there exists a real number m such that mxx S;for allsuch a number m is called a lower bound for S. (iii) Bounded if it is bounded above and below. Note that if M is an upper bound for S, then any number greater than M is also an upper bound for S, and if m is a lower bound for S, than any number less than m is also a lower bound for S.Examples The intervals (, 2] and (, 2) are both bounded above by 2 (and by every number greater than 2), but these sets are not bounded below. The set of positive integers {1, 2, 3, . . .} is bounded below by 1 (and by every number less than 1), but the set is not bounded above; there being no number M greater than or equal to all positive integers, the set has no upper bound. All nite sets of numbers are bounded (bounded below by the least element and bounded above by the greatest). Finally, the set of all integers, { , 3, 2, 1, 0, 1, 2, 3, }, is unbounded in both directions; it is unbounded above and unbounded below. FactorialsLet n be a positive integer. By n factorial, denoted n!, we mean the product of the integers from n down to 1: n! = n(n 1)(n 2) 3 2 1. In particular 1! = 1, 2! = 2 1 = 2, 3! = 3 2 1 = 6, 4! = 4 3 2 1 = 24, and so on. For convenience we dene 0! = 1.Algebra Powers and Roots p factorslaws of exponents a real, q odd a nonnegative, q even notation rational exponentsa = a,a = a a aa=0:a real, p a positive integera 0 = 1,1pa p+q = a p a q ,a p = 1/a pa pq = a p a q ,(a q ) p = a pqa 1/q , called the qth root of a, is the number b such that bq = a a 1/q is the nonnegative number b such that bq = a a 1/q can be written q a (a 1/2 is written a) a p/q = (a 1/q ) p In dening bounded above, bounded below, and bounded we used the conditional if, not iff. We could have used iff, but that would have been unnecessary. Denitions are by their very nature iff statements.715:42 32. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20068 CHAPTER 1 PRECALCULUS REVIEW Examples 20 = 1, 21 = 1, 22 = 2 2 = 4, 23 = 2 2 2 = 8, and so on 25+3 = 25 23 = 32 8 = 256, 235 = 22 = 1/22 = 1/4 (22 )3 = 232 = 26 = 64, (23 )2 = 223 = 26 = 64 81/3 = 2, (8)1/3 = 2, 161/2 = 16 = 4, 161/4 = 2 85/3 = (81/3 )5 = 25 = 32, 85/3 = (81/3 )5 = 25 = 1/25 = 1/32Basic Formulas(a + b)2 = a 2 + 2ab + b2 (a b)2 = a 2 2ab + b2 (a + b)3 = a 3 + 3a 2 b + 3ab2 + b3 (a b)3 = a 3 3a 2 b + 3ab2 b3 a 2 b2 = (a b)(a + b) a 3 b3 = (a b)(a 2 + ab + b2 ) a 4 b4 = (a b)(a 3 + a 2 b + ab2 + b3 ) More generally: a n bn = (a b)(a n1 + a n2 b + + abn2 + bn1 ) Quadratic EquationsThe roots of a quadratic equation ax 2 + bx + c = 0with a = 0are given by the general quadratic formula b b2 4ac . r= 2a If b2 4ac > 0, the equation has two real roots; if b2 4ac = 0, the equation has one real root; if b2 4ac < 0, the equation has no real roots, but it has two complex roots.Geometry Elementary FiguresTriangleEquilateral Triangle sh barea =1 bh 2area =1 4ss3 s215:42 33. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.2 REVIEW OF ELEMENTARY MATHEMATICSRectangleRectangular Solidwhwllarea = lw perimeter = + 2w 2l diagonal = l 2 + w 2volume = lwh surface area = 2lw + 2lh + 2w hSquareCubexx xxxarea = x 2 perimeter = 4x diagonal = x 2volume = x 3 surface area = 6x 2CircleSpherer rarea = r 2 circumference = 2rvolume = 4 r 3 3 surface area = 4r 2Sector of a Circle: radius r, central angle measured in radians (see Section 1.6). arc length rrarc length = r area = 1 r 2 2 (Table continues)915:42 34. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200610 CHAPTER 1 PRECALCULUS REVIEW Right Circular CylinderRight Circular Conerhhrvolume = r h lateral area = 2r h total surface area = 2r 2 + 2r h 21 2 r h 3volume = slant height = r 2 + h 2 lateral area = r r 2 + h 2 total surface area = r 2 + r r 2 + h 2EXERCISES 1.2 Exercises 110. Is the number rational or irrational? 2. 6. 3. 2.131313 . . . = 2.13. 4. 2 3. 5. 0. 6. 2. 8. 0.125. 7. 3 8. 10. ( 2 3) ( 2 + 3) 9. 9. Exercises 1116. Replace the symbol by , or = to make the statement true. 12. 0.33 1 . 11. 3 0.75. 4 3 13. 2 1.414. 14. 4 16. 16. 22 . 15. 2 0.285714. 7 7 Exercises 1723. Evaluate 17. 6 . 18. 4 . 19. 3 7 . 20. 5 8 . 21. 5 + 8 . 22. 2 . 23. 5 5 . Exercises 2433. Indicate on a number line the numbers x that satisfy the condition. 24. x 3 25. x 3 . 2 1.17 . 726. 2 x 3. 27. x 2 < 16. 29. x 0. 28. x 2 16. 31. x 4 2. 30. x 2 0. 32. x + 1 > 3. 33. x + 3 0. Exercises 3440. Sketch the set on a number line. 34. [3, ). 35. (, 2). 36. (4, 3]. 37. [2, 3] [1, 5]. 39. (, 1) (2, ). 38. 3, 3 3 , 5 . 2 2 2 40. (, 2) [3, ). Exercises 4147. State whether the set is bounded above, bounded below, bounded. If a set is bounded above, give an upper bound; if it is bounded below, give a lower bound; if it is bounded, give an upper bound and a lower bound.41. {0, 1, 2, 3, 4}. 43. The set of even integers. 44. {x : x 4}. 46. n1 : n = 1, 2, 3 . . . . n42. {0, 1, 2, 3, . . .}. 45. {x : x 2 > 3}. 47. The set of rational numbers less than 2. Exercises 4850. 48. Order the following numbers and place them on a number line: 3 , 2 , 2, 3 , 3 . 3 17 + 2xn1 49. Let x0 = 2 and dene xn = for n = 2 3xn1 1, 2, 3, 4, . . . Find at least ve values for xn . Is the set S = {x0 , x1 , x2 , . . . , xn , . . .} bounded above, bounded below, bounded? If so, give a lower bound and/or an upper bound for S. If n is a large positive integer, what is the approximate value of xn ? 5 231 + 4xn1 50. Rework Exercise 49 with x0 = 3 and xn = 4 5xn1 Exercises 5156. Write the expression in factored form. 52. 9x 2 4. 51. x 2 10x + 25. 6 54. 27x 3 8. 53. 8x + 64. 56. 4x 4 + 4x 2 + 1. 55. 4x 2 + 12x + 9. Exercises 5764. Find the real roots of the equation. 58. x 2 9 = 0. 57. x 2 x 2 = 0. 2 60. 2x 2 5x 3 = 0. 59. x 6x + 9 = 0. 62. x 2 + 8x + 16 = 0. 61. x 2 2x + 2 = 0. 2 64. x 2 2x + 5 = 0. 63. x + 4x + 13 = 0. Exercises 6569. Evaluate. 5! 65. 5!. 66. . 8! 8! 9! 67. . 68. . 3!5! 3!6! 7! 69. . 0!7!15:42 35. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.3 REVIEW OF INEQUALITIES70. Show that the sum of two rational numbers is a rational number. 71. Show that the sum of a rational number and an irrational number is irrational. 72. Show that the product of two rational numbers is a rational number. 73. Is the product of a rational number and an irrational number necessarily rational? necessarily irrational? 74. Show by example that the sum of two irrational numbers (a) can be rational; (b) can be irrational. Do the same for the product of two irrational numbers. 75. Prove that 2 is irrational. HINT: Assume that 2 = p/q with the fraction written in lowest terms. Square both sides of this equation and argue that both p and q must be divisible by 2. 76. Prove that 3 is irrational. 77. Let S be the set of all rectangles with perimeter P. Show that the square is the element of S with largest area. 78. Show that if a circle and a square have the same perimeter, then the circle has the larger area. Given that a circle and a rectangle have the same perimeter, which has the larger area?(a phony one): Let a and b be real numbers, both different from 0. Suppose now that a = b. ThenPROOFab = b2 ab a 2 = b2 a 2 a(b a) = (b + a)(b a) a = b + a. Since a = b, we have a = 2a. Division by a, which by assumption is not 0, gives 1 = 2. What is wrong with this argument?All our work with inequalities is based on the order properties of the real numbers given in Section 1.2. In this section we work with the type if inequalities that arise frequently in calculus, inequalities that involve a variable. To solve an inequality in x is to nd the numbers x that satisfy the inequality. These numbers constitute a set, called the solution set of the inequality. We solve inequalities much as we solve an equation, but there is one important difference. We can maintain an inequality by adding the same number to both sides, or by subtracting the same number from both sides, or by multiplying or dividing both sides by the same positive number. But if we multiply or divide by a negative number, then the inequality is reversed: givesx < 6, 1 x iff a >or a < .0 a > Example 7 Solve the inequality 2x + 3 > 5. SOLUTION In general a >iffa>ora< .So here 2x + 3 > 5or2x + 3 < 5.The rst possibility gives 2x > 2 and thus x > 1.11515:42 40. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200616 CHAPTER 1 PRECALCULUS REVIEW The second possibility gives 2x < 8 and thus x < 44The total solution is therefore the union (, 4) (1, ). 41We come now to one of the fundamental inequalities of calculus: for all real numbers a and b, a + b a + b .(1.3.6)This is called the triangle inequality in analogy with the geometric observation that in any triangle the length of each side is less than or equal to the sum of the lengths of the other two sides. PROOF OF THE TRIANGLE INEQUALITY The key here is to think of x as x 2 . Note rst that (a + b)2 = a 2 + 2ab + b2 a 2 + 2 a b + b 2 = ( a + b )2 . Comparing the extremes of the inequality and taking square roots, we have (a + b)2 a + b .(Exercise 51)The result follows from observing that (a + b)2 = a + b .Here is a variant of the triangle inequality that also comes up in calculus: for all real numbers a and b,(1.3.7) a b a b .The proof is left to you as an exercise. EXERCISES 1.3 Exercises 120. Solve the inequality and mark the solution set on a number line. 1. 2 + 3x < 5.2. 1 (2x + 3) < 6. 23. 16x + 64 16.4. 3x + 5 > 1 (x 2). 45. 1 (1 + x) < 1 (1 x). 2 36. 3x 2 1 + 6x.7. 9. 11. 13. 15. 17. 19.x 1 < 0. x 2 x 6 0. 2x 2 + x 1 0. x(x 1)(x 2) > 0. x 3 2x 2 + x 0. x 3 (x 2)(x + 3)2 < 0. x 2 (x 2)(x + 6) > 0. 28. 10. 12. 14. 16. 18. 20.x 2 + 9x + 20 < 0. x 2 4x 5 > 0. 3x 2 + 4x 4 0. x(2x 1)(3x 5) 0. x 2 4x + 4 0. x 2 (x 3)(x + 4)2 > 0. 7x(x 4)2 < 0.Exercises 2136. Solve the inequality and express the solution set as an interval or as the union of intervals. 21. x < 2. 22. x 1. 23. x > 3. 24. x 1 < 1. 26. x 1 < 2. 25. x 2 < 1 . 2 2 27. 0 < x < 1.28. 0 < x < 1 . 229. 0 < x 2 < 1 . 30. 0 < x 1 < 2. 2 2 31. 0 < x 3 < 8. 32. 3x 5 < 3. 34. 5x 3 < 1 . 33. 2x + 1 < 1 . 4 2 35. 2x + 5 > 3. 36. 3x + 1 > 5. Exercises 3742. Each of the following sets is the solution of an inequality of the form x c < . Find c and .15:42 41. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.4 COORDINATE PLANE; ANALYTIC GEOMETRY37. (3, 3). 38. (2, 2). 39. (3, 7). 40. (0, 4). 41. (7, 3). 42. (a, b). Exercises 4346. Determine all numbers A > 0 for which the statement is true. 43. If x 2 < 1, then 2x 4 < A. 44. If x 2 < A, then 2x 4 < 3. 45. If x + 1 < A, then 3x + 3 < 4. 46. If x + 1 < 2, then 3x + 3 < A. 47. Arrange the following in order :1, x, x, 1/x, 1/ x, given that: (a) x > 1; (b) 0 < x < 1. 48. Given that x > 0, compare x x +1andx +1 . x +249. Suppose that ab > 0. Show that if a < b, then 1/b < 1/a. 50. Given that a > 0 and b > 0, show that if a 2 b2 , then a b. 51. Show that if 0 a b, then a b.1752. Show that a b a + b for all real numbers a and b. 53. Show that a b a b for all real numbers a and b. 2HINT: Calculate a b . 54. Show that a + b = a + b iff ab 0. 55. Show that if0 a b,b a . 1+a 1+bthen56. Let a, b, c be nonnegative numbers. Show that ifa b + c,thenb c a + . 1+a 1+b 1+c57. Show that if a and b are real numbers and a < b, then a < (a + b)/2 < b. The number (a + b)/2 is called the arithmetic mean of a and b. 58. Given that 0 a b, show that a The number a+b ab b. 2 ab is called the geometric mean of a and b. 1.4 COORDINATE PLANE; ANALYTIC GEOMETRY Rectangular Coordinates The one-to-one correspondence between real numbers and points on a line can be used to construct a coordinate system for the plane. In the plane, we draw two number lines that are mutually perpendicular and intersect at their origins. Let O be the point of intersection. We set one of the lines horizontally with the positive numbers to the right of O and the other vertically with the positive numbers above O. The point O is called the origin, and the number lines are called the coordinate axes. The horizontal axis is usually labeled the x-axis and the vertical axis is usually labeled the y-axis. The coordinate axes separate four regions, which are called quadrants. The quadrants are numbered I, II, III, IV in the counterclockwise direction starting with the upper right quadrant. See Figure 1.4.1. Rectangular coordinates are assigned to points of the plane as follows (see Figure 1.4.2.). The point on the x-axis with line coordinate a is assigned rectangular coordinates (a, 0). The point on the y-axis with line coordinate b is assigned rectangular coordinates (0, b). Thus the origin is assigned coordinates (0, 0). A point P not on one of the coordinate axes is assigned coordinates (a, b) provided that the line l1 that passes through P and is parallel to the y-axis intersects the x-axis at the point with coordinates (a, 0), and the l2 that passes through P and is parallel to the x-axis intersects the y-axis at the point with coordinates (0, b). This procedure assigns an ordered pair of real numbers to each point of the plane. Moreover, the procedure is reversible. Given any ordered pair (a, b) of real numbers, there is a unique point P in the plane with coordinates (a, b). To indicate P with coordinates (a, b) we write P(a, b). The number a is called the x-coordinate (the abscissa); the number b is called the y-coordinate (the ordinate). The coordinate system that we have dened is called a rectangular coordinate system. It is often referred to as a Cartesian coordinate system after the French mathematician Ren Descartes (15961650). eyIIIxOIIIIVFigure 1.4.1y P bca xO l1 dl2Q(c, d)Figure 1.4.215:42 42. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200618 CHAPTER 1 PRECALCULUS REVIEW Distance and Midpoint Formulas Let P0 (x0 , y0 ) and P1 (x1 , y1 ) be points in the plane. The formula for the distance d(P0 , P1 ) between P0 and P1 follows from the Pythagorean theorem: d(P0 , P1 ) = x1 x0 2 + y1 y0 2 = (x1 x0 )2 + (y1 y0 )2 . a 2 = a 2(Figure 1.4.3)y P1(x1, y1) y1 y0 P0(x0, y0) P(x1, y0)x1 x0xDistance : d(P0 , P1 ) =(x1 x0 )2 + (y1 y0 )2Figure 1.4.3yLet M(x, y) be the midpoint of the line segment P0 P1 . ThatP1(x1, y1) M (x, y)x=P0(x0, y0)x0 + x1 2andy=y0 + y1 2follows from the congruence of the triangles shown in Figure 1.4.4 xMidpoint: M =x0 +x1 y0 +y1 , 2 2Figure 1.4.4Lines Let l be the line determined by P0 (x0 , y0 ) and P1 (x1 , y1 ). If l is not vertical, then x1 = x0 and the slope of l is given by the formula(i) Slopem=y1 y0 . x1 x0(Figure 1.4.5)With (as indicated in the gure) measured counterclockwise from the x-axis, m = tan . The angle is called the inclination of l. If l is vertical, then = /2 and the slope of l is not dened. (ii) InterceptsIf a line intersects the x-axis, it does so at some point (a, 0). We call a the x-intercept. If a line intersects the y-axis, it does so at some point (0, b). We call b the y-intercept. Intercepts are shown in Figure 1.4.6. The trigonometric functions are reviewed in Section 1.6.15:42 43. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 2006yy191.4 COORDINATE PLANE; ANALYTIC GEOMETRY yP1(x1, y1)P0(x0, y0)P0(x0, y0) P1(x1, y1)P1(x1, y1)P0(x0, y0) xO m=y1 y0 x1 x00< 0m=y1 y0 x1 x0=0xO m= 2=02y1 y0 x1 x0 aFigure 1.4.10 Parabola yy xx y = ax 2, a > 0y = ax 2, a < 0y = ax 2 + bx + c , a > 0y = ax 2 + bx + c, a < 0Figure 1.4.11 Hyperbola yyxx2 a2y2 b2xy2=1a2Figure 1.4.12x2 b2=115:42 47. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.4 COORDINATE PLANE; ANALYTIC GEOMETRY23Remark The circle, the ellipse, the parabola, and the hyperbola are known as the conic sections because each of these congurations can be obtained by slicing a double right circular cone by a suitably inclined plane. (See Figure 1.4.13.) circleellipseparabolahyperbolaFigure 1.4.13EXERCISES 1.4 Exercises 14. Find the distance between the points. 1. P0 (0, 5),P1 (6, 3).2. P0 (2, 2),P1 (5, 5).4. P0 (2, 7), P1 (4, 7). 3. P0 (5, 2), P1 (3, 2). Exercises 58. Find the midpoint of the line segment P0 P1 . 6. P0 (3, 1), P1 (1, 5). 5. P0 (2, 4), P1 (6, 8). 8. P0 (a, 3), P1 (3, a). 7. P0 (2, 3), P1 (7, 3). Exercises 914. Find the slope of the line through the points. 10. P0 (4, 3), P1 (2, 7). 9. P0 (2, 5), P1 (4, 1). 11. P(a, b), Q(b, a). 12. P(4, 1), Q(3, 1). 14. O(0, 0), P(x0 , y0 ). 13. P(x0 , 0), Q(0, y0 ). Exercises 1520. Find the slope and y-intercept. 15. y = 2x 4. 16. 6 5x = 0. 17. 3y = x + 6. 18. 6y 3x + 8 = 0. 19. 7x 3y + 4 = 0. 20. y = 3. Exercises 2124. Write an equation for the line with 21. slope 5 and y-intercept 2. 22. slope 5 and y-intercept 2. 23. slope 5 and y-intercept 2. 24. slope 5 and y-intercept 2. Exercises 2526. Write an equation for the horizontal line 3 units 25. above the x-axis.26. below the x-axis. Exercises 2728. Write an equation for the vertical line 3 units 27. to the left of the y-axis. 28. to the right of the y-axis. Exercises 2934. Find an equation for the line that passes through the point P(2, 7) and is 29. parallel to the x-axis. 30. parallel to the y-axis. 31. parallel to the line 3y 2x + 6 = 0. 32. perpendicular to the line y 2x + 5 = 0. 33. perpendicular to the line 3y 2x + 6 = 0. 34. parallel to the line y 2x + 5 = 0. Exercises 3538. Determine the point(s) where the line intersects the circle. 35. y = x, x 2 + y 2 = 1. 36. y = mx,x 2 + y 2 = 4.37. 4x + 3y = 24, x 2 + y 2 = 25. 38. y = mx + b, x 2 + y 2 = b2 . Exercises 3942. Find the point where the lines intersect. 39. l1 : 4x y 3 = 0, l2 : 3x 4y + 1 = 0. 40. l1 : 3x + y 5 = 0, l2 : 7x 10y + 27 = 0. 41. l1 : 4x y + 2 = 0, l2 : 19x + y = 0.15:42 48. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200624 CHAPTER 1 PRECALCULUS REVIEW 42. l1 : 5x 6y + 1 = 0, l2 : 8x + 5y + 2 = 0. 43. Find the area of the triangle with vertices (1, 2), (1, 3), (2, 4). 44. Find the area of the triangle with vertices (1, 1), (3, 2), ( 2, 1). 45. Determine the slope of the line that intersects the circle x 2 + y 2 = 169 only at the point (5, 12). 46. Find an equation for the line which is tangent to the circle x 2 + y 2 2x + 6y 15 = 0 at the point (4, 1). HINT: A line is tangent to a circle at a point P iff it is perpendicular to the radius at P. 47. The point P(1, 1) is on a circle centered at C(1, 3). Find an equation for the line tangent to the circle at P. Exercises 4851. Estimate the point(s) of intersection. 48. l1 : 3x 4y = 7, l2 : 5x + 2y = 11. 49. l1 : 2.41x + 3.29y = 5, l2 : 5.13x 4.27y = 13. 50. l1 : 2x 3y = 5, circle : x 2 + y 2 = 4. 51. circle : x 2 + y 2 = 9, parabola : y = x 2 4x + 5. Exercises 5253. The perpendicular bisector of the line segment P Q is the line which is perpendicular to P Q and passes through the midpoint of P Q. Find an equation for the perpendicular bisector of the line segment that joins the two points. 52. P(1, 3), Q(3, 4). 53. P(1, 4), Q(4, 9). Exercises 5456. The points are the vertices of a triangle. State whether the triangle is isosceles (two sides of equal length), a right triangle, both of these, or neither of these. 54. P0 (4, 3), P1 (4, 1), P2 (2, 1). 55. P0 (2, 5), P1 (1, 3), P2 (1, 0). 56. P0 (1, 2), P1 (1, 3), P2 (4, 1). 57. Show that the distance from the origin to the line Ax + By + C = 0 is given by the formula d(0, l) = C A2 + B 2.58. An equilateral triangle is a triangle the three sides of which have the same length. Given that two of the vertices of an equilateral triangle are (0, 0) and (4, 3), nd all possible locations for a third vertex. How many such triangles are there? 59. Show that the midpoint M of the hypotenuse of a right triangle is equidistant from the three vertices of the triangle. HINT: Introduce a coordinate system in which the sides of the triangle are on the coordinate axes; see the gure.y (0, b) Mx(a, 0)60. A median of a triangle is a line segment from a vertex to the midpoint of the opposite side. Find the lengths of the medians of the triangle with vertices (1, 2), (2, 1), (4, 3). 61. The vertices of a triangle are (1, 0), (3, 4), (1, 6). Find the point(s) where the medians of this triangle intersect. 62. Show that the medians of a triangle intersect in a single point (called the centroid of the triangle). HINT: Introduce a coordinate system such that one vertex is at the origin and one side is on the positive x-axis; see the gure. y (a, b)(c, 0)x63. Prove that each diagonal of a parallelogram bisects the other. HINT: Introduce a coordinate system with one vertex at the origin and one side on the positive x-axis. 64. P1 (x1 , y1 ), P2 (x2 , y2 ), P3 (x3 , y3 ), P4 (x4 , y4 ) are the vertices of a quadrilateral. Show that the quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram. 65. Except in scientic work, temperature is usually measured in degrees Fahrenheit (F) or in degrees Celsius (C). The relation between F and C is linear. (In the equation that relates F to C, both F and C appear to the rst degree.) The freezing point of water in the Fahrenheit scale is 32 F; in the Celsius scale it is 0 C. The boiling point of water in the Fahrenheit scale is 212 F; in the Celsius scale it is 100 C. Find an equation that gives the Fahrenheit temperature F in terms of the Celsius temperature C. Is there a temperature at which the Fahrenheit and Celsius readings are equal? If so, nd it. 66. In scientic work, temperature is measured on an absolute scale, called the Kelvin scale (after Lord Kelvin, who initiated this mode of temperature measurement). The relation between Fahrenheit temperature F and absolute temperature K is linear. Given that K = 273 when F = 32 , and K = 373 when F = 212 , express K in terms of F. Then use your result in Exercise 65 to determine the connection between Celsius temperature and absolute temperature. 1.5 FUNCTIONS The fundamental processes of calculus (called differentiation and integration) are processes applied to functions. To understand these processes and to be able to carry them out, you have to be comfortable working with functions. Here we review some of the basic ideas and the nomenclature. We assume that you are familiar with all of this.15:42 49. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.5 FUNCTIONSFunctions can be applied in a very general setting. At this stage, and throughout the rst thirteen chapters of this text, we will be working with what are called real-valued functions of a real variable, functions that assign real numbers to real numbers.Domain and Range Lets suppose that D is some set of real numbers and that f is a function dened on D. Then f assigns a unique number f (x) to each number x in D. The number f (x) is called the value of f at x, or the image of x under f. The set D, the set on which the function is dened, is called the domain of f, and the set of values taken on by f is called the range of f. In set notation dom ( f ) = D,range ( f ) = { f (x) : x D}.We can specify the function f by indicating exactly what f (x) is for each x in D. Some examples. We begin with the squaring function f (x) = x 2 ,for all real numbers x.The domain of f is explicitly given as the set of real numbers. Particular values taken on by f can be found by assigning particular values to x. In this case, for example, f (4) = 42 = 16,f (3) = (3)2 = 9,f (0) = 02 = 0.As x runs through the real numbers, x 2 runs through all the nonnegative numbers. Thus the range of f is [0, ). In abbreviated form, we can write dom ( f ) = (, ),range ( f ) = [0, )and we can say that f maps (, ) onto [0, ). Now lets look at the function g dened by g(x) = 2x + 4, x [0, 6]. The domain of g is given as the closed interval [0, 6]. At x = 0, g takes on the value 2: g(0) = 2 0 + 4 = 4 = 2; at x = 6, g has the value 4: g(6) = 2 6 + 4 = 16 = 4.As x runs through the numbers in [0, 6], g(x) runs through the numbers from 2 to 4. Therefore, the range of g is the closed interval [2, 4]. The function g maps [0, 6] onto [2, 4]. Some functions are dened piecewise. As an example, take the function h, dened by setting h(x) =2x + 1, x 2,if x < 0 if x 0.As explicitly stated, the domain of h is the set of real numbers. As you can verify, the range of h is also the set of real numbers. Thus the function h maps (, ) onto (, ). A more familiar example is the absolute value function f (x) = x . Here f (x) =x, x,if x 0 if x < 0.The domain of this function is (, ) and the range is [0, ). Remark Functions are often given by equations of the form y = f (x) with x restricted to some set D, the domain of f. In this setup x is called the independent variable (or the argument of the function) and y, clearly dependent on x, is called the dependent variable. 2515:42 50. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200626 CHAPTER 1 PRECALCULUS REVIEW The Graph of a Function If f is a function with domain D, then the graph of f is the set of all points P(x, f (x)) with x in D. Thus the graph of f is the graph of the equation y = f (x) with x restricted to D; namely the graph of f = {(x, y) : x D, y = f (x)}. The most elementary way to sketch the graph of a function is to plot points. We plot enough points so that we can see what the graph may look like and then connect the points with a curve. Of course, if we can identify the curve in advance (for example, if we know that the graph is a straight line, a parabola, or some other familiar curve), then it is much easier to draw the graph. The graph of the squaring function f (x) = x 2 ,x (, )is the parabola shown in Figure 1.5.1. The points that we plotted are indicated in the table and marked on the graph. The graph of the function x [0, 6] g(x) = 2x + 4, is the arc shown in Figure 1.5.2 y 6 (2, 4) y x1 40 28 2.84 612 3.5 41 41 2P(x, x2)01 2xx20(2, 4)(1, 1)(1, 1) 21122x + 4 42(6, 4) P(x, 2x + 4) (4, 12)2(2, 8 )2xFigure 1.5.146xFigure 1.5.2The graph of the function h(x) =2x + 1, x 2,if x < 0 if x 0and the graph of the absolute value function are shown in Figures 1.5.3 and 1.5.4. Although the graph of a function is a curve in the plane, not every curve in the plane is the graph of a function. This raises a question: How can we tell whether a curve is the graph of a function? A curve C which intersects each vertical line at most once is the graph of a function: for each P(x, y) C, dene f (x) = y. A curve C which intersects some vertical line more than once is not the graph of a function: If P(x, y1 ) and P(x, y2 ) are both on C, then how can we decide what f (x) is? Is it y1 ; or is it y2 ? These observations lead to what is called the vertical line test: a curve C in the plane is the graph of a function iff no vertical line intersects C at more than one point. Thus circles, ellipses, hyperbolas are not the graphs of functions. The curve shown in Figure 1.5.5 is the graph of a function, but the curve shown in Figure 1.5.6 is not the graph of a function.15:42 51. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.5 FUNCTIONS x 0 1 2 2023(2, 4)0 1 41yh(x)2xx 0 1 10 1 1y(0, 1)112(1, 1)(1, 1)x1 h(x) ={2x + 1, x < 0 x2, x 0x =(2, 3)Figure 1.5.31{xx, x 0 x, x < 0Figure 1.5.4yyxxFigure 1.5.5Figure 1.5.6Graphing calculators and computer algebra systems (CAS) are valuable aids to graphing, but, used mindlessly, they can detract from the understanding necessary for more advanced work. We will not attempt to teach the use of graphing calculators or the ins and outs of computer software, but technology-oriented exercises will appear throughout the text.Even Functions, Odd Functions; Symmetry For even integers n, (x)n = x n ; for odd integers n, (x)n = x n . These simple observations prompt the following denitions: A function f is said to be even if f (x) = f (x)for allx dom ( f );a function f is said to be odd if f (x) = f (x)for allx dom ( f ).The graph of an even function is symmetric about the y-axis, and the graph of an odd function is symmetric about the origin. (Figures 1.5.7 and 1.5.8.) The absolute value function is even: f (x) = x = x = f (x).2715:42 52. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200628 CHAPTER 1 PRECALCULUS REVIEW y y (x, f (x))(x, f (x))(x, f (x))x x xxxOdd functionxEven function(x, f (x))Figure 1.5.7y 3y = 4x x3Its graph is symmetric about the y-axis. (See Figure 1.5.4.) The function f (x) = 4x x 3 is odd:2f (x) = 4(x) (x 3 ) = 4x + x 3 = (4x x 3 ) = f (x).1 21Figure 1.5.8The graph, shown in Figure 1.5.9, is symmetric about the origin. 1 23Figure 1.5.9xConvention on Domains If the domain of a function f is not explicitly given, then by convention we take as domain the maximal set of real numbers x for which f (x) is a real number. For the function f (x) = x 3 + 1, we take as domain the set of real numbers. For g(x) = x, we take as domain the set of nonnegative numbers. For 1 x 2 we take as domain the set of all real numbers x = 2. In interval notation. h(x) =dom( f ) = (, ),dom(g) = [0, ),anddom(h) = (, 2) (2, ).The graphs of the three functions are shown in Figure 1.5.10. yyy P(x, x)P x,1 x2(0, 1) xxx10, 2P(x, x3 + 1) f (x) = x3 + 1)g(x) = xh(x) =1 x2Figure 1.5.10Example 1 Give the domain of each function: x +1 4 x2 , (a) f (x) = 2 (b) g(x) = . x +x 6 x 1 SOLUTION (a) You can see that f (x) is a real number iff x 2 + x 6 = 0. Sincex 2 + x 6 = (x + 3)(x 2),15:42 53. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.5 FUNCTIONSthe domain of f is the set of real numbers other than 3 and 2. This set can be expressed as (, 3) (3, 2) (2, ). (b) For g(x) to be a real number, we need 4 x2 0andx = 1.Since 4 x 0 iff x 4 iff 2 x 2, the domain of g is the set of all numbers x in the closed interval [2, 2] other than x = 1. This set can be expressed as the union of two half-open intervals: 22[2, 1) (1, 2]. Example 2 Give the domain and range of the function: 1 + 5. f (x) = 2x SOLUTION First we look for the domain. Since 2 x is a real number iff 2 x 0, we need x 2. But at x = 2, 2 x = 0 and its reciprocal is not dened. We must therefore restrict x to x < 2. The domain is (, 2). Now we look for the range. As x runs through (, 2), 2 x takes on all positive values and so does its reciprocal. The range of f is therefore (5, ). The function f maps (, 2) onto (5, ). Functions are used in applications to show how variable quantities are related. The domain of a function that appears in an application is dictated by the requirements of the application.Example 3 U.S. Postal Service regulations require that the length plus the girth (the perimeter of a cross section) of a package for mailing cannot exceed 108 inches. A rectangular box with a square end is designed to meet the regulation exactly (see Figure 1.5.11). Express the volume V of the box as a function of the edge length of the square end and give the domain of the function.GirththngLeSOLUTION Let x denote the edge length of the square end and let h denote the lengthof the box. The girth is the perimeter of the square, or 4x. Since the box meets the regulations exactly, 4x + h = 108and thereforeFigure 1.5.11h = 108 4x.The volume of the box is given by V = x 2 h and so it follows that V (x) = x 2 (108 4x) = 108x 2 4x 3 . Since neither the edge length of the square end nor the length of the box can be negative, we have x 0andh = 108 4x 0.The second condition requires x 27. The full requirement on x, 0 x 27, gives dom(V ) = [0, 27]. Example 4 A soft-drink manufacturer wants to fabricate cylindrical cans. (See Figure 1.5.12.) The can is to have a volume of 12 uid ounces, which we take to be approximately 22 cubic inches. Express the total surface area S of the can as a function of the radius and give the domain of the function.rhFigure 1.5.122915:42 54. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200630 CHAPTER 1 PRECALCULUS REVIEW SOLUTION Let r be the radius of the can and h the height. The total surface area (top, bottom, and lateral area) of a right circular cylinder is given by the formulaS = 2r 2 + 2 rh. Since the volume V = r 2 h is to be 22 cubic inches, we have 22 and h= r 2 h = 22 r 2 and therefore 22 44 = 2r 2 + . S(r ) = 2r 2 + 2r 2 r r(square inches)Since r can take on any positive value, dom (S) = (0, ). EXERCISES 1.5 Exercises 16. Calculate (a) f (0), (b) f (1), (c) f (2), (d) f (3/2). 2x 1 1. f (x) = 2x 2 3x + 2. 2. f (x) = 2 . x +4 3. f (x) = x 2 + 2x. 4. f (x) = x + 3 5x. 2x 1 5. f (x) = . 6. f (x) = 1 . x + 2 + x 2 (x + 1)2 Exercises 710. Calculate (a) f (x), (b) f (1/x), (c) f (a + b). x 7. f (x) = x 2 2x. 8. f (x) = 2 x +1 x 9. f (x) = 1 + x 2 . 10. f (x) = 2 . x 1 Exercises 11 and 12. Calculate f (a + h) and [ f (a + h) f (a)]/ h for h = 0. 1 12. f (x) = 11. f (x) = 2x 2 3x. . x 2 Exercises 1318. Find the number(s) x, if any, where f takes on the value 1. 13. f (x) = 2 x . 14. f (x) = 1 + x. 16. f (x) = 4 + 10x x 2 . 15. f (x) = x 2 + 4x + 5. x 2 . 18. f (x) = 17. f (x) = . 25 x x Exercises 1930. Give the domain and range of the function. 19. f (x) = x . 20. g(x) = x 2 1. 21. f (x) = 2x 3. 22. g(x) = x + 5. 4 1 24. g(x) = . 23. f (x) = 2 . x x 26. g(x) = x 3. 25. f (x) = 1 x. 28. g(x) = x 1 1. 27. f (x) = 7 x 1. 1 1 . 30. g(x) = 29. f (x) = . 2x 4 x2 Exercises 3140. Give the domain of the function and sketch the graph. 31. f (x) = 1. 32. f (x) = 1.33. f (x) = 2x. 34. f (x) = 2x + 1. 36. f (x) = 1 x 3. 35. f (x) = 1 x + 2. 2 2 37. f (x) = 4 x 2 . 38. f (x) = 9 x 2 . 40. f (x) = x 1 . 39. f (x) = x 2 x 6. Exercises 4144. Sketch the graph and give the domain and range of the function. 1, x < 0 41. f (x) = 1, x > 0. 42. f (x) =x 2, 1 x,x 0 x > 0. 1 + x, 0 x 1 x, 1 < x < 2 43. f (x) = 1 x + 1, 2 x. 2 x 2, x < 0 44. f (x) = 1, 0 < x < 2 x, 2 < x. Exercises 4548. State whether the curve is the graph of a function. If it is, give the domain and the range. 45. y 3 2 1 3 2 11 2 3123x15:42 55. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.5 FUNCTIONS46.3 2 1 1 1123xB2 A347.31(b) Find the zero(s) of f (the values of x such that f (x) = 0) accurate to three decimal places. (c) Find the coordinates of the points marked A and B, accurate to three decimal places. 58. The graph of f (x) = x 4 + 8x 2 + x 1 looks something like this:y3 2y 3 2 1 3 2 11123x2 348.y 3 2 1 3 2 1123x1 2 3Exercises 4954. State whether the function is odd, even, or neither. 50. f (x) = x 2 + 1. 49. f (x) = x 3 . 51. g(x) = x(x 1). 52. g(x) = x(x 2 + 1). 1 x2 . 54. F(x) = x + . 53. f (x) = 1 x x x 55. f (x) = 2 56. f (x) = 5 x x 3 . x 9 57. The graph of f (x) = 1 x 3 + 1 x 2 12x 6 looks some3 2 thing like this: AB(a) Use a graphing utility to sketch an accurate graph of f.(a) Use a graphing utility to sketch an accurate graph of f. (b) Find the zero(s) of f, if any. Use three decimal place accuracy. (c) Find the coordinates of the points marked A and B, accurate to three decimal places. Exercises 59 and 60. Use a graphing utility to draw several views of the graph of the function. Select the one that most accurately shows the important features of the graph. Give the domain and range of the function. 59. f (x) = x 3 3x 2 24x + 4 . 60. f (x) = x 3 8. 61. Determine the range of y = x 2 4x 5: (a) by writing y in the form (x a)2 + b. (b) by rst solving the equation for x. 2x : 62. Determine the range of y = 4x b (a) by writing y in the form a + . 4x (b) by rst solving the equation for x. 63. Express the area of a circle as a function of the circumference. 64. Express the volume of a sphere as a function of the surface area. 65. Express the volume of a cube as a function of the area of one of the faces. 66. Express the volume of a cube as a function of the total surface area. 67. Express the surface area of a cube as a function of the length of the diagonal of a face. 68. Express the volume of a cube as a function of one of the diagonals. 69. Express the area of an equilateral triangle as a function of the length of a side. 70. A right triangle with hypotenuse c is revolved about one of its legs to form a cone. (See the gure.) Given that x is the length of the other leg, express the volume of the cone as a function of x.15:42 56. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200632 CHAPTER 1 PRECALCULUS REVIEW y (2, 5) c (x, 0) x71. A Norman window is a window in the shape of a rectangle surmounted by a semicircle. (See the gure.) Given that the perimeter of the window is 15 feet, express the area as a function of the width x.x75. A string 28 inches long is to be cut into two pieces, one piece to form a square and the other to form a circle. Express the total area enclosed by the square and circle as a function of the perimeter of the square. 76. A tank in the shape of an inverted cone is being lled with water. (See the gure.) Express the volume of water in the tank as a function of the depth h.x10r 20x72. A window has the shape of a rectangle surmounted by an equilateral triangle. Given that the perimeter of the window is 15 feet, express the area as a function of the length of one side of the equilateral triangle. 73. Express the area of the rectangle shown in the accompanying gure as a function of the x-coordinate of the point P. yh77. Suppose that a cylindrical mailing container exactly meets the U.S. Postal Service regulations given in Example 3. (See the gure.) Express the volume of the container as a function of the radius of an end.(0, b) P(x, y)(a, 0)x r74. A right triangle is formed by the coordinate axes and a line through the point (2,5). (See the gure.) Express the area of the triangle as a function of the x-intercept.thngLe 1.6 THE ELEMENTARY FUNCTIONS The functions that gure most prominently in single-variable calculus are the polynomials, the rational functions, the trigonometric functions, the exponential functions, and the logarithm functions. These functions are generally known as the elementary functions. Here we review polynomials, rational functions, and trigonometric functions. Exponential and logarithm functions are introduced in Chapter 7.Polynomials We begin with a nonnegative integer n. A function of the form P(x) = an x n + an1 x n1 + + a1 x + a0for all real x,15:42 57. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.6 THE ELEMENTARY FUNCTIONSwhere the coefcients an , an1 , . . . , a1 , a0 are real numbers and an = 0 is called a (real) polynomial of degree n. If n = 0, the polynomial is simply a constant function: P(x) = a0for all real x.Nonzero constant functions are polynomials of degree 0. The function P(x) = 0 for all real x is also a polynomial, but we assign no degree to it. Polynomials satisfy a condition known as the factor theorem: if P is a polynomial and r is a real number, then P(r ) = 0iff(x r ) is a factor of P(x).The real numbers r at which P(x) = 0 are called the zeros of the polynomial. The linear functions P(x) = ax + b,a=0are the polynomials of degree 1. Such a polynomial has only one zero: r = b/a. The graph is the straight line y = ax + b. The quadratic functions P(x) = ax 2 + bx + c,a=0are the polynomials of degree 2. The graph of such a polynomial is the parabola y = ax 2 + bx + c. If a > 0, the vertex is the lowest point on the curve; the curve opens up. If a < 0, the vertex is the highest point on the curve. (See Figure 1.6.1.) vertexvertexa>0a 0.x xx b2 4ac > 0 two real rootsb2 4ac = 0 one real rootb2 4ac < 0 no real rootsFigure 1.6.2Polynomials of degree 3 have the form P(x) = ax 3 + bx 2 + cx + d, a = 0. These functions are called cubics. In general, the graph of a cubic has one of the two following shapes, again determined by the sign of a (Figure 1.6.3). Note that we have not tried to3315:42 58. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200634 CHAPTER 1 PRECALCULUS REVIEW locate these graphs with respect to the coordinate axes. Our purpose here is simply to indicate the two typical shapes. You can see, however, that for a cubic there are three possibilities: three real roots, two real roots, one real root. (Each cubic has at least one real root.)a>0Cubicsa 0, in the clockwise direction if < 0. If = 0, there is no movement; every point remains in place. In degree measure a full turn is effected over the course of 360 . In radian measure, a full turn is effected during the course of 2 radians. (The circumference of a circle of radius 1 is 2 .) Thus0Figure 1.6.62 radians = 360 degrees one radian = 360/2 degrees 57.30 = 0.0175 radians. one degree = 2/360 radians = The following table gives some common angles (rotations) measured both in degrees and in radians. degrees030456090120135150180270360radians01 61 41 31 22 33 45 63 22Cosine and Sine In Figure 1.6.7 you can see a circle of radius 1 centered at the origin of a coordinate plane. We call this the unit circle. On the circle we have marked the point A (1, 0). Now let be any real number. The rotation takes A (1, 0) to some point P, also on the unit circle. The coordinates of P are completely determined by and have namesA(1, 0)Figure 1.6.715:42 60. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200636 CHAPTER 1 PRECALCULUS REVIEW y P(cos , sin ) A(1, 0) xrelated to . The second coordinate of P is called the sine of (we write sin ) and the rst coordinate of P is called the cosine of (we write cos ). Figure 1.6.8 illustrates the idea. To simplify the diagram, we have taken from 0 to 2 . For each real , the rotation and the rotation + 2 take the point A to exactly the same point P. It follows that for each , sin( + 2 ) = sin ,cos( + 2) = cos .In Figure 1.6.9 we consider two rotations: a positive rotation and its negative counterpart . From the gure, you can see that Figure 1.6.8sin( ) = sin ,cos( ) = cos .The sine function is an odd function and the cosine function is an even function. In Figure 1.6.10 we have marked the effect of consecutive rotations of 1 radians: 2 (a, b) (b, a) (a, b) (b, a). In each case, (x, y) (y, x). Thus, sin( + 1 ) = cos , 2cos( + 1 ) = sin . 2A rotation of radians takes each point to the point antipodal to it: (x, y) (x, y). Thus sin( + ) = sin ,cos( + ) = cos .yy (b, a) P(cos , sin )(a, b) xx(a, b)Q(cos( ), sin( )) (b, a)Figure 1.6.9Figure 1.6.10Tangent, Cotangent, Secant, CosecantThere are four other trigonometric functions: the tangent, the cotangent, the secant, the cosecant. These are obtained as follows: tan =sin , cos cos , sin cot =sec =1 , cos csc =1 . sin The most important of these functions is the tangent. Note that the tangent function is an odd function tan() = sin sin( ) = = tan cos( ) cos and repeats itself every radians: tan( + ) =sin( + ) sin = = tan . cos( + ) cos 15:42 61. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.6 THE ELEMENTARY FUNCTIONS Particular ValuesThe values of the sine, cosine, and tangent at angles (rotations) frequently encountered are given in the following table. 01 6sin 0cos 1tan 01 2 1 3 2 1 3 31 4 1 2 2 1 2 211 3 1 3 2 1 2 31 21 0 2 3 1 3 2 1 2 33 4 1 2 2 1 2 25 63 221 2 1 2 3 1 3 31 01010100The (approximate) values of the trigonometric functions for any angle can be obtained with a hand calculator or from a table of values. IdentitiesBelow we list the basic trigonometric identities. Some are obvious; some have just been veried; the rest are derived in the exercises. (i) unit circle sin2 + cos2 = 1,tan2 + 1 = sec2 ,1 + cot2 = csc2 .(the rst identity is obvious; the other two follow from the rst)(ii) periodicity sin( + 2) = sin ,cos( + 2 ) = cos ,tan( + ) = tan (iii) odd and even sin( ) = sin ,cos( ) = cos ,tan() = tan .(the sine and tangent are odd functions; the cosine is even)(iv) sines and cosines sin( + ) = sin , cos( + ) = cos , sin( + 1 ) = cos , cos( + 1 ) = sin , 2 2 sin( 1 ) = cos , cos( 1 ) = sin . 2 2 (only the third pair of identities still has to be veried)(v) addition formulas sin( + ) = sin cos + cos sin , sin( ) = sin cos cos sin , cos( + ) = cos cos sin sin , cos( ) = cos cos + sin sin . (taken up in the exercises)(vi) double-angle formulas sin 2 = 2 sin cos ,cos 2 = cos2 sin2 = 2 cos2 1 = 1 2 sin2 .(follow from the addition formulas) A function f with an unbounded domain is said to be periodic if there exists a number p > 0 such that, if is in the domain of f, then + p is in the domain and f ( + p) = f ( ). The least number p with this property (if there is a least one) is called the period of the function. The sine and cosine have period 2 . Their reciprocals, the cosecant and secant, also have period 2. The tangent and cotangent have period .3715:42 62. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200638 CHAPTER 1 PRECALCULUS REVIEW (vii) half-angle formulas sin2 = 1 (1 cos 2 ), 2cos2 = 1 (1 + cos 2) 2(follow from the double-angle formulas)sin =adjacent sidecsc =hypotenuse , opposite sideadjacent side , hypotenusesec =hypotenuse , adjacent sidetan =opposite side , hypotenusecos =opposite sidehypotenuseIn Terms of a Right Triangle For angles between 0 and /2, the trigonometric functions can also be dened as ratios of the sides of a right triangle. (See Figure 1.6.11.)opposite side , adjacent sidecot =adjacent side . opposite side(Exercise 81)Figure 1.6.11 Arbitrary TrianglesLet a, b, c be the sides of a triangle and let A, B, C be the opposite angles. (See Figure 1.6.12.) areaAblaw of sinesc C BaFigure 1.6.121 ab 2sin C = 1 ac sin B = 1 bc sin A. 2 2sin A sin B sin C = = . a b c(taken up in the exercises)law of cosines a 2 = b2 + c2 2bc cos A, b2 = a 2 + c2 2ac cos B, c2 = a 2 + b2 2ab cos C.Usually we work with functions y = f (x) and graph them in the x y-plane. To bring the graphs of the trigonometric functions into harmony with this convention, we replace by x and write y = sin x, y = cos x, y = tan x. (These are the only functions that we are going to graph here.) The functions have not changed, only the symbols: x is the rotation that takes A(1, 0) to the point P(cos x, sin x). The graphs of the sine, cosine, and tangent appear in Figure 1.6.13. The graphs of sine and cosine are waves that repeat themselves on every interval of length 2. These waves appear to chase each other. They do chase each other. In the chase the cosine wave remains 1 units behind the sine wave: 2Graphscos x = sin(x + 1 ). 2 Changing perspective, we see that the sine wave remains 3 units behind the cosine 2 wave: sin x = cos(x + 3 ). 2 All these waves crest at y = 1, drop down to y = 1, and then head up again. The graph of the tangent function consists of identical pieces separated every units by asymptotes that mark the points x where cos x = 0.15:42 63. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.6 THE ELEMENTARY FUNCTIONS39y y=1sine 5 2 2 3 1 21 2 23 225 2x y = 1y = sin x period 2 ycosiney=1 5 2 23 2 1 21 2 3 222x5 2x y = 1y = cos x period 2 y tangent2y = tan x period 1vertical asymptotes x = (n + ), n an integer 2Figure 1.6.13EXERCISES 1.6 Exercises 110. State whether the function is a polynomial, a rational function (but not a polynomial), or neither a polynomial nor a rational function. If the function is a polynomial, give the degree. 1. f (x) = 3.2. f (x) = 1 + 1 x. 21 . 4. x x 3 3x 2/2 + 2x . 6. 5. F(x) = x2 1 7. f (x) = x( x + 1). 8. x2 + 1 9. f (x) = 2 . 10. x 1 3. g(x) =h(x) =x2 4 . 2f (x) = 5x 4 x 2 + 1 . 2 x 2 2x 8 . x +2 ( x + 2)( x 2) h(x) = . x2 + 4 g(x) =Exercises 1116. Determine the domain of the function and sketch the graph. 1 . 12. f (x) = 11. f (x) = 3x 1 . 2 x +1 13. g(x) = x 2 x 6. 14. F(x) = x 3 x. 1 1 15. f (x) = 2 . 16. g(x) = x + . x 4 x Exercises 1722. Convert the degree measure into radian measure. 18. 210 . 17. 225 . 20. 450 . 19. 300 . 22. 3 . 21. 15 . Exercises 2328. Convert the radian measure into degree measure. 23. 3/2. 24. 5/4.15:42 64. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200640 CHAPTER 1 PRECALCULUS REVIEW 25. 5/3. 26. 11/6. 27. 2. 28. 3. 29. Show that in a circle of radius r, a central angle of radians subtends an arc of length r . 30. Show that in a circular disk of radius r, a sector with a central angle of radians has area 1 r 2 . Take between 0 and 2. 2 HINT: The area of the circle is r 2 . Exercises 3138. Find the number(s) x in the interval [0, 2] which satisfy the equation. 31. sin x = 1/2. 32. cos x = 1/2. 33. tan x/2 = 1. 35. cos x = 2/2. 37. cos 2x = 0.34. sin x = 1. 36. sin 2x = 3/2. 38. tan x = 3. 2Exercises 3944. Evaluate to four decimal place accuracy. 39. sin 51 . 40. cos 17 . 41. sin(2.352). 42. cos(13.461). 44. cot(7.311). 43. tan 72.4 . Exercises 4552. Find the solutions x that are in the interval [0, 2]. Express your answers in radians and use four decimal place accuracy. 45. sin x = 0.5231. 46. cos x = 0.8243. 47. tan x = 6.7192. 48. cot x = 3.0649. 49. sec x = 4.4073. 50. csc x = 10.260. Exercises 5152. Solve the equation f (x) = y0 for x in [0, 2 ] by using a graphing utility. Display the graph of f and the line y = y0 in one gure; then use the trace function to nd the point(s) of intersection. 51. f (x) = sin 3x; y0 = 1/ 2. 52. f (x) = cos 1 x; y0 = 3 2 4 Exercises 5358. Give the domain and range of the function. 53. f (x) = sin x 54. g(x) = sin2 x + cos2 x. 55. f (x) = 2 cos 3x. 56. F(x) = 1 + sin x. 2 58. h(x) = cos2 x. 57. f (x) = 1 + tan x. Exercises 5962. Determine the period. (The least positive number p for which f (x + p) = f (x) for all x.) 59. f (x) = sin x. 60. f (x) = cos 2x. 1 62. f (x) = sin 1 x. 61. f (x) = cos 3 x. 2 Exercises 6368. Sketch the graph of the function. 63. f (x) = 3 sin 2x. 64. f (x) = 1 + sin x. 65. g(x) = 1 cos x. 66. F(x) = tan 1 x. 2 67. f (x) =2sin x.68. g(x) = 2 cos x.Exercises 6974. State whether the function is odd, even, or neither. 69. f (x) = sin 3x. 70. g(x) = tan x. 71. f (x) = 1 + cos 2x. 72. g(x) = sec x.cos x . x2 + 1 75. Suppose that l1 and l2 are two nonvertical lines. If m 1 m 2 = 1, then l1 and l2 intersect at right angles. Show that if l1 and l2 do not intersect at right angles, then the angle between l1 and l2 (see Section 1.4) is given by the formula 73. f (x) = x 3 + sin x.74. h(x) =tan =m1 m2 . 1 + m1m2HINT: Derive the identity tan(1 2 ) =tan 1 tan 2 1 + tan 1 tan 2by expressing the right side in terms of sines and cosines. Exercises 7679. Find the point where the lines intersect and determine the angle between the lines. 76. l1 : 4x y 3 = 0, l2 : 3x 4y + 1 = 0. 77. l1 : 3x + y 5 = 0, l2 : 7x 10y + 27 = 0. 78. l1 : 4x y + 2 = 0, l2 : 19x + y = 0. 79. l1 : 5x 6y + 1 = 0, l2 : 8x + 5y + 2 = 0. 1, x rational 80. Show that the function f (x) = is periodic 0, x irrational but has no period. 81. Verify that, for angles between 0 and /2, the denition of the trigonometric functions in terms of the unit circle and the denitions in terms of a right triangle are in agreement. HINT: Set the triangle as in the gure. y (cos , sin ) (1, 0)xThe setting for Exercises 82, 83, 84 is a triangle with sides a, b, c and opposite angles A, B, C. 82. Show that the area of the triangle is given by the formula A = 1 ab sin C. 2 83. Conrm the law of sines: sin B sin C sin A = = . a b c HINT: Drop a perpendicular from one vertex to the opposite side and use the two right triangles formed. 84. Conrm the law of cosines: a 2 = b2 + c2 2bc cos A. HINT: Drop a perpendicular from angle B to side b and use the two right triangles formed.15:42 65. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.7 COMBINATIONS OF FUNCTIONS85. Verify the identity cos( ) = cos cos + sin sin . HINT: With P and Q as in the accompanying gure, calculate the length of P Q by applying the law of cosines. O41Conjecture a property shared by the graphs of all polynomials of the form P(x) = x 4 + ax 3 + bx 2 + cx + d. Make an analogous conjecture for polynomials of the form. Q(x) = x 4 + ax 3 + bx 2 + cx + d.Q (cos , sin ) P(cos , sin )92. (a) Use a graphing utility to graph the polynomials.A(1, 0)f (x) = x 5 7x 3 + 6x + 2, g(x) = x 5 + 5x 3 3x 3.86. Use Exercise 85 to show that cos( + ) = cos cos sin sin .(b) Based on your graphs in part (a), make a conjecture about the general shape of the graph of a polynomial of degree 5. (c) Now graph87. Verify the following identities: sin( 1 ) = cos , 2P(x) = x 5 + ax 4 + bx 3 + cx 2 + d x + e cos( 1 ) = sin . 288. Verify that sin( + ) = sin cos + cos sin . HINT: sin( + ) = cos[( 1 ) ]. 2 89. Use Exercise 88 to show that sin( ) = sin cos cos sin . 90. It has been said that all of trigonometry lies in the undulations of the sine wave. Explain. 91. (a) Use a graphing utility to graph the polynomials f (x) = x 4 + 2x 3 5x 2 3x + 1, g(x) = x 4 + x 3 + 4x 2 3x + 2. (b) Based on your graphs in part (a), make a conjecture about the general shape of the graphs of polynomials of degree 4. (c) Test your conjecture by graphing f (x) = x 4 4x 2 + 4x + 2andg(x) = x 4 .for several choices of a, b, c, d, e. (For example, try a = b = c = d = e = 0.) How do these graphs compare with your graph of f from part (a)? 93. (a) Use a graphing utility to graph f A (x) = A cos x for several values of A; use both positive and negative values. Compare your graphs with the graph of f (x) = cos x. (b) Now graph f B (x) = cos Bx for several values of B. Since the cosine function is even, it is sufcient to use only positive values for B. Use some values between 0 and 1 and some values greater than 1. Again, compare your graphs with the graph of f (x) = cos x. (c) Describe the effects that the coefcients A and B have on the graph of the cosine function. 94. Let f n (x) = x n , n = 1, 2, 3 . . . . (a) Using a graphing utility, draw the graphs of f n for n = 2, 4, 6 in one gure, and in another gure draw the graphs of f n for n = 1, 3, 5. (b) Based on your results in part (a), make a general sketch of the graph of f n for even n and for odd n. (c) Given a positive integer k, compare the graphs of f k and f k+1 on [0, 1] and on (1, ). 1.7 COMBINATIONS OF FUNCTIONS In this section we review the elementary ways of combining functions.Algebraic Combinations of Functions Here we discuss with some precision ideas that were used earlier without comment. On the intersection of their domains, functions can be added and subtracted: ( f + g)(x) = f (x) + g(x),( f g)(x) = f (x) g(x);they can be multiplied: ( f g)(x) = f (x)g(x);15:42 66. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200642 CHAPTER 1 PRECALCULUS REVIEW and, at the points where g(x) = 0, we can form the quotient: f g(x) =f (x) , g(x)a special case of which is the reciprocal: 1 1 (x) = . g g(x)Example 1 Let f (x) =x +3andg(x) = 5 x 2.(a) Give the domain of f and of g. (b) Determine the domain of f + g and specify ( f + g)(x). (c) Determine the domain of f /g and specify ( f /g)(x). SOLUTION (a) We can form x + 3 iff x + 3 0, which holds iff x 3. Thus dom( f ) = [3, ). We can form 5 x 2 iff 5 x 0, which holds iff x 5. Thus dom(g) = (, 5]. (b) dom( f + g) = dom( f ) dom(g) = [3, ) (, 5] = [3, 5], ( f + g)(x) = f (x) + g(x) = x + 3 + 5 x 2. (c) To obtain the domain of the quotient, we must exclude from [ 3, 5] the numbers x at which g(x) = 0. There is only one such number: x = 1. Therefore domf g= {x [3, 5] : x = 1} = [3, 1) (1, 5], f g x +3 f (x) . (x) = = g(x) 5x 2We can multiply functions f by real numbers and form what are called scalar multiples of f : ( f )(x) = f (x). With functions f and g and real numbers and , we can form linear combinations: ( f + g)(x) = f (x) + g(x). These are just specic instances of the products and sums that we dened at the beginning of the section. You have seen all these algebraic operations many times before: (i) The polynomials are simply nite linear combinations of powers x n , each of which is a nite product of identity functions f (x) = x. (Here we are taking the point of view that x 0 = 1.) (ii) The rational functions are quotients of polynomials. (iii) The secant and cosecant are reciprocals of the cosine and the sine. (iv) The tangent and cotangent are quotients of sine and cosine. Vertical Translations (Vertical Shifts) Adding a positive constant c to a function raises the graph by c units. Subtracting a positive constant c from a function lowers the graph by c units. (Figure 1.7.1.)15:42 67. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.7 COMBINATIONS OF FUNCTIONS yy = cos x + 2 x y = cos x y = cos x 2Figure 1.7.1Composition of Functions You have seen how to combine functions algebraically. There is another (probably less familiar) way to combine functions, called composition. To describe it, we begin with two functions, f and g, and a number x in the domain of g. By applying g to x, we get the number g(x). If g(x) is in the domain of f, then we can apply f to g(x) and thereby obtain the number f (g(x)). What is f (g(x))? It is the result of rst applying g to x and then applying f to g(x). The idea is illustrated in Figure 1.7.2. This new functionit takes x in the domain of g to g(x) in the domain of f, and assigns to it the value f (g(x))is called the composition of f with g and is denoted by f g. (See Figure 1.7.3.) The symbol f g is read f circle g. g gff g(x) f (g(x))xg(x) f (g(x)) xf gFigure 1.7.2Figure 1.7.3DEFINITION 1.7.1 COMPOSITIONLet f and g be functions. For those x in the domain of g for which g(x) is in the domain of f, we dene the composition of f with g, denoted f g, by setting ( f g)(x) = f (g(x)).In set notation, dom( f g) = {x dom(g) : g(x) dom( f )}Example 2 Suppose that g(x) = x 2(the squaring function)4315:42 68. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200644 CHAPTER 1 PRECALCULUS REVIEW and f (x) = x + 3.(the function that adds 3)Then ( f g)(x) = f (g(x)) = g(x) + 3 = x 2 + 3. Thus, f g is the function that rst squares and then adds 3. On the other hand, the composition of g with f gives (g f )(x) = g( f (x)) = (x + 3)2 . Thus, g f is the function that rst adds 3 and then squares. Since f and g are everywhere dened, both f g and g f are also everywhere dened. Note that g f is not the same as f g. Example 3 Let f (x) = x 2 1 and g(x) = 3 x.The domain of g is (, 3]. Since f is everywhere dened, the domain of f g is also (, 3]. On that interval 2 3 x 1 = (3 x) 1 = 2 x. ( f g)(x) = f (g(x)) = Since g( f (x)) = 3 f (x), we can form g( f (x)) only for those x in the domain of f for which f (x) 3. As you can verify, this is the set [2, 2]. On [2, 2] 3 (x 2 1) =(g f )(x) = g( f (x)) =4 x 2. Horizontal Translations (Horizontal Shifts)Adding a positive constant c to the argument of a function shifts the graph c units left: the function g(x) = f (x + c) takes on at x the value that f takes on at x + c. Subtracting a positive constant c from the argument of a function shifts the graph c units to the right: the function h(x) = f (x c) takes on at x the value that f takes on at x c. (See Figure 1.7.4.) y gf2 g (x)2= f (x + 2)h(x)hx= f (x 2)Figure 1.7.4We can form the composition of more than two functions. For example, the triple composition f g h consists of rst h, then g, and then f : ( f g h)(x) = f [g(h(x))].15:42 69. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.7 COMBINATIONS OF FUNCTIONS45We can go on in this manner with as many functions as we like.Example 4 If f (x) =1 , xg(x) = x 2 + 1,( f g h)(x) = f [g(h(x))] =then=cos21 . x +1h(x) = cos x, 1 1 = g(h(x)) [h(x)]2 + 1Example 5 Find functions f and g such that f g = F given that F(x) = (x + 1)5 . A SOLUTION The function consists of rst adding 1 and then taking the fth power. We can therefore set g(x) = x + 1 (adding 1)andf (x) = x 5 .(taking the fth power)As you can see, ( f g)(x) = f (g(x)) = [g(x)]5 = (x + 1)5 . Example 6 Find three functions f, g, h such that f g h = F given that 1 . F(x) = x + 3 A SOLUTION F takes the absolute value, adds 3, and then inverts. Let h take theabsolute value: seth(x) = x .Let g add 3: setg(x) = x + 3.Let f do the inverting: setf (x) =1 . xWith this choice of f, g, h, we have ( f g h)(x) = f [g(h(x))] =1 1 1 = = . g(h(x)) h(x) + 3 x + 3EXERCISES 1.7 Exercises 18. Set f (x) = 2x 2 3x + 1 and g(x) = x 2 + 1/x. Calculate the indicated value. 1. ( f + g)(2). 3. ( f g)(2). 5. (2 f 3g)( 1 ). 22. ( f g)(1). f (1). 4. g f + 2g 6. (1). f7. ( f g)(1). 8. (g f )(1). Exercises 912. Determine f + g, f g, f g, f /g, and give the domain of each 9. f (x) = 2x 3, g(x) = 2 x. 10. f (x) = x 2 1, g(x) = x + 1/x. 11. f (x) = x 1, g(x) = x x + 1. 12. f (x) = sin2 x, g(x) = cos 2x.15:42 70. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200646 CHAPTER 1 PRECALCULUS REVIEW 13. Given that f (x) = x + 1/ x and g(x) = x 2 x, nd (a) 6 f + 3g, (b) f g, (c) f /g. 14. Given that 1 x, 2x 1,f (x) =x 1 and g(x) = x > 1,0, 1,x 1.How is f dened for x < 0 if (a) f is even? (b) f is odd? For x 0, f (x) = x 2 x. How is f dened for x < 0 if (a) f is even? (b) f is odd? Given that f is dened for all real numbers, show that the function g(x) = f (x) + f (x) is an even function. Given that f is dened for all real numbers, show that the function h(x) = f (x) f (x) is an odd function. Show that every function dened for all real numbers can be written as the sum of an even function and an odd function. For x = 0, 1, dene 1 , f 3 (x) = 1 x, x 1 x 1 x , f 5 (x) = , f 6 (x) = . f 4 (x) = 1x x x 1 f 1 (x) = x,f 2 (x) =This family of functions is closed under composition; that is, the composition of any two of these functions is again one of these functions. Tabulate the results of composing these functions one with the other by lling in the table shown in the gure. To indicate that f i f j = f k , write f k in the ith row, jth column. We have already made two entries in the table. Check out these two entries and then ll in the rest of the table.15:42 71. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.8 A NOTE ON MATHEMATICAL PROOF; MATHEMATICAL INDUCTION f1f2f3f4f5f6f1 f2 f3f6 f2f4 f5 f63x Exercises 6162. Set f (x) = x 2 4, g(x) = , h(x) = 2x 2x x + 4, and k(x) = . Use a CAS to nd the indicated 3+x composition. 61. (a) f g; (b) g k; (c) f k g. 62. (a) g f ; (b) k g; (c) g f k. Exercises 63 and 64. Set f (x) = x 2 and F(x) = (x a)2 + b. 63. (a) Choose a value for a and, using a graphing utility, graph F for several different values of b. Be sure to choose47both positive and negative values. Compare your graphs with the graph of f, and describe the effect that varying b has on the graph of F. (b) Now x a value of b and graph F for several values of a; again, use both positive and negative values. Compare your graphs with the graph of f, and describe the effect that varying a has on the graph of F. (c) Choose values for a and b, and graph F. What effect does changing the sign of F have on the graph? 64. For all values of a and b, the graph of F is a parabola which opens upward. Find values for a and b such that the parabola will have x-intercepts at 3 and 2. Verify your result alge2 braically. Exercises 6566. Set f (x) = sin x. 65. (a) Using a graphing utility, graph cf for c = 3, 2, 1, 2, 3. Compare your graphs with the graph of f. (b) Now graph g(x) = f (cx) for c = 3, 2, 1 , 1 , 1 , 2, 3. 2 3 2 Compare your graphs with the graph of f. 66. (a) Using a graphing utility, graph g(x) = f (x c) for c = 1 , 1 , 1 , 1 , , 2. Compare your graphs 2 4 3 2 with the graph of f. (b) Now graph g(x) = a f (bx c) for several values of a, b, c. Describe the effect of a, the effect of b, the effect of c. 1.8 A NOTE ON MATHEMATICAL PROOF; MATHEMATICAL INDUCTION Mathematical Proof The notion of proof goes back to Euclids Elements, and the rules of proof have changed little since they were formulated by Aristotle. We work in a deductive system where truth is argued on the basis of assumptions, denitions, and previously proved results. We cannot claim that such and such is true without clearly stating the basis on which we make that claim. A theorem is an implication; it consists of a hypothesis and a conclusion: if (hypothesis) . . . , then (conclusion) . . . . Here is an example: If a and b are positive numbers, then ab is positive. A common mistake is to ignore the hypothesis and persist with the conclusion: to insist, for example, that ab > 0 just because a and b are numbers. Another common mistake is to confuse a theorem if A, then B with its converse if B, then A. The fact that a theorem is true does not mean that its converse is true: While it is true that if a and b are positive numbers, then ab is positive,15:42 72. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200648 CHAPTER 1 PRECALCULUS REVIEW it is not true that if ab is positive, then a and b are positive numbers; [(2)(3) is positive but 2 and 3 are not positive]. A third, more subtle mistake is to assume that the hypothesis of a theorem represents the only condition under which the conclusion is true. There may well be other conditions under which the conclusion is true. Thus, for example, not only is it true that if a and b are positive numbers, then ab is positive but it is also true that if a and b are negative numbers, then ab is positive. In the event that a theorem if A, then B and its converse if B, then A are both true, then we can write A if and only if Bor more brieyA iff B.We know, for example, that ifx 0,then x = x;if x = x,we also know that then x 0.We can summarize this by writing x 0iff x = x.Remark Well use iff frequently in this text but not in denitions. As stated earlier in a footnote, denitions are by their very nature iff statements. For example, we can say that a number r is called a zero of P if P(r ) = 0; we dont have to say a number r is called a zero of P iff P(r ) = 0. In this situation, the only if part is taken for granted. A nal point. One way of proving if A, then B is to assume that (1)A holds and B does not holdand then arrive at a contradiction. The contradiction is taken to indicate that (1) is a false statement and therefore if A holds, then B must hold. Some of the theorems of calculus are proved by this method. Calculus provides procedures for solving a wide range of problems in the physical and social sciences. The fact that these procedures give us answers that seem to make sense is comforting, but it is only because we can prove our theorems that we can have condence in the mathematics that is being applied. Accordingly, the study of calculus should include the study of some proofs.15:42 73. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.8 A NOTE ON MATHEMATICAL PROOF; MATHEMATICAL INDUCTIONMathematical Induction Mathematical induction is a method of proof which can be used to show that certain propositions are true for all positive integers n. The method is based on the following axiom:1.8.1 AXIOM OF INDUCTIONLet S be a set of positive integers. If (A) 1 S, and (B) k S implies that k + 1 S, then all the positive integers are in S.You can think of the axiom of induction as a kind of domino theory. If the rst domino falls (Figure 1.8.1), and if each domino that falls causes the next one to fall, then, according to the axiom of induction, each domino will fall.domino theoryFigure 1.8.1While we cannot prove that this axiom is valid (axioms are by their very nature assumptions and therefore not subject to proof), we can argue that it is plausible. Lets assume that we have a set S that satises conditions (A) and (B). Now lets choose a positive integer m and argue that m S. From (A) we know that 1 S. Since 1 S, we know that 1 + 1 S, and thus that (1 + 1) + 1 S, and so on. Since m can be obtained from 1 by adding 1 successively (m 1) times, it seems clear that m S. To prove that a given proposition is true for all positive integers n, we let S be the set of positive integers for which the proposition is true. We prove rst that 1 S; that is, that the proposition is true for n = 1. Next we assume that the proposition is true for some positive integer k, and show that it is true for k + 1; that is, we show that k S implies that k + 1 S. Then by the axiom of induction, we conclude that S contains the set of positive integers and therefore the proposition is true for all positive integers.Example 1 Well show that 1 + 2 + 3 + + n =n(n + 1) 2for all positive integers n.4915:42 74. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200650 CHAPTER 1 PRECALCULUS REVIEW SOLUTION Let S be the set of positive integers n for which1 + 2 + 3 + + n =n(n + 1) . 2Then 1 S since 1(1 + 1) . 2 Next, we assume that k S; that is, we assume that 1=1 + 2 + 3 + k =k(k + 1) . 2Adding up the rst k + 1 integers, we have 1 + 2 + 3 + + k + (k + 1) = [1 + 2 + 3 + + k] + (k + 1) k(k + 1) + (k + 1) (by the induction hypothesis) 2 k(k + 1) + 2(k + 1) = 2 (k + 1)(k + 2) , = 2 and so k + 1 S. Thus, by the axiom of induction, we can conclude that all positive integers are in S; that is, we can conclude that =1 + 2 + 3 + + n =n(n + 1) 2for all positive integers n. Example 2 Well show that, if x 1, then (1 + x)n 1 + nxfor all positive integers n.SOLUTION We take x 1 and let S be the set of positive integers n for which(1 + x)n 1 + nx. Since (1 + x)1 = 1 + 1 x, we have 1 S. We now assume that k S. By the denition of S, (1 + x)k 1 + kx. Since (1 + x)k+1 = (1 + x)k (1 + x) (1 + kx)(1 + x) and (1 + kx)(1 + x) = 1 + (k + 1)x + kx 2 1 + (k + 1)x, we can conclude that (1 + x)k+1 1 + (k + 1)x and thus that k + 1 S. We have shown that 1Sand thatkSimpliesk + 1 S.(explain)15:42 75. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 20061.8 A NOTE ON MATHEMATICAL PROOF; MATHEMATICAL INDUCTIONBy the axiom of induction, all positive integers are in S.51Remark An induction does not have to begin with the integer 1. If, for example, you want to show that some proposition is true for all integers n 3, all you have to do is show that it is true for n = 3, and that, if it is true for n = k, then it is true for n = k + 1. (Now you are starting the chain reaction by pushing on the third domino.) EXERCISES 1.8 Exercises 110. Show that the statement holds for all positive integers n. 2. 1 + 2n 3n . 1. 2n 2n . 0 1 2 3 n1 = 2n 1. 3. 2 + 2 + 2 + 2 + + 2 4. 1 + 3 + 5 + + (2n 1) = n 2 .13. Find a simplifying expression for the product 116. 13 + 23 + 33 + + n 3 = (1 + 2 + 3 + + n)2 . HINT: Use Example 1. 15.8. 12 + 22 + + (n 1)2 < 1 n 3 < 12 + 22 + + n 2 . 3 1 1 1 1 9. + + + + > n. n 1 2 3 1 1 1 n 1 + + + + = . 10. 12 23 34 n(n + 1) n+1 11. For what integers n is 3 +2 divisible by 7? Prove that your answer is correct. 12. For what integers n is 9n 8n 1 divisible by 64? Prove that your answer is correct. 2n+111 1 1 3 nand verify its validity for all integers n 2. 14. Find a simplifying expression for the product5. 12 + 22 + 32 + + n 2 = 1 n(n + 1)(2n + 1). 67. 13 + 23 + + (n 1)3 < 1 n 4 < 13 + 23 + + n 3 . 41 216. 17. 18.n+219.1 2211 32 1 1 n2and verify its validity for all integers n 2. Prove that an N -sided convex polygon has 1 N (N 3) di2 agonals. Take N > 3. Prove that the sum of the interior angles in an N -sided convex polygon is (N 2)180 . Take N > 2. Prove that all sets with n elements have 2n subsets. Count the empty set and the whole set as subsets. Show that, given a unit length, for each positive integer n, a line segment of length n can be constructed by straight edge and compass. Find the rst integer n for which n 2 n + 41 is not a prime number. CHAPTER 1. REVIEW EXERCISES Exercises 14. Is the number rational or irrational? 1. 1.25. 2. 16/9. 4. 1.001001001 . . . . 3. 5 + 1. Exercises 58. State whether the set is bounded above, bounded below, bounded. If the set is bounded above, give an upper bound; if it is bounded below, give a lower bound; if it is bounded, give an upper bound and a lower bound. 5. S = {1, 3, 5, 7, }. 6. S = {x : x 1}. 7. S = {x : x + 2 < 3}. 8. S = {(1/n)n : n = 1, 2, 3, }. Exercises 912. Find the real roots of the equation. 10. x 2 + 2x + 5 = 0. 9. 2x 2 + x 1 = 0. 2 12. 9x 3 x = 0. 11. x 10x + 25 = 0.Exercises 1322. Solve the inequality. Express the solution as an interval or as the union of intervals. Mark the solution on a number line. 13. 5x 2 < 0. 14. 3x + 5 < 1 (4 x). 2 15. x 2 x 6 0. 16. x(x 2 3x + 2) 0. x 2 4x + 4 x +1 > 0. 18. 2 0. 17. (x + 2)(x 2) x 2x 3 19. x 2 < 1. 20. 3x 2 4. 2 5 21. >2 22. < 1. x +4 x +1 Exercises 2324. (a) Find the distance between the points P, Q. (b) Find the midpoint of the line segment P Q. 23. P(2, 3), Q(1, 4). 24. P(3, 4), Q(1, 6).15:42 76. P1: PBU/OVYP2: PBU/OVYJWDD023-01JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 5, 200652 CHAPTER 1 PRECALCULUS REVIEW Exercises 2528. Find an equation for the line that passes through the point (2, 3) and is 25. parallel to the y-axis. 26. parallel to the line y = 1. 27. perpendicular to the line 2x 3y = 6. 28. parallel to the line 3x + 4y = 12. Exercises 2930. Find the point where the lines intersect. l2 : 3x + 4y = 3. 29. l1 : x 2y = 4, l2 : 3x + 2y = 0. 30. l1 : 4x y = 2, 31. Find the point(s) where the line y = 8x 6 intersects the parabola y = 2x 2 . 32. Find an equation for the line tangent to the circle x 2 + y 2 + 2x 6y 3 = 0 at the point (2, 1). Exercises 3338. Give the domain and range of the function. 34. f (x) = 3x 2. 33. f (x) = 4 x 2 . 36. f (x) = 1 1 4x 2 . 35. f (x) = x 4. 2 37. f (x) = 1 + 4x 2 . 38. f (x) = 2x + 1 . Exercises 3940. Sketch the graph and give the domain and range of the function. 4 2x, x 2 39. f (x) = . x 2, x > 2 x + 2, x 0 . 2 x 2, x > 0 Exercises 4144. Find the number(s) x in the interval [0, 2] which satisfy the equation. 42. cos 2x = 1 . 41. sin x = 1 . 2 2 43. tan(x/2) = 1. 44. sin 3x = 0. Exercises 4548. Sketch the graph of the function. 45. f (x) = cos 2x. 46. f (x) = cos 2x. 47. f (x) = 3 cos 2x. 48. f (x) = 1 cos 2x. 3 Exercises 4951. Form the combinations f + g, f g, f g, f /g and specify the domain of combination. 49. f (x) = 3x + 2, g(x) = x 2 1. 50. f (x) = x 2 4, g(x) = x + 1/x. 51. f (x) = cos2 x, g(x) = sin 2x, for x [0, 2 ]. 40. f (x) =2Exercises 5254. Form the compositions f g and g f , and specify the domain of each of these combinations. 52. f (x) = x 2 2x, g(x) = x + 1. 53. f (x) = x + 1, g(x) = x 2 5. 54. f (x) = 1 x 2 , g(x) = sin 2x. 55. (a) Write an equation in x and y for an arbitrary line l that passes through the origin. (b) Verify that if P(a, b) lies on l and is a real number, then the point Q(a, b) also lies on l. (c) What additional conclusion can you draw if > 0? if < 0? 56. The roots of a quadratic equation. You can nd the roots of a quadratic equation by resorting to the quadratic formula. The approach outlined below is more illuminating. Since division by the leading coefcient does not alter the roots of the equation, we can make the coefcient 1 and work with the equation x 2 + ax + b = 0. (a) Show that the equation x 2 + ax + b = 0 can be written as (x )2 2 = 0,or(x ) = 0,or2(x )2 + 2 = 0. HINT: Set = a/2, complete the square, and go on from there. (b) What are the roots of the equation (x )2 2 = 0? (c) What are the roots of the equation (x )2 = 0? (d) Show that the equation (x )2 + 2 = 0 has no real roots. 57. Knowing that a + b a + b for all real a, b show that a b a b for all real a, b. 58. (a) Express the perimeter of a semicircle as a function of the diameter. (b) Express the area of a semicircle as a function of the diameter.15:42 77. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER2LIMITS AND CONTINUITY 2.1 THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION) We could begin by saying that limits are important in calculus, but that would be a major understatement. Without limits, calculus would not exist. Every single notion of calculus is a limit in one sense or another. For example,yT P QWhat is the slope of a curve? It is the limit of slopes of secant lines. (Figure 2.1.1.) What is the length of a curve? It is the limit of the lengths of polygonal paths inscribed in the curve. (Figure 2.1.2) What is the area of a region bounded by a curve? It is the limit of the sum of areas of approximating rectangles. (Figure 2.1.3)xFigure 2.1.1 yyyyR x region Rxy1 approximating rectanglexy8 approximating rectanglesx3 approximating rectanglesxFigure 2.1.2y14 approximating rectanglesxxFigure 2.1.35313:24 78. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200654 CHAPTER 2 LIMITS AND CONTINUITY The Idea of Limit Technically there are several limit processes, but they are all very similar. Once you master one of them, the others will pose few difculties. The limit process that we start with is the one that leads to the notion of continuity and the notion of differentiability. At this stage our approach is completely informal. All we are trying to do here is lay an intuitive foundation for the mathematics that begins in Section 2.2 We start with a number c and a function f dened at all numbers x near c but not necessarily at c itself. In any case, whether or not f is dened at c and, if so, how is totally irrelevant. Now let L be some real number. We say that the limit of f (x) as x tends to c is L and write lim f (x) = Lxcprovided that (roughly speaking) as x approaches c, f(x) approaches L or (somewhat more precisely) provided that f (x) is close to L for all x = c which are close to c. Lets look at a few functions and try to apply this limit idea. Remember, our work at this stage is entirely intuitive.Example 1 Set f (x) = 4x + 5 and take c = 2. As x approaches 2, 4x approaches 8 and 4x + 5 approaches 8 + 5 = 13. We conclude that lim f (x) = 13. x2 Example 2 Set f (x) = 1 x and take c = 8. As x approaches 8, 1 x approaches 9 and 1 x approaches 3. We conclude that lim f (x) = 3.x8If for that same function we try to calculate lim f (x),x2 we run into a problem. The function f (x) = 1 x is dened only for x 1. It is therefore not dened for x near 2, and the idea of taking the limit as x approaches 2 makes no sense at all: lim f (x)does not exist.lim5 x 3 2x + 4 = . 2+1 x 2x2Example 3 x3First we work with the numerator: as x approaches 3, x 3 approaches 27, 2x approaches 6, and x 3 2x + 4 approaches 27 6 + 4 = 25. Now for the denominator: as x approaches 3, x 2 + 1 approaches 10. The quotient (it would seem) approaches 25/10 = 5/2. 13:24 79. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.1 THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION)The curve in Figure 2.1.4 represents the graph of a function f. The number c is on the x-axis and the limit L is on the y-axis. As x approaches c along the x-axis, f (x) approaches L along the y-axis.y ff (x) L f (x)xcxxFigure 2.1.4As we have tried to emphasize, in taking the limit of a function f as x tends to c, it does not matter whether f is dened at c and, if so, how it is dened there. The only thing that matters is the values taken on by f at numbers x near c. Take a look at the three cases depicted in Figure 2.1.5. In the rst case, f (c) = L. In the second case, f is not dened at c. In the third case, f is dened at c, but f (c) = L. However, in each case lim f (x) = Lxcbecause, as suggested in the gures, as x approaches c, f (x) approaches L.y y yf (c) f (x)ff (x) ff (x)fLL L f (x)f (x)f (x)xcxxx(a)cx(b)xx (c)Figure 2.1.5x2 9 and let c = 3. Note that the function f is not dened x 3 at 3: at 3, both numerator and denominator are 0. But that doesnt matter. For x = 3, and therefore for all x near 3,Example 4 Set f (x) =(x 3)(x + 3) x2 9 = = x + 3. x 3 x 3cxx5513:24 80. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200656 CHAPTER 2 LIMITS AND CONTINUITY yTherefore, if x is close to 3, then f (x)fthat6x2 9 = x + 3 is close to 3 + 3 = 6. We conclude x 3x2 9 = lim (x + 3) = 6. x3 x 3 x3 The graph of f is shown in Figure 2.1.6. limf (x)x3xxExample 5Figure 2.1.6limx2x3 8 = 12. x 2x 8 is undened at x = 2. But, as we said before, that doesnt x 2 3The function f (x) = matter. For all x = 2,x3 8 (x 2)(x 2 + 2x + 4) = = x 2 + 2x + 4. x 2 x 2 Therefore, limx2x3 8 = lim (x 2 + 2x + 4) = 12. x2 x 2Example 6 If f (x) =3x 4, 10,x =0 x = 0,then lim f (x) = 4. x0It does not matter that f (0) = 10. For x = 0, and thus for all x near 0, f (x) = 3x 4and therefore lim f (x) = lim (3x 4) = 4. x0x0One-Sided Limits Numbers x near c fall into two natural categories: those that lie to the left of c and those that lie to the right of c. We write lim f (x) = Lxc[The left-hand limit of f(x) as x tends to c is L.]to indicate that as x approaches c from the left, f(x) approaches L. We write lim f (x) = Lxc+[The right-hand limit of f(x) as x tends to c is L.]to indicate that as x approaches c from the right, f(x) approaches L.The left-hand limit is sometimes written lim f (x) and the right-hand limit, lim f (x). xcxc13:24 81. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.1 THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION)As an example, take the function indicated in Figure 2.1.7. As x approaches 5 from the left, f (x) approaches 2; thereforef (x)lim f (x) = 2.574x5As x approaches 5 from the right, f (x) approaches 4; thereforey f2 f (x)lim+ f (x) = 4.5xx5xxFigure 2.1.7The full limit, lim f (x), does not exist: consideration of x < 5 would force the limit to x5be 2, but consideration of x > 5 would force the limit to be 4. For a full limit to exist, both one-sided limits have to exist and they have to be equal.Example 7 For the function f indicated in Figure 2.1.8, limx(2)f (x) = 5andlimx(2)+yf (x) = 5.7 5In this case lim f (x) = 5.2fx2It does not matter that f (2) = 3. Examining the graph of f near x = 4, we nd that lim f (x) = 7x442lim+ f (x) = 2.whereasxFigure 2.1.8x4Since these one-sided limits are different, lim f (x)x4does not exist. Example 8 Set f (x) = x/ x . Note that f (x) = 1 for x > 0, and f (x) = 1 for x < 0: 1, if x > 0 f (x) = (Figure 2.1.9) 1, if x < 0. Lets try to apply the limit process at different numbers c. If c < 0, then for all x sufciently close to c, x < 0 and f (x) = 1. It follows that for c < 0 lim f (x) = lim (1) = 1.xcxcIf c > 0, then for all x sufciently close to c, x > 0 and f (x) = 1. It follows that for c 1, f (x) = 1/(x 1). Therefore, as x 1+ , f (x) . The function has no numerical limit as x 1: lim f (x)does not exist.x1We now assert that ylim f (x) = 2.x1.52To see this, note that for x close to 1.5, x > 1 and therefore f (x) = 1/(x 1). It follows that lim f (x) = limx1.5x1.511 1 = = 2. x 1 0.5Figure 2.1.12See Figure 2.1.12. Example 12 Here we set f (x) = sin (/x) and show that the function can have no limit as x 0. y y = sin (/x)111x1Figure 2.1.13The function is not dened at x = 0, but, as you know, thats irrelevant. What keeps f from having a limit as x 0 is indicated in Figure 2.1.13. As x 0, f (x) keeps oscillating between y = 1 and y = 1 and therefore cannot remain close to any one number L. In our nal example we rely on a calculator and deduce a limit from numerical calculation. We can approach x = 0 by numbers an =1 1.5 22 4n + 1andby numbers bn = 2 , 4n + 1n = 0, 1, 2, 3, . . . . As you can check, f (an ) = 1 and f (bn ) = 1. This conrms the oscillatory behavior of f near x = 0.x13:24 84. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200660 CHAPTER 2 LIMITS AND CONTINUITY Example 13 Let f (x) = (sin x)/x. If we try to evaluate f at 0, we get the meaningless ratio 0/0; f is not dened at x = 0. However, f is dened for all x = 0, and so we can consider limx0sin x . xWe select numbers that approach 0 closely from the left and numbers that approach 0 closely from the right. Using a calculator, we evaluate f at these numbers. The results are tabulated in Table 2.1.1. Table 2.1.1(Left side)(Right side)x (radians)sin x xx (radians)sin x x1 0.5 0.1 0.01 0.0010.84147 0.95885 0.99833 0.99998 0.999991 0.5 0.1 0.01 0.0010.84147 0.95885 0.99833 0.99998 0.99999These calculations suggest that limx0sin x =1 xandlim+x0sin x =1 xand therefore that limx0sin x = 1. xThe graph of f, shown in Figure 2.1.14, supports this conclusion. A proof that this limit is indeed 1 is given in Section 2.5. y 10.5xFigure 2.1.14If you have found all this to be imprecise, you are absolutely right. Our work so far has been imprecise. In Section 2.2 we will work with limits in a more coherent manner.13:24 85. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.1 THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION)EXERCISES 2.1 Exercises 110. You are given a number c and the graph of a function f. Use the graph to nd (a) lim f (x) (b) lim+ f (x) (c) lim f (x) (d) f (c) xc4. c = 4.xcxcy1. c = 2.y = f (x) y y = f (x)1 41 2xx5. c = 2. 2. c = 3. y y y = f (x) y = f (x) 1 12 3xx6. c = 1.3. c = 3.yyy = f (x)y = f (x)11 3x1x6113:24 86. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200662 CHAPTER 2 LIMITS AND CONTINUITY 7. c = 1.Exercises 1112. Give the values of c for which lim f (x) does xcynot exist. 11. y = f (x)y1 1x1xy = f (x)8. c = 1. y y = f (x)12. y1 2x 1xy = f (x)9. c = 2.Exercises 1349. Decide on intuitive grounds whether or not the indicated limit exists; evaluate the limit if it does exist. 14. lim (2 5x). 13. lim (2x 1). x0 x1 16. lim x 2 + 2x + 1. 15. lim (x 2 2x + 4).yy = f (x)x2117. lim ( x 2). x34x3 . x1 x + 1 2 lim . x1 x + 1 2x 6 . lim x3 x 3 x 3 lim 2 . x3 x 6x + 9 x 2 lim . x2 x 2 3x + 2 1 lim x + . x0 x 2x 5x 2 lim . x0 x x2 1 lim . x1 x 1 x3 1 lim . x1 x + 1x4x019. lim20.21.22.23. 10. c = 3. 25.y27. y = f (x)29.1 3x31. 33. 35.1 . x 4 lim . x1 x + 1 1 lim . x2 3x 6 x 2 6x + 9 lim . x3 x 3 x 2 3x + 2 lim . x2 x 2 x 2 lim . x1 x 2 3x + 2 1 lim x + . x1 x x 3 lim . x3 6 2x x3 1 lim . x1 x 1 x2 + 1 lim 2 . x1 x 118. lim24. 26. 28. 30. 32. 34. 36..13:24 87. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.1 THE LIMIT PROCESS (AN INTUITIVE INTRODUCTION)37. lim f (x); x0f (x) =1, 3,x =0 x = 0.38. lim f (x);f (x) =3x, 3,39. lim f (x);f (x) =x 2, 0,f (x) =x 2 , x 2,41. lim f (x);f (x) =x 2, 1 + x,42. lim f (x);f (x) =2x, x 2 + 1,43. lim f (x);f (x) =3x, x + 2,x 1.x1x4x0x0x1x244. lim f (x); x0x4x 0.2x 2 11x + 12 (i) f (x) = ; x 4 2x 2 11x + 12 (ii) f (x) = 2 . x 8x + 16 (b) Use a CAS to nd each of the limits in part (a).x 0. x 1.2, 2,2x, 46. lim f (x); f (x) = 2, x1 2+1 x 2 . 47. lim x1 x 1 x 2 + 5 30 . 48. lim x5 x 5 x2 + 1 49. lim . x1 2x + 2 256. (a) Use a graphing utility to estimate lim f (x): x4(i) f (x) =5x 2 26x + 24 . 4x 2 11x 20 (b) Use a CAS to nd each of the limits in part (a). 57. (a) Use a graphing utility to estimate lim f (x):x rational x irrational.x0f (x) = 6x x (i) f (x) = ; x 2(i) f (x) =1 cos x x0 x(radian measure)by evaluating the quotient at x = 1, 0.1, 0.01, 0.001. 51. Estimate tan 2x lim x0 x(radian measure)by evaluating the quotient at x = 1, 0.1, 0.01, 0.001. 52. Estimate x sin x x3(radian measure)after evaluating the quotient at x = 1, 0.1, 0.01, 0.001, 0.0001. 53. Estimate limx1(ii) f (x) =x 2 4x + 4 . x 6xx2limx0x2(b) Use a CAS to nd each of the limits in part (a). 58. (a) Use a graphing utility to estimate lim f (x)Exercises 5054. After estimating the limit using the prescribed values of x, validate or improve your estimate by using a graphing utility. 50. Estimatelim3x 10x 8 ; 5x 2 + 16x 16 2(ii) f (x) =x rational x irrational.45. lim f (x);63x 3/2 1 x 1by evaluating the quotient at x = 0.9, 0.99, 0.99, 0.9999 and at x = 1.1, 1.01, 1.001, 1.0001.2x 18 x ; 4 x2 2 2x . (ii) f (x) = 8x 4(b) Use a CAS to nd each of the limits in part (a). Exercises 5962. Use a graphing utility to nd at least one number c at which lim f (x) does not exist. xcx +1 . 59. f (x) = 3 x + 1 6x 2 x 35 60. f (x) = . 2x 5 x . + + + 26x 2 + 36x + 72 5x 3 22x 2 + 15x + 18 62. f (x) = 3 . x 9x 2 + 27x 27 63. Use a graphing utility to draw the graphs of 61. f (x) =x5f (x) =2x 413x 31 sin x xandg(x) = x sin1 xfor x = 0 between /2 and /2. Describe the behavior of f (x) and g(x) for x close to 0. 64. Use a graphing utility to draw the graphs of f (x) =1 tan x xandg(x) = x tan1 xfor x = 0 between /2 and /2. Describe the behavior of f (x) and g(x) for x close to 0.13:24 88. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200664 CHAPTER 2 LIMITS AND CONTINUITY 2.2 DEFINITION OF LIMIT In Section 2.1 we tried to give you an intuitive feeling for the limit process. However, our description was too vague to be called mathematics. We relied on statements such as as x approaches c, f (x) approaches L and f (x) is close to L for all x = c which are close to c. But what exactly do these statements mean? What are we saying by stating that f (x) approaches L? How close is close? In this section we formulate the limit process in a coherent manner and, by so doing, establish a foundation for more advanced work. As before, in taking the limit of f (x) as x approaches c, we dont require that f be dened at c, but we do require that f be dened at least on an open interval (c p, c + p) except possibly at c itself. cpc+pcxTo say that lim f (x) = Lxcis to say that f (x) L can be made as small as we choose, less than any > 0 we choose, by restricting x to a sufciently small set of the form (c , c) (c, c + ), by restricting x by an inequality of the form 0 < x c < with > 0 sufciently small. Phrasing this idea precisely, we have the following denition.DEFINITION 2.2.1 THE LIMIT OF A FUNCTIONLet f be a function dened at least on an open interval (c p, c + p) except possibly at c itself. We say that lim f (x) = Lxc> 0, there exists a > 0 such thatif for each0 < x c < ,ifythen f (x) L < .LcxFigures 2.2.1 and 2.2.2 illustrate this denition.Figure 2.2.1 )) LFor each > 0xcx )cxthere exists > 0 such that,f (x)L ((Lcif 0 < x c< ,Figure 2.2.2( yx)Lyyc(ythen f (x) L < .x13:24 89. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.2 DEFINITION OF LIMIT65Except in the case of a constant function, the choice of depends on the previous choice of . We do not require that there exists a number which works for all , but rather, that for each there exists a which works for that particular . In Figure 2.2.3, we give two choices of and for each we display a suitable . For a to be suitable, all points within of c (with the possible exception of c itself) must be taken by the function f to within of L. In part (b) of the gure, we began with a smaller and had to use a smaller . yyfL+ f L+ L LL Lc c +cc c c +x(a)x(b)Figure 2.2.3The of Figure 2.2.4 is too large for the given . In particular, the points marked x1 and x2 in the gure are not taken by f to within of L. y ff (x2) L+ L L f (x1)c x1cx2 c + xFigure 2.2.4As these illustrations suggest, the limit process can be described entirely in terms of open intervals. (See Figure 2.2.5.)y)L+ (2.2.2)L L(Let f be dened at least on an open interval (c p, c + p) except possibly at c itself. We say that lim f (x) = Lxcif for each open interval (L , L + ) there is an open interval (c , c + ) such that all the numbers in (c , c + ), with the possible exception of c itself, are mapped by f into (L , L + ). Next we apply the , denition of limit to a variety of functions. At rst you may nd the , arguments confusing. It usually takes a little while for the , idea to take hold.c () cc+Figure 2.2.5x13:24 90. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200666 CHAPTER 2 LIMITS AND CONTINUITY yExample 1 Show that f (x) = 2x 1lim (2x 1) = 3.(Figure 2.2.6)x23+Finding a . Let3 3> 0. We seek a number > 0 such that 0 < x 2 < ,if (2x 1) 3 < .thenWhat we have to do rst is establish a connection between 2 2 2+ (2x 1) 3 xlim (2x 1) = 3and x 2 .The connection is evident:x2 (2x 1) 3 = 2x 4 = 2 x 2 .()Figure 2.2.6To make (2x 1) 3 less than , we need to make 2 x 2 < , which we can accomplish by making x 2 < /2. This suggests that we choose = 1 . 2 Showing that the works. If 0 < x 2 < 1 , then 2 x 2 < and, by 2 (), (2x 1) 3 < . Remark In Example 1 we chose = 1 , but we could have chosen any positive 2 number less than 1 . In general, if a certain works for a given , then any less 2 than will also work. Example 2 Show thatylim (2 3x) = 5.5+(Figure 2.2.7)x1Finding a . Let5 5if> 0. We seek a number > 0 such that0 < x (1) < ,then (2 3x) 5 < .To nd a connection between x (1) and (2 3x) 5 ,we simplify both expressions: x (1) = x + 1 f (x) = 2 3x x1 1 1 + lim (2 3x) = 5x 1Figure 2.2.7and (2 3x) 5 = 3x 3 = 3 x + 1 = 3 x + 1 . We can conclude that () (2 3x) 5 = 3 x (1) .We can make the expression on the left less than by making x (1) less than /3. This suggests that we set = 1 . 3 Showing that the works. If 0 < x (1) < 1 , then 3 x (1) < and, by 3 (), (2 3x) 5 < . Three Basic Limits Here we apply the , method to conrm three basic limits that are intuitively obvious. (If the , method did not conrm these limits, then the method would have been thrown out a long time ago.)13:24 91. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.2 DEFINITION OF LIMITExample 3 For each number c,67y f (x) = x c+lim x = c.(2.2.3)ifc c> 0. We must nd a > 0 such thatLet c be a real number and letPROOF(Figure 2.2.8)xc0 < x c < ,cxlim x = c x c < .thenc+cxcObviously we can choose = . Figure 2.2.8Example 4 For each real number c ylim x = c .(2.2.4)0 < x c < ,iff (x) = x> 0. We seek a > 0 such thatLet c be a real number and letPROOFc+ c c(Figure 2.2.9)xcc c c+ x c < .thenlim x = cxxcSinceFigure 2.2.9 [(1.3.7)] x c x c , we can choose = , for 0 < x c < ,ifthen x c < . yExample 5 For each constant kk+ f (x) = kk klim k = k.(2.2.5)(Figure 2.2.10)xcccc+lim k = k xcPROOFHere we are dealing with the constant functionFigure 2.2.10f (x) = k. Let> 0. We must nd a > 0 such that if0 < x c < ,then k k < .Since k k = 0, we always have k k < no matter how is chosen; in short, any positive number will do for . Usually , arguments are carried out in two stages. First we do a little scratch work, labeled nding a in Examples 1 and 2. This scratch work involves working backward from f (x) L < to nd a > 0 sufciently small so that we can begin with the inequality 0 < x c < and arrive at f (x) L < . This rst stage isx13:24 92. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200668 CHAPTER 2 LIMITS AND CONTINUITY just preliminary, but it shows us how to proceed in the second stage. The second stage consists of showing that the works by verifying that, for our choice of , it is true that 0 < x c < ,if f (x) L < .thenThe next two examples will give you a better feeling for this idea of working backward to nd a . yExample 6 lim x 2 = 92f (x) = x(Figure 2.2.11)x39+Finding a . Let9 9if> 0. We seek a > 0 such that 0 < x 3 < , x 2 9 < .thenThe connection between x 3 and x 2 9 can be found by factoring: x 2 9 = (x + 3)(x 3), 33x 3+and thus, x 2 9 = x + 3 x 3 .()Figure 2.2.11At this point, we need to get an estimate for the size of x + 3 for x close to 3. For convenience, well take x within one unit of 3. If x 3 < 1, then 2 < x < 4 and x + 3 x + 3 = x + 3 < 7. Therefore, by (), ()if x 3 < 1, x 2 9 < 7 x 3 .thenIf, in addition, x 3 < /7, then it will follow that x 2 9 < 7( /7) = . This means that we can let = the minimum of 1 and /7. Showing that the works. Let > 0. Choose = min{1, /7} and assume that 0 < x 3 < . Then x 3 < 1and x 3 < /7.By (), x 2 9 < 7 x 3 , and since x 3 < /7, we have y 2+ 2 2 x 2 9 < 7( /7) = .f (x) = xExample 7 4 4 4+ lim x = 2 x4Figure 2.2.12limxx4Finding a . Let ifx = 2.> 0. We seek a > 0 such that 0 < x 4 < ,then(Figure 2.2.12) x 2 < .13:24 93. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.2 DEFINITION OF LIMIT69 To be able to form x, we need to have x 0. To ensure this, we must have 4. (Explain.) Remembering that we must have 4, lets move on to nd a connection between x 4 and x 2 . With x 0, we can form x and write x 4 = ( x)2 22 = ( x + 2)( x 2). Taking absolute values, we have x 4 = x + 2 x 2 . Since x + 2 2 > 1, it follows that x 2 < x 4 .This last inequality suggests that we can simply set . But remember the requirement 4. We can meet both requirements on by setting = the minimum of 4 and . Showing that the works. Let > 0. Choose = min{4, } and assume that 0 < x 4 < . Since 4, we have x 0, and so x is dened. Now, as shown above, x 4 = x + 2 x 2 . Since x + 2 2 > 1, we can conclude that x 2 < x 4 . Since x 4 < and , it does follow that x 2 < .There are several different ways of formulating the same limit statement. Sometimes one formulation is more convenient, sometimes another. In particular, it is useful to recognize that the following four statements are equivalent:(i) lim f (x) = L (2.2.6)xc(iii) lim ( f (x) L) = 0 xc(ii) lim f (c + h) = Ly(iv) lim f (x) L = 0.Lh0 xch0cFigure 2.2.13xx13:24 94. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200670 CHAPTER 2 LIMITS AND CONTINUITY DEFINITION 2.2.7 LEFT-HAND LIMITLet f be a function dened at least on an open interval of the form (c p, c). We say that lim f (x) = Lxcif for each> 0 there exists a > 0 such that ifc < x < c, f (x) L < .thenDEFINITION 2.2.8 RIGHT-HAND LIMITLet f be a function dened at least on an open interval of the form (c, c + p). We say that lim f (x) = Lxc+if for each> 0 there exists a > 0 such that ifc < x 0,2(Figure 2.2.14)lim f (x) does not exist.x0PROOFyThe left- and right-hand limits at 0 are as follows: lim f (x) = lim (2x + 1) = 1,x03lim f (x) = lim+ (x 2 x) = 0.x0+x0x0Since these one-sided limits are different, lim f (x) does not exist. x02 1g12Figure 2.2.153xExample 10 For the function dened by setting x 1, lim g(x) = 2.x1(Figure 2.2.15)13:24 95. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.2 DEFINITION OF LIMIT71The left- and right-hand limits at 1 are as follows:PROOFlim g(x) = lim (1 + x 2 ) = 2,x1lim g(x) = lim+ (4 2x) = 2.x1+x1x1Thus, lim g(x) = 2. NOTE: It does not matter that g(1) = 2. x1At an endpoint of the domain of a function we cant take a (full) limit and we cant take a one-sided limit from the side on which the function is not dened, but we can try to take a limit from the side on which the function is dened. For example, it makes no sense to write or lim x. lim x x0yf (x) = xx0But it does make sense to try to nd limx0+x.x(Figure 2.2.16) lim x = 0 x 0+As you probably suspect, this one-sided limit exists and is 0.Figure 2.2.16EXERCISES 2.2 Exercises 120. Decide in the manner of Section 2.1 whether or not the indicated limit exists. Evaluate the limits that do exist. x . x1 x + 1 x(1 + x) lim . x0 2x 2 x4 1 lim . x1 x 1 x . lim x0 x x lim . x2 x x +3 . lim x3+ x 2 7x + 12 x 1 . lim x1+ xx 2 (1 + x) . x0 2x x lim . x4 x +1 1x lim . x1 x + 1 x2 1 lim 2 . x1 x 2x + 1 x 3 lim . x9 x 3 x lim . x0 x lim 9 x 2 .1. lim4.y2. lim3.21. Which of the s displayed in the gure works for the given ?5. 7. 9. 11. 13.15. lim+ f (x) if f (x) = x26. 8. 10. 12. 14. 2x 1, x 2 x, 1, x + 2,16. lim f (x) if f (x) = x1L+ L Lc 3L + 3 L + 2 L + 1 L L 1 L 2 L 3x 1 x > 1.19. lim f (x) if f (x) =3, 1,x an integer otherwise.20. lim f (x) if f (x) =x 2, 5x,18. lim f (x) if f (x) = x3x2x2x 2, 7, 2x + 3,x 3.x 1 x > 1.xyx 2 x > 2.x an integer otherwise.x2c c + 2 c 1 c + 1 c + 322. For which of the s given in the gure does the specied work?x33, 1, 17. lim f (x) if f (x) =c 2c c c+ xExercises 2326. Find the largest that works for the given . 24. lim 5x = 20; = 0.5. 23. lim 2x = 2; = 0.1. x125.lim 1 x x2 2x4= 1;= 0.01.26. lim 1 x = 2 ; 5 5 x2= 0.1.13:24 96. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200672 CHAPTER 2 LIMITS AND CONTINUITY 27. The graphs of f (x) = x and the horizontal lines y = 1.5 and y = 2.5 are shown in the gure. Use a graphing utility to nd a > 0 which is such that if 0 < x 4 < , then x 2 < 0.5. y 2.50 < x 3 < 1,ifthen f (x) 5 < 0.1.Which of the following statements are necessarily true? (a) If x 3 < 1, then f (x) 5 < 0.1. (b) If x 2.5 < 0.3, then f (x) 5 < 0.1. (c) lim f (x) = 5. x32 1.54x28. The graphs of f (x) = 2x 2 and the horizontal lines y = 1 and y = 3 are shown in the gure. Use a graphing utility to nd a > 0 which is such that if41. Let f be some function for which you know only that0 < x + 1 < , 2x 2 2 < 1.then y(d) (e) (f) (g) (h) (i)If 0 < x 3 < 2, then f (x) 5 < 0.1. If 0 < x 3 < 0.5, then f (x) 5 < 0.1. If 0 < x 3 < 1 , then f (x) 5 < 1 (0.1). 4 4 If 0 < x 3 < 1, then f (x) 5 < 0.2. If 0 < x 3 < 1, then f (x) 4.95 < 0.05. If lim f (x) = L, then 4.9 L 5.1. x342. Suppose that A B < for each > 0. Prove that A = B. HINT: Suppose that A = B and set = 1 A B . 2 Exercises 4344. Give the four limit statements displayed in (2.2.6), taking x 1 , c=3 44. f (x) = 2 , c = 1. 43. f (x) = x 1 x +2 45. Prove that lim f (x) = 0,(2.2.10)iffxclim f (x) = 0.xc3 246. (a) Prove that1if1xExercises 2934. For each of the limits stated and the s given, use a graphing utility to nd a > 0 which is such that if 0 < x c < , then f (x) L < . Draw the graph of f together with the vertical lines x = c , x = c + and the horizontal lines y = L , y = L + . + x + 1 = 4;= 0.5,= 0.25.30. lim (x 3 + 4x + 2) = 2;= 0.5,= 0.25.1 2 x x2 429. limx2x 1 31. lim = 2; x1 x 1= 0.5,= 0.25.1 3x = 2; = 0.5, = 0.1. 2x + 4 sin 3x = 3; = 0.25, = 0.1. 33. lim x0 x 34. lim tan( x/4) = 1; = 0.5, = 0.1.32. limx1x1Give an , proof for the following statements. 35. lim (2x 5) = 3. 36. lim (3x 1) = 5. x4x237. lim (6x 7) = 11.40. lim x 2 = 0.x3 x2lim f (x) = L lim f (x) = L .xclim f (x) = M = L ,andxcxcand then give an example where lim f (x) xcexistsbutlim f (x)xcdoes not exist.47. Give an , proof that statement (i) in (2.2.6) is equivalent to (ii). 48. Give an , proof of (2.2.9). 49. (a) Show that lim x = c for each c > 0. xcHINT: If x and c are positive, then x c 1 0 x c = < x c . x+ c c (b) Show that lim+ x = 0. x0Give an , proof for the following statements. 51. lim x 3 = 1. 50. lim x 2 = 4. x2 x1 53. lim 3 x = 0. 52. lim x + 1 = 2. x3x354. Prove that, for the function g(x) =x0 x2then(b) Show that the converse is false. Give an example where38. lim (2 5x) = 2.39. lim 1 3x = 5.lim f (x) = L ,xclim g(x) = 0.x0x, 0,x rational x irrational,13:24 97. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.3 SOME LIMIT THEOREMS(b) Prove that if L < 0, then f (x) < 0 for all x = c in an interval of the form (c , c + ). 60. Prove or give a counterexample: if f (c) > 0 and lim f (x)55. The function 1, 0,f (x) =x rational x irrationalxcis called the Dirichlet function. Prove that for no number c does lim f (x) exist. xcProve the limit statement. 56. lim f (x) = L iff lim f (c h ) = L. h0xc57. lim+ f (x) = Liffxc58. lim f (x) = L xciffexists, then f (x) > 0 for all x in an interval of the form (c , c + ). 61. Suppose that f (x) g(x) for all x (c p, c + p), except possibly at c itself. (a) Prove that lim f (x) lim g(x), provided each of these xclim f (c + h ) = L.h0lim [ f (x) L] = 0.xcxclim g(x)?xc(a) Prove that if L > 0, then f (x) > 0 for all x = c in an interval of the form (c , c + ). HINT: Use an , argument, setting = L.xc62. Prove that if lim f (x) = L, then there are positive numbers xc and B such that if 0 < x c < , then f (x) < B. 2.3 SOME LIMIT THEOREMS As you probably gathered by working through the previous section, it can become rather tedious to apply the , denition of limit time and time again. By proving some general theorems, we can avoid some of this repetitive work. Of course, the theorems themselves (at least the rst ones) will have to be proved by , methods. We begin by showing that if a limit exists, it is unique.THEOREM 2.3.1 THE UNIQUENESS OF A LIMITlim f (x) = Lxcandlim f (x) = M,xcthenL = M.We show L = M by proving that the assumption L = M leads to the false conclusion thatPROOF L M < L M . Assume that L = M. Then L M /2 > 0. Since lim f (x) = L, we know that xcthere exists a 1 > 0 such that (1)if0 < x c < 1 ,then f (x) L < L M /2. (Here we are using L M /2 as .)Since lim f (x) = M, we know that there exists a 2 > 0 such that xc(2)if0 < x c < 2 ,then f (x) L < L M /2. (Again, we are using L M /2 as .)Now let x1 be a number that satises the inequality 0 < x1 c < minimum of 1 and 2 . Then, by (1) and (2), f (x1 ) L 0. To prove (i), we must show that there exists a > 0 such thatifPROOF0 < x c < ,then [ f (x) + g(x)] [L + M] < .Note that [ f (x) + g(x)] [L + M] = [ f (x) L] + [g(x) M] f (x) L + g(x) M .()We can make [ f (x) + g(x)] [L + M] less than by making f (x) L and g(x) M each less than 1 . Since > 0, we know that 1 > 0. Since 2 2 lim f (x) = Landxclim g(x) = M,xcwe know that there exist positive numbers 1 and 2 such that if0 < x c < 1 ,then f (x) L 0 such that if0 < x c < 1 ,then g(x) M M 2so that1 2 < g(x) M and thus 1 2 1 g(x) M 2 g(x) M = 2 g(x) M . = g(x) M g(x) M M 2 M Now let> 0 and choose 2 > 0 such that if0 < x c < 2 ,then g(x) M 0. xcThis makes the square-root function continuous at each positive number. What happens at c = 0, we discuss later. With f and g continuous at c, we have lim f (x) = f (c)xclim g(x) = g(c)xcand thus, by the limit theorems, lim [ f (x) + g(x)] = f (c) + g(c),xclim [ f (x)] = f (c)xcand, if g(c) = 0,for each real lim [ f (x) g(x)] = f (c) g(c)xclim [ f (x)g(x)] = f (c)g(c)xclim [ f (x)/g(x)] = f (c)/g(c).xcFigure 2.4.4x13:24 108. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200684 CHAPTER 2 LIMITS AND CONTINUITY We summarize all this in a theorem. THEOREM 2.4.2If f and g are continuous at c, then (i) (ii) (iii) (iv) (v)f + g is continuous at c; f g is continuous at c; f is continuous at c for each real ; f g is continuous at c; f /g is continuous at c provided g(c) = 0.These results can be combined and extended to any nite number of functions. x3 x + 4 is continuous at all real x 2 5x + 6 numbers other than 2 and 3. You can see this by noting thatExample 1 The function F(x) = 3 x +F = 3 f + g/ h + k where f (x) = x ,g(x) = x 3 x,h(x) = x 2 5x + 6,k(x) = 4.Since f, g, h, k are everywhere continuous, F is continuous except at 2 and 3, the numbers at which h takes on the value 0. (At those numbers F is not dened.) Our next topic is the continuity of composite functions. Before getting into this, however, lets take a look at continuity in terms of , . A direct translation of lim f (x) = f (c)xcinto , terms reads like this: for each if0 < x c < ,> 0, there exists a > 0 such that then f (x) f (c) < .Here the restriction 0 < x c is unnecessary. We can allow x c = 0 because then x = c, f (x) = f (c), and thus f (x) f (c) = 0. Being 0, f (x) f (c) is certainly less than . Thus, an , characterization of continuity at c reads as follows:(2.4.3)f is continuous at c iffor each > 0 there exists a > 0 such that if x c < , then f (x) f (c) < .In intuitive terms f is continuous at ciffor x close to c,f (x) is close to f (c).We are now ready to take up the continuity of composite functions. Remember the dening formula: ( f g)(x) = f (g(x)). (You may wish to review Section 1.7.)13:24 109. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.4 CONTINUITYTHEOREM 2.4.4If g is continuous at c and f is continuous at g(c), then the composition f g is continuous at c. The idea here is as follows: with g continuous at c, we know that for x close to c, g(x) is close to g(c); from the continuity of f at g(c), we know that with g(x) close to g(c), f (g(x)) is close to f (g(c)). In summary, with x close to c, f (g(x)) is close to f (g(c)). The argument we just gave is too vague to be a proof. Here, in contrast, is a proof. We begin with > 0. We must show that there exists a number > 0 such that x c < ,if f (g(x)) f (g(c)) < .thenIn the rst place, we observe that, since f is continuous at g(c), there does exist a number 1 > 0 such that (1) t g(c) < 1 ,if f (t) f (g(c)) < .thenWith 1 > 0, we know from the continuity of g at c that there exists a number > 0 such that (2)if x c < , g(x) g(c) < 1 .thenCombining (2) and (1), we have what we want: by (2), if x c < , g(x) g(c) < 1thenso that by (1) f (g(x)) f (g(c)) < . This proof is illustrated in Figure 2.4.5. The numbers within of c are taken by g to within 1 of g(c), and then by f to within of f (g(c)). fg f gf (g(c)) 11 g(c) cFigure 2.4.5Its time to look at some examples.8513:24 110. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200686 CHAPTER 2 LIMITS AND CONTINUITY x2 + 1 is continuous at all numbers greater x 3 than 3. To see this, note that F = f g, whereExample 2 The function F(x) = x2 + 1 . x 3 Now, take any c > 3. Since g is a rational function and g is dened at c, g is continuous at c. Also, since g(c) is positive and f is continuous at each positive number, f is continuous at g(c). By Theorem 2.4.4, F is continuous at c. f (x) =xg(x) =andThe continuity of composites holds for any nite number of functions. The only requirement is that each function be continuous where it is applied.Example 3 The function F(x) =1is continuous everywhere except 5 x 2 + 16 at x = 3, where it is not dened. To see this, note that F = f g k h, where 1 g(x) = 5 x, k(x) = x, h(x) = x 2 + 16, f (x) = , x and observe that each of these functions is being evaluated only where it is continuous. In particular, g and h are continuous everywhere, f is being evaluated only at nonzero numbers, and k is being evaluated only at positive numbers. Just as we considered one-sided limits, we can consider one-sided continuity.DEFINITION 2.4.5 ONE-SIDED CONTINUITYA function f is called continuous from the left at ciflim f (x) = f (c).xcIt is called continuous from the right at ciflim f (x) = f (c).xc+The function of Figure 2.4.6 is continuous from the right at 0; the function of Figure 2.4.7 is continuous from the left at 1. y yfgxf (x) = xFigure 2.4.61xg(x) = 1 xFigure 2.4.7It follows from (2.2.9) that a function is continuous at c iff it is continuous from both sides at c. Thus13:24 111. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.4 CONTINUITY87f is continuous at c iff f (c), lim f (x), lim+ f (x) xc(2.4.6)xcall exist and are equal.Example 4 Determine the discontinuities, if any, of the following function: x 0 2x + 1, (Figure 2.4.8) 1, 0 1.yf3 2SOLUTION Clearly f is continuous at each point in the open intervals (, 0), (0, 1), (1, ). (On each of these intervals f is a polynomial.) Thus, we have to check the behavior of f at x = 0 and x = 1. The gure suggests that f is continuous at 0 and discontinuous at 1. Indeed, that is the case:f (0) = 1,lim f (x) = lim (2x + 1) = 1,x0lim f (x) = lim (1) = 1andx1Figure 2.4.8lim f (x) = lim+ (x 2 + 1) = 2.x1+x1Example 5 Determine the discontinuities, if any, of the following function: x 1 x 3, 2 x 2, 1 < x < 1 6 x, 1x 1.89 1, x < 1 x 3 , 1 x 1 f (x) = 1, 1 < x. 1, x 2 1 x, 2 < x < 4 g(x) = 2 x, 4 x. 1, x 0 2 x , 0 2.13:24 114. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200690 CHAPTER 2 LIMITS AND CONTINUITY 37. Give necessary and sufcient conditions on A and B for the function Ax B, x 1 3x, 1 < x < 2 f (x) = Bx 2 A, 2 x to be continuous at x = 1 but discontinuous at x = 2. 38. Give necessary and sufcient conditions on A and B for the function in Exercise 37 to be continuous at x = 2 but discontinuous at x = 1. 1 + cx, x < 2 Find a value of c that makes c x, x 2. f continuous on (, ). Use a graphing utility to verify your result. 1 cx + d x 2 , x 1 x 2 + x, 1 < x < 2 Find values of 40. Set f (x) = 2 cx + d x + 4, x 2. c and d that make f continuous on (, ). Use a graphing utility to verify your result. Exercises 4144. Dene the function at 5 so that it becomes continuous at 5. x +43 x +43 . . 42. f (x) = 41. f (x) = x 5 x 5 2x 1 3 43. f (x) = . x 5 x 2 7x + 16 6 . 44. f (x) = (x 5) x + 1 39. Set f (x) =Exercises 4547. At what points (if any) is the function continuous? 1, x rational 45. f (x) = 0, x irrational. 46. g(x) =x, 0,x rational x irrational.2x, x an integer x 2 , otherwise. 48. The following functions are important in science and engineering: 0, x < c 1. The Heaviside function Hc (x) = 1, x c. 2. The unit pulse function 1[Hc (x) Hc+ (x)].(a) Graph Hc and P ,c . (b) Determine where each of the functions is continuous. (c) Find lim Hc (x) and lim+ Hc (x). What can you say xcxcabout lim H (x)? xc49. (Important) Prove that f is continuous at c f (x) f (c) B x c for all x in the interval (c p, c + p). Prove that f is continuous at c. 54. Suppose the function f has the property that f (x) f (t) x t for each pair of points x, t in the interval (a, b). Prove that f is continuous on (a, b). 55. Prove that if f (c + h) f (c) lim h0 h exists, then f is continuous at c. 56. Suppose that the function f is continuous on (, ). Show that f can be written f = fe + f0 ,47. h(x) =P ,c (x) =50. (Important) Let f and g be continuous at c. Prove that if: (a) f (c) > 0, then there exists > 0 such that f (x) > 0 for all x (c , c + ). (b) f (c) < 0, then there exists > 0 such that f (x) < 0 for all x (c , c + ). (c) f (c) < g(c), then there exists > 0 such that f (x) < g(x) for all x (c , c + ). 51. Suppose that f has an essential discontinuity at c. Change the value of f as you choose at any nite number of points x1 , x2 , . . ., xn and call the resulting function g. Show that g also has an essential discontinuity at c. 52. (a) Prove that if f is continuous everywhere, then f is continuous everywhere. (b) Give an example to show that the continuity of f does not imply the continuity of f. (c) Give an example of a function f such that f is continuous nowhere, but f is continuous everywhere. 53. Suppose the function f has the property that there exists a number B such thatifflim f (c + h) = f (c).h0where f e is an even function which is continuous on (, ) and f 0 is an odd function which is continuous on (, ). Exercises 5760. The function f is not dened at x = 0. Use a graphing utility to graph f. Zoom in to determine whether there is a number k such that the function F(x) =f (x), k,x =0 x =0is continuous at x = 0. If so, what is k? Support your conclusion by calculating the limit using a CAS. sin 5x . 57. f (x) = sin 2x 2 x . 58. f (x) = 1 cos 2x sin x 59. f (x) = . x x sin 2x 60. f (x) = . sin x 213:24 115. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS 2.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS Figure 2.5.1 shows the graphs of three functions f, g, h. Suppose that, as suggested by the gure, for x close to c, f is trapped between g and h. (The values of these functions at c itself are irrelevant.) If, as x tends to c, both g(x) and h(x) tend to the same limit L, then f (x) also tends to L. This idea is made precise in what we call the pinching theorem. f y ggf L h cx hFigure 2.5.1THEOREM 2.5.1 THE PINCHING THEOREMLet p > 0. Suppose that, for all x such that 0 < x c < p, h(x) f (x) g(x). If lim h(x) = Landxclim g(x) = L ,xcthen lim f (x) = L .xcPROOF> 0. Let p > 0 be such thatLetif0 < x c < p,thenh(x) f (x) g(x).Choose 1 > 0 such that if0 < x c < 1 ,thenL < h(x) < L + .thenL < g(x) < L + .Choose 2 > 0 such that if0 < x c < 2 ,Let = min{ p, 1 , 2 }. For x satisfying 0 < x c < , we have L < h(x) f (x) g(x) < L + , and thus f (x) L < . 9113:24 116. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200692 CHAPTER 2 LIMITS AND CONTINUITY Remark With straightforward modications, the pinching theorem holds for onesided limits. We do not spell out the details here because throughout this section we will be working with two-sided limits. We come now to some trigonometric limits. All calculations are based on radian measure. As our rst application of the pinching theorem, we prove thatlim sin x = 0.(2.5.2)x0To follow the argument, see Figure 2.5.2. For small x = 0PROOF0 < sin x = length of B P < length of A P < length of AP = x . Thus, for such x 0 < sin x < x .1PBO sin xOAxxsin xx BxA1 x0PFigure 2.5.2Since lim 0 = 0andx0lim x = 0,x0we know from the pinching theorem that lim sin x = 0x0and thereforelim sin x = 0. x0From this it follows readily that(2.5.3)PROOFlim cos x = 1.x0In general, cos2 x + sin2 x = 1. For x close to 0, the cosine is positive and wehave cos x =1 sin2 x.As x tends to 0, sin x tends to 0, sin2 x tends to 0, and therefore cos x tends to 1. Recall that in a circle of radius 1, a central angle of x radians subtends an arc of length x .13:24 117. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITS93Next we show that the sine and cosine functions are everywhere continuous; which is to say, for all real numbers c,lim sin x = sin c(2.5.4)PROOFandxclim cos x = cos c.xcTake any real number c. By (2.2.6) we can write lim sin xxclim sin (c + h).ash0This form of the limit suggests that we use the addition formula sin (c + h) = sin c cos h + cos c sin h. Since sin c and cos c are constants, we have lim sin (c + h) = (sin c)(lim cos h) + (cos c)(lim sin h)h0h0h0= (sin c)(1) + (cos c)(0) = sin c. The proof that lim cos x = cos c is left to you. xcThe remaining trigonometric functions cos x 1 1 sin x , cot x = , sec x = , csc x = cos x sin x cos x sin x are all continuous where dened. Justication? They are all quotients of continuous functions. We turn now to two limits, the importance of which will become clear in Chapter 3: tan x =(2.5.5)limx0sin x =1 xandlimx01 cos x = 0. xRemark These limits were investigated by numerical methods in Section 2.1, the rst in the text, the second in the exercises. PROOFWe show thatsin x =1 x0 x by using some simple geometry and the pinching theorem. For any x satisfying 0 < x /2 (see Figure 2.5.3), length of PB = sin x, length of OB = cos x, and length OA = 1. Since triangle OAQ is a right triangle, tan x = QA/1 = QA. Now 1 1 area of triangle OAP = (1) sin x = sin x 2 2 1 1 2 area of sector OAP = (1) x = x 2 2 1 sin x 1 . area of triangle OAQ = (1) tan x = 2 2 cos x limy Q(0, 1) Ptan x sin x x OB Acos x 1Figure 2.5.3x13:24 118. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200694 CHAPTER 2 LIMITS AND CONTINUITY Since triangle OAP sector OAP triangle OAQ (and these are all proper containments), we have 1 1 sin x 1 sin x < x < 2 2 2 cos x 1 x < 1< sin x cos x sin x < 1. cos x < x This inequality was derived for x > 0, but since cos (x) = cos x sin x sin x sin (x) = = , x x xandthis inequality also holds for x < 0. We can now apply the pinching theorem. Since lim cos x = 1lim 1 = 1,andx0x0we can conclude that limx0sin x = 1. xNow lets show that 1 cos x = 0. x0 x limFor small x = 0, cos x is close to 1 and so cos x = 1. Therefore, we can write 1 cos x = x1 cos x x=1 cos2 x x(1 + cos x)=1 + cos x 1 + cos xsin2 x x(1 + cos x)=sin x xsin x 1 + cos x.Since limx0sin x =1 xandlimx0sin x 0 = = 0, 1 + cos x 2it follows that limx01 cos x = 0. x This trick is a fairly common procedure with trigonometric expressions. It is much like using conjugates to revise algebraic expressions: 3 3 4 2 3(4 2) . = = 14 4+ 2 4+ 2 4 213:24 119. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.5 THE PINCHING THEOREM; TRIGONOMETRIC LIMITSRemarkThe limit statements in (2.5.5) can be generalized as follows: For each number a = 0 1 cos ax = 0. x0 axsin ax =1 x0 ax(2.5.6)limlimExercise 38.We are now in a position to evaluate a variety of trigonometric limits. sin 4x x0 3xExample 1 Find lim1 cos 2x . x0 5xandlimSOLUTION To calculate the rst limit, we pair off sin 4x with 4x and use (2.5.6):sin 4x 4 sin 4x 4 sin 4x = = . 3x 4 3x 3 4x Therefore, limx04 sin 4x sin 4x = lim x0 3 3x 4x=4 4 4 sin 4x lim = (1) = . 3 x0 4x 3 3The second limit can be obtained the same way: 1 cos 2x 2 1 cos 2x 1 cos 2x 2 2 = lim = lim = (0) = 0. x0 x0 5 5x 2x 5 x0 2x 5 limExample 2 Find lim x cot 3x. x0SOLUTION We begin by writingx cot 3x = xcos 3x 1 = sin 3x 33x sin 3x(cos 3x).Since sin 3x =1 x0 3x lim3x = 1, x0 sin 3xgiveslimand lim cos 3x = cos 0 = 1, we see that x0lim x cot 3x =x0Example 3 Find lim1 3x lim x0 sin 3x 3x/4SOLUTIONsin x 1 4 (x 1 )2 4sin x 1 4 x 1 4 =2lim (cos 3x) =x0.sin x 1 4 (x 1 ) 41 . x 1 4We know that limx/4sin x 1 4 x 1 4= 1.1 1 (1)(1) = . 3 39513:24 120. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200696 CHAPTER 2 LIMITS AND CONTINUITY Since limx/4x 1 = 0, you can see by Theorem 2.3.10 that 4 limsin x 1 4x/4x 1 4does not exist. 2x2 . x0 sec x 1Example 4 Find limSOLUTION The evaluation of this limit requires a little imagination. Since both the numerator and denominator tend to zero as x tends to zero, it is not clear what happens to the fraction. However, we can rewrite the fraction in a more amenable form by multiplying both numerator and denominator by sec x + 1.x2 x2 = sec x 1 sec x 1 =sec x + 1 sec x + 1x 2 (sec x + 1) x 2 (sec x + 1) = sec2 x 1 tan2 x 1x 2 cos2 x(sec x + 1) sin2 x x 2 = (cos2 x)(sec x + 1). sin x Since each of these factors has a limit as x tends to 0, the fraction we began with has a limit: x2 x 2 = lim lim cos2 x lim (sec x + 1) = (1)(1)(2) = 2. lim x0 sec x 1 x0 sin x x0 x0 =EXERCISES 2.5 Exercises 132. Evaluate the limits that exist. 1. 3. 5. 7. 9. 11. 13.sin 3x lim . x0 x 3x . lim x0 sin 5x sin 4x . lim x0 sin 2x sin x 2 lim . x0 x sin x lim . x0 x 2 sin2 3x lim . x0 5x 2 2x lim . x0 tan 3x15. lim x csc x. x0x2 17. lim . x0 1 cos 2x2. 4. 6. 8. 10. 12. 14. 16. 18.2x lim . x0 sin x sin 3x lim . x0 2x sin 3x lim . x0 5x sin x 2 . lim x0 x 2 sin2 x 2 lim . x0 x2 tan2 3x lim . x0 4x 2 4x lim . x0 cot 3x cos x 1 lim . x0 2x x 2 2x lim . x0 sin 3x1 sec2 2x . x0 x219. lim2x 2 + x . x0 sin x tan 3x 23. lim 2 . x0 2x + 5x sec x 1 . 25. lim x0 x sec x 21. limsin x . x/4 x27. lim 29. limx/2cos x . x 1 231. limsin (x + 1 ) 1 432. limsin (x + 1 ) 1 3x/4x 1 420. limx01 . 2x csc x1 cos 4x . x0 9x 222. lim24. lim x 2 (1 + cot2 3x). x026. lim1 cos x . x28. limsin2 x . x(1 cos x)30. limsin x . x x/4x0x. HINT: x + 1 = x 1 + 1 . 4 4 2. x 1 6 33. Show that lim cos x = cos c for all real numbers c. x/6xc13:24 121. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.6 TWO BASIC THEOREMSExercises 3437. Evaluate the limit, taking a and b as nonzero constants. sin ax 1 cos ax . 35. lim . 34. lim x0 bx x0 bx sin ax cos ax . 37. lim . 36. lim x0 sin bx x0 cos bx 38. Show that if lim f (x) = L , x0lim f (ax) = L for each a = 0.thenx0HINT: Let > 0. If 1 > 0 works for the rst limit, then = 1 / a works for the second limit. Exercises 3942. Evaluate lim [ f (c + h) f (c)]/ h.x0theorem. 44. Show that lim (x ) cos2 [1/(x )] = 0. x45. Show that lim x 1 sin x = 0. x1x0special cases of this general result. 48. Prove that if there is a number B such that f (x)/x B for all x = 0, then lim f (x) = 0. x049. Prove that if there is a number B such that f (x) L / x c B for all x = c, then lim f (x) = L. xc50. Given that lim f (x) = 0 and g(x) B for all x in an inxcterval (c p, c + p), prove thatf (x) =lim f (x)g(x) = 0.xcExercises 5152. Use the limit utility in a CAS to evaluate the limit. 20x 15x 2 51. lim . x0 sin 2x tan x 52. lim 2 . x0 x x 53. Use a graphing utility to plot f (x) = on [0.2, 0.2]. tan 3x Estimate lim f (x); use the zoom function if necessary. x0Verify your result analytically.46. Let f be the Dirichlet function 1, 0,9747. Prove that if there is a number B such that f (x) B for all x = 0, then lim x f (x) = 0. NOTE: Exercises 4346 areh039. f (x) = sin x, c = /4. HINT: Use the addition formula for the sine function. 40. f (x) = cos x, c = /3. 41. f (x) = cos 2x, c = /6. 42. f (x) = sin 3x, c = /2. 43. Show that lim x sin(1/x) = 0. HINT: Use the pinchingx rational x irrational.Show that lim x f (x) = 0. x0tan x on tan x + x [0.2, 0.2]. Estimate lim f (x); use the zoom function if54. Use a graphing utility to plot f (x) = x0necessary. Verify your result analytically. 2.6 TWO BASIC THEOREMS A function which is continuous on an interval does not skip any values, and thus its graph is an unbroken curve. There are no holes in it and no jumps. This idea is expressed coherently by the intermediate-value theorem.THEOREM 2.6.1 THE INTERMEDIATE-VALUE THEOREMy f (b)K f (a) aIf f is continuous on [a, b] and K is any number between f (a) and f (b), then there is at least one number c in the interval (a, b) such that f (c) = K .We illustrate the theorem in Figure 2.6.1. What can happen in the discontinuous case is illustrated in Figure 2.6.2. There the number K has been skipped. Its a small step from the intermediate-value theorem to the following observation: continuous functions map intervals onto intervals. A proof of the intermediate-value theorem is given in Appendix B. We will assume the result and illustrate its usefulness.bxbcxFigure 2.6.1y f (b)K f (a) aFigure 2.6.213:24 122. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 200698 CHAPTER 2 LIMITS AND CONTINUITY Here we apply the theorem to the problem of locating the zeros of a function. In particular, suppose that the function f is continuous on [a, b] and that either f (a) < 0 < f (b)orf (b) < 0 < f (a)(Figure 2.6.3)Then, by the intermediate-value theorem, we know that the equation f (x) = 0 has at least one root between a and b. ffaxbabxFigure 2.6.3Example 1 We set f (x) = x 2 2. Since f (1) = 1 < 0 and f (2) = 2 > 0, there exists a number c between 1 and 2 such that f (c) = 0. Since f increases on [1, 2], there is only one such number. This is the number we call 2. So far we have shown only that 2 lies between 1 and 2. We can locate 2 more precisely by evaluating f at 1.5, the midpoint of the interval [1, 2]. Since f (1.5) = 0.25 > 0 and f (1) < 0, we know that 2 lies between 1 and 1.5. We now evaluate f at 1.25, the midpoint of [1, 1.5]. Since f (1.25) 0.438 < 0 and f (1.5) > 0, we know = that 2 lies between 1.25 and 1.5. Our next step is to evaluate f at 1.375, the midpoint of [1.25, 1.5]. Since f (1.375) 0.109 < 0 and f (1.5) > 0, we know that 2 lies = between 1.375 and 1.5. We now evaluate f at 1.4375, the midpoint of [1.375, 1.5]. Since f (1.4375) 0.066 > 0 and f (1.375) < 0, we know that 2 lies between 1.375 = and 1.4375. The average these two numbers, rounded off to two decimal places, is of 1.41. A calculator gives 2 1.4142. So we are not far off. = The method used in Example 1 is called the bisection method. It can be used to locate the roots of a wide variety of equations. The more bisections we carry out, the more accuracy we obtain. As you will see in the exercise section, the intermediate-value theorem gives us results that are otherwise elusive, but, as our next example makes clear, the theorem has to be applied with some care.Example 2 The function f (x) = 2/x takes on the value 2 at x = 1 and it takes on the value 2 at x = 1. Certainly 0 lies between 2 and 2. Does it follow that f takes on the value 0 somewhere between 1 and 1? No: the function is not continuous on [1, 1], and therefore it can and does skip the number 0. yBoundedness; Extreme Values xA function f is said to be bounded or unbounded on a set I in the sense in which the set of values taken on by f on the set I is bounded or unbounded. For example, the sine and cosine functions are bounded on (, ): 1 sin x 1 y = tan xFigure 2.6.4and 1 cos x 1for all x (, ).Both functions map (, ) onto [1, 1]. The situation is markedly different in the case of the tangent. (See Figure 2.6.4.) The tangent function is bounded on [0, /4]; on [0, /2) it is bounded below but13:24 123. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20062.6 TWO BASIC THEOREMSnot bounded above; on (/2, 0] it is bounded above but not bounded below; on (/2, /2) it is unbounded both below and above.99y2Example 3 Let g(x) =1/x 2 , 0,x >0 x = 0.1 g(Figure 2.6.5)It is clear that g is unbounded on [0, ). (It is unbounded above.) However, it is bounded on [1, ). The function maps [0, ) onto [0, ), and it maps [1, ) onto (0, 1]. 12Figure 2.6.5A function may take on a maximum value; it may take on a minimum value; it may take on both a maximum value and a minimum value; it may take on neither. Here are some simple examples: f (x) =1, 0,x rational x irrationaltakes on both a maximum value (the number 1) and a minimum value (the number 0) on every interval of the real line. The function f (x) = x 2x (0, 5]takes on a maximum value (the number 25), but it has no minimum value. The function 1 x (0, ) f (x) = , x has no maximum value and no minimum value. For a function continuous on a bounded closed interval, the existence of both a maximum value and a minimum value is guaranteed. The following theorem is fundamental. THEOREM 2.6.2 THE EXTREME-VALUE THEOREMIf f is continuous on a bounded closed interval [a, b], then on that interval f takes on both a maximum value M and a minimum value m. For obvious reasons, M and m are called the extreme values of the function. The result is illustrated in Figure 2.6.6. The maximum value M is taken on at the point marked d, and the minimum value m is taken on at the point marked c.M m acdbxFigure 2.6.6In Theorem 2.6.2, the full hypothesis is needed. If the interval is not bounded, the result need not hold: the cubing function f (x) = x 3 has no maximum on the interval [0, ). If the interval is not closed, the result need not hold: the identity function3x13:24 124. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006100 CHAPTER 2 LIMITS AND CONTINUITY f (x) = x has no maximum and no minimum on (0, 2). If the function is not continuous, the result need not hold. As an example, take the function x =1 3, 1 1 3 + x, x < 2 5, x = 2 . if f (x) = x 2 3, x > 2if f (x) =29. lim f (x) x230. lim f (x) x231. Let f be some function for which you know only that if0 < x 2 < 1,then f (x) 4 < 0.1.Which of the following statements are necessarily true? (a) If 0 x 2 < 1, then f (x) 4 0.1. (b) If 0 < x 2 < 1 , then f (x) 4 < 0.05. 2 (c) If 0 x 2.5 < 0.2, then f (x) 4 < 0.1. (d) If 0 < x 1.5 < 1, then f (x) 4 < 0.1. (e) If lim f (x) = L, then 3.9 L 4.1. x22x 2 3ax + x a 1 exx3 x 2 2x 3 ists and then evaluate the limit.32. Find a number a for which lim33. (a) Sketch the graph of 3x + 4, 2x 2, f (x) = 2x, x 2,x < 1 1 < x < 2 x >2 x = 1, 2(b) Evaluate the limits that exist. (i) lim f (x). x1(ii) lim + f (x). x1(iii) lim f (x). x1(iv) lim f (x). x2(v) lim+ f (x). x2(vi) lim f (x). x2(c) (i) Is f continuous from the left at 1? Is f continuous from the right at 1? (ii) Is f continuous from the left at 2? Is f continuous from the right at 2? 1 cos x exist? If so, what is the limit? 34. (a) Does lim cos x0 2x sin x exist? If so, what is the limit? (b) Does lim cos x0 2x 2x 2 1, x < 2 A, x = 2 For what values of A and 35. Set f (x) = 3 x 2Bx, x > 2. B is f continuous at 2? 36. Give necessary and sufcient conditions on A and B for the function x < 1 Ax + B, 2x, 1 x 2 f (x) = 2Bx A, 2 0 and f (b) < 0, then there must exist at least one number c in (a, b) for which f (c) = 0.13:24 128. P1: PBU/OVYP2: PBU/OVYJWDD023-02JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006104 CHAPTER 2 LIMITS AND CONTINUITY (b) If f is continuous on [a, b] with f (a) < 0 and f (b) > 0, then there must exist at least one number c in (a, b) for which f (c) = 0. (c) If f is continuous on (a, b) with f (a) > 0 and f (b) < 0, then there must exist at least one number c in (a, b), for which f (c) = 0. (d) If f is continuous on [a, b] with f (c) = 0 for some number c in (a, b), then f (a) and f (b) have opposite signs. Exercises 4243. Use the intermediate-value theorem to show that the equation has a solution in the interval specied. 42. x 3 3x 4 = 0, [2, 3]. 43. 2 cos x x + 1, [1, 2]. Exercises 4446. Give an , proof for each statement. 44. lim (5x 4) = 6. x245. lim 2x + 5 = 3. x446. limx9x 5 = 2.47. Prove that if lim [ f (x)/x] exists, then lim f (x) = 0. x0x048. Prove that if lim g(x) = l and if f is continuous at l, then xclim f (g(x)) = f (l). HINT: See Theorem 2.4.4.xc49. A function f is continuous at all x 0. Can f take on the value zero at and only (a) at the positive integers? (b) at the reciprocals of the positive integers? If the answer is yes, sketch a gure that supports your answer; if the answer is no, prove it. 50. Two functions f and g are everywhere dened. Can they both be everywhere continuous (a) if they differ only at a nite number of points? (b) if they differ only on a bounded closed interval [a, b]? (c) if they differ only on a bounded open interval (a, b)? Justify your answers.13:24 129. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER3THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 3.1 THE DERIVATIVE Introductionsecant lineWe begin with a function f . On the graph of f we choose a point (x, f (x)) and a nearby point (x + h, f (x + h)). (See Figure 3.1.1.) Through these two points we draw a line. We call this line a secant line because it cuts through the graph of f . The gure shows this secant line rst with h > 0 and then with h < 0.secant line(x, f (x))(x + h, f (x + h))(x, f (x))(x + h, f (x + h))h>0 h>0Figure 3.1.1Whether h is positive or negative, the slope of the secant line is the difference quotient f (x + h) f (x) . h (Check this out.)The word secant comes from the Latin secare, to cut.10513:54 130. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006106 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION If we let h tend to zero (from one side or the other), then ideally the point (x + h, f (x + h)) slides along the curve toward (x, f (x)), x + h tends to x, f (x + h) tends to f (x), and the slope of the secant f (x + h) f (x) h()tends to a limit that we denote by f (x) . While () represents the slope of the approaching secant, the number f (x) represents the slope of the graph at the point (x, f (x)). What we call differential calculus is the implementation of this idea.Derivatives and Differentiation DEFINITION 3.1.1A function f is said to be differentiable at x if f (x + h) f (x) exists. h0 h If this limit exists, it is called the derivative of f at x and is denoted by f (x). limAs indicated in the introduction, the derivative f (x) = limh0f (x + h) f (x) hrepresents slope of the graph of f at the point (x, f (x)). The line that passes through the point (x, f (x)) with slope f (x) is called the tangent line at the point (x, f (x)). (This line is marked by dashes in Figure 3.1.1.)Example 1 We begin with a linear function f (x) = mx + b . The graph of this function is the line y = mx + b, a line with constant slope m. We therefore expect f (x) to be constantly m. Indeed it is: for h = 0, f (x + h) f (x) [m(x + h) + b] [mx + b] mh = = =m h h h and therefore f (x + h) f (x) = lim m = m. f (x) = lim h0 h0 hy2xExample 2 Now we look at the squaring functionslopef (x) = x 2 . To nd f (x), we form the difference quotient(x, x 2)xFigure 3.1.2(Figure 3.1.2)(x + h)2 x 2 f (x + h) f (x) = h h This prime notation goes back to the French mathematician Joseph-Louis Lagrange (17361813). Other notations are introduced later.13:54 131. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20061073.1 THE DERIVATIVEand take the limit as h 0. Since (x + h)2 x 2 (x 2 + 2xh + h 2 ) x 2 2xh + h 2 = = = 2x + h, h h h we have f (x + h) f (x) = 2x + h. h Therefore f (x + h) f (x) = lim (2x + h) = 2x. h0 h The slope of the graph changes with x. For x < 0, the slope is negative and the curve tends down; at x = 0, the slope is 0 and the tangent line is horizontal; for x > 0, the slope is positive and the curve tends up. f (x) = limh0Example 3 Here we look for f (x) for the square-root function f (x) = x, x 0.ySince f (x) is a two-sided limit, we can expect a derivative at most for x > 0. We take x > 0 and form the difference quotient f (x + h) f (x) x +h x = . h h We simplify this expression by multiplying both numerator and denominator by x + h + x. This gives us f (x + h) f (x) x +h x x +h+ x = h h x +h+ x =tgentan(Figure 3.1.3) 1peslox 2(x, x) x square-root functionFigure 3.1.3(x + h) x 1 = . h( x + h + x) x +h+ xIt follows that f (x) = limh0f (x + h) f (x) 1 1 = lim = . h0 h 2 x x +h+ xAt each point of the graph to the right of the origin the slope is positive. As x increases, the slope diminishes and the graph attens out. The derivative f is a function, its value at x being f (x). However, this function f is dened only at those numbers x where f is differentiable. As you just saw in Example 3, while the square-root function is dened on [0, ), its derivative f is dened only on (0, ): 1 f (x) = x for all x 0; f (x) = only for x > 0. 2 xy(x, 1 ) xTo differentiate a function f is to nd its derivative f .Example 4 Lets differentiate the reciprocal function 1 f (x) = . x We begin by forming the difference quotient 1 1 f (x + h) f (x) x +h x. = h hx slope 1x2tangent(Figure 3.1.4) y=1 xFigure 3.1.413:54 132. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006108 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Now we simplify: x h x +h 1 1 1 x +h x = x(x + h) x(x + h) = x(x + h) = . h h h x(x + h) It follows that f (x) = limh0f (x + h) f (x) 1 1 = lim = 2. h0 x(x + h) h xThe graph of the function consists of two curves. On each curve the slope, 1/x 2 , is negative: large negative for x close to 0 (each curve steepens as x approaches 0 and tends toward the vertical) and small negative for x far from 0 (each curve attens out as x moves away from 0 and tends toward the horizontal). Evaluating Derivatives Example 5 We take f (x) = 1 x 2 and calculate f (2). We can rst nd f (x): f (x) = limh0f (x + h) f (x) h[1 (x + h)2 ] [1 x 2 ] 2xh h 2 = lim = lim (2x h) = 2x h0 h0 h0 h h and then substitute 2 for x: = limf (2) = 2(2) = 4. We can also evaluate f (2) directly: f (2) = limh0f (2 + h) f (2) h[1 (2 + h)2 ] [1 (2)2 ] 4h h 2 = lim = lim (4 h) = 4. h0 h0 h0 h h= limExample 6 Lets nd f (3) and f (1) given that f (x) =x 2, 2x 1,x 1 x > 1.By denition, f (3) = limh0f (3 + h) f (3) . hFor all x sufciently close to 3, f (x) = x 2 . Thus (3 + h)2 (3)2 (9 6h + h 2 ) 9 = lim = lim (6 + h) = 6. h0 h0 h0 h hf (3) = lim Now lets ndf (1) = limh0f (1 + h) f (1) . hSince f is not dened by the same formula on both sides of 1, we will evaluate this limit by taking one-sided limits. Note that f (1) = 12 = 1.13:54 133. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.1 THE DERIVATIVETo the left of 1, f (x) = x 2 . Thus limh0f (1 + h) f (1) (1 + h)2 1 = lim h0 h h = lim h0(1 + 2h + h 2 ) 1 = lim (2 + h) = 2. h0 hTo the right of 1, f (x) = 2x 1. Thus f (1 + h) f (1) [2(1 + h) 1] 1 = lim+ = lim+ 2 = 2. h0 h0 h0 h h The limit of the difference quotient exists and is 2: lim+f (1) = limh0f (1 + h) f (1) = 2. hTangent Lines If f is differentiable at c, the line that passes through the point (c, f (c)) with slope f (c) is the tangent line at that point. As an equation for this line we can write y f (c) = f (c)(x c).(3.1.2)(point-slope form)This is the line through (c, f (c)) that best approximates the graph of f near the point (c, f (c)).Example 7 We go back to the square-root function f (x) = x and write an equation for the tangent line at the point (4, 2). As we showed in Example 3, for x > 0 1 f (x) = . 2 x Thus f (4) = 1 . The equation for the tangent line at the point (4, 2) can be written 4 y 2 = 1 (x 4). 4Example 8 We differentiate the function f (x) = x 3 12x and seek the points of the graph where the tangent line is horizontal. Then we write an equation for the tangent line at the point of the graph where x = 3. First we calculate the difference quotient: [(x + h)3 12(x + h)] [x 3 12x] f (x + h) f (x) = h h =x 3 + 3x 2 h + 3xh 2 + h 3 12x 12h x 3 + 12x h=3x 2 h + 3xh 2 + h 3 12h = 3x 2 + 3xh + h 2 12. h10913:54 134. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006110 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Now we take the limit as h 0: f (x + h) f (x) = lim (3x 2 + 3xh + h 2 12) = 3x 2 12. f (x) = lim h0 h0 hy (2, 16)The function has a horizontal tangent at the points (x, f (x)) where f (x) = 0. In this case 22f (x) = 0xiff3x 2 12 = 0iffx = 2.The graph has a horizontal tangent at the pointsf (x) = x3 12x(2, f (2)) = (2, 16)(2, 16)(2, f (2)) = (2, 16).andThe graph of f and the horizontal tangents are shown in Figure 3.1.5. The point on the graph where x = 3 is the point (3, f (3)) = (3, 9). The slope at this point is f (3) = 15, and the equation of the tangent line at this point can be writtenFigure 3.1.5y + 9 = 15(x 3). A Note on Vertical TangentsThe cube-root function f (x) = x 1/3 is everywhere continuous, but as we show below, it is not differentiable at x = 0. The difference quotient at x = 0, yxf (x) = x1/3Figure 3.1.6f (0 + h) f (0) h 1/3 0 1 = = 2/3 , h h h increases without bound as h 0. In the notation established in Section 2.1, f (0 + h) f (0) as h 0, . h Thus f is not differentiable at x = 0. The behavior of f at x = 0 is depicted in Figure 3.1.6. For reasons geometrically evident, we say that the graph of f has a vertical tangent at the origin.Differentiability and Continuity A function can be continuous at a number x without being differentiable there. Viewed geometrically, this can happen for only one of two reasons: either the tangent line at (x, f (x)) is vertical (you just saw an example of this), or there is no tangent line at (x, f (x)). The lack of a tangent line at a point of continuity is illustrated below. The graph of the absolute value function f (x) = x is shown in Figure 3.1.7. The function is continuous at 0 (it is continuous everywhere), but it is not differentiable at 0: 0 + h 0 h f (0 + h) f (0) 1, h < 0 = = = 1, h > 0 h h hyso that limh0f (0 + h) f (0) = 1, hlimh0+f (0 + h) f (0) =1 hand thus xlimno derivative at 0h0f (0 + h) f (0) hdoes not exist.Figure 3.1.7 Vertical tangents will be considered in detail in Section 4.7.13:54 135. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.1 THE DERIVATIVEThe lack of differentiability at 0 is evident geometrically. At x = 0 the graph changes direction abruptly and there is no tangent line. Another example of this sort of behavior is offered by the function f (x) =1 x 2x 2, + 1, 2x 1 x > 1.limf (1 + h) f (1) (1 + h)2 1 h 2 + 2h = lim = lim = lim (h + 2) = 2, h0 h0 h0 h h hlim+1 (1 + h) + f (1 + h) f (1) = lim+ 2 h0 h hh01 21= lim+ h0111y(Figure 3.1.8)As you can check, the function is everywhere continuous, but at the point (1, 1) the graph has an abrupt change of direction. The calculation below conrms that f is not differentiable at x = 1: h01 2=1xno derivative at 1Figure 3.1.81 . 2Since these one-sided limits are different, limh0f (1 + h) f (1) hdoes not exist.1 x 5, 0 y 10For our last example, f (x) = x 3 6x 2 + 8x + 3,(Figure 3.1.9)we used a graphing utility. So doing, it appeared that f is differentiable except, possibly, at x = 0, at x = 2, and at x = 4. There abrupt changes in direction seem to occur. By zooming in near the point (2, f (2)), we conrmed that the left-hand limits and righthand limits of the difference quotient both exist at x = 2 but are not equal. See Figure 3.1.10. A similar situation was seen at x = 0 and x = 4. From the look of it, f fails to be differentiable at x = 0, at x = 2, and at x = 4. Although not every continuous function is differentiable, every differentiable function is continuous.Figure 3.1.9 1.997 x 2.003, 2.996 y 3.006THEOREM 3.1.3If f is differentiable at x, then f is continuous at x.PROOFFor h = 0 and x + h in the domain of f , f (x + h) f (x) =f (x + h) f (x) h . hWith f differentiable at x, limh0f (x + h) f (x) = f (x). hSince lim h = 0, we have h0lim [ f (x + h) f (x)] = limh0h0f (x + h) f (x) lim h = f (x) 0 = 0. h0 hIt wasnt necessary to use a graphing utility here, but we gured that the use of it might make for a pleasant change of paceFigure 3.1.1013:54 136. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006112 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION It follows that lim f (x + h) = f (x),h0and thus f is continuous at x.(2.2.6)EXERCISES 3.1 Exercises 110. Differentiate the function by forming the difference quotient22. Exercise 21 for the function f graphed below. yf (x + h) f (x) h2and taking the limit as h tends to 0. 1. 3. 5. 7. 9.f (x) = 2 3x. f (x) = 5x x 2 . f (x) = x 4 . f (x) = x 1. f (x) = 1/x 2 .2. 4. 6. 8. 10.f (x) = k, k constant. f (x) = 2x 3 + 1. f (x) = 1/(x + 3). f (x) = x 3 4x. f (x) = 1/ x.f1211324xExercises 1116. Find f (c) by forming the difference quotient f (c + h) f (c) hExercises 2328. Draw the graph of f ; indicate where f is not differentiable. 23. f (x) = x + 1 . 24. f (x) = 2x 5 . 26. f (x) = x 2 4 . 25. f (x) = x .and taking the limit as h 0. 12. f (x) = 7x x 2 ; c = 2 11. f (x) = x 2 4x; c = 3. 14. f (x) = 5 x 4 ; c = 1. 13. f (x) = 2x 3 + 1; c = 1. 8 15. f (x) = ; c = 2. 16. f (x) = 6 x; c = 2. x +427. f (x) =Exercises 1720. Write an equation for the tangent line at (c, f (c)). 18. f (x) = x; c = 4. 17. f (x) = 5x x 2 ; c = 4. 20. f (x) = 5 x 3 ; c = 2. 19. f (x) = 1/x 2 ; c = 2. 21. The graph of f is shown below.x 2, 2 x,x 1 x > 1.28. f (x) =30. f (x) =3x 2 , 2x + 1,x 1 x > 1;31. f (x) =x + 1, (x + 1)2 ,x 1 x > 1;32. f (x) = 1 x 2, 2 3x,22f112341(a) At which numbers c is f discontinuous? Which of the discontinuities is removable? (b) At which numbers c is f continuous but not differentiable?x 2 x > 2.Exercises 2932. Find f (c) if it exists. 4x, x < 1 c = 1. 29. f (x) = 2x 2 + 2, x 1;y1x 2 1, 3,x3x 0.(a) Determine f (x) for x = 0. (b) Show that f is not differentiable at x = 0. 41. Find A and B given that the function f (x) =x 3, Ax + B,x 1 x > 1.is differentiable at x = 1. 42. Find A and B given that the function35. yf (x) =(2, 2)x (2, 2)36. yx2x 2 2, Bx + Ax, 2x 2 x >2is differentiable at x = 2. Exercises 4348. Give an example of a function f that is dened for all real numbers and satises the given conditions. 43. f (x) = 0 for all real x. 44. f (x) = 0 for all x = 0; f (0) does not exist. 45. f (x) exists for all x = 1; f (1) does not exist. 46. f (x) exists for all x = 1; neither f (1) nor f (1) exists. 47. f (1) = 2 and f (1) = 7. 48. f (x) = 1 for x < 0 and f (x) = 1 for x > 0. x 2 x, x 2 49. Set f (x) = 2x 2, x > 2. (a) Show that f is continuous at 2. (b) Is f differentiable at 2? 50. Let f (x) = x x, x 0. Calculate f (x) for each x > 0. 1 x 2, x 0 x 2 , x > 0. (a) Is f differentiable at 0? (b) Sketch the graph of f . 52. Set x, x rational f (x) = g(x) = 0, x irrational,51. Set f (x) =37. y1 x 1x 2, 0,x rational x irrational.(a) Show that f is not differentiable at 0. (b) Show that g is differentiable at 0 and give g (0). (Normal lines) If the graph of f has a tangent line at (c, f (c)), then38. y (2, 2)(2, 2)x39. Show that f (x) =(x + 1)2 , (x 1)2 ,x 2, 2x,is not differentiable at x = 1.x 1 x >1(3.1.4)the line through (c, f (c)) that is perpendicular to the tangent line is called the normal line.53. Write an equation for the normal line at (c, f (c)) given that the tangent line at this point (a) is horizontal; (b) has slope f (c) = 0; (c) is vertical. 54. All the normals through a circular arc pass through one point. What is this point? 55. As you saw in Example 7, the line y 2 = 1 (x 4) is tangent 4 to the graph of the square-root function at the point (4, 2). Write an equation for the normal line through this point.13:54 138. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006114 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 56. (A follow-up to Exercise 55) Sketch the graph of the squareroot function displaying both the tangent and the normal at the point (4, 2). 57. The lines tangent and normal to the graph of the squaring function at the point (3, 9) intersect the x-axis at points s units apart. What is s? 58. The graph of the function f (x) = 1 x 2 is the upper half of the unit circle. On that curve (see the gure below) we have marked a point P(x, y). y(a) Show that f and g are both continuous at 0. (b) Show that f is not differentiable at 0. (c) Show that g is differentiable at 0 and give g (0). (Important). By denition f (c + h) f (c) h provided this limit exists. Setting x = c + h, we can write f (c) = limh0(3.1.5)f (c) = limxcnormal P(x, y) tangent(1, 0)x(1, 0)(a) What is the slope of the normal at P? Express your answer in terms of x and y. (b) Deduce from (a) the slope of the tangent at P. Express your answer in terms of x and y. (c) Conrm your answer in (b) by calculating 1 (x + h)2 1 x 2 . f (x) = lim h0 h HINT: First rationalize the numerator of the difference quotient by multiplying both numerator and denominator by 1 (x + h)2 + 1 x 2 . x sin(1/x), x = 0 and g(x) = x f (x). 0, x = 0 The graphs of f and g are indicated in the gures below.59. Letf (x) =y y=xf xy = xyy = x2g xy = x 2f (x) f (c) x cThis is an alternative denition of derivative which has advantages in certain situations. Convince yourself of the equivalence of both denitions by calculating f (c) by both methods. 61. f (x) = x 2 3x; c = 1. 60. f (x) = x 3 + 1; c = 2. 63. f (x) = x 1/3 ; c = 1. 62. f (x) = 1 + x; c = 3. 1 64. f (x) = ; c = 0. x +2 65. Set f (x) = x 5/2 and consider the difference quotient f (2 + h) f (2) . h (a) Use a graphing utility to graph D for h = 0. Estimate f (2) to three decimal places from the graph. (b) Create a table of values to estimate lim D(h). Estimate D(h) =h0f (2) to three decimal places from the table. (c) Compare your results from (a) and (b). 66. Exercise 65 with f (x) = x 2/3 . 67. Use the denition of the derivative with a CAS to nd f (x) in general and f (c) in particular. (a) f (x) = 5x 4; c = 3. (b) f (x) = 2 x 2 + 4x 4 x 6 ; c = 2. 3 2x (c) f (x) = ; c = 1. 2 + 3x 68. Use a CAS to evaluate, if possible, f (c + h) f (c) . h (a) f (x) = x 1 + 2; c = 1. (b) f (x) = (x + 2)5/3 1; c = 2. (c) f (x) = (x 3)2/3 + 3; c = 3. 69. Let f (x) = 5x 2 7x 3 on [1, 1]. (a) Use a graphing utility to draw the graph of f . (b) Use the trace function to approximate the points on the graph where the tangent line is horizontal. (c) Use a CAS to nd f (x). (d) Use a solver to solve the equation f (x) = 0 and compare what you nd to what you found in (b). 70. Exercise 69 with f (x) = x 3 + x 2 4x + 3 on [2, 2]. 71. Set f (x) = 4x x 3 . (a) Use a CAS to nd f ( 3 ). Then nd equations for the 2 tangent T and the normal N at the point ( 3 , f ( 3 )). 2 2 f (c) = limh013:54 139. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.2 SOME DIFFERENTIATION FORMULAS(b) Use a graphing utility to display N , T , and the graph of f in one gure. (c) Note that T is a good approximation to the graph of f for x close to 3 . Determine the interval on which the 2 vertical separation between T and the graph of f is of absolute value less than 0.01. 72. If f (x) = x, then f (x) = 1 x 0 = 1. If f (x) = x 2 , then f (x) = 2x 1 = 2x.iff (x) = x 3 ,f (x) = x nf (x) = , any constant,f (x) =then6x 2 1 x 4 + 5x + 1f (x) = 0for all x,and the identity function f (x) = x has constant derivative 1: (3.2.2)PROOFiff (x) = x,thenf (x) = 1for all x.For f (x) = , f (x) = limh0f (x + h) f (x) = lim = lim 0 = 0. h0 h0 h hFor f (x) = x, f (x) = limh0has derivativef (x) = nx n1 .(x + h)k+1 x k+1 = x(x + h)k x x k +h(x + h)k .f (x + h) f (x) h and then taking the limit as h tends to 0 is somewhat laborious. Here we derive some general formulas that enable us to calculate such derivatives quite quickly and easily. We begin by pointing out that constant functions have derivative identically 0: iff (x) = 3x 2 .HINT:by forming the appropriate difference quotient(3.2.1)then(b) Prove by induction that for each positive integer n,Calculating the derivative of or115(a) Show that 3.2 SOME DIFFERENTIATION FORMULAS f (x) = (x 3 + 2x 3)(4x 2 + 1)f (x + h) f (x) (x + h) x h = lim = lim = lim 1 = 1. h0 h0 h h0 h hRemark These results can be veried geometrically. The graph of a constant function f (x) = is a horizontal line, and the slope of a horizontal line is 0. The graph of the identity function f (x) = x is the graph of the line y = x. The line has slope 1. THEOREM 3.2.3 DERIVATIVES OF SUMS AND SCALAR MULTIPLESLet be a real number. If f and g are differentiable at x, then f + g and f are differentiable at x. Moreover, and ( f ) (x) = f (x). ( f + g) (x) = f (x) + g (x)13:54 140. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006116 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION To verify the rst formula, note thatPROOF( f + g)(x + h) ( f + g)(x) [ f (x + h) + g(x + h)] [ f (x) + g(x)] = h h f (x + h) f (x) g(x + h) g(x) + . = h h By denition, limh0f (x + h) f (x) = f (x) handlimh0g(x + h) g(x) = g (x). hThus limh0( f + g)(x + h) ( f + g)(x) = f (x) + g (x), hwhich means that ( f + g) (x) = f (x) + g (x). To verify the second formula, we must show that ( f )(x + h) ( f )(x) = f (x). h0 h limThis follows directly from the fact that ( f )(x + h) ( f )(x) f (x + h) f (x) = = h hf (x + h) f (x) . hRemark In this section and in the next few sections we will derive formulas for calculating derivatives. It will be to your advantage to commit these formulas to memory. You may nd it useful to put these formulas into words. According to Theorem 3.2.3, the derivative of a sum is the sum of the derivatives and the derivative of a scalar multiple is the scalar multiple of the derivative. Since f g = f + (g), it follows that if f and g are differentiable at x, then f g is differentiable at x, and(3.2.4)( f g) (x) = f (x) g (x). The derivative of a difference is the difference of the derivatives. These results can be extended by induction to any nite collection of functions: if f 1 , f 2 , . . . , f n are differentiable at x, and 1 , 2 , . . . , n are numbers, then the linear combination 1 f 1 + 2 f 2 + . . . + n f n is differentiable at x and(3.2.5)(1 f 1 + 2 f 2 + + n f n ) (x) = 1 f 1 (x) + 2 f 2 (x) + + n f n (x).The derivative of a linear combination is the linear combination of the derivatives. 13:54 141. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.2 SOME DIFFERENTIATION FORMULASTHEOREM 3.2.6 THE PRODUCT RULEIf f and g are differentiable at x, then so is their product, and ( f g) (x) = f (x)g (x) + g(x) f (x).The derivative of a product is the rst function times the derivative of the second plus the second function times the derivative of the rst. PROOFWe form the difference quotientf (x + h)g(x + h) f (x)g(x) ( f g)(x + h) ( f g)(x) = h h f (x + h)g(x + h) f (x + h)g(x) + f (x + h)g(x) f (x)g(x) = h and rewrite it as f (x h) f (x) g(x + h) g(x) + g(x) . f (x + h) h h [Here we have added and subtracted f (x + h)g(x) in the numerator and then regrouped the terms so as to display the difference quotients for f and g.] Since f is differentiable at x, we know that f is continuous at x (Theorem 3.1.5) and thus lim f (x + h) = f (x).(Exercise 49, Section 2.4)h0Since limh0g(x + h) g(x) = g (x) handlimh0f (x + h) f (x) = f (x), hwe obtain limh0( f g)(x + h) ( f g)(x) = hlim f (x + h) limh0h0g(x + h) g(x) + g(x) lim h0 h= f (x)g (x) + g(x) f (x).f (x + h) f (x) hUsing the product rule, it is not hard to show that(3.2.7)for each positive integer n p(x) = x nhas derivativep (x) = nx n1 .In particular, has derivativep (x) = 1 = 1 x 0 ,p(x) = x 2has derivativep (x) = 2x,p(x) = x3has derivativep (x) = 3x 2 ,p(x) = x 4has derivativep (x) = 4x 3 ,p(x) = xand so on. In this setting we are following the convention that x 0 is identically 1 even though in itself 00 is meaningless.11713:54 142. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006118 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION PROOF OF (3.2.7)We proceed by induction on n. If n = 1, then we have the identityfunction p(x) = x, which we know has derivative p (x) = 1 = 1 x 0 . This means that the formula holds for n = 1. We assume now that the result holds for n = k; that is, we assume that if p(x) = x k , then p (x) = kx k1 , and go on to show that it holds for n = k + 1. We let p(x) = x k+1 and note that p(x) = x x k . Applying the product rule (Theorem 3.2.6) and our induction hypothesis, we obtain p (x) = x kx k1 + x k 1 = (k + 1)x k . This shows that the formula holds for n = k + 1. By the axiom of induction, the formula holds for all positive integers n. Remark quotientFormula (3.2.7) can be obtained without induction. From the difference p(x + h) p(x) (x + h)n x n = , h happly the formula a n bn = (a b)(a n1 + a n2 b + + abn2 + bn1 ),(Section 1.2)and youll see that the difference quotient becomes the sum of n terms, each of which tends to x n1 as h tends to zero. The formula for differentiating polynomials follows from (3.2.5) and (3.2.7):IfP(x) = an x n + an1 x n1 + + a2 x 2 + a1 x + a0 ,thenP (x) = nan x n1 + (n 1)an1 x n2 + + 2a2 x + a1 .(3.2.8)For example, P(x) = 12x 3 6x 2has derivativeP (x) = 36x 2 6Q(x) = 1 x 4 2x 2 + x + 5 4has derivativeQ (x) = x 3 4x + 1.andExample 1 Differentiate F(x) = (x 3 2x + 3)(4x 2 + 1) and nd F (1). SOLUTION We have a product F(x) = f (x)g(x) withf (x) = x 3 2x + 3andg(x) = 4x 2 + 1.13:54 143. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.2 SOME DIFFERENTIATION FORMULASThe product rule gives F (x) = f (x)g (x) + g(x) f (x) = (x 3 2x + 3)(8x) + (4x 2 + 1)(3x 2 2) = 8x 4 16x 2 + 24x + 12x 4 5x 2 2 = 20x 4 21x 2 + 24x 2. Setting x = 1, we have F (1) = 20(1)4 21(1)2 + 24(1) 2 = 20 21 24 2 = 27.Example 2 Differentiate F(x) = (ax + b)(cx + d), where a, b, c, d are constants. SOLUTION We have a product F(x) = f (x)g(x) withf (x) = ax + bg(x) = cx + d.andAgain we use the product rule F (x) = f (x)g (x) + g(x) f (x). In this case F (x) = (ax + b)c + (cx + d)a = 2acx + bc + ad. We can also do this problem without using the product rule by rst carrying out the multiplication F(x) = acx 2 + bcx + ad x + bd and then differentiating F (x) = 2acx + bc + ad.Example 3 Suppose that g is differentiable at each x F(x) = (x 3 5x)g(x). Find F (2) given that g(2) = 3 and g (2) = 1.andthatSOLUTION Applying the product rule, we haveF (x) = [(x 3 5x)g(x)] = (x 3 5x)g (x) + g(x)(3x 2 5). Therefore, F (2) = (2)g (2) + (7)g(2) = (2)(1) + (7)(3) = 23. We come now to reciprocals.THEOREM 3.2.9 THE RECIPROCAL RULEIf g is differentiable at x and g(x) = 0, then 1/g is differentiable at x and 1 g(x) = g (x) . [g(x)]2Since g is differentiable at x, g is continuous at x. (Theorem 3.1.5) Since g(x) = 0, we know that 1/g is continuous at x and thus that 1 1 = . lim h0 g(x + h) g(x)PROOF11913:54 144. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006120 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION For h different from 0 and sufciently small, g(x + h) = 0. The continuity of g at x and the fact that g(x) = 0 guarantee this. (Exercise 50, Section 2.4) The difference quotient for 1/g can be written 1 1 1 g(x) g(x + h) 1 = h g(x + h) g(x) h g(x + h)g(x) 1 g(x + h) g(x) . h g(x + h)g(x)=As h tends to zero, the right-hand side (and thus the left) tends to g (x) . [g(x)]2 Using the reciprocal rule, we can show that Formula (3.2.7) also holds for negative integers: for each negative integer n,(3.2.10)p(x) = x nhas derivativep (x) = nx n1 .This formula holds at all x except, of course, at x = 0, where no negative power is even dened. In particular, for x = 0, p(x) = x 1has derivativep (x) = (1)x 2 = x 2 ,p(x) = x 2has derivativep (x) = 2x 3 ,p(x) = x 3has derivativep (x) = 3x 4 ,and so on. PROOF OF (3.2.10)Note that 1 where p(x) = g(x)g(x) = x n and n is a positive integer.The rule for reciprocals gives p (x) = g (x) (nx n1 ) = = (nx n1 )x 2n = nx n1 . [g(x)]2 x 2nExample 4 Differentiate f (x) =5 6 and nd f ( 1 ). 2 x2 xSOLUTION To apply (3.2.10), we writef (x) = 5x 2 6x 1 . Differentiation gives f (x) = 10x 3 + 6x 2 . Back in fractional notation, f (x) = 10 6 + 2. x3 xSetting x = 1 , we have 2 f1 2=10 1 3 2+6 1 2 2= 80 + 24 = 56. 13:54 145. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.2 SOME DIFFERENTIATION FORMULASExample 5 Differentiate f (x) =1 , where a, b, c are constants. ax 2 + bx + cSOLUTION Here we have a reciprocal f (x) = 1/g(x) withg(x) = ax 2 + bx + c. The reciprocal rule (Theorem 3.2.9) gives f (x) = g (x) 2ax + b = . 2 2 + bx + c]2 [g(x)] [axFinally we come to quotients in general.THEOREM 3.2.11 THE QUOTIENT RULEIf f and g are differentiable at x and g(x) = 0, then the quotient f /g is differentiable at x and f g(x) =g(x) f (x) f (x)g (x) [g(x)]2The derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Since f /g = f (1/g), the quotient rule can be obtained from the product and reciprocal rules. The proof of the quotient rule is left to you as an exercise. Finally, note that the reciprocal rule is just a special case of the quotient rule. [Take f (x) = 1.] From the quotient rule you can see that all rational functions (quotients of polynomials) are differentiable wherever they are dened.Example 6 Differentiate F(x) =6x 2 1 . x 4 + 5x + 1SOLUTION Here we are dealing with a quotient F(x) = f (x)/g(x). The quotient rule,F (x) =g(x) f (x) f (x)g (x) , [g(x)]2gives F (x) = =(x 4 + 5x + 1)(12x) (6x 2 1)(4x 3 + 5) (x 4 + 5x + 1)2 12x 5 + 4x 3 + 30x 2 + 12x + 5 . (x 4 + 5x + 1)2Example 7 Find equations for the tangent and normal lines to the graph of 3x f (x) = 1 2x at the point (2, f (2)) = (2, 2).12113:54 146. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006122 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION SOLUTION We need to nd f (2). Using the quotient rule, we getf (x) =(1 2x)(3) 3x(2) 3 = . 2 (1 2x) (1 2x)2This gives f (2) =3 3 1 = = . 2 2 [1 2(2)] (3) 3As an equation for the tangent, we can write y (2) = 1 (x 2), 3which simplies toy + 2 = 1 (x 2). 3The equation for the normal line can be written y + 2 = 3(x 2).Example 8 Find the points on the graph of 4x f (x) = 2 x +4 where the tangent line is horizontal. SOLUTION The quotient rule givesf (x) =(x 2 + 4)(4) 4x(2x) 16 4x 2 = 2 . (x 2 + 4)2 (x + 4)2The tangent line is horizontal only at the points (x, f (x)) where f (x) = 0. Therefore, we set f (x) = 0 and solve for x: y 1 212Figure 3.2.1x16 4x 2 =0 (x 2 + 4)216 4x 2 = 0iffiffx = 2.The tangent line is horizontal at the points where x = 2 or x = 2. These are the points (2, f (2)) = (2, 1) and (2, f (2)) = (2, 1). See Figure 3.2.1. Remark Some expressions are easier to differentiate if we rewrite them in more convenient form. For example, we can differentiate f (x) =x4 2 x 5 2x = x2 xby the quotient rule, or we can write f (x) = (x 4 2)x 1 and use the product rule; even better, we can write f (x) = x 3 2x 1 and proceed from there: f (x) = 3x 2 + 2x 2 .EXERCISES 3.2 Exercises 120. Differentiate. 1. F(x) = 1 x.2. F(x) = 2(1 + x). 3 4. F(x) = 2 . 3. F(x) = 11x 6x + 8. x 5. F(x) = ax 2 + bx + c; a, b, c constant. 53x3 x2 x x4 + . 4 3 2 1 (x 2 + 2) 1 8. F(x) = . 7. F(x) = 2 . x x3 6. F(x) =9. G(x) = (x 2 1)(x 3).13:54 147. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.2 SOME DIFFERENTIATION FORMULAS1 . x ax b 12. F(x) = ; cx d x2 1 13. G(x) = . 2x + 3 10. F(x) = x 15. 17. 19. 20.11. G(x) =x3 . 1xa, b, c, d constant.7x 4 + 11 . x +1 x 3 + 3x G(x) = (x 3 2x)(2x + 5). 16. G(x) = 2 . x 1 6 1/x 1 + x4 G(x) = . . 18. G(x) = x 2 x2 1 G(x) = (9x 8 8x 9 ) x + . x 1 1 G(x) = 1 + 1+ 2 . x x 14. G(x) =Exercises 2126. Find f (0) and f (1). 1 21. f (x) = . 22. f (x) = x 2 (x + 1). x 2 1 x2 2x 2 + x + 1 23. f (x) = . 24. f (x) = 2 . 1 + x2 x + 2x + 1 ax + b ; a, b, c, d constant. 25. f (x) = cx + d ax 2 + bx + c 26. f (x) = 2 ; a, b, c constant. cx + bx + a Exercises 2730. Find f (0) given that h(0) = 3 and h (0) = 2. 27. f (x) = xh(x). 28. f (x) = 3x 2 h(x) 5x. 1 x 29. f (x) = h(x) . 30. f (x) = h(x) + . h(x) h(x) Exercises 3134. Find an equation for the tangent line at the point (c, f (c)). x ; c = 4. 31. f (x) = x +2 32. f (x) = (x 3 2x + 1)(4x 5); c = 2. 33. f (x) = (x 2 3)(5x x 3 ); c = 1. 10 34. f (x) = x 2 ; c = 2. x12344. f (x) = x 3 3x is perpendicular to the line 5y 3x = 8. Exercises 4548. Find a function f with the given derivative. 46. f (x) = 4x 3 2x + 4. 45. f (x) = 3x 2 + 2x + 1. 1 1 47. f (x) = 2x 2 3x 2 . 48. f (x) = x 4 + 2x 3 + . x 2 x 49. Find A and B given that the derivative of f (x) =Ax 3 + Bx + 2,x 2Bx A,x >22is everywhere continuous. HINT: First of all, f must be continuous. 50. Find A and B given that the derivative of f (x) =51.52.53.54.55.56.Exercises 3538. Find the point(s) where the tangent line is horizontal. 35. f (x) = (x 2)(x 2 x 11). 5x 16 36. f (x) = x 2 . 37. f (x) = 2 . x x +1 38. f (x) = (x + 2)(x 2 2x 8).57.Exercises 3942. Find all x at which (a) f (x) = 0; (b) f (x) > 0; (c) f (x) < 0. 40. f (x) = 3x 4 4x 3 2. 39. f (x) = x 4 8x 2 + 3. x 2 2x + 4 4 42. f (x) = . 41. f (x) = x + 2 . x x2 + 459.Exercises 4344. Find the points where the tangent to the graph of 43. f (x) = x 2 6 is parallel to the line y = 4x 1.61.58.60.Ax 2 + B,x < 1Bx + Ax + 4,x 15is everywhere continuous. Find the area of the triangle formed by the x-axis, the tangent to the graph of f (x) = 6x x 2 at the point (5, 5), and the normal through this point (the line through this point that is perpendicular to the tangent). Find the area of the triangle formed by the x-axis and the lines tangent and normal to the graph of f (x) = 9 x 2 at the point (2, 5). Find A, B, C such that the graph of f (x) = Ax 2 + Bx + C passes through the point (1, 3) and is tangent to the line 4x + y = 8 at the point (2, 0). Find A, B, C, D such that the graph of f (x) = Ax 3 + Bx 2 + C x + D is tangent to the line y = 3x 3 at the point (1, 0) and is tangent to the line y = 18x 27 at the point (2, 9). Find the point where the line tangent to the graph of the quadratic function f (x) = ax 2 + bx + c is horizontal. NOTE: This gives a way to nd the vertex of the parabola y = ax 2 + bx + c. Find conditions on a, b, c, d which guarantee that the graph of the cubic p(x) = ax 3 + bx 2 + cx + d has: (a) exactly two horizontal tangents. (b) exactly one horizontal tangent. (c) no horizontal tangents. Find the points (c, f (c)) where the line tangent to the graph of f (x) = x 3 x is parallel to the secant line that passes through the points (1, f (1)) and (2, f (2)). Find the points (c, f (c)) where the line tangent to the graph of f (x) = x/(x + 1) is parallel to the secant line that passes through the points (1, f (1)) and (3, f (3)). Let f (x) = 1/x, x > 0. Show that the triangle that is formed by each line tangent to the graph of f and the coordinate axes has an area of 2 square units. Find two lines through the point (2, 8) that are tangent to the graph of f (x) = x 3 . Find equations for all the lines tangent to the graph of f (x) = x 3 x that pass through the point (2, 2).13:54 148. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006124 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 62. Set f (x) = x 3 . (a) Find an equation for the line tangent to the graph of f at (c, f (c)), c = 0. (b) Determine whether the tangent line found in (a) intersects the graph of f at a point other than (c, c3 ). If it does, nd the x-coordinate of the second point of intersection. 63. Given two functions f and g, show that if f and f + g are differentiable, then g is differentiable. Give an example to show that the differentiability of f + g does not imply that f and g are each differentiable. 64. We are given two functions f and g, with f and f g differentiable. Does it follow that g is differentiable? If not, nd a condition that guarantees that g is differentiable if both f and f g are differentiable. 65. Prove the validity of the quotient rule. 66. Verify that, if f, g, h are differentiable, then ( f gh) (x) = f (x)g(x)h(x) + f (x)g (x)h(x) + f (x)g(x)h (x). HINT: Apply the product rule to [ f (x)g(x)]h(x). 67. Use the result in Exercise 66 to nd the derivative of F(x) = (x 2 + 1)[1 + (1/x)](2x 3 x + 1). 68. Use the result in Exercise 66 to nd the derivative of G(x) = x[1/(1 + 2x)](x 2 + x 1). 69. Use the product rule to show that if f is differentiable, then g(x) = [ f (x)]2has derivativeg (x) = 2 f (x) f (x).70. Use the result in Exercise 69 to nd the derivative of g(x) = (x 3 2x 2 + x + 2)2 . Exercises 7174. Use a CAS to nd where f (x) = 0, f (x) > 0, f (x) < 0. Verify your results with a graphing utility. x2 . 71. f (x) = x +1 72. f (x) = 8x 5 60x 4 + 150x 3 125x 2 . x3 + 1 x 4 16 . 74. f (x) = . 73. f (x) = x2 x4 75. Set f (x) = sin x. (a) Estimate f (x) at x = 0, x = /6, x = /4, x = /3, and x = /2 using the difference quotient f (x + h) f (x) h taking h = 0.001. (b) Compare the estimated values of f (x) found in (a) with the values of cos x at each of these points. (c) Use your results in (b) to guess the derivative of the sine function. 76. Let f (x) = x 4 + x 3 5x 2 + 2. (a) Use a graphing utility to graph f on the interval [4, 4] and estimate the x-coordinates of the points where the tangent line to the graph of f is horizontal. (b) Use a graphing utility to graph f . Are there any points where f is not differentiable? If so, estimate the numbers where f fails to be differentiable. 3.3 THE d/dx NOTATION; DERIVATIVES OF HIGHER ORDER The d/dx Notation So far we have indicated the derivative by a prime. There are, however, other notations that are widely used, particularly in science and in engineering. The most popular of these is the double-d notation of Leibniz. In the Leibniz notation, the derivative of a function y is indicated by writing dy dx dy dt dy dzifyis a function of x,ifyis a function of t,ifyis a function of z,and so on. Thus, if y = x 3 ,dy = 3x 2 ; dxif y =1 dy 2 , = 3; 2 dt t tif y =z,dy 1 = dz 2 z Gottfried Wilhelm Leibniz (16461716), the German mathematician whose role in the creation of calculus was outlined on page 3.13:54 149. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.3 THE d/dx NOTATION; DERIVATIVES OF HIGHER ORDERThe symbols d d d , , , and so forth dx dt dz are also used as prexes before expressions to be differentiated. For example, d 2 d 5 d 3 (x 4x) = 3x 2 4, (t + 3t + 1) = 2t + 3, (z 1) = 5z 4 . dx dt dz In the Leibniz notation the differentiation formulas read: d d d d d [ f (x) + g(x)] = [ f (x)] + [g(x)], [ f (x)] = [ f (x)], dx dx dx dx dx d d d [ f (x)g(x)] = f (x) [g(x)] + g(x) [ f (x)], dx dx dx d dx d dxd 1 1 = [g(x)], 2 dx g(x) [g(x)]d d g(x) [ f (x)] f (x) [g(x)] f (x) dx dx = . g(x) [g(x)]2Often functions f and g are replaced by u and v and the x is left out altogether. Then the formulas look like this: du dv d du d (u + v) = + , (u) = , dx dx dx dx dx d du du (uv) = u +v , dx dx dx dv du u v 1 dv d u d 1 = 2 , = dx 2 dx . dx v v dx dx v v The only way to develop a feeling for this notation is to use it. Below we work out some examples.Example 1 Finddy 3x 1 for y = . dx 5x + 2SOLUTION We use the quotient rule:d d (5x + 2) (3x 1) (3x 1) (5x + 2) dy dx dx = dx (5x + 2)2 =(5x + 2)(3) (3x 1)(5) 11 = . (5x + 2)2 (5x + 2)2dy for y = (x 3 + 1)(3x 5 + 2x 1). dx SOLUTION Here we use the product rule:Example 2 Finddy d d = (x 3 + 1) (3x 5 + 2x 1) + (3x 5 + 2x 1) (x 3 + 1) dx dx dx = (x 3 + 1)(15x 4 + 2) + (3x 5 + 2x 1)(3x 2 ) = (15x 7 + 15x 4 + 2x 3 + 2) + (9x 7 + 6x 3 3x 2 ) = 24x 7 + 15x 4 + 8x 3 3x 2 + 2.12513:54 150. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006126 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Example 3 Findd dtt3 t2t . 1SOLUTIONd dtt3 t2t 1=d 3 d (t ) dt dt= 3t 2 Example 4 Finddu dxfort2t 1(t 2 1)(1) t(2t) t2 + 1 . = 3t 2 + 2 (t 2 1)2 (t 1)2u = x(x + 1)(x + 2).SOLUTION You can think of u as[x(x + 1)](x + 2)or asx[(x + 1)(x + 2)].From the rst point of view, d du = [x(x + 1)](1) + (x + 2) [x(x + 1)] dx dx = x(x + 1) + (x + 2)[x(1) + (x + 1)(1)] ()= x(x + 1) + (x + 2)(2x + 1).From the second point of view, d du = x [(x + 1)(x + 2)] + (x + 1)(x + 2)(1) dx dx = x[(x + 1)(1) + (x + 2)(1)] + (x + 1)(x + 2) ()= x(2x + 3) + (x + 1)(x + 2).Both () and () can be multiplied out to give du = 3x 2 + 6x + 2. dx Alternatively, this same result can be obtained by rst carrying out the multiplication and then differentiating u = x(x + 1)(x + 2) = x(x 2 + 3x + 2) = x 3 + 3x 2 + 2x so that du = 3x 2 + 6x + 2. dxExample 5 Evaluate dy/d x at x = 0 and x = 1 given that y = SOLUTIONAt x = 0, Remarkx2 . 4x28x (x 2 4)2x x 2 (2x) dy = 2 . = 2 4)2 dx (x (x 4)2 80 dy = 2 = 0; dx (0 4)2at x = 1,The notation dy dxx=a81 dy 8 = 2 = . 2 dx (1 4) 913:54 151. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.3 THE d/dx NOTATION; DERIVATIVES OF HIGHER ORDERis sometimes used to emphasize the fact that we are evaluating the derivative dy/d x at x = a. Thus, in Example 5, we have dy dx=0andx=0dy dx= x=18 9Derivatives of Higher Order As we noted in Section 3.1, when we differentiate a function f we get a new function f , the derivative of f . Now suppose that f can be differentiated. If we calculate ( f ) , we get the second derivative of f. This is denoted f . So long as we have differentiability, we can continue in this manner, forming the third derivative of f, written f , and so on. The prime notation is not used beyond the third derivative. For the fourth derivative of f, we write f (4) and more generally, for the nth derivative we write f (n) . The functions f , f , f , f (4) , . . . , f (n) are called the derivatives of f of orders 1, 2, 3, 4, . . . , n, respectively. For example, if f (x) = x 5 , then f (x) = 5x 4 ,f (x) = 20x 3 ,f (x) = 60x 2 ,f (4) (x) = 120x,f (5) (x) = 120.In this case, all derivatives of orders higher than ve are identically zero. As a variant of this notation, you can write y = x 5 and then y = 5x 4 ,y = 20x 3 ,y = 60x 2 ,and so on.Since each polynomial P has a derivative P that is in turn a polynomial, and each rational function Q has a derivative Q that is in turn a rational function, polynomials and rational functions have derivatives of all orders. In the case of a polynomial of degree n, derivatives of order greater than n are all identically zero. (Explain.) In the Leibniz notation the derivatives of higher order are written d d2 y = dx2 dxdy dx,d3 y d = dx3 dxd2 y dx2, . . . , and so on.or d d2 [ f (x)] = 2 dx dxd [ f (x)] , dxd d3 [ f (x)] = 3 dx dxd2 [ f (x)] , . . . , and so on. dx2Below we work out some examples.Example 6 If f (x) = x 4 3x 1 + 5, then f (x) = 4x 3 + 3x 2andf (x) = 12x 2 6x 3 .Example 7 d 5 (x 4x 3 + 7x) = 5x 4 12x 2 + 7, dx d2 5 d (5x 4 12x 2 + 7) = 20x 3 24x, (x 4x 3 + 7x) = 2 dx dx d3 5 d (20x 3 24x) = 60x 2 24. (x 4x 3 + 7x) = 3 dx dxExample 8 Finally, we consider y = x 1 . In the Leibniz notation dy = x 2 , dxd2 y = 2x 3 , dx2d3 y = 6x 4 , dx3d4 y = 24x 5 , . . . . dx412713:54 152. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006128 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION On the basis of these calculations, we are led to the general result dn y = (1)n n!x n1 . dxn[Recall that n! = n(n 1)(n 2) 3 2 1.]In Exercise 61 you are asked to prove this result. In the prime notation we have y = x 2 ,y = 2x 3 ,y = 6x 4 ,y (4) = 24x 5 , . . . .In general y (n) = (1)n n!x n1 . EXERCISES 3.3 Exercises 110. Find dy/d x. 2. y = x 2 + 2x 4 .1. y = 3x 4 x 2 + 1. 3. y = x 5. y =4. y =2x . 1x7. y =1 . xx2 . 1xx . 1 + x26. y = x(x 2)(x + 1). 8. y =x 1+x9. y =2x . 3x +1 . x3 1 3d dt10. y =4t . 2t 3 12x . (1 + x)16.d dt2t + 1 . t4 3217.d du2u . 1 2u18.d du19.d duu u . u1 u+120.d 2 [u (1 u 2 )(1 u 3 )]. du21.d dxx3 + x2 + x + 1 . x3 x2 + x 122.d dxu . +1u3x3 + x2 + x 1 . x3 x2 + x + 1Exercises 2326. Evaluate dy/d x at x = 2. 23. y = (x + 1)(x + 2)(x + 3). 24. y = (x + 1)(x 2 + 2)(x 3 + 3). 25. y =(x 1)(x 2) . (x + 2)1 x2 3 . 30. f (x) = x 2 2 . x x 31. f (x) = (x 2 2)(x 2 + 2).29. f (x) =32. f (x) = (2x 3)2x + 3 . xExercises 3338. Find d 3 y/d x 3 .Exercises 1122. Find the indicated derivative. d d 11. (2x 5). 12. (5x + 2). dx dx d 13. [(3x 2 x 1 )(2x + 5)]. dx d 14. [(2x 2 + 3x 1 )(2x 3x 2 )]. dx 15.Exercises 2732. Find the second derivative. 27. f (x) = 7x 3 6x 5 . 28. f (x) = 2x 5 6x 4 + 2x 1.26. y =(x 2 + 1)(x 2 2) . x2 + 233. y = 1 x 3 + 1 x 2 + x + 1. 3 2 34. y = (1 + 5x)2 .35. y = (2x 5)2 .36. y = 1 x 3 1 x 2 + x 3. 6 4 37. y = x 3 1 . x338. y =x4 + 2 . xExercises 3944. Find the indicated derivative. d d x (x x 2 ) . 39. dx dx 40.d d2 (x 2 3x) (x + x 1 ) . dx2 dxd4 [3x x 4 ]. dx4 d5 [ax 4 + bx 3 + cx 2 + d x + e]. 42. dx5 d2 d2 (1 + 2x) 2 (5 x 3 ) . 43. 2 dx dx 41.44.d3 dx31 d2 4 (x 5x 2 ) . x dx2Exercises 4548. Find a function y = f (x) for which: 2 46. y = x 3 + 3. 45. y = 4x 3 x 2 + 4x. x 4 dy 5 dy 48. = 5x 4 + 5 . = 4x 5 4 2. 47. dx x dx x13:54 153. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.3 THE d/dx NOTATION; DERIVATIVES OF HIGHER ORDER49. Find a quadratic polynomial p with p(1) = 3, p (1) = 2, and p (1) = 4. 50. Find a cubic polynomial p with p(1) = 0, p (1) = 3, p (1) = 2, and p (1) = 6. 51. Set f (x) = x n , n a positive integer. (a) Find f (k) (x) for k = n. (b) Find f (k) (x) for k > n. (c) Find f (k) (x) for k < n. 52. Let p be an arbitrary polynomial p(x) = an x n + an1 x n1 + + a1 x + a0 , an = 0. (a) Find (d n /d x n )[ p(x)]. (b) What is (d k /d x k )[ p(x)] for k > n? 53. Set f (x) = (a) (b) (c) (d)x 2, 0,x 0 . x 0, (c) f (x) < 0. 58. f (x) = x 4 . 57. f (x) = x 3 . 4 3 2 59. f (x) = x + 2x 12x . 60. f (x) = x 4 + 3x 3 6x 2 x. 61. Prove by induction that ify = x 1 ,thendn y = (1)n n!x n1 . dxn62. Calculate y , y , y for y = 1/x 2 . Use these results to guess a formula for y (n) for each positive integer n, and then prove the validity of your conjecture by induction. 63. Let u, , w be differentiable functions of x. Express the derivative of the product uw in terms of the functions u, , w, and their derivatives. dn 64. (a) Find n (x n ) for n = 1, 2, 3, 4, 5. Give the general fordx mula.129d n+1 n (x ). d x n+1 dn 1 [ f (x)]. . Find a formula for 65. Set f (x) = 1x dxn 1x . Use a CAS to nd a formula for 66. Set f (x) = 1+x n d [ f (x)]. dxn (b) Give the general formula for67. Set f (x) = x 3 x. (a) Use a graphing utility to display in one gure the graph of f and the line l : x 2y + 12 = 0. (b) Find the points on the graph of f where the tangent is parallel to l. (c) Verify the results you obtained in (b) by adding these tangents to your previous drawing. 68. Set f (x) = x 4 x 2 . (a) Use a graphing utility to display in one gure the graph of f and the line l : x 2y 4 = 0. (b) Find the points on the graph of f where the normal is perpendicular to l. (c) Verify the results you obtained in (b) by adding these normals to your previous drawing. 69. Set f (x) = x 3 + x 2 4x + 1. (a) Calculate f (x). (b) Use a graphing utility to display in one gure the graphs of f and f . If possible, graph f and f in different colors. (c) What can you say about the graph of f where f (x) < 0? What can you say about the graph of f where f (x) > 0? 70. Set f (x) = x 4 x 3 5x 2 x 2. (a) Calculate f (x). (b) Use a graphing utility to display in one gure the graphs of f and f . If possible, graph f and f in different colors. (c) What can you say about the graph of f where f (x) < 0? What can you say about the graph of f where f (x) > 0? 71. Set f (x) = 1 x 3 3x 2 + 3x + 3. 2 (a) Calculate f (x). (b) Use a graphing utility to display in one gure the graphs of f and f . If possible, graph f and f in different colors. (c) What can you say about the graph of f where f (x) = 0? (d) Find the x-coordinate of each point where the tangent to the graph of f is horizontal by nding the zeros of f to three decimal places. 72. Set f (x) = 1 x 3 3x 2 + 4x + 1. 2 (a) Calculate f (x). (b) Use a graphing utility to display in one gure the graphs of f and f . If possible, graph f and f in different colors. (c) What can you say about the graph of f where f (x) = 0? (d) Find the x-coordinate of each point where the tangent to the graph of f is horizontal by nding the zeros of f to three decimal places.13:54 154. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006130 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 3.4 THE DERIVATIVE AS A RATE OF CHANGEyIn the case of a linear function y = mx + b, the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x. As x changes from x0 to x1 , y changes m times as much:y1 y1 y0 y0y1 y0 = m(x1 x0 )x1 x0(Figure 3.4.1)Thus the slope m gives the change in y per unit change in x. In the more general case of a differentiable function x0x1xy = f (x)y1 y0 = m (x1 x 0)Figure 3.4.1the graph is a curve. The slope dy = f (x) dx still gives the rate of change of y with respect to x, but this rate of change can vary from point to point. At x = x1 (see Figure 3.4.2) the rate of change of y with respect to x is f (x1 ); the steepness of the graph is that of a line of slope f (x1 ). At x = x2 , the rate of change of y with respect to x is f (x2 ); the steepness of the graph is that of a line of slope f (x2 ). At x = x3 , the rate of change of y with respect to x is f (x3 ); the steepness of the graph is that of a line of slope f (x3 ). m3 = f ' (x3) m1 = f ' (x1)m2 = f ' (x2) x1x2x3Figure 3.4.2The derivative as a rate of change is one of the fundamental ideas of calculus. Keep it in mind whenever you see a derivative. This section is only introductory. Well develop the idea further as we go on.Example 1 The area of a square is given by the formula A = x 2 where x is the length of a side. As x changes, A changes. The rate of change of A with respect to x is the derivative d 2 dA = (x ) = 2x. dx dx When x = 1 , this rate of change is 1 : the area is changing at half the rate of x. When 4 2 x = 1 , the rate of change of A with respect to x is 1: the area is changing at the same rate 2 as x. When x = 1, the rate of change of A with respect to x is 2 : the area is changing at twice the rate of x. In Figure 3.4.3 we have plotted A against x. The rate of change of A with respect to x at each of the indicated points appears as the slope of the tangent line. Example 2 An equilateral triangle of side x has area A = 1 3x 2 . 4(Check this out.)13:54 155. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.4 THE DERIVATIVE AS A RATE OF CHANGE Am=21m=1m=1 41 211 2xA = x 2, x > 0Figure 3.4.3The rate of change of A with respect to x is the derivative dA = 1 3x. 2 dx When x = 2 3, the of change of A with respect to x is 3. In other words, when rate the side has length 2 3, the area is changing three times as fast as the length of the side. x 2 . x2 (a) Find the rate of change of y with respect to x at x = 2. (b) Find the value(s) of x at which the rate of change of y with respect to x is 0.Example 3 Set y =SOLUTION The rate of change of y with respect to x is given by the derivative, dy/d x:x 2 + 4x 4x x 2 (1) (x 2)(2x) dy = = . = 4 dx x x4 x3 (a) At x = 2, 1 42 dy = . = 3 dx 2 4 4x dy = 0, we have = 0, and therefore x = 4. The rate of change of y dx x3 with respect to x at x = 4 is 0. (b) SettingExample 4 Suppose that we have a right circular cylinder of changing dimensions. (Figure 3.4.4.) When the base radius is r and the height is h, the cylinder has volume V = r 2 h.rhFigure 3.4.413113:54 156. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006132 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION If r remains constant while h changes, then V can be viewed as a function of h. The rate of change of V with respect to h is the derivative dV = r 2 . dh If h remains constant while r changes, then V can be viewed as a function of r. The rate of change of V with respect to r is the derivative dV = 2r h. dh Suppose now that r changes but V is kept constant. How does h change with respect to r? To answer this, we express h in terms of r and V : h=V V = r 2 . r 2 Since V is held constant, h is now a function of r. The rate of change of h with respect to r is the derivative dh 2(r 2 h) 3 2h 2V = r 3 = r = . dr rEXERCISES 3.4 1. Find the rate of change of the area of a circle with respect to the radius r. What is the rate when r = 2? 2. Find the rate of change of the volume of a cube with respect to the length s of a side. What is the rate when s = 4? 3. Find the rate of change of the area of a square with respect to the length z of a diagonal. What is the rate when z = 4? 4. Find the rate of change of y = 1/x with respect to x at x = 1. 5. Find the rate of change of y = [x(x + 1)]1 with respect to x at x = 2. 6. Find the values of x at which the rate of change of y = x 3 12x 2 + 45x 1 with respect to x is zero. 7. Find the rate of change of the volume of a sphere with respect to the radius r. 8. Find the rate of change of the surface area of a sphere with respect to the radius r. What is this rate of change when r = r0 ? How must r0 be chosen so that the rate of change is 1? 9. Find x0 given that the rate of change of y = 2x 2 + x 1 with respect to x at x = x0 is 4. 10. Find the rate of change of the area A of a circle with respect to (a) the diameter d; (b) the circumference C. 11. Find the rate of change of the volume V of a cube with respect to (a) the length w of a diagonal on one of the faces. (b) the length z of one of the diagonals of the cube. 12. The dimensions of a rectangle are changing in such a way that the area of the rectangle remains constant. Find the rate of change of the height h with respect to the base b.13. The area of a sector in a circle is given by the formula A = 1 r 2 where r is the radius and is the central angle 2 measured in radians. (a) Find the rate of change of A with respect to if r remains constant. (b) Find the rate of change of A with respect to r if remains constant. (c) Find the rate of change of with respect to r if A remains constant. 14. The total surface area of a right circular cylinder is given by the formula A = 2r (r + h) where r is the radius and h is the height. (a) Find the rate of change of A with respect to h if r remains constant. (b) Find the rate of change of A with respect to r if h remains constant. (c) Find the rate of change of h with respect to r if A remains constant. 15. For what value of x is the rate of change of y = ax 2 + bx + c with respect to x the same as the rate of change of z = bx 2 + ax + c with respect to x? Assume that a, b, c are constant with a = b. 16. Find the rate of change of the product f (x)g(x)h(x) with respect to x at x = 1 given that f (1) = 0,g(1) = 2,h(1) = 2,f (1) = 1,g (1) = 1,h (1) = 013:54 157. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.5 THE CHAIN RULE 3.5 THE CHAIN RULE In this section we take up the differentiation of composite functions. Until we get to Theorem 3.5.6, our approach is completely intuitiveno real denitions, no proofs, just informal discussion. Our purpose is to give you some experience with the standard computational procedures and some insight into why these procedures work. Theorem 3.5.6 puts this all on a sound footing. Suppose that y is a differentiable function of u and u in turn is a differentiable function of x. Then y is a composite function of x. Does y have a derivative with respect to x? Yes it does, and dy/d x is given by a formula that is easy to remember: dy dy du = . dx du d x(3.5.1)This formula, known as the chain rule, says that the rate of change of y with respect to x is the rate of change of y with respect to u times the rate of change of u with respect to x. Plausible as all this sounds, remember that we have proved nothing. All we have done is assert that the composition of differentiable functions is differentiable and given you a formulaa formula that needs justication and is justied at the end of this section. Before using the chain rule in elaborate computations, lets conrm its validity in some simple instances. If y = 2u and u = 3x, then y = 6x. Clearly dy du dy =6=23= , dx du d x and so, in this case, the chain rule is conrmed: dy du dy = . dx du d x If y = u 3 and u = x 2 , then y = (x 2 )3 = x 6 . This time dy = 6x 5 , dxdy = 3u 2 = 3(x 2 )2 = 3x 4 , dudu = 2x dxand once again dy du dy = 6x 5 = 3x 4 2x = . dx du d xExample 1 Find dy/d x by the chain rule given that y=u1 u+1andu = x 2.SOLUTIONdy (u + 1)(1) (u 1)(1) 2 = = du (u + 1)2 (u + 1)2anddu = 2x dxso that dy dy dy 2 4x = = 2x = 2 . 2 dx du d x (u + 1) (x + 1)213313:54 158. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006134 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Remark We would have obtained the same result without the chain rule by rst writing y as a function of x and then differentiating: withy=u1 u+1andu = x 2,we havey=x2 1 x2 + 1and dy (x 2 + 1)2x (x 2 1)2x 4x = = 2 . 2 + 1)2 dx (x (x + 1)2Suppose now that you were asked to calculate d [(x 2 1)100 ]. dx You could expand (x 2 1)100 into a polynomial by using the binomial theorem (thats assuming that you are familiar with the theorem and are adept at applying it) or you could try repeated multiplication, but in either case you would have a terrible mess on your hands: (x 2 1)100 has 101 terms. Using the chain rule, we can derive a formula that will render such calculations almost trivial. By the chain rule, we can show that, if u is a differentiable function of x and n is a positive or negative integer, then d n du (u ) = nu n1 . dx dx(3.5.2)If n is a positive integer, the formula holds without restriction. If n is negative, the formula is valid except at those numbers where u(x) = 0. PROOFSet y = u n . In this case, dy dy du = dx du d xgives du d n du d n (u ) = (u ) = nu n1 . dx du dx dxTo calculate d [(x 2 1)100 ], dx we set u = x 2 1. Then by our formula d d [(x 2 1)100 ] = 100(x 2 1)99 (x 2 1) = 100(x 2 1)99 2x = 200x(x 2 1)99 . dx dx Remark While it is clear that (3.5.2) is the only practical way to calculate the derivative of y = (x 2 1)100 , you do have a choice when differentiating a similar, but simpler, function such as y = (x 2 1)4 . By (3.5.2) d d [(x 2 1)4 ] = 4(x 2 1)3 (x 2 1) = 4(x 2 1)3 2x = 8x(x 2 1)3 . dx dx On the other hand, if we were to rst expand the expression (x 2 1)4 , we would get y = x 8 4x 6 + 6x 4 4x 2 + 113:54 159. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.5 THE CHAIN RULEand then dy = 8x 7 24x 5 + 24x 3 8x. dx As a nal answer, this is correct but somewhat unwieldy. To reconcile the two results, note that 8x is a factor of dy/d x: dy = 8x(x 6 3x 4 + 3x 2 1), dx and the expression in parentheses is (x 2 1)3 multiplied out. Thus, dy = 8x(x 2 1)3 , dx as we saw above. However, (3.5.2) gave us this neat, compact result much more efciently. Here are additional examples of a similar sort.Example 2 d dxx+31 x= 3 x +1 x41 d x+ dx x= 3 x +1 x411 . x2Example 3 d d [1 + (2 + 3x)5 ]3 = 3[1 + (2 + 3x)5 ]2 [1 + (2 + 3x)5 ]. dx dx Since d d [1 + (2 + 3x)5 ] = 5(2 + 3x)4 (2 + 3x) = 5(2 + 3x)4 (3) = 15(2 + 3x)4 , dx dx we have d [1 + (2 + 3x)5 ]3 = 3[1 + (2 + 3x)5 ]2 [15(2 + 3x)4 ] dx = 45(2 + 3x)4 [1 + (2 + 3x)5 ]2 .Example 4 Calculate the derivative of f (x) = 2x 3 (x 2 3)4 . SOLUTION Here we need to use the product rule and the chain rule:d d d [2x 3 (x 2 3)4 ] = 2x 3 [(x 2 3)4 ] + (x 2 3)4 (2x 3 ) dx dx dx = 2x 3 [4(x 2 3)3 (2x)] + (x 2 3)4 (6x 2 ) = 16x 4 (x 2 3)3 + 6x 2 (x 2 3)4 = 2x 2 (x 2 3)3 (11x 2 9).The formula dy du dy = dx du d x can be extended to more variables. For example, if x itself depends on s, then we have (3.5.3)dy du d x dy = . ds du d x ds13513:54 160. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006136 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION If, in addition, s depends on t, then dy du d x ds dy = , dt du d x ds dt(3.5.4)and so on. Each new dependence adds a new link to the chain.Example 5 Find dy/ds given that y = 3u + 1, u = x 2 , x = 1 s. SOLUTIONdy = 3, dudu = 2x 3 , dxdx = 1. dsTherefore dy du d x dy = = (3)(2x 3 )(1) = 6x 3 = 6(1 s)3 . ds du d x dsExample 6 Find dy/dt at t = 9 given that u+2 , u = (3s 7)2 , y= u1s= t.SOLUTION As you can check,dy 3 = , du (u 1)2ds 1 = . dt 2 tdu = 6(3s 7), dsAt t = 9, we have s = 3 and u = 4, so that 1 3 dy = , = 2 du (4 1) 3du = 6(9 7) = 12, dsds 1 1 = = . dt 6 2 9Thus, at t = 9, dy dy du ds 1 1 = = (12) dt du ds dt 3 62 = . 3Example 7 Gravel is being poured by a conveyor onto a conical pile at the constant rate of 60 cubic feet per minute. Frictional forces within the pile are such that the height is always two-thirds of the radius. How fast is the radius of the pile changing at the instant the radius is 5 feet? SOLUTION The formula for the volume V of a right circular cone of radius r and height h isV = 1 r 2 h. 3 However, in this case we are told that h = 2 r , and so we have 3 ()V = 2 r 3 . 9Since gravel is being poured onto the pile, the volume, and hence the radius, are functions of time t. We are given that d V /dt = 60 and we want to nd dr/dt at the13:54 161. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.5 THE CHAIN RULEinstant r = 5. Differentiating () with respect to t by the chain rule, we get dV dr d V dr = = 2 r 2 . 3 dt dr dt dt Solving for dr/dt and using the fact that d V /dt = 60 , we nd that 180 90 dr = = 2. dt 2r 2 r When r = 5, 90 90 dr = = 3.6. = dt (5)2 25 Thus, the radius is increasing at the rate of 3.6 feet per minute at the instant the radius is 5 feet. So far we have worked entirely in Leibnizs notation. What does the chain rule look like in prime notation? Lets go back to the beginning. Once again, let y be a differentiable function of u: say y = f (u). Let u be a differentiable function of x: say u = g(x). Then y = f (u) = f (g(x)) = ( f g)(x) and, according to the chain rule (as yet unproved), dy du dy = . dx du d x Since dy d = [( f g)(x)] = ( f g) (x), dx dx the chain rule can be written(3.5.5)dy = f (u) = f (g(x)), dudu = g (x), dx( f g) (x) = f (g(x)) g (x).The chain rule in prime notation says that the derivative of a composition f g at x is the derivative of f at g(x) times the derivative of g at x. In Leibnizs notation the chain rule appears seductively simple, to some even obvious. After all, to prove it, all you have to do is cancel the dus: dy dy du = . dx du d x Of course, this is just nonsense. What would one cancel from ( f g) (x) = f (g(x))g (x)? Although Leibnizs notation is useful for routine calculations, mathematicians generally turn to prime notation where precision is required.13713:54 162. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006138 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION It is time for us to be precise. How do we know that the composition of differentiable functions is differentiable? What assumptions do we need? Under what circumstances is it true that ( f g) (x) = f (g(x))g (x)? The following theorem provides the denitive answer.THEOREM 3.5.6 THE CHAIN-RULE THEOREMIf g is differentiable at x and f is differentiable at g(x), then the composition f g is differentiable at x and ( f g) (x) = f (g(x))g (x).A proof of this theorem appears in the supplement to this section. The argument is not as easy as canceling the dus. One nal point. The statement ( f g) (x) = f (g(x))g (x) is often written d [ f (g(x))] = f (g(x))g (x). dx EXERCISES 3.5 Exercises 16. Differentiate the function: (a) by expanding before differentiation, (b) by using the chain rule. Then reconcile your results. 1. y = (x 2 + 1)2 . 3. y = (2x + 1)3 . 5. y = (x + x 1 )2 .2. y = (x 3 1)2 . 4. y = (x 2 + 1)3 . 6. y = (3x 2 2x)2 .Exercises 720. Differentiate the function. 8. f (x) = (1 + 2x)5 . 7. f (x) = (1 2x)1 . 9. f (x) = (x 5 x 10 )20 .10. f (x) = x 2 +13. f (x) = (x x 3 + x 5 )4 . 15. f (t) = (t 1 + t 2 )4 .16. f (x) =.4 1 . 1+t 14. f (t) = (t t 2 )3 .11. f (x) = x 17. f (x) =1 x3x x2 + 1431 x212. f (t) =.4.18. f (x) = [(2x + 1)2 + (x + 1)2 ]3 . 1x2 x x3 . + + 3 2 1 20. f (x) = [(6x + x 5 )1 + x]2 . 19. f (x) =4x + 3 5x 23.Exercises 2124. Find dy/d x at x = 0. 1 , u = 2x + 1. 21. y = 1 + u2 1 22. y = u + , u = (3x + 1)4 . u 2u 23. y = , u = (5x 2 + 1)4 . 1 4u 1x 24. y = u 3 u + 1, u = . 1+x Exercises 2526. Find dy/dt. 1 7u 25. y = , u = 1 + x 2 , x = 2t 5. 1 + u2 1 7x 26. y = 1 + u 2 , u = , x = 5t + 2. 1 + x2 Exercises 2728. Find dy/d x at x = 2. 27. y = (s + 3)2 , s = t 3, t = x 2 . 1+s 1 28. y = , s = t , t = x. 1s t Exercises 2938. Evaluate the following, given that f (0) = 1, f (0) = 2, f (1) = 0, f (1) = 1, f (2) = 1, f (2) = 1, g(0) = 2, g (0) = 1, g(1) = 1, g (1) = 0, g(2) = 1, g (2) = 1, h(0) = 1, h (0) = 2, h(1) = 2, h (1) = 1, h(2) = 0, h (2) = 2,13:54 163. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.5 THE CHAIN RULE29. 31. 33. 35. 37.( f g) (0). ( f g) (2). (g f ) (1). ( f h) (0). (g f h) (2).30. 32. 34. 36. 38.( f g) (1). (g f ) (0). (g f ) (2). ( f h g) (1). (g h f ) (0).Exercises 3942. Find f (x). 39. f (x) = (x 3 + x)4 . 40. f (x) = (x 2 5x + 2)10 .13963. The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if P(x) = (x a)2 q(x)andq(a) = 0.Prove that if a is a double zero of P, then a is a zero of both P and P , and P (a) = 0. 64. The number a is called a triple zero (or a zero of multiplicity 3) of the polynomial P if P(x) = (x a)3 q(x)andq(a) = 0.341. f (x) =x 1x42. f (x) =x 2 + 1 recall that. d 1 [ x] = . dx 2 xExercises 4346. Express the derivative in prime notation. d x 1 d 44. [ f (x 2 + 1)]. f . 43. dx dx x +1 d d f (x) 1 45. 46. [[ f (x)]2 + 1]. . dx d x f (x) + 1 Exercises 4750. Determine the values of x for which (a) f (x) = 0; (b) f (x) > 0; (c) f (x) < 0. 48. f (x) = (1 x 2 )2 . 47. f (x) = (1 + x 2 )2 . 50. f (x) = x(1 x 2 )3 . 49. f (x) = x(1 + x 2 )1 . Exercises 5153. Find a formula for the nth derivative. x 1 . 52. y = . 51. y = 1x 1+x 53. y = (a + bx)n ; n a positive integer, a, b constants. a 54. y = , a, b, c constants. bx + c Exercises 5558. Find a function y = f (x) with the given derivative. Check your answer by differentiation. 56. y = 2x(x 2 1). 55. y = 3(x 2 + 1)2 (2x). dy dy = 2(x 3 2)(3x 2 ). = 3x 2 (x 3 + 2)2 . 58. 57. dx dx 59. A function L has the property that L (x) = 1/x for x = 0. Determine the derivative with respect to x of L(x 2 + 1). 60. Let f and g be differentiable functions such that f (x) = g(x) and g (x) = f (x), and let H (x) = [ f (x)]2 [g(x)]2 . Find H (x). 61. Let f and g be differentiable functions such that f (x) = g(x) and g (x) = f (x), and let T (x) = [ f (x)]2 + [g(x)]2 . Find T (x). 62. Let f be a differentiable function. Use the chain rule to show that: (a) if f is even, then f is odd. (b) if f is odd, then f is even.Prove that if a is a triple zero of P, then a is a zero of P, P , and P , and P (a) = 0. 65. The number a is called a zero of multiplicity k of the polynomial P if P(x) = (x a)k q(x)andq(a) = 0.Use the results in Exercises 63 and 64 to state a theorem about a zero of multiplicity k. 66. An equilateral triangle of side length x and altitude h has area A given by 3 2 2 3 where x= x h. A= 4 3 Find the rate of change of A with respect to h and determine this rate of change when h = 2 3. 67. As air is pumped into a spherical balloon, the radius increases at the constant rate of 2 centimeters per second. What is the rate of change of the balloons volume when the radius is 10 centimeters? (The volume V of a sphere of radius r is 4 r 3 .) 3 68. Air is pumped into a spherical balloon at the constant rate of 200 cubic centimeters per second. How fast is the surface area of the balloon changing when the radius is 5 centimeters? (The surface area S of a sphere of radius r is 4r 2 .) 69. Newtons law of gravitational attraction states that if two bodies are at a distance r apart, then the force F exerted by one body on the other is given by F(r ) = k r2where k is a positive constant. Suppose that, as a function of time, the distance between the two bodies is given by r (t) = 49t 4.9t 2 ,0 t 10.(a) Find the rate of change of F with respect to t. (b) Show that (F r ) (3) = (F r ) (7). 70. Set f (x) = 3 1 x. (a) Use a CAS to nd f (9). Then nd an equation for the line l tangent to the graph of f at the point (9, f (9)). (b) Use a graphing utility to display l and the graph of f in one gure. (c) Note that l is a good approximation to the graph of f for x close to 9. Determine the interval on which the vertical separation between l and the graph of f is of absolute value less than 0.01.13:54 164. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006140 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 1 . 1 + x2 (a) Use a CAS to nd f (1). Then nd an equation for the line l tangent to the graph of f at the point (1, f (1)). (b) Use a graphing utility to display l and the graph of f in one gure. (c) Note that l is a good approximation to the graph of f for x close to 1. Determine the interval on which the vertical separation between l and the graph of f is of absolute value less than 0.01.73. Use a CAS to express the following derivatives in f notation. d 1 d x2 1 (a) f , (b) f , dx x dx x2 + 1 d f (x) (c) . d x 1 + f (x) 74. Use a CAS to nd the following derivatives: d d [u 1 (u 2 (x))], [u 1 (u 2 (u 3 (x)))], (a) (b) dx dx d [u 1 (u 2 (u 3 (u 4 (x))))]. (c) dx d2 75. Use a CAS to nd a formula for [ f (g(x))]. dx271. Set f (x) =72. Use a CAS to ndd4 d x 2 4 (x 2 + 1)4 . dx dx*SUPPLEMENT TO SECTION 3.5 To prove Theorem 3.5.6, it is convenient to use a slightly different formulation of derivative.THEOREM 3.5.7 The function f is differentiable at x iff limtxf (t) f (x) exists. txIf this limit exists, it is f (x).PROOFFix x. For each t = x in the domain of f , dene G(t) =f (t) f (x) . txNote that G(x + h) =f (x + h) f (x) hand therefore f is differentiable at xifflim G(x + h) exists.h0The result follows from observing that lim G(x + h) = Lh0ifflim G(t) = L .txFor the equivalence of these two limits we refer you to (2.2.6). PROOF OF THEOREM 3.5.6By Theorem 3.5.7 it is enough to show that limtxf (g(t)) f (g(x)) = f (g(x))g (x). txWe begin by dening an auxiliary function F on the domain of f by setting f (y) f (g(x)) , y = g(x) y g(x) F(y) = f (g(x)), y = g(x)13:54 165. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.6 THE CHAIN RULE141F is continuous at g(x) since lim F(y) = limyg(x)yg(x)f (y) f (g(x)) , y g(x)and the right-hand side is (by Theorem 3.5.7) f (g(x)), which is the value of F at g(x). For t = x, (1)f (g(t)) f (g(x)) g(t) g(x) = F(g(t)) . tx txTo see this we note that, if g(t) = g(x), then both sides are 0. If g(t) = g(x), then F(g(t)) =f (g(t)) f (g(x)) , g(t) g(x)so that again we have equality. Since g, being differentiable at x, is continuous at x and since F is continuous at g(x), we know that the composition F g is continuous at x. Thus lim F(g(t)) = F(g(x)) = f (g(x)). by our denition of F txThis, together with (1), gives limtxf (g(t)) f (g(x)) = f (g(x))g (x). tx PROJECT 3.5 ON THE DERIVATIVE OF u n If n is a positive or negative integer and the function u is differentiable at x, then by the chain rule d d [u(x)]n = n[u(x)]n1 [u(x)], dx dx except that, if n is negative, the formula fails at those numbers x where u(x) = 0. We can obtain this result without appealing to the chain rule by using the product rule and carrying out an induction on n. Let u be a differentiable function of x. Then d d [u(x)]2 = [u(x) u(x)] dx dx d d = u(x) [u(x)] + u(x) [u(x)] dx dx d = 2u(x) [u(x)]; dx d d [u(x)]3 = [u(x) [u(x)]2 ] dx dx d d = u(x) [u(x)]2 + [u(x)]2 [u(x)] dx dxd d [u(x)] + [u(x)]2 [u(x)] dx dx d = 3[u(x)]2 [u(x)]. dx = 2[u(x)]2Problem 1. Show that d d [u(x)]4 = 4[u(x)]3 [u(x)]. dx dx Problem 2. Show by induction that d d [u(x)]n = n[u(x)]n1 [u(x)] dx dxfor all positive integers n.Problem 3. Show that if n is a negative integer, then d d [u(x)]n = n[u(x)]n1 [u(x)] dx dx except at those numbers x where u(x) = 0. HINT: Problem 2 and the reciprocal rule.13:54 166. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006142 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 3.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONS An outline review of trigonometrydenitions, identities, and graphsappears in Chapter 1. As indicated there, the calculus of the trigonometric functions is simplied by the use of radian measure. We will use radian measure throughout our work and refer to degree measure only in passing. The derivative of the sine function is the cosine function: d (sin x) = cos x. dx(3.6.1)PROOFFix any number x. For h = 0,sin(x + h) sin x [sin x cos h + cos x sin h] [sin x] = h h sin h cos h 1 + cos x . = sin x h h Now, as shown in Section 2.5 cos h 1 sin h lim =0 and lim = 1. h0 h0 h h Since x is xed, sin x and cos x remain constant as h approaches zero. It follows that sin h cos h 1 sin(x + h) sin x = lim sin x + cos x h0 h0 h h h cos h 1 sin h + cos x lim = sin x lim h0 h0 h h limThus limh0sin(x + h) sin x = (sin x)(0) + (cos x)(1) = cos x. hThe derivative of the cosine function is the negative of the sine function: d (cos x) = sin x. dx(3.6.2)PROOFFix any number x. For h = 0, cos(x + h) = cos x cos h sin x sin h.Therefore cos(x + h) cos x [cos x cos h sin x sin h] [cos x] = lim h0 h0 h h lim= cos xcos h 1 h0 h lim sin xsin h h0 h lim= sin x. Example 1 To differentiate f (x) = cos x sin x, we use the product rule: d d (sin x) + sin x (cos x) dx dx = cos x(cos x) + sin x( sin x) = cos2 x sin2 x. f (x) = cos x.13:54 167. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONSWe come now to the tangent function. Since tan x = sin x/ cos x, we have d d cos x (sin x) sin x (cos x) cos2 x + sin2 x 1 d dx dx (tan x) = = = = sec2 x. 2x 2x dx cos cos cos2 x The derivative of the tangent function is the secant squared: d (tan x) = sec2 x. dx(3.6.3)The derivatives of the other trigonometric functions are as follows: d (cot x) = csc2 x, dx d (sec x) = sec x tan x, dx d (csc x) = csc x cot x. dx(3.6.4)The verication of these formulas is left as an exercise. It is time for some sample problems.Example 2 Find f (/4) for f (x) = x cot x. SOLUTION We rst nd f (x). By the product rule,d d (cot x) + cot x (x) = x csc2 x + cot x. dx dx Now we evaluate f at /4: f (/4) = ( 2)2 + 1 = 1 . 4 2 f (x) = xExample 3 Findd dx1 sec x . tan xSOLUTION By the quotient rule,1 sec x tan xd dx= = (sec x tan x = 1) 22d d (1 sec x) (1 sec x) (tan x) dx dx tan2 x tan x( sec x tan x) (1 sec x)(sec2 x) tan2 x 2 sec x(sec x tan2 x) sec2 x tan2 x 2 sec x sec x sec x(1 sec x) = . 2x tan tan2 x tan xExample 4 Find an equation for the line tangent to the curve y = cos x at the point where x = /3.14313:54 168. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006144 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION SOLUTION Since cos /3 = 1/2, the point of tangency is (/3, 1/2). To nd theslope of the tangent line, we evaluate the derivative dy = sin x dx at x = /3. This gives m = 3/2. The equation for the tangent line can be written 3 1 x . y = 2 2 3Example 5 Set f (x) = x + 2 sin x. Find the numbers x in the open interval (0, 2) at which (a) f (x) = 0, (b) f (x) > 0, (c) f (x) < 0. SOLUTION The derivative of f is the functionf (x) = 1 + 2 cos x. The only numbers in (0, 2 ) at which f (x) = 0 are the numbers at which cos x = 1 : 2 x = 2/3 and x = 4/3. These numbers separate the interval (0, 2) into three open subintervals (0, 2/3), (2/3, 4/3), (4/3, 2 ). On each of these subintervals f keeps a constant sign. The sign of f is recorded below: sign of f + + + + + + + + 0 0 + + + + + + + + 02/34/32Answers: (a) f (x) = 0 at x = 2/3 and x = 4/3. (b) f (x) > 0 on (0, 2/3) (4/3, 2 ). (c) f (x) < 0 on (2/3, 4/3). The Chain Rule Applied to the Trigonometric Functions If f is a differentiable function of u and u is a differentiable function of x, then, as you saw in Section 3.5, du d du d [ f (x)] = [ f (u)] = f (u) . dx du dx dx Written in this form, the derivatives of the six trigonometric functions appear as follows:(3.6.5)d du (sin u) = cos u , dx dx du d (tan u) = sec2 u , dx dx du d (sec u) = sec u tan u , dx dxExample 6d du (cos u) = sin u , dx dx du d (cot u) = csc2 u , dx dx d du (csc u) = csc u cot u . dx dxd d (cos 2x) = sin 2x (2x) = 2 sin 2x. dx dxExample 7 d d [sec(x 2 + 1)] = sec(x 2 + 1) tan(x 2 + 1) (x 2 + 1) dx dx = 2x sec(x 2 + 1) tan(x 2 + 1). 13:54 169. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.6 DIFFERENTIATING THE TRIGONOMETRIC FUNCTIONS145Example 8 d d (sin3 x) = (sin x)3 dx dx d = 3(sin x)2 (sin x) dx d = 3(sin x)2 cos x ( x) dx = 3(sin x)2 cos x( ) = 3 sin2 x cos x. Our treatment of the trigonometric functions has been based entirely on radian measure. When degrees are used, the derivatives of the trigonometric functions contain 1 the extra factor 180 0.0175. =Example 9 Findd (sin x ). dxSOLUTION Since x =1 x 180radians,d d 1 (sin x ) = (sin 180 x) = dx dx1 1801 cos 180 x =1 180cos x .1 The extra factor 180 is a disadvantage, particularly in problems where it occurs repeatedly. This tends to discourage the use of degree measure in theoretical work.EXERCISES 3.6 Exercises 112. Differentiate the function. 1. y = 3 cos x 4 sec x. 2. y = x 2 sec x. 3 4. y = sin2 x. 3. y = x csc x. 6. y = 3t 2 tan t. 5. y = cos2 t. 4 8. y = u csc u 2 . 7. y = sin u. 10. y = cos x. 9. y = tan x 2 . 12. y = [x 2 sec 2x]3 . 11. y = [x + cot x]4 . Exercises 1324. Find the second derivative. 13. y = sin x. 14. y = cos x. cos x . 16. y = tan3 2 x. 15. y = 1 + sin x 17. y = cos3 2u. 18. y = sin5 3t. 19. y = tan 2t. 20. y = cot 4u. sin x . 22. y = 21. y = x 2 sin 3x. 1 cos x 23. y = sin2 x + cos2 x. 24. y = sec2 x tan2 x. Exercises 2530. Find the indicated derivative. d4 d4 25. (sin x). 26. (cos x). dx4 dx4 2 d 2d d d 27. (t cos 3t) . 28. t t (cos t 2 ) . 2 dt dt dt dt d d [ f (sin 3x)]. 30. [sin( f (3x))]. 29. dx dxExercises 3136. Find an equation for the line curve at the point with x coordinate a. 31. y = sin x; a = 0. 32. y = tan x; 33. y = cot x; a = /6. 34. y = cos x; 35. y = sec x; a = /4. 36. y = csc x;tangent to the a = /6. a = 0. a = /3.Exercises 3746. Determine the numbers x between 0 and 2 where the line tangent to the curve is horizontal. 37. y = cos x. 38. y = sin x. 40. y = cos x 3 sin x. 39. y = sin x + 3 cos x. 42. y = cos2 x. 41. y = sin2 x. 43. y = tan x 2x. 44. y = 3 cot x + 4x. 45. y = 2 sec x + tan x. 46. y = cot x 2 csc x. Exercises 4750. Find all x in (0, 2) at which (a) f (x) = 0; (b) f (x) > 0; (c) f (x) < 0. 47. f (x) = x + 2 cos x. 48. f (x) = x 2 sin x. 49. f (x) = sin x + cos x. 50. f (x) = sin x cos x. Exercises 5154. Find dy/dt (a) by the chain rule and (b) by writing y as a function of t and then differentiating. 51. y = u 2 1, u = sec x, x = t. 52. y = [ 1 (1 + u)]3 , u = cos x, x = 2t. 2 53. y = [ 1 (1 u)]4 , u = cos x, x = 2t. 2 54. y = 1 u 2 , u = csc x, x = 3t.13:54 170. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006146 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION 55. It can be shown by induction that the nth derivative of the sine function is given by the formula dn (sin x) = dxn(1)(n1)/2 cos x, n odd (1)n/2 sin x, n even.Persuade yourself that this formula is correct and obtain a similar formula for the nth derivative of the cosine function. 56. Verify the following differentiation formulas: d (cot x) = csc2 x. (a) dx d (b) (sec x) = sec x tan x. dx d (csc x) = csc x cot x. (c) dx 57. Use the identities cos x = sin x and sin x = cos x 2 269. Set g(x) =(a) For what values of a and b is g differentiable at 2/3? (b) Using the values of a and b you found in part (a), sketch the graph of g. 1 + a cos x, x /3 70. Set f (x) = b + sin(x/2), x > /3. (a) For what values of a and b is f differentiable at /3? (b) Using the values of a and b you found in part (a), sketch the graph of f . 71. Let y = A sin t + B cos t where A, B, are constants. Show that y satises the equation d2 y + 2 y = 0. dt 2 72. A simple pendulum consists of a mass m swinging at the end of a rod or wire of negligible mass. The gure shows a simple pendulum of length L. The angular displacement at time t is given by a trigonometric expression:to give an alternative proof of (3.6.2). 58. The double-angle formula for the sine function takes the form: sin 2x = 2 sin x cos x. Differentiate this formula to obtain a double-angle formula for the cosine function. 59. Set f (x) = sin x. Show that nding f (0) from the denition of derivative amounts to nding sin x . lim x0 xsin x, 0 x 2/3 ax + b, 2/3 < x 2. (t) = A sin(t + ) where A, , are constants.L (see Section 2.5)60. Set f (x) = cos x. Show that nding f (0) from the denition of derivative amounts to nding cos x 1 . x0 x Exercises 6166. Find a function f with the given derivative. Check your answer by differentiation. 61. f (x) = 2 cos x 3 sin x. 62. f (x) = sec2 x csc2 x. 63. f (x) = 2 cos 2x + sec x tan x. 64. f (x) = sin 3x csc 2x cot 2x. 65. f (x) = 2x cos(x 2 ) 2 sin 2x. 66. f (x) = x 2 sec2 (x 3 ) + 2 sec 2x tan 2x. x sin(1/x), x = 0 67. Set f (x) = and g(x) = x f (x). 0, x = 0. In Exercise 62, Section 3.1, you were asked to show that f is continuous at 0 but not differentiable there, and that g is differentiable at 0. Both f and g are differentiable at each x = 0. (a) Find f (x) and g (x) for x = 0. (b) Show that g is not continuous at 0. cos x, x 0 68. Set f (x) = ax + b, x < 0.(a) Show that the function satises the equationlim(a) For what values of a and b is f differentiable at 0? (b) Using the values of a and b you found in part (a), sketch the graph of f .d 2 + 2 = 0. dt 2 (Except for notation, this is the equation of Exercise 71.) (b) Show that can be written in the form (t) = A sin t + B cos t 73.74.75.76.where A, B, are constants. An isosceles triangle has two sides of length c. The angle between them is x radians. Express the area A of the triangle as a function of x and nd the rate of change of A with respect to x. A triangle has sides of length a and b, and the angle between them is x radians. Given that a and b are kept constant, nd the rate of change of the third side c with respect to x. HINT: Use the law of cosines. Let f (x) = cos kx, k a positive integer. Use a CAS to nd dn (a) [ f (x)], dxn (b) all positive integers m for which y = f (x) is a solution of the equation y + my = 0. Use a CAS to show that y = A cos 2x + B sin 2x is a solution of the equation y + 2y = 0. Find A and B given that y(0) = 2 and y (0) = 3. Verify your results analytically.13:54 171. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS77. Let f (x) = sin x cos 2x for 0 x 2. (a) Use a graphing utility to estimate the points on the graph where the tangent is horizontal. (b) Use a CAS to estimate the numbers x at which f (x) = 0. (c) Reconcile your results in (a) and (b). 78. Exercise 77 with f (x) = sin x sin2 x for 0 x 2.147Exercises 7980. Find an equation for the line l tangent to the graph of f at the point with x-coordinate c. Use a graphing utility to display l and the graph of f in one gure. Note that l is a good approximation to the graph of f for x close to c. Determine the interval on which the vertical separation between l and the graph of f is of absolute value less than 0.01. 79. f (x) = sin x; c = 0. 80. f (x) = tan x; c = /4. 3.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS Up to this point we have been differentiating functions dened explicitly in terms of an independent variable. We can also differentiate functions not explicitly given in terms of an independent variable. Suppose we know that y is a differentiable function of x and satises a particular equation in x and y. If we nd it difcult to obtain the derivative of y, either because the calculations are burdensome or because we are unable to express y explicitly in terms of x, we may still be able to obtain dy/d x by a process called implicit differentiation. This process is based on differentiating both sides of the equation satised by x and y. Example 1 We know that the function y = 1 x 2 (Figure 3.7.1) satises the equation x 2 + y 2 = 1. (Figure 3.7.2) We can obtain dy/d x by carrying out the differentiation in the usual manner, or we can do it more simply by working with the equation x 2 + y 2 = 1. Differentiating both sides of the equation with respect to x (remembering that y is a differentiable function of x), we have d 2 d 2 d (x ) + (y ) = (1) dx dx dx dy =0 2x + 2y dx (by the chain rule) dy x = . dx y We have obtained dy/d x in terms of x and y. Usually this is as far as we can Here go. we can go further since we have y explicitly in terms of x. The relation y = 1 x 2 gives x dy . = dx 1 x2 Verify this result by differentiating y = 1 x 2 . in the usual manner. Example 2 Assume that y is a differentiable function of x which satises the given equation. Use implicit differentiation to express dy/d x in terms of x and y. (a) 2x 2 y y 3 + 1 = x + 2y.(b) cos(x y) = (2x + 1)3 y.SOLUTION(a) Differentiating both sides of the equation with respect to x, we have dy dy dy =1+2 + 4x y 3y 2 2x 2 dx dx dx (by the chain rule) (by the product rule) (2x 2 3y 2 2)dy = 1 4x y. dxy1 x1 y = 1 x2Figure 3.7.1 y(1, 0)(1, 0) xx2 + y2 = 1Figure 3.7.213:54 172. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006148 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Therefore dy 1 4x y = 2 . dx 2x 3y 2 2 (b) We differentiate both sides of the equation with respect to x: dy dx sin(x y) 1 = (2x + 1)3dy + 3(2x + 1)2 (2)y dx(by the chain rule) [sin(x y) (2x + 1)3 ]dy = 6(2x + 1)2 y + sin(x y). dxThus 6(2x + 1)2 y + sin(x y) dy = . dx sin(x y) (2x + 1)3y2321Example 3 Figure 3.7.3 shows the curve 2x 3 + 2y 3 = 9x y and the tangent line at the point (1, 2). What is the slope of the tangent line at that point?(1, 2)123xSOLUTION We want dy/d x where x = 1 and y = 2. We proceed by implicit differ-entiation: 2 2x3 + 2y3 = 9xyFigure 3.7.3dy dy = 9x + 9y dx dx dy dy 2x 2 + 2y 2 = 3x + 3y. dx dx Setting x = 1 and y = 2, we have 6x 2 + 6y 2dy dy dy dy 4 =3 + 6, 5 = 4, = . dx dx dx dx 5 The slope of the tangent line at the point (1, 2) is 4/5. 2+8We can also nd higher derivatives by implicit differentiation.Example 4 The function y = (4 + x 2 )1/3 satises the equation y 3 x 2 = 4. Use implicit differentiation to express d 2 y/d x 2 in terms of x and y. SOLUTION Differentiation with respect to x gives()3y 2dy 2x = 0. dxDifferentiating again, we have 3y 2d dxdy dx+dy dx d2 y 3y 2 2 + 6y dx(by the product rule) d (3y 2 ) 2 = 0 dx dy dxSince () gives dy 2x = 2, dx 3y2 2 = 0.13:54 173. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERSwe have 3y 2d2 y + 6y dx22x 3y 22 2 = 0.As you can check, this gives d2 y 6y 3 8x 2 = . dx2 9y 5 RemarkIf we differentiate x 2 + y 2 = 1 implicitly, we nd that 2x + 2ydy =0 dxand thereforedy x = . dx yHowever, the result is meaningless. It is meaningless because there is no real-valued function y of x that satises the equation x 2 + y 2 = 1. Implicit differentiation can be applied meaningfully to an equation in x and y only if there is a differentiable function y of x that satises the equation. Rational Powers You have seen that the formula d n (x ) = nx n1 dx holds for all real x if n is a positive integer and for all x = 0 if n is a negative integer. For x = 0, we can stretch the formula to n = 0 (and it is a bit of a stretch) by writing d 0 d (x ) = (1) = 0 = 0x 1 . dx dx The formula can then be extended to all rational exponents p/q: d p/q p (x ) = x ( p/q)1 . dx q(3.7.1)The formula applies to all x = 0 where x p/q is dened. We operate under the assumption that the function y = x 1/q is differentiable at all x where x 1/q is dened. (This assumption is readily veried from considerations explained in Section 7.1.) From y = x 1/q we get DERIVATION OF (3.7.1)y q = x. Implicit differentiation with respect to x gives qy q1dy =1 dxand therefore dy 1 1 1 = y 1q = x (1q)/q = x (1/q)1 . dx q q q So far we have shown that d 1/q 1 (x ) = x (1/q)1 . dx q The function y = x p/q is a composite function: y = x p/q = (x 1/q ) p .14913:54 174. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006150 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION Applying the chain rule, we have d 1 p dy = p(x 1/q ) p1 (x 1/q ) = px ( p1)/q x (1/q)1 = x ( p/q)1 dx dx q q as asserted. Here are some simple examples: d 5/2 d 7/9 d 2/3 (x ) = 2 x 1/3 , (x ) = 5 x 3/2 , (x ) = 7 x 16/9 . 3 2 9 dx dx dx If u is a differentiable function of x, then, by the chain rule p d p/q du (u ) = u ( p/q)1 . dx q dx(3.7.2)The verication of this is left to you. The result holds on every open x-interval where u ( p/q)1 is dened.Example 5 d [(1 + x 2 )1/5 ] = 1 (1 + x 2 )4/5 (2x) = 2 x(1 + x 2 )4/5 . 5 5 dx d (b) [(1 x 2 )2/3 ] = 2 (1 x 2 )1/3 (2x) = 4 x(1 x 2 )1/3 . 3 3 dx d (c) [(1 x 2 )1/4 ] = 1 (1 x 2 )3/4 (2x) = 1 x(1 x 2 )3/4 . 4 2 dx The rst statement holds for all real x, the second for all x = 1, and the third only for x (1, 1). (a)Example 6 1/21 2x 1 + x21 2x 1 + x21 = 2x 1 + x2= =d dx1 + x2 x= The result holds for all x > 0.1/21/21/2d dxx 1 + x2(1 + x 2 )(1) x(2x) (1 + x 2 )21 x2 (1 + x 2 )21 x2 . 2x 1/2 (1 + x 2 )3/2EXERCISES 3.7 Preliminary note. In many of the exercises below you are asked to use implicit differentiation. We assure you that in each case there is a function y = y(x) that satises the indicated equation and has the requisite derivative(s). Exercises 110. Use implicit differentiation to express dy/d x in terms of x and y.1. 3. 5. 7. 9.x 2 + y 2 = 4. 4x 2 + 9y 2 = 36. x 4 + 4x 3 y + y 4 = 1. (x y)2 y = 0. sin (x + y) = x y.2. 4. 6. 8. 10.x 3 + y 3 3x y = 0. x + y = 4. x 2 x 2 y + x y 2 + y 2 = 1. (y + 3x)2 4x = 0. tan x y = x y.13:54 175. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERSExercises 1116. Express d 2 y/d x 2 in terms of x and y. 12. x 2 2x y + 4y 2 = 3. 11. y 2 + 2x y = 16. 2 2 14. x 2 3x y = 18. 13. y + x y x = 9. 16. sin2 x + cos2 y = 1. 15. 4 tan y = x 3 . Exercises 1720. Evaluate dy/d x and d 2 y/d x 2 at the point indicated. 17. x 2 4y 2 = 9; (5, 2). 18. x 2 + 4x y + y 3 + 5 = 0; (2, 1). 19. cos(x + 2y) = 0; (/6, /6). 20. x = sin2 y; ( 1 , /4). 2 Exercises 2126. Find equations for the tangent and normal lines at the point indicated. 21. 2x + 3y = 5; (2, 3). 22. 9x 2 + 4y 2 = 72; (2, 3). 23. x 2 + x y + 2y 2 = 28; (2, 3). 24. x 3 ax y + 3ay 2 = 3a 3 ; (a, a). 1 , . 25. x = cos y; 2 3 26. tan x y = x; 1, . 4 Exercises 2732. Find dy/d x. 28. y = (x + 1)1/3 . 27. y = (x 3 + 1)1/2 . 4 30. y = (x + 1)1/3 (x + 2)2/3 . 29. y = 2x 2 + 1. 31. y = 2 x 2 3 x 2 . 32. y = (x 4 x + 1)3 . Exercises 3336. Carry out the differentiation. 33.d dx35.d dx1 x+ . x x x2+134.36..d dx d dx3x + 1 . 2x + 5 x2 + 1 . x37. (Important) Show the general form of the graph. (a) f (x) = x 1/n , n a positive even integer. (b) f (x) = x 1/n , n a positive odd integer. (c) f (x) = x 2/n , n an odd integer greater than 1. Exercises 3842. Find the second derivative. 39. y = 3 a + bx. 38. y = a 2 + x 2 . 41. y = x tan x. 40. y = x a 2 x 2 . 42. y = x sin x. 43. Show that all normals to the circle x 2 + y 2 = r 2 pass through the center of the circle. 44. Determine the x-intercept of the tangent to the parabola y 2 = x at the point where x = a. The angle between two curves is the angle between their tangent lines at the point of intersection. If the slopes are m 1 and m 2 , then the angle of intersection can be obtained from the formula tan =m2 m1 . 1 + m1m245. At what angles do the parabolas y 2 = 2 px + p 2 and y 2 = p 2 2 px intersect?15146. At what angles does the line y = 2x intersect the curve x 2 x y + 2y 2 = 28? 47. The curves y = x 2 and x = y 3 intersect at the points (1, 1) and (0, 0). Find the angle between the curves at each of these points. 48. Find the angles at which the circles (x 1)2 + y 2 = 10 and x 2 + (y 2)2 = 5 intersect. Two curves are said to be orthogonal iff, at each point of intersection, the angle between them is a right angle. Show that the curves given in Exercises 49 and 50 are orthogonal. 49. The hyperbola x 2 y 2 = 5 and the ellipse 4x 2 + 9y 2 = 72. 50. The ellipse 3x 2 + 2y 2 = 5 and y 3 = x 2 . HINT: The curves intersect at (1, 1) and (1, 1) Two families of curves are said to be orthogonal trajectories (of each other) if each member of one family is orthogonal to each member of the other family. Show that the families of curves given in Exercises 51 and 52 are orthogonal trajectories. 51. The family of circles x 2 + y 2 = r 2 and the family of lines y = mx. 52. The family of parabolas x = ay 2 and the family of ellipses x 2 + 1 y 2 = b. 2 53. Find equations for the lines tangent to the ellipse 4x 2 + y 2 = 72 that are perpendicular to the line x + 2y + 3 = 0. 54. Find equations for the lines normal to the hyperbola 4x 2 y 2 = 36 that are parallel to the line 2x + 5y 4 = 0. 55. The curve (x 2 + y 2 )2 = x 2 y 2 is called a lemniscate. The curve is shown in the gure. Find the four points of the curve at which the tangent line is horizontal. y(1, 0)(1, 0)x56. The curve x 2/3 + y 2/3 = a 2/3 is called an astroid. The curve is shown in the gure.y (0, a)(a, 0)(a, 0)(0, a)x13:54 176. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006152 CHAPTER 3 THE DERIVATIVE; THE PROCESS OF DIFFERENTIATION (a) Find the slope of the graph at an arbitrary point (x1 , y1 ), which is not a vertex. (b) At what points of the curve is the slope of the tangent line 0, 1, 1? 57. Show that the sum of the x- and y-intercepts of any line tangent to the graph of x 1/2 + y 1/2 = c1/2 is constant and equal to c. 58. A circle of radius 1 with center on the y-axis is inscribed in the parabola y = 2x 2 . See the gure. Find the points of contact. y1 (0, a)y = 2x 2 x 59. Set f (x) = 3 3 x. Use a CAS to f (h) f (0) . (a) Find d(h) = h (b) Find lim d(h) and lim+ d(h). h0h0(c) Is there a tangent line at (0, 0)? Explain. (d) Use a graphing utility to draw the graph of f on [2, 2]. 3 60. Exercise 59 with f (x) = 3 x 2 . Exercises 61 and 62. Use a graphing utility to determine where (a) f (x) = 0; 3(b) f (x) > 0;(c) f (x) < 0,x2 + 1 . x 63. A graphing utility in parametric mode can be used to graph some equations in x and y. Draw the graph of the equation x 2 + y 2 = 4 rst by setting x = t, y = 4 t 2 and then by setting x = t, y = 4 t 2 . Exercises 6467. Use a CAS to nd the slope of the line tangent to the curve at the given point. Use a graphing utility to draw the curve and the tangent line together in one gure. 64. 3x 2 + 4y 2 = 16; P(2, 1). 65. 4x 2 y 2 = 20; P(3, 4). 66. 2 sin y cos x = 0; P(0, /6). 3 67. x 2 + 3 y 2 = 4; P(1, 3 3). 68. (a) Use a graphing utility to draw the graph of the equation x 3 + y 3 = 6x y. (b) Use a CAS to nd equations for the lines tangent to the curve at the points where x = 3. (c) Draw the graph of the equation and the tangent lines in one gure. 69. (a) Use a graphing utility to draw the gure-eight curve 61. f (x) = x x 2 + 1.62. f (x) =x 4 = x 2 y2. (b) Find the x-coordinates of the points of the graph where the tangent line is horizontal. 70. Use a graphing utility to draw the curve (2 x)y 2 = x 3 . Such a curve is called a cissoid. CHAPTER 3. REVIEW EXERCISES Exercises 14. Differentiate by taking the limit of the appropriate difference quotient. 2. f (x) = 1 + 2x. 1. f (x) = x 3 4x + 3. 1 3. g(x) = . 4. F(x) = x sin x. x 2 Exercises 522. Find the derivative. 6. y = 2x 3/4 4x 1/4 . 5. y = x 2/3 72/3 . 1 + 2x + x 2 7. y = . 8. f (t) = (2 3t 2 )3 . x3 1 1 b 2 9. f (x) = . 10. y = a . x a2 x 2 b 3 11. y = a + 2 . 12. y = x 2 + 3x. x 14. g(x) = x 2 cos(2x 1). 13. y = tan 2x + 1. a2 + x 2 15. F(x) = (x + 2)2 x 2 + 2. 16. y = 2 . a x217. h(t) = t sec t 2 + 2t 3 . 2 3t 19. s = 3 . 2 + 3t 21. f ( ) = cot(3 + ).sin 2x . 1 + cos x 20. r = 2 3 4 . 18. y =22. y =x sin 2x . 1 + x2Exercises 2326. Find f (c). 23. f (x) = 3 x + x; c = 64. 24. f (x) = x 8 x 2 ; c = 2. 1 25. f (x) = x 2 sin2 x; c = . 6 1 26. f (x) = cot 3x; c = . 9 Exercises 2730. Find equations for the lines tangent and normal to the graph of f at the point indicated. 27. f (x) = 2x 3 x 2 + 3; (1, 4). 2x 3 28. f (x) = ; (1, 5). 3x + 413:54 177. P1: PBU/OVYP2: PBU/OVYJWDD023-03JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20063.7 IMPLICIT DIFFERENTIATION; RATIONAL POWERS29. f (x) = (x + 1) sin 2x; (0, 0). 30. f (x) = x 1 + x 2 ; (1, 2). Exercises 3134. Find the second derivative. 31. f (x) = cos (2 x). 32. f (x) = (x 2 + 4)3/2 . 33. y = x sin x. 34. g(u) = tan2 u. Exercises 3536. Find a formula for the nth derivative. a 36. y = . 35. y = (a bx)n . bx + c Exercises 3740. Use implicit differentiation to express dy/d x in terms of x and y. 38. tan (x + 2y) = x 2 y. 37. x 3 y + x y 3 = 2. 3 40. x 2 + 3x y = 1 + x/y. 39. 2x + 3x cos y = 2x y. Exercises 4142. Find equations for the lines tangent and normal to the curve at the point indicated. 41. x 2 + 2x y 3y 2 = 9; (3, 2). 42. y sin 2x x sin y = 1 ; ( 1 , 1 ). 4 4 2 Exercises 4344. Find all x at which (a) f (x) = 0; (b) f (x) > 0; (c) f (x) < 0. 43. f (x) = x 3 9x 2 + 24x + 3. 2x . 44. f (x) = 1 + 2x 2 Exercises 4546. Find all x in (0, 2) at which (a) f (x) = 0; (b) f (x) > 0; (c) f (x) < 0. 45. f (x) = x + sin 2x 46. f (x) = 3x 2 cos x. 47. Find the points on the curve y = 2 x 3/2 where the inclination 3 of the tangent line is (a) /4, (b) 60 , (c) /6. 48. Find equations for all tangents to the curve y = x 3 that pass through the point (0, 2). 49. Find equations for all tangents to the curve y = x 3 x that pass through the point (2, 2). 50. Find A, B, C given that the curve y = Ax 2 + Bx + C passes through the point (1, 3) and is tangent to the line x y + 1 = 0 at the point (2, 3). 51. Find A, B, C, D given that the curve y = Ax 3 + Bx 2 + C x + D is tangent to the line y = 5x 4 at the point (1, 1) and is tangent to the line y = 9x at the point (1, 9).15352. Show that d/d x(x n ) = n/x n+1 for all positive integers n by showing that limh01 1 1 n h (x + h)n x=n . x n+1Exercises 5357. Evaluate the following limits. HINT: Apply either Denition 3.1.1 or (3.1.5). (1 + h)2 2(1 + h) + 1 . h0 h sin ( 1 + h) 1 9+h3 6 2 54. lim . 55. lim . h0 h0 h h x 5 32 sin x 56. lim . 57. lim . x x x2 x 2 53. lim58. The gure is intended to depict a function f which is continuous on [x0 , ) and differentiable on (x0 , ). fabcdexFor each x (x0 , ) dene M(x) = maximum value of f on [x0 , x]. m(x) = minimum value of f on [x0 , x]. a. Sketch the graph of M and specify the number(s) at which M fails to be differentiable. b. Sketch the graph of m and specify the number(s) at which m fails to be differentiable.13:54 178. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER4THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 4.1 THE MEAN-VALUE THEOREM We come now to the mean-value theorem. From this theorem ow most of the results that give power to the process of differentiation.THEOREM 4.1.1 THE MEAN-VALUE THEOREMIf f is differentiable on the open interval (a, b) and continuous on the closed interval [a, b], then there is at least one number c in (a, b) for which f (c) =f (b) f (a) . baNote that for this number f (b) f (a) = f (c)(b a). The quotient f (b) f (a) ba is the slope of the line l that passes through the points (a, f (a)) and (b, f (b)). To say that there is at least one number c for which f (c) =f (b) f (a) bais to say that the graph of f has at least one point (c, f (c)) at which the tangent line is parallel to the line l. See Figure 4.1.1. 154The theorem was rst stated and proved by the French mathematician Joseph-Louis Lagrange (17361813).13:26 179. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.1 THE MEAN-VALUE THEOREM enttangl (b, f (b))(a, f (a)) acbFigure 4.1.1We will prove the mean-value theorem in steps. First we will show that if a function f has a nonzero derivative at some point x0 , then, for x close to x0 , f (x) is greater than f (x0 ) on one side of x0 and less than f (x0 ) on the other side of x0 . THEOREM 4.1.2Suppose that f is differentiable at x0 . If f (x0 ) > 0, then f (x0 h) < f (x0 ) < f (x0 + h) for all positive h sufciently small. If f (x0 ) < 0, then f (x0 h) > f (x0 ) > f (x0 + h) for all positive h sufciently small. We take the case f (x0 ) > 0 and leave the other case to you. By the denition of the derivative,PROOFf (x0 + k) f (x0 ) = f (x0 ). k With f (x0 ) > 0 we can use f (x0 ) itself as and conclude that there exists > 0 such that f (x0 + k) f (x0 ) if 0 < k < , then f (x0 ) < f (x0 ). k limk0For such k we have f (x0 ) 0. k We have shown that () holds for all numbers k which satisfy the condition 0 < k < . If 0 < h < , then 0 < h < and 0 < h < . Consequently,()f (x0 + h) f (x0 ) >0 handf (x0 h) f (x0 ) > 0. h15513:26 180. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006156 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES The rst inequality shows that f (x0 + h) f (x0 ) > 0and thereforef (x0 ) < f (x0 + h).The second inequality shows that f (x0 h) f (x0 ) < 0 tangentacFigure 4.1.2bxand thereforef (x0 h) < f (x0 ).Next we prove a special case of the mean-value theorem, known as Rolles theorem [after the French mathematician Michel Rolle (16521719), who rst announced the result in 1691]. In Rolles theorem we make the additional assumption that f (a) and f (b) are both 0. (See Figure 4.1.2.) In this case the line through (a, f (a)) and (b, f (b)) is horizontal. (It is the x-axis.) The conclusion is that there is a point (c, f (c)) at which the tangent line is horizontal.THEOREM 4.1.3 ROLLE'S THEOREMSuppose that f is differentiable on the open interval (a, b) and continuous on the closed interval [a, b]. If f (a) and f (b) are both 0, then there is at least one number c in (a, b) for which f (c) = 0. If f is constantly 0 on [a, b], then f (c) = 0 for all c in (a, b). If f is not constantly 0 on [a, b], then f takes on either some positive values or some negative values. We assume the former and leave the other case to you. Since f is continuous on [a, b], f must take on a maximum value at some point c of [a, b] (Theorem 2.6). This maximum value, f (c), must be positive. Since f (a) and f (b) are both 0, c cannot be a and it cannot be b. This means that c must lie in the open interval (a, b) and therefore f (c) exists. Now f (c) cannot be greater than 0 and it cannot be less than 0 because in either case f would have to take on values greater than f (c). (This follows from Theorem 4.1.2.) We can conclude therefore that f (c) = 0. PROOFRemarkRolles theorem is sometimes formulated as follows:Suppose that g is differentiable on the open interval (a, b) and continuous on the closed interval [a, b]. If g(a) = g(b), then there is at least one number c in (a, b) for which g (c) = 0. That these two formulations are equivalent is readily seen by setting f (x) = g(x) g(a) (Exercise 44). Rolles theorem is not just a stepping stone toward the mean-value theorem. It is in itself a useful tool.Example 1 We use Rolles theorem to show that p(x) = 2x 3 + 5x 1 has exactly one real zero.13:26 181. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.1 THE MEAN-VALUE THEOREM157SOLUTION Since p is a cubic, we know that p has at least one real zero (Exercise 29,Section 2.6). Suppose that p has more than one real zero. In particular, suppose that p(a) = p(b) = 0 where a and b are real numbers and a = b. Without loss of generality, we can assume that a < b. Since every polynomial is everywhere differentiable, p is differentiable on (a, b) and continuous on [a, b]. Thus, by Rolles theorem, there is a number c in (a, b) for which p (c) = 0. But p (x) = 6x 2 + 5 5for all x,and p (c) cannot be 0. The assumption that p has more than one real zero has led to a contradiction. We can conclude therefore that p has only one real zero. We are now ready to give a proof of the mean-value theorem. PROOF OF THE MEAN-VALUE THEOREMWe create a function g that satises the conditions of Rolles theorem and is so related to f that the conclusion g (c) = 0 leads to the conclusion f (b) f (a) . f (c) = baThe functionf g(x)(a, f (a))g(x) = f (x) f (b) f (a) (x a) + f (a) bais exactly such a function. A geometric view of g(x) is given in Figure 4.1.3. The line that passes through (a, f (a)) and (b, f (b)) has equation y=f (b) f (a) (x a) + f (a). ba[This is not hard to verify. The slope is right, and, at x = a, y = f (a).] The difference g(x) = f (x) f (b) f (a) (x a) + f (a) bais simply the vertical separation between the graph of f and the line featured in the gure. If f is differentiable on (a, b) and continuous on [a, b], then so is g. As you can check, g(a) and g(b) are both 0. Therefore, by Rolles theorem, there is at least one number c in (a, b) for which g (c) = 0. Since g (x) = f (x) f (b) f (a) , bag (c) = f (c) f (b) f (a) . bawe haveSince g (c) = 0, f (c) =f (b) f (a) . baExample 2 The function f (x) = 1 x,1 x 1a(b, f (b))xFigure 4.1.3b13:26 182. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006158 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES satises the conditions of the mean-value theorem: it is differentiable on (1, 1) and continuous on [1, 1]. Thus, we know that there exists a number c between 1 and 1 at which f (1) f (1) = 1 2. f (c) = 2 1 (1) f (1) = 0, f (1) =2What is c in this case? To answer this, we differentiate f. By the chain rule, 1 . f (x) = 2 1xy The condition f (c) = 1 2 gives 211 1 = 2. 2 2 1c 11xFigure 4.1.4Solve this equation for c and youll nd that c = 1 . 2 The tangent line at ( 1 , f ( 1 )) = ( 1 , 1 2) is parallel to the secant line that passes 2 2 2 2 through the endpoints of the graph. (Figure 4.1.4) Example 3 Suppose that f is differentiable on (1, 4), continuous on [1, 4], and f (1) = 2. Given that 2 f (x) 3 for all x in (1,4), what is the least value that f can take on at 4? What is the greatest value that f can take on at 4? SOLUTION By the mean-value theorem, there is at least one number c between 1 and 4 at whichf (4) f (1) = f (c)(4 1) = 3 f (c). Solving this equation for f (4), we have f (4) = f (1) + 3 f (c). Since f (x) 2 for every x in (1, 4), we know that f (c) 2. It follows that f (4) 2 + 3(2) = 8. Similarly, since f (x) 3 for every x in (1, 4), we know that f (c) 3, and therefore f (4) 2 + 3(3) = 11. We have shown that f (4) is at least 8 and no more than 11. Functions which do not satisfy the hypotheses of the mean-value theorem (differentiability on (a, b), continuity on [a, b]) may fail to satisfy the conclusion of the theorem. This is demonstrated in the Exercises.EXERCISES 4.1 Exercises 14. Show that f satises the conditions of Rolles theorem on the indicated interval and nd all numbers c on the interval for which f (c) = 0. 1. 2. 3. 4.f (x) = x 3 x; [0, 1]. f (x) = x 4 2x 2 8; [2, 2]. f (x) = sin 2x; [0, 2]. f (x) = x 2/3 2x 1/3 ; [0, 8].Exercises 510. Verify that f satises the conditions of the mean-value theorem on the indicated interval and nd all numbers c that satisfy the conclusion of the theorem. 5. 6. 7. 8.f (x) = x 2 ; [1, 2]. f (x) = 3 x 4x; [1, 4]. f (x) = x 3 ; [1, 3]. f (x) = x 2/3 ; [1, 8].13:26 183. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.1 THE MEAN-VALUE THEOREM 9. f (x) = 1 x 2 ; [0, 1]. 10. f (x) = x 3 3x; [1, 1]. 11. Determine whether the function f (x) = 1 x 2 /(3 + x 2 ) satises the conditions of Rolles theorem on the interval [1, 1]. If so, nd the numbers c for which f (c) = 0. 12. The function f (x) = x 2/3 1 has zeros at x = 1 and at x = 1. (a) Show that f has no zeros in (1, 1). (b) Show that this does not contradict Rolles theorem. 13. Does there exist a differentiable function f with f (0) = 2, f (2) = 5, and f (x) 1 for all x in (0, 2)? If not, why not? 14. Does there exist a differentiable function f with f (x) = 1 only at x = 0, 2, 3, and f (x) = 0 only at x = 1, 3/4, 3/2? If not, why not? 15. Suppose that f is differentiable on (2, 6) and continuous on [2, 6]. Given that 1 f (x) 3 for all x in (2, 6), show that 4 f (6) f (2) 12. 16. Find a point on the graph of f (x) = x 2 + x + 3, x between 1 and 2, where the tangent line is parallel to the line through (1, 3) and (2, 9). 17. Sketch the graph of f (x) =2x + 2, x 3 x,x 1 x > 1and nd the derivative. Determine whether f satises the conditions of the mean-value theorem on the interval [3, 2] and, if so, nd the numbers c that satisfy the conclusion of the theorem. 18. Sketch the graph of f (x) =2 + x 3, 3x,x 1 x >1and nd the derivative. Determine whether f satises the conditions of the mean-value theorem on the interval [1, 2] and, if so, nd the numbers c that satisfy the conclusion of the theorem. 19. Set f (x) = Ax 2 + Bx + C. Show that, for any interval [a, b], the number c that satises the conclusion of the meanvalue theorem is (a + b)/2, the midpoint of the interval. 20. Set f (x) = x 1 , a = 1, b = 1. Verify that there is no number c for which f (c) =f (b) f (a) . baExplain how this does not violate the mean-value theorem. 21. Exercise 20 with f (x) = x . 22. Graph the function f (x) = 2x 1 3. Verify that f (1) = 0 = f (2) and yet f (x) is never 0. Explain how this does not violate Rolles theorem.15923. Show that the equation 6x 4 7x + 1 = 0 does not have more than two distinct real roots. 24. Show that the equation 6x 5 + 13x + 1 = 0 has exactly one real root. 25. Show that the equation x 3 + 9x 2 + 33x 8 = 0 has exactly one real root. 26. (a) Let f be differentiable on (a, b). Prove that if f (x) = 0 for each x (a, b), then f has at most one zero in (a, b). (b) Let f be twice differentiable on (a, b). Prove that if f (x) = 0 for each x (a, b), then f has at most two zeros in (a, b). 27. Let P(x) = an x n + + a1 x + a0 be a nonconstant polynomial. Show that between any two consecutive roots of the equation P (x) = 0 there is at most one root of the equation P(x) = 0. 28. Let f be twice differentiable. Show that, if the equation f (x) = 0 has n distinct real roots, then the equation f (x) = 0 has at least n 1 distinct real roots and the equation f (x) = 0 has at least n 2 distinct real roots. 29. A number c is called a xed point of f if f (c) = c. Prove that if f is differentiable on an interval I and f (x) < 1 for all x I , then f has at most one xed point in I. HINT: Form g(x) = f (x) x. 30. Show that the equation x 3 + ax + b = 0 has exactly one real root if a 0 and at most one real root between 1 3 a 3 and 1 3 a if a < 0. 3 31. Set f (x) = x 3 3x + b. (a) Show that f (x) = 0 for at most one number x in [1, 1]. (b) Determine the values of b which guarantee that f (x) = 0 for some number x in [1, 1]. 32. Set f (x) = x 3 3a 2 x + b, a > 0. Show that f (x) = 0 for at most one number x in [a, a]. 33. Show that the equation x n + ax + b = 0, n an even positive integer, has at most two distinct real roots. 34. Show that the equation x n + ax + b = 0, n an odd positive integer, has at most three distinct real roots. 35. Given that f (x) 1 for all real numbers x, show that f (x1 ) f (x2 ) x1 x2 for all real numbers x1 and x2 . 36. Let f be differentiable on an open interval I. Prove that, if f (x) = 0 for all x in I, then f is constant on I. 37. Let f be differentiable on (a, b) with f (a) = f (b) = 0 and f (c) = 0 for some c in (a, b). Show by example that f need not be continuous on [a, b]. 38. Prove that for all real x and y (a) cos x cos y x y . (b) sin x sin y x y . 39. Let f be differentiable on (a, b) and continuous on [a, b]. (a) Prove that if there is a constant M such that f (x) M for all x (a, b), then f (b) f (a) + M(b a).13:26 184. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006160 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES (b) Prove that if there is a constant m such that f (x) m for all x (a, b), thenThis result is known as the Cauchy mean-value theorem. It reduces to the mean-value theorem if g(x) = x. HINT: To prove the result, setf (b) f (a) + m(b a). (c) Parts (a) and (b) together imply that if there exists a constant K such that f (x) K on (a, b), thenCf (a) K (b a) f (b) f (a) + K (b a). Show that this is the case. 40. Suppose that f and g are differentiable functions and f (x)g (x) g(x) f (x) has no zeros on some interval I. Assume that there are numbers a, b in I with a < b for which f (a) = f (b) = 0, and that f has no zeros in (a, b). Prove that if g(a) = 0 and g(b) = 0, then g has exactly one zero in (a, b). HINT: Suppose that g has no zeros in (a, b) and consider h = f /g. Then consider k = g/ f . 41. Suppose that f and g are nonconstant, everywhere differentiable functions and that f = g and g = f . Show that between any two consecutive zeros of f there is exactly one zero of g and between any two consecutive zeros of g there is exactly one zero of f . 42. (Important) Use the mean-value theorem to show that if f is continuous at x and at x + h and is differentiable between these two numbers, thenCC Cf (x + h) f (x) = f (x + h)h for some number between 0 and 1. (In some texts this is how the mean-value theorem is stated.) 43. Let h > 0. Suppose f is continuous on [x0 h, x0 + h] and differentiable on (x0 h, x0 + h). Show that if lim f (x) = L ,xx0then f is differentiable at x0 and f (x0 ) = L. HINT: Exercise 42. 44. Suppose that g is differentiable on (a, b) and continuous on [a, b]. Without appealing to the mean-value theorem, show that if g(a) = g(b), then there is at least one number c in (a, b) for which g (c) = 0. HINT: Figure out a way to use Rolles theorem. 45. (Generalization of the mean-value theorem) Suppose that f and g both satisfy the hypotheses of the mean-value theorem. Prove that if g has no zeros in (a, b), then there is at least one number c in (a, b) for which f (c) f (b) f (a) = . g(b) g(a) g (c)CCF(x) = [ f (b) f (a)]g(x) [g(b) g(a)] f (x). Exercises 4647. Show that the given function satises the hypotheses of Rolles theorem on the indicated interval. Use a graphing utility to graph f and estimate the number(s) c where f (c) = 0. Round off your estimates to three decimal places. 46. f (x) = 2x 3 + 3x 2 3x 2; [2, 1]. 47. f (x) = 1 x 3 cos ( x/2); [0, 1]. 48. Set f (x) = x 4 x 3 + x 3 x. Find a number b, if possible, such that Rolles theorem is satised on [0, b]. If such a number b exists, nd a number c that conrms Rolles theorem on [0, b] and use a graphing utility to draw the graph of f together with the line y = f (c). 49. Exercise 49 with f (x) = x 4 + x 3 + x 2 x. Exercises 5052. Use a CAS. Find the x-intercepts of the graph. Between each pair of intercepts, nd, if possible, a number c that conrms Rolles theorem. x2 x 50. f (x) = 2 . x + 2x + 2 x 4 16 51. f (x) = 2 . x +4 52. f (x) = 125x 7 300x 6 760x 5 + 2336x 4 + 80x 3 4288x 2 + 3840x 1024. Suppose that the function f satises the hypotheses of the meanvalue theorem on an interval [a, b]. We can nd the numbers c that satisfy the conclusion of the mean-value theorem by nding the zeros of the function f (b) f (a) . g(x) = f (x) ba Exercises 5354. Use a graphing utility to graph the function g that corresponds to the given f on the indicated interval. Estimate the zeros of g to three decimal places. For each zero c in the interval, graph the line tangent to the graph of f at (c, f (c)), and graph the line through (a, f (a)) and (b, f (b)). Verify that these lines are parallel. 53. f (x) = x 4 7x 2 + 2; [1, 3]. 54. f (x) = x cos x + 4 sin x; [/2, /2]. Exercises 5556. The function f satises the hypotheses of the mean-value theorem on the given interval [a, b]. Use a CAS to nd the number(s) c that satisfy the conclusion of the theorem. Then graph the function, the line through the endpoints (a, f (a)) and (b, f (b)), and the tangent line(s) at (c, f (c)). 55. f (x) = x 3 x 2 + x 1; [1, 4]. 56. f (x) = x 4 2x 3 x 2 x + 1; [2, 3]. 4.2 INCREASING AND DECREASING FUNCTIONS We are going to talk about functions increasing or decreasing on an interval. To place our discussion on a solid footing, we will dene these terms.13:26 185. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20061614.2 INCREASING AND DECREASING FUNCTIONS y(i) increase on the interval I if for every two numbers x1 , x2 in I, x1 < x2f (x1 ) < f (x2 );implies thatinc rsing rea decA function f is said toeas ingDEFINITION 4.2.1(ii) decrease on the interval I if for every two numbers x1 , x2 in I, x1 < x2xf (x1 ) > f (x2 ).implies thatFigure 4.2.1 yPreliminary Examples (a) The squaring function (Figure 4.2.1)f (x) =1, x,x 0 for all x in I, then f increases on I. (ii) If f (x) < 0 for all x in I, then f decreases on I. (iii) If f (x) = 0 for all x in I, then f is constant on I.Dirichlet functionFigure 4.2.4 PROOF Choose any two numbers x 1 and x 2 in I with x 1 < x 2 . Since f is differentiable on I, it is continuous on I. Therefore we know that f is differentiable on (x1 , x2 ) and13:26 186. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006162 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES continuous on [x1 , x2 ]. By the mean-value theorem there is a number c in (x1 , x2 ) for which f (x2 ) f (x1 ) . f (c) = x2 x1 In (i), f (x) > 0 for all x. Therefore, f (c) > 0 and we have f (x2 ) f (x1 ) > 0, x2 x1which implies thatf (x1 ) < f (x2 ).In (ii), f (x) < 0 for all x. Therefore, f (c) < 0 and we have f (x2 ) f (x1 ) < 0, x2 x1which implies thatf (x1 ) > f (x2 ).In (iii), f (x) = 0 for all x. Therefore, f (c) = 0 and we have f (x2 ) f (x1 ) = 0, x2 x1which implies thatf (x1 ) = f (x2 ).Remark In Section 3.2 we showed that if f is constant on an open interval I, then f (x) = 0 for all x I . Part (iii) of Theorem 4.2.2 gives the converse: if f (x) = 0 for all x in an open interval I, then f is constant on I. Combining these two statements, we can assert thatif I is an open interval, then f is constant on Iifff (x) = 0 for all x I. Theorem 4.2.2 is useful but doesnt tell the complete story. Look, for example, at the function f (x) = x 2 . The derivative f (x) = 2x is negative for x in (, 0), zero at x = 0, and positive for x in (0, ). Theorem 4.2.2 assures us thatf decreases on (, 0) and increases on (0, ), but actually f decreases on (, 0] and increases on [0, ). To get these stronger results, we need a theorem that applies to closed intervals. To extend Theorem 4.2.2 so that it works for an arbitrary interval I, the only additional condition we need is continuity at the endpoint(s).THEOREM 4.2.3Suppose that f is differentiable on the interior of an interval I and continuous on all of I. (i) If f (x) > 0 for all x in the interior of I, then f increases on all of I. (ii) If f (x) < 0 for all x in the interior of I, then f decreases on all of I. (iii) If f (x) = 0 for all x in the interior of I, then f is constant on all of I.The proof of this theorem is a simple modication of the proof of Theorem 4.2.2. It is time for examples.13:26 187. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006Example 2 The function f (x) = 1/x is dened for all x = 0. The derivative f (x) = 1/x 2 is negative for all x = 0. Thus the function f decreases on (, 0) and on (0, ). (See Figure 4.2.6.) Note that we did not say that f decreases on (, 0) (0, ); it does not. If x1 < 0 < x2 , then f (x1 ) < f (x2 ). inc rey fing as11decr(1, 3),f(3, ).singxdec reag sinThe derivative g takes on the value 0 at 2, at 1, and at 3. These numbers determine four intervals on which g keeps a constant sign: (2, 1),yeag (x) = 4x 4 12x 3 12x 2 + 44x 24 = 4(x 4 3x 3 3x 2 + 11x 6) = 4(x + 2)(x 1)2 (x 3).xFigure 4.2.5Example 3 The function g(x) = 4 x 5 3x 4 4x 3 + 22x 2 24x + 6 is a polyno5 mial. It is therefore everywhere continuous and everywhere differentiable. Differentiation gives(, 2),dec reing as Example 1 The function f (x) = 1 x 2 has derivative f (x) = x/ 1 x 2 . Since f (x) > 0 for all x in (1, 0) and f is continuous on [1, 0], f increases on [1, 0]. Since f (x) < 0 for all x in (0, 1) and f is continuous on [0, 1], f decreases on [0, 1]. The graph of f is the semicircle shown in Figure 4.2.5. 1634.2 INCREASING AND DECREASING FUNCTIONSFigure 4.2.6The sign of g on these intervals and the consequences for g are as follows: y++++++++++++0 0 0+++++++sign of g': behavior of g:increases2decreases1decreases3 increasesx(2, 100.4)Since g is everywhere continuous, g increases on (, 2], decreases on [2, 3], and increases on [3, ). (See Figure 4.2.7.) gExample 4 Let f (x) = x 2 sin x, 0 x 2 . Find the intervals on which f increases and the intervals on which f decreases.(1, 2.2) xSOLUTION In this case f (x) = 1 2 cos x. Setting f (x) = 0, we have (3, 24.6)1 2 cos x = 0and thereforecos x = 1 . 2The only numbers in [0, 2 ] at which the cosine takes on the value 1/2 are x = /3 and x = 5/3. It follows that on the intervals (0, /3), (/3, 5/3), (5/3, 2 ), the derivative f keeps a constant sign. The sign of f and the behavior of f are recorded below. 0+++++++++++++++++++++++++++0 sign of f': 0 behavior of f :decreases 3 increases5 32xdecreasesSince f is continuous throughout, f decreases on [0, /3], increases on [/3, 5/3], and decreases on [5/3, 2 ]. (See Figure 4.2.8.) Figure 4.2.713:26 188. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006164 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 5 5 ( 3 , 3 + 3)yf 35 3( 3 , 3 3)2xFigure 4.2.8While the theorems we have proven have wide applicability, they do not tell the whole story.y 5 2Example 5 The function 12f (x) =x1 x 2x 3, + 2,x 0 on (, 0) and f is continuous on (, 0], f increases on (, 0] (Theorem 4.2.3). Since f (x) > 0 on (0, 1) and is continuous on [0, 1), f increases on [0, 1) (Theorem 4.2.3). Since f increases on (, 0] and on [0, 1), f increases on (, 1). (We dont need a theorem to tell us that.) Since f (x) > 0 on (1, ) and f is continuous on [1, ), f increases on [1, ). (Theorem 4.2.3) That f increases on (, ) is not derivable from the theorems weve stated but is obvious by inspection. yExample 6 The function5 2g(x) =11 x 2+ 2, x 3,x 0 and the intervals on which f (x) < 0. 41. yExercises 3336. Find the intervals on which f increases and the intervals on which f decreases. x < 3 x + 7, 33. f (x) = x + 1 , 3 x < 1 5 2x, 1 x. 2 (x 1) x < 1 5 x, 1 x < 3 34. f (x) = 7 2x, 3 x. 2 4x , x 0 for all x, f (0) = 1, and f (x) < 0 for all x. 44. f (1) = 1, f (x) < 0 for all x = 1, and f (1) = 0. 45. f (1) = 4, f (2) = 2, and f (x) > 0 for all x. 46. f (x) = 0 only at x = 1 and at x = 2, f (3) = 4, f (5) = 1. Exercises 4750. Either prove the assertion or show that the assertion is not valid by giving a counterexample. A pictorial counterexample sufces. 47. (a) If f increases on [a, b] and increases on [b, c], then f increases on [a, c]. (b) If f increases on [a, b] and increases on (b, c], then f increases on [a, c]. 48. (a) If f decreases on [a, b] and decreases on [b, c], then f decreases on [a, c]. (b) If f decreases on [a, b) and decreases on [b, c], then f decreases on [a, c]. 49. (a) If f increases on (a, b), then there is no number x in (a, b) at which f (x) < 0. (b) If f increases on (a, b), then there is no number x in (a, b) at which f (x) = 0. 50. If f (x) = 0 at x = 1, x = 2, x = 3, then f cannot possibly increase on [0, 4]. 51. Set f (x) = x sin x. (a) Show that f increases on (, ). (b) Use the result in part (a) to show that sin x < x on (0, ) and sin x > x on (, 0). 52. Prove Theorem 4.2.4.13:26 191. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.3 LOCAL EXTREME VALUES56.57. 58. 59. 60. 61.andg (x) = f (x).(a) Show that f 2 (x) + g 2 (x) = C for some constant C. (b) Suppose that f (0) = 0 and g(0) = 1. What is C? (c) Give an example of a pair of functions that satisfy parts (a) and (b). Assume that f and g are differentiable on the interval (c, c) and f (0) = g(0). (a) Show that if f (x) > g (x) for all x (0, c), then f (x) > g(x) for all x (0, c). (b) Show that if f (x) > g (x) for all x (c, 0), then f (x) < g(x) for all x (c, 0). Show that tan x > x for all x (0, /2). Show that 1 x 2 /2 < cos x for all x (0, ). Let n be an integer greater than 1. Show that (1 + x)n > 1 + nx for all x < 0. Show that x x 3 /6 < sin x for all x > 0. It follows from Exercises 51 and 60 that x 1 x 3 < sin x < x 6for all x > 0.1 1 x 2 < cos x < 1 1 x 2 + 2 2 CCCfor all x > 0.f (x) = sin x sin (x + 2) sin2 (x + 1). From the graph, what do you conclude about f and f ? Conrm your conclusions by calculating f .In many problems in economics, engineering, and physics it is important to determine how large or how small a certain quantity can be. If the problem admits a mathematical formulation, it is often reducible to the problem of nding the maximum or minimum value of some function. Suppose that f is a function dened at some number c. We call c an interior point of the domain of f provided f is dened not only at c but at all numbers within an open interval (c , c + ). This being the case, f is dened at all numbers x within of c. DEFINITION 4.3.1 LOCAL EXTREME VALUESSuppose that f is a function and c is an interior point of the domain. The function f is said to have a local maximum at c provided that for all x sufciently close to c.The function f is said to have a local minimum at c provided that f (c) f (x)1 4 x 24Use this result to estimate cos 6 . (The x above is in radians.) Exercises 6366. Use a graphing utility to graph f and its derivative f on the indicated interval. Estimate the zeros of f to three decimal places. Estimate the subintervals on which f increases and the subintervals on which f decreases. 63. f (x) = 3x 4 10x 3 4x 2 + 10x + 9; [2, 5]. 64. f (x) = 2x 3 x 2 13x 6; [3, 4]. 65. f (x) = x cos x 3 sin 2x; [0, 6]. 66. f (x) = x 4 + 3x 3 2x 2 + 4x + 4; [5, 3]. Exercises 6770. Use a CAS to nd the numbers x at which (a) f (x) = 0, (b) f (x) > 0, (c) f (x) < 0. 67. f (x) = cos3 x, 0 x 2. x x2 1 68. f (x) = . 69. f (x) = 2 . x +1 x2 + 4 70. f (x) = 8x 5 36x 4 + 6x 3 + 73x 2 + 48x + 9. 71. Use a graphing utility to draw the graph of 4.3 LOCAL EXTREME VALUESf (c) f (x)167Use this result to estimate sin 4 . (The x above is in radians). 1 62. (a) Show that cos x < 1 1 x 2 + 24 x 4 for all x > 0. 2 (b) It follows from part (a) and Exercise 58 that53. Set f (x) = sec2 x and g(x) = tan2 x on the interval ( , ). Show that f (x) = g (x) for all x in ( , ). 2 2 2 2 54. Having carried out Exercise 53, you know from Theorem 4.2.4 that there exists a constant C such that f (x) g(x) = C for all x in (/2, /2). What is C? 55. Suppose that for all real x f (x) = g(x)for all x sufciently close to c.The local maxima and minima of f comprise the local extreme values of f.We illustrate these notions in Figure 4.3.1. A careful look at the gure suggests that local maxima and minima occur only at points where the tangent is horizontal [ f (c) = 0] or where there is no tangent line [ f (c) does not exist]. This is indeed the case.13:26 192. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006168 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES y flocal maximumlocal maximumlocal minimum x1x2x3local minimum x4xFigure 4.3.1THEOREM 4.3.2Suppose that c is an interior point of the domain of f. If f has a local maximum or local minimum at c, then f (c) = 0orf (c) does not exist.PROOF Lets suppose that f has a local extreme value at c, and lets suppose that f (c) exists. If f (c) > 0 or f (c) < 0, then, by Theorem 4.1.2, there must be points x1 and x2 arbitrarily close to c for whichf (x1 ) < f (c) < f (x2 ). This makes it impossible for a local maximum or a local minimum to occur at c. Therefore, if f (c) exists, it must have the value 0. The only other possibility is that f (c) does not exist. On the basis of this result, we make the following denition (an important one):DEFINITION 4.3.3 CRITICAL POINTThe interior points c of the domain of f for which f (c) = 0yorf (c) does not existare called the critical points for f.local maximum = 3x f (x) = 3 x 2Figure 4.3.2As a consequence of Theorem 4.3.2, in searching for local maxima and local minima, the only points we need to consider are the critical points. We illustrate the technique for nding local maxima and minima by some examples. In each case the rst step is to nd the critical points.Example 1 For f (x) = 3 x 2 ,(Figure 4.3.2) Also called the critical numbers for f. We prefer the term critical point because it is more in consonance with the term used in the study of functions of several variables.13:26 193. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.3 LOCAL EXTREME VALUES169ythe derivative f (x) = 2x exists everywhere. Since f (x) = 0 only at x = 0, the number 0 is the only critical point. The number f (0) = 3 is a local maximum. local minimum = 2Example 2 In the case of1f (x) = x + 1 + 2 = differentiation givesx + 1, x + 3,x < 1 x 1,xf (x) = x + 1 + 2(Figure 4.3.3)Figure 4.3.3 1, f (x) = does not exist, 1,x < 1 x = 1 x > 1.yThis derivative is never 0. It fails to exist only at 1. The number 1 is the only critical point. The value f (1) = 2 is a local minimum. 11 . Example 3 Figure 4.3.4 shows the graph of the function f (x) = x 1 The domain is (, 1) (1, ). The derivative 1 f (x) = (x 1)2f (x) =x1 x1Figure 4.3.4exists throughout the domain of f and is never 0. Thus there are no critical points. In particular, 1 is not a critical point for f because 1 is not in the domain of f. Since f has no critical points, there are no local extreme values. yThe fact that c is a critical point for f does not gurantee that f (c) is a local extreme value. This is made clear by the next two examples. CAUTION1xExample 4 In the case of the function f (x) = x 3 ,(Figure 4.3.5)the derivative f (x) = 3x is 0 at 0, but f (0) = 0 is not a local extreme value. The function is everywhere increasing. 2f (x) = x3Figure 4.3.5 yExample 5 The function f (x) =2x + 5, 1 x + 2, 2x 0 for all x in (c , c) and f (x) < 0 for all x in (c, c + ), then (Figures 4.3.7 and 4.3.8) f (c) is a local maximum. (ii) f (x) < 0 for all x in (c , c) and f (x) > 0 for all x in (c, c + ), then (Figures 4.3.9 and 4.3.10) f (c) is a local minimum. (iii) f (x) keeps constant sign on (c , c) (c, c + ), then f (c) is not a local extreme value.0f'local maximumc>0f'local maximum cxf ' (c) = 0xf ' (c) does not existFigure 4.3.7Figure 4.3.8f' >000f'f' 0f' 1(Figure 4.3.13)has no derivative at x = 1. Therefore 1 is a critical point. While it is true that f (x) > 0 for x < 1 and f (x) < 0 for x > 1, it does not follow that f (1) is a local maximum. The function is discontinuous at x = 1 and the rst-derivative test does not apply. There are cases where it is difcult to determine the sign of f on both sides of a critical point. If f is twice differentiable, then the following test may be easier to apply.5 4 3 2 1 12f (x) =THEOREM 4.3.5 THE SECOND-DERIVATIVE TEST(i) If f (c) > 0, then f (c) is a local minimum. (ii) If f (c) < 0, then f (c) is a local maximum. (Note that no conclusion is drawn if f (c) = 0.) We handle the case f (c) > 0. The other is left as an exercise. (Exercise 32) Since f is the derivative of f , we see from Theorem 4.1.2 that there exists a > 0 such that, ifPROOFc < x1 < c < x2 < c + , then f (x1 ) < f (c) < f (x2 ). Since f (c) = 0, we have for x in (c , c)andf (x) > 0By the rst-derivative test, f (c) is a local minimum.for x in (c, c + ).Example 8 For f (x) = 2x 3 3x 2 12x + 5 we have f (x) = 6x 2 6x 12 = 6(x 2 x 2) = 6(x 2)(x + 1) and f (x) = 12x 6.4 51 + 2x, x 1 5 x, x > 1Figure 4.3.13Suppose that f (c) = 0 and f (c) exists.f (x) < 03x13:26 196. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006172 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES The critical points are 2 and 1; the rst derivative is 0 at each of these points. Since f (2) = 18 > 0 and f (1) = 18 < 0, we can conclude from the second-derivative test that f (2) = 15 is a local minimum and f (1) = 12 is a local maximum. Comparing the First- and Second-Derivative Tests The rst-derivative test is more general than the second-derivative test. The rstderivative test can be applied at a critical point c even if f is not differentiable at c (provided of course that f is continuous at c). In contrast, the second-derivative test can be applied at c only if f is twice differentiable at c, and, even then, the test gives us information only if f (c) = 0.Example 9 Set f (x) = x 4/3 . Here f (x) = 4 x 1/3 so that 3 f (0) = 0,f (x) < 0x < 0,forf (x) > 0x > 0.forBy the rst-derivative test, f (0) = 0 is a local minimum. We cannot get this information from the second-derivative test because f (x) = 4 x 2/3 is not dened at x = 0. 9Example 10 To show what can happen if the second derivative is zero at a critical point c, we examine the functions f (x) = x 3 ,g(x) = x 4 ,yh(x) = x 4 .(Figure 4.3.14)yyf (x) = x 3g(x) = x 4xxx h(x) = x 4Figure 4.3.14In each case x = 0 is a critical point: f (x) = 3x 2 ,g (x) = 4x 3 ,h (x) = 4x 3 ,f (0) = 0,g (0) = 0,h (0) = 0.In each case the second derivative is zero at x = 0: f (x) = 6x,g (x) = 12x 2 ,h (0) = 12x 2 ,f (0) = 0,g (0) = 0,h (0) = 0.The rst function, f (x) = x 3 , has neither a local maximum nor a local minimum at x = 0. The second function, g(x) = x 4 , has derivative g (x) = 4x 3 . Since g (x) < 0forx < 0,g (x) > 0forx > 0,g(0) is a local minimum. (The rst-derivative test.) The last function, being the negative of g, has a local maximum at x = 0. 13:26 197. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.3 LOCAL EXTREME VALUES173EXERCISES 4.3 Exercises 128. Find the critical points and the local extreme values. 1. f (x) = x 3 + 3x 2. 1 3. f (x) = x + . x 5. f (x) = x 2 (1 x). 1+x 7. f (x) = . 1x 2 9. f (x) = . x(x + 1)2. f (x) = 2x 4 4x 2 + 6. 3 4. f (x) = x 2 2 . x 6. f (x) = (1 x)2 (1 + x). 2 3x 8. f (x) = . 2+x31. Let f and g be the differentiable functions, with graphs shown below. The point c is the point in the interval [a, b] where the vertical separation between the two curves is greatest. Show that the line tangent to the graph of f at x = c is parallel to the line tangent to the graph of g at x = c.f10. f (x) = x 2 16 . x 2 x +2311. f (x) = x 3 (1 x)2 .12. f (x) =13. f (x) = (1 2x)(x 1)3 .14. f (x) = (1 x)(1 + x)3 . 16. f (x) = x 3 1 x.x2 . 1+x 17. f (x) = x 2 3 2 + x. 15. f (x) =18. f (x) =.1 1 . x +1 x 219. f (x) = x 3 + 2x + 1 . 20. f (x) = x 7/3 7x 1/3 . 21. f (x) = x 2/3 + 2x 1/3 . 23. f (x) = sin x + cos x,22. f (x) =x3 . x +10 < x < 2 .24. f (x) = x + cos 2x, 0 < x < . 25. f (x) = sin2 x 3 sin x, 0 < x < . 26. f (x) = sin2 x,0 < x < 2 .27. f (x) = sin x cos x 3 sin x + 2x,0 < x < 2.28. f (x) = 2 sin x 3 sin x, 0 < x < . Exercises 2930. The graph of f is given. (a) Find the intervals on which f increases and the intervals on which f decreases. (b) Find the local maximum(s) and the local minimum(s) of f. Sketch the graph of f given that f (0) = 1.ga1Show that the equation P(x) = 0 has exactly two real roots, both positive. 36. A function f has derivative f (x) = x 3 (x 1)2 (x + 1)(x 2).123 x37.38. y30.39. 40. 1xP(x) = x 4 8x 3 + 22x 2 24x + 4.y2b32. Prove the validity of the second-derivative test in the case that f (c) < 0. 33. Let f (x) = ax 2 + bx + c, a = 0. Show that f has a local maximum at x = b/(2a) if a < 0 and a local minimum there if a > 0. 34. Let f (x) = ax 3 + bx 2 + cx + d, a = 0. Under what conditions on a, b, c will f have: (1) two local extreme values, (2) only one local extreme value, (3) no local extreme values? 35. Find the critical points and the local extreme values of the polynomial329.c12xAt what numbers x, if any, does f have a local maximum? A local minimum? Suppose that p(x) = an x n + an1 x n1 + + a1 x + a0 has critical points 1, 1, 2, 3, and corresponding values p(1) = 6, p(1) = 1, p(2) = 3, p(3) = 1. Sketch a possible graph for p if: (a) n is odd, (b) n is even. Suppose that f (x) = Ax 2 + Bx + C has a local minimum at x = 2 and the graph passes through the points (1, 3) and (3, 1). Find A, B, C. Find a and b given that f (x) = ax/(x 2 + b2 ) has a local minimum at x = 2 and f (0) = 1. Let f (x) = x p (1 x)q , p and q integers greater than or equal to 2. (a) Show that the critical points of f are 0, p/( p + q), 1. (b) Show that if p is even, then f has a local minimum at 0.13:26 198. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006174 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES (c) Show that if q is even, then f has a local minimum at 1. (d) Show that f has a local maximum at p/( p + q) for all p and q under consideration. 41. Prove that a polynomial of degree n has at most n 1 local extreme values. 42. Let y = f (x) be differentiable and suppose that the graph of f does not pass through the origin. The distance D from the origin to a point P(x, f (x)) of the graph is given by D=43. 44.C45.46.CExercises 4749. Use a graphing utility to graph the function on the indicated interval. (a) Use the graph to estimate the critical points and local extreme values. (b) Estimate the intervals on which the function increases and the intervals on which the function decreases. Round off your estimates to three decimal places. 47. f (x) = 3x 3 7x 2 14x + 24; [3, 4]. 48. f (x) = 3x 3 + x 2 10x + 2 + 3x; [4, 4]. 8 sin 2x 49. f (x) = ; [3, 3]. 1 + 1 x2 2CExercises 5052. Find the local extreme values of f by using a graphing utility to draw the graph of f and noting the numbers x at which f (x) = 0. 50. f (x) = x 5 + 13x 4 67x 3 + 171x 2 216x + 108. 51. f (x) = x 2 3x 2. 52. f (x) = cos2 2x.CExercises 5354. The derivative f of a function f is given. Use a graphing utility to graph f on the indicated interval. Estimate the critical points of f and determine at each such point whether f has a local maximum, a local minimum, or neither. Round off your estimates to three decimal places. 53. f (x) = sin2 x + 2 sin 2x; [2, 2]. 54. f (x) = 2x 3 + x 2 4x + 3; [4, 4].x 2 + [ f (x)]2 .Show that if D has a local extreme value at c, then the line through (0, 0) and (c, f (c)) is perpendicular to the line tangent to the graph of f at (c, f (c)). Show that f (x) = x 4 7x 2 8x 3 has exactly one critical point c in the interval (2, 3). Show that f (x) = sin x + 1 x 2 2x has exactly one critical 2 point c in the interval (2, 3). ax 2 + b , d = 0. Use a CAS to show that f has Set f (x) = 2 cx + d a local minimum at x = 0 if ad bc > 0 and a local maximum at x = 0 if ad bc < 0. Conrm this by calculating ad bc for each of the functions given below and using a graphing utility to draw the graph. 3 2x 2 2x 2 + 3 . . (b) f (x) = 2 (a) f (x) = 2 4x x +2 Set f (x) =Earlier we stated that f is differentiable at 0 and that f (0) = 0. Show that f has neither a local maximum nor a local minimum at x = 0.x 2 sin (1/x), 0,x = 0. x = 0. 4.4 ENDPOINT EXTREME VALUES; ABSOLUTE EXTREME VALUES We will work with functions dened on an interval or on an interval with a nite number of points removed. A number c is called the left endpoint of the domain of f if f is dened at c but undened to the left of c. We call c the right endpoint of the domain of f if f is dened at c but undened to the right of c. The assumptions made on the structure of the domain guarantee that if c is the left endpoint of the domain, then f is dened at least on an interval [c, c + ), and if c is the right endpoint, then f is dened at least on an interval (c , c]. Endpoints of the domain can give rise to what are called endpoint extreme values. Endpoint extreme values (illustrated in Figures 4.4.14.4.4) are dened below.endpoint maximumendpoint minimumccFigure 4.4.1Figure 4.4.213:26 199. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.4 ENDPOINT EXTREME VALUES; ABSOLUTE EXTREME VALUESendpoint minimumendpoint maximumccFigure 4.4.3Figure 4.4.4DEFINITION 4.4.1 ENDPOINT EXTREME VALUESIf c is an endpoint of the domain of f, then f is said to have an endpoint maximum at c provided that f (c) f (x)for all x in the domain of f sufciently close to c.It is said to have an endpoint minimum at c provided that f (c) f (x)for all x in the domain of f sufciently close to c.Endpoints in the domain of a continuous function which is differentiable at all points of the domain near that endpoint can be tested by examining the sign of the rst derivative at nearby points and then reasoning as we did in Section 4.3. Suppose, for example, that c is a left endpoint and that f is continuous from the right at c. If f (x) < 0 at all nearby x > c, then f decreases on an interval [c, c + ) and f (c) is an endpoint maximum. (Figure 4.4.1) If, on the other hand, f (x) > 0 at all nearby x > c, then f increases on an interval [c, c + ) and f (c) is an endpoint minimum. (Figure 4.4.2) Similar reasoning can be applied to right endpoints.Absolute Maxima and Absolute Minima Whether or not a function f has a local extreme value or an endpoint extreme value at some point c depends entirely on the behavior of f at c and at points close to c. Absolute extreme values, which we dene below, depend on the behavior of the function on its entire domain. We begin with a number d in the domain of f. Here d can be an interior point or an endpoint.DEFINITION 4.4.2 ABSOLUTE EXTREME VALUESThe function f is said to have an absolute maximum at d provided that f (d) f (x)for all x in the domain of f ;f is said to have an absolute minimum at d provided that f (d) f (x)for all x in the domain of f.A function can be continuous on an interval (or even differentiable there) without taking on an absolute maximum or an absolute minimum. All we can say in general is that if f takes on an absolute extreme value, then it does so at a critical point or at an endpoint.17513:26 200. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006176 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES There are, however, special conditions that guarantee the existence of absolute extreme values. From Section 2.6 we know that continuous functions map bounded closed intervals [a, b] onto bounded closed intervals [m, M]; M is the maximum value taken on by f on [a, b] and m is the minimum value. If [a, b] constitutes the entire domain of f , then, clearly, M is the absolute maximum and m is the absolute minimum. For a function continuous on a bounded closed interval [a, b], the absolute extreme values can be found as indicated below. Step 1. Find the critical points c1 , c2 , . . . . (These are the numbers in the open interval (a, b) at which the derivative is zero or does not exist.) Step 2. Calculate f (a), f (c1 ), f (c2 ), . . . , f (b). Step 3. The greatest of these numbers is the absolute maximum value of f and the least of these numbers is the absolute minimum.Example 1 Find the critical points of the function f (x) = 1 + 4x 2 1 x 4 , 2x [1, 3].Then nd and classify all the extreme values. SOLUTION Since f is continuous and the entire domain is the bounded closed interval [1, 3], we know that f has an absolute maximum and an absolute minimum. To nd the critical points of f, we differentiate:f (x) = 8x 2x 3 = 2x(4 x 2 ) = 2x(2 x)(2 + x). The numbers x in (1, 3) at which f (x) = 0 are x = 0 and x = 2. Thus, 0 and 2 are the critical points. The sign of f and the behavior of f are as follows: sign of f' behavior of f 0 + + + + + + + + 0 1 decreases 09f (1) = 1 + 4(1)2 1 (1)4 = 29 2xis an endpoint maximum;f (0) = 1is a local minimum;f (2) = 1 + 4(2)2 1 (24 ) = 9 29 2is a local maximum;f (3) = 1 + 4(3) 21 2 decreases 3Taking the endpoints into consideration, we have:y1increases27 2Figure 4.4.53x1 (3)4 2=7 2is an endpoint minimum.The least of these extremes, f (3) = 7 , is the absolute minimum; the greatest of these 2 extremes, f (2) = 9, is the absolute maximum. The graph of the function is shown in Figure 4.4.5. Example 2 Find the critical points of the function f (x) =x 2 + 2x + 2,1 x < 0 2x 2 2x + 2,0 x 2.Then nd and classify all the extreme values. SOLUTION Since f is continuous on its entire domain, which is the bounded closed interval [ 1 , 2], we know that f has an absolute maximum and an absolute minimum. 2 Differentiating f, we see that f (x) is2x + 2 on ( 1 , 0), 2nonexistent at x = 0,2x 2 on (0, 2).13:26 201. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.4 ENDPOINT EXTREME VALUES; ABSOLUTE EXTREME VALUES177This makes x = 0 a critical point. Since f (x) = 0 at x = 1, 1 is a critical point. The sign of f and the behavior of f are as follows: + + + + + + dne 0 + + + + + + +sign of f ': behavior of f : 1 increases 0 2decreases1 increases3 2yx(2, 2)(0, 2)Therefore f ( 1 ) = 21+2=1 45 4is an endpoint minimum;f (0) = 2 f (1) = 1 2 + 2 = 1( , )is a local maximum; is a local minimum;f (2) = 21 2is an endpoint maximum.The least of these extremes, f (1) = 1, is the absolute minimum; the greatest of these extremes, f (0) = f (2) = 2, is the absolute maximum. The graph of the function is shown in Figure 4.4.6. Behavior of f (x) as x and as x . We now state four denitions. Once you grasp the rst one, the others become transparent. To say that x ,asf (x) is to say that, as x increases without bound, f (x) becomes arbitrarily large. More precisely, given any positive number M, there exists a positive number K such that x K,ifthenf (x) M.For example, as x , x 2 ,1 + x 2 ,tan 1 2 2 xTo say that asx ,f (x) is to say that, as x increases without bound, f (x) becomes arbitrarily large negative: given any negative number M, there exists a positive number K such that x K,ifthenf (x) M.For example, as x , x 4 ,1x ,tan1 2 x 2 .To say that asx ,f (x) is to say that, as x decreases without bound, f (x) becomes arbitrarily large: given any positive number M, there exists a negative number K such that x K,ifthenf (x) M.For example, as x , 1 x ,x 2 ,tan 1 2 2 xFinally, to say that asx ,f (x) .1 25 4(1, 1)1Figure 4.4.62x13:26 202. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006178 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES is to say that, as x decreases without bound, f(x) becomes arbitrarily large negative: given any negative number M, there exists a negative number K such that, x K,iff (x) M.thenFor example, as x , x 3 , Remark 1 x ,tan1 2 x 2 .As you can readily see, f (x) iff f (x) . Suppose now that P is a nonconstant polynomial: P(x) = an x n + an1 x n1 + + a1 x + a0 (an = 0, n 1). For large x that is, for large positive x and for large negative x the leading term an x n dominates. Thus, what happens to P(x) as x depends entirely on what happens to an x n . (You are asked to conrm this in Exercise 43.)Example 3 (a) As x , 3x 4 100x 3 + 2x 5 since 3x 4 . (b) As x , 5x 3 + 12x 2 + 80 since 5x 3 . Finally, we point out that if f (x) , then f cannot have an absolute maximum value, and if f (x) , then f cannot have an absolute minimum value.Example 4 Find the critical points of the function f (x) = 6 x x x. Then nd and classify all the extreme values. SOLUTION The domain is [0, ). To simplify the differentiation, we use fractional exponents:f (x) = 6x 1/2 x 3/2 . On (0, ) 3 3(2 x) f (x) = 3x 1/2 x 1/2 = . 2 2 x(Verify this.)Since f (x) = 0 at x = 2, we see that 2 is a critical point. The sign of f and the behavior of f are as follows:y(2, 42(sign of f ': behavior of f : 0+++++++0 increases1decreasesxTherefore,2Figure 4.4.76xf (0) = 0 is an endpoint minimum; f (2) = 6 2 2 2 = 4 2 is a local maximum. Since f (x) = x(6 x) as x , the function has no absolute minimum value. Since f increases on [0, 2] and decreases on [2, ), the local maximum is the absolute maximum. The graph of f appears in Figure 4.4.7. 13:26 203. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.4 ENDPOINT EXTREME VALUES; ABSOLUTE EXTREME VALUESA Summary for Finding All the Extreme Values (Local, Endpoint, and Absolute) of a Continuous Function f Step 1. Find the critical points the interior points c at which f (c) = 0 or f (c) does not exist. Step 2. Test each endpoint of the domain by examining the sign of the rst derivative at nearby points. Step 3. Test each critical point c by examining the sign of the rst derivative on both sides of c (the rst-derivative test) or by checking the sign of the second derivative at c itself (second-derivative test). Step 4. If the domain is unbounded on the right, determine the behavior of f (x) as x ; if unbounded on the left, check the behavior of f (x) as x . Step 5. Determine whether any of the endpoint extremes and local extremes are absolute extremes.Example 5 Find the critical points of the function f (x) = 1 (x 3 3 x 2 6x + 2), 4 2x [2, ).The nd and classify all the extreme values. SOLUTION To nd the critical points, we differentiate:f (x) = 1 (3x 2 3x 6) = 3 (x + 1)(x 2). 4 4 Since f (x) = 0 at x = 1 and x = 2, the numbers 1 and 2 are critical points. The sign of f and the behavior of f are as follows: +++++++0 0+++++++sign of f ': behavior of f :2increases 1decreases2increasesxWe can see from the sign of f that f (2) = 1 (8 6 + 12 + 2) = 0 4is an endpoint minimum;f (1) = 1 (1 411 8is a local maximum;f (2) = 1 (8 6 12 + 2) = 2 4is a local minimum.3 2+ 6 + 2) =The function takes on no absolute maximum value since f (x) as x ; f (2) = 2 is the absolute minimum value. The graph of f is shown in Figure 4.4.8. y 2 (1,11 ) 81211212(2, 2)Figure 4.4.8x17913:26 204. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006180 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES Example 6 Find the critical points of the function f (x) = sin x sin2 x,x [0, 2].Then nd and classify all the extreme values. SOLUTION On the interval (0, 2 )f (x) = cos x 2 sin x cos x = cos x(1 2 sin x). Setting f (x) = 0, we have cos x(1 2 sin x) = 0. This equation is satised by the numbers x at which cos x = 0 and the numbers x at which sin x = 1 . On (0, 2 ), the cosine is 0 only at x = /2 and x = 3/2, and the 2 sine is 1 only at x = /6 and x = 5/6. The critical points, listed in order, are 2 /6,/2,5/6,3/2.The sign of f and the behavior of f are as follows: + + + 0 0 + + + + + + 0 0 + + + + + + + + +sign of f ': 2 60behavior of f : increases decreases5 6increases23 2decreasesxincreasesTherefore f (0) = 0f (/6) =is an endpoint minimum;1 4f (/2) = 0is a local minimum;f (5/6) =f (3/2) = 2is a local minimum;f (2 ) = 0is a local maximum; 1 4is a local maximum; is an endpoint maximum.Note that f (/6) = f (5/6) = 1 is the absolute maximum and f (3/2) = 2 is the 4 absolute minimum. The graph of the function is shown in Figure 4.4.9. y0.25 6 25 63 2x212Figure 4.4.9EXERCISES 4.4 Exercises 130. Find the critical points. Then nd and classify all the extreme values. 1. f (x) = x + 2. 2. f (x) = (x 1)(x 2).3. f (x) = x 2 4x + 1, 4. f (x) = 2x 2 + 5x 1, 1 5. f (x) = x 2 + . xx [0, 3]. x [2, 0]. 6. f (x) = x +1 . x213:26 205. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.4 ENDPOINT EXTREME VALUES; ABSOLUTE EXTREME VALUES1 1 , x 10 , 2 . x 1 8. f (x) = x + 2 , x[1, 2]. x 9. f (x) = (x 1)(x 2), x [0, 2].7. f (x) = x 2 +10. f (x) = (x 1)2 (x 2)2 , x [0, 4]. x , x [3, 1]. 11. f (x) = 4 + x2 12. 13. 15. 17.x2 f (x) = , x [1, 2]. 1 + x2 f (x) = (x x)2 . 14. f (x) = x 4 x 2 . 1 f (x) = x 3 x. 16. f (x) = x . x 3 f (x) = 1 x 1.18. f (x) = (4x 1)1/3 (2x 1)2/3 . 19. f (x) = sin2 x 3 cos x, 0 x . 20. f (x) = cot x + x,0 x 2. 321. f (x) = 2 cos3 x + 3 cos x, 22. f (x) = sin 2x x, 23. f (x) = tan x x,0 x .24. f (x) = sin4 x sin2 x, 0 x 2 . 3 2x, 0 x < 1 25. f (x) = x 3, 1 x 4 5 x, 4 < x 7. x + 9, 8 x < 3 26. f (x) = x 2 + x, 3 x 2 5x 4, 2 < x < 5. x 2 + 1, 2 x < 1 27. f (x) = 5 + 2x x 2 , 1 x 3 x 1, 3 < x < 6. 2 2x x 2 , 2 x 0 x 2 , 0 0 for all x. 34. f (x) = 0 at each integer x; f has no extreme values. 35. Show that the cubic p(x) = x 3 + ax 2 + bx + c has extreme values iff a 2 > 3b. 36. Let r be a rational number, r > 1, and set f (x) = (1 + x)r (1 + r x)37.38.39. 40.0 x .1 x 1. 3 241.42.43.forShow that 0 is a critical point for f and show that f (0) = 0 is the absolute minimum value. Suppose that c is a critical point for f and f (x) > 0 for x = c. Show that if f (c) is a local maximum, then f is not continuous at c. What can you conclude about a function f continuous on [a, b], if for some c in (a, b), f (c) is both a local maximum and a local minimum? Suppose that f is continuous on [a, b] and f (a) = f (b). Show that f has at least one critical point in (a, b). Suppose that c1 < c2 and that f takes on local maxima at c1 and c2 . Prove that if f is continuous on [c1 , c2 ], then there is at least one point c in (c1 , c2 ) at which f takes on a local minimum. Give an example of a nonconstant function that takes on both its absolute maximum and absolute minimum on every interval. Give an example of a nonconstant function that has an innite number of distinct local maxima and an innite number of distinct local minima. Let P be a polynomial with positive leading coefcient: P(x) = an x n + an1 x n1 + + a1 x + a0 ,44. 45.46. 47.x 1.n 1.Clearly, as x , an x n . Show that, as x , P(x) by showing that, given any positive number M, there exists a positive number K such that, if x K , then P(x) M. Show that of all rectangles with diagonal of length c, the square has the largest area. Let p and q be positive rational numbers and set f (x) = x p (1 x)q , 0 x 1. Find the absolute maximum value of f. The sum of two numbers is 16. Find the numbers given that the sum of their cubes is an absolute minimum. If the angle of elevation of a cannon is and a projectile is red with muzzle velocity v ft/sec, then the range of the projectile is given by the formula R=v 2 sin 2 feet. 32What angle of elevation maximizes the range? 48. A piece of wire of length L is to be cut into two pieces, one piece to form a square and the other piece to form an equilateral triangle. How should the wire be cut so as to (a) maximize the sum of the areas of the square and the triangle?13:26 206. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006182 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVESCC(b) minimize the sum of the areas of the square and the triangle? Exercises 4952. Use a graphing utility to graph the function on the indicated interval. Estimate the critical points of the function and classify the extreme values. Round off your estimates to three decimal places. 49. f (x) = x 3 4x + 2x sin x; [2.5, 3]. 50. f (x) = x 4 7x 2 + 10x + 3; [3, 3]. 51. f (x) = x cos 2x cos2 x; [, ]. 52. f (x) = 5x 2/3 + 3x 5/3 + 1; [3, 1]. Exercises 5355. Use a graphing utility to determine whether the function satises the hypothesis of the extreme-value theo-rem on [a, b] (Theorem 2.6.2). If the hypothesis is satised, nd the absolute maximum value M and the absolute minimum value m. If the hypothesis is not satised, nd M and m if they exist. 1 2 x, if 1 x 2 53. f (x) = [a, b] = [1, 3]. 1 x 2, if 2 < x 3; 11 x 419 , 4if 0 x 354. f (x) =x 3 + 2,if 3 < x 4;55. f (x) =1 x + 1, 2 x 3 + 2,if 1 x < 4 if 4 x 6;[a, b] = [0, 4]. [a, b] = [1, 6]. 4.5 SOME MAX-MIN PROBLEMS The techniques of the preceding two sections can be brought to bear on a wide variety of max-min problems. The key idea is to express the quantity to be maximized or minimized as a function of one variable. If the function is differentiable, we can differentiate and analyze the results. We begin with a geometric example. 12Example 1 An isosceles triangle has a base of 6 units and a height of 12 units. Find the maximum possible area for a rectangle that is inscribed in the triangle and has one side resting on the base of the triangle. What are the dimensions of the rectangle(s) of maximum area? SOLUTION Figure 4.5.1 shows the isosceles triangle and a rectangle inscribed in the specied manner. In Figure 4.5.2 we have introduced a rectangular coordinate system. With x and y as in the gure, the area of the rectangle is given by the product6Figure 4.5.1A = 2x y. yThis is the quantity we want to maximize. To do this we have to express A as a function of only one variable. Since the point (x, y) lies on the line that passes through (0, 12) and (3, 0),(0, 12)y = 12 4x. The area of the rectangle can now be expressed entirely in terms of x:(x, y)(x, y)A(x) = 2x(12 4x) = 24x 8x 2 .yx(Verify this.)x(3, 0)Figure 4.5.2xSince x and y represent lengths, x and y cannot be negative. As you can check, this restricts x to the closed interval [0, 3]. Our problem can now be formulated as follows: nd the absolute maximum of the function A(x) = 24x 8x 2 ,x [0, 3].The derivative A (x) = 24 16x is dened for all x (0, 3). Setting A (x) = 0, we have 24 16x = 0which impliesx = 3. 213:26 207. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.5 SOME MAX-MIN PROBLEMSThe only critical point is x = 3 . Evaluating A at the endpoints of the interval and at the 2 critical point, we have: A(0) = 24(0) 8(0)2 = 0, A3 2= 243 283 2 2= 18,A(3) = 24(3) 8(3)2 = 0. The function has an absolute maximum of 18, and this value is taken on at x = 3 . At 2 x = 3 , the base 2x = 3 and the height y = 12 4x = 6. 2 The maximum possible area is 18 square units. The rectangle that produces this area has a base of 3 units and a height of 6 units. The example we just considered suggests a basic strategy for solving max-min problems.Strategy Step 1. Draw a representative gure and assign labels to the relevant quantities. Step 2. Identify the quantity to maximized or minimized and nd a formula for it. Step 3. Express the quantity to be maximized or minimized in terms of a single variable; use the conditions given in the problem to eliminate the other variable(s). Step 4. Determine the domain of the function generated by Step 3. Step 5. Apply the techniques of the preceding sections to nd the extreme value(s). rExample 2 A paint manufacturer wants cylindrical cans for its specialty enamels. The can is to have a volume of 12 uid ounces, which is approximately 22 cubic inches. Find the dimensions of the can that will require the least amount of material. See Figure 4.5.3. SOLUTION Let r be the radius of the can and h the height. The total surface area(top, bottom, lateral area) of a circular cylinder of radius r and height h is given by the formula S = 2r 2 + 2r h. This is the quantity that we want to minimize. Since the volume V = r 2 h is to be 22 cubic inches, we require that r 2 h = 22and thush=22 . r 2It follows from these equations that r and h must both be positive. Thus, we want to minimize the function S(r ) = 2r 2 + 2r22 r 2= 2r 2 +44 , rr (0, ).Differentiation gives dS 44 4r 3 44 4(r 3 11) = 4r 2 = = . dr r r2 r2hFigure 4.5.318313:26 208. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006184 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES The derivative is 0 where r 3 11 = 0, which is the point r0 = (11/)1/3 . Since r < r0 negative for dS 0 at r = r0 is positive for dr r >r , 0S decreases on (0, r0 ] and increases on [r0 , ). Therefore, the function S is minimized by setting r = r0 = (11/ )1/3 . The dimensions of the can that will require the least amount of material are as follows: 22 = 2(11/)1/3 radius r = (11/ )1/3 1.5 inches, height h = = (11/)2/3 3 inches. = The can should be as wide as it is tall.xExample 3 A window in the shape of rectangle capped by a semicircle is to have perimeter p. Choose the radius of the semicircular part so that the window admits the most light.xySOLUTION We take the point of view that the window which admits the most light is the one with maximum area. As in Figure 4.5.4, we let x be the radius of the semicircular part and y be the height of the rectangular part. We want to express the areaA = 1 x 2 + 2x y 2 Figure 4.5.4as a function of x alone. To do this, we must express y in terms of x. Since the perimeter is p, we have p = 2x + 2y + x and thus y = 1 [ p (2 + )x]. 2 Since x and y represent lengths, these variables must be nonnegative. For both x and y to be nonnegative, we must have 0 x p/(2 + ). The area can now be expressed in terms of x alone: A(x) = 1 x 2 + 2x y 2 = 1 x 2 + 2x 21 [p 2 (2 + )x]= 1 x 2 + px (2 + )x 2 = px 2 + 1 x 2 . 2 2 We want to maximize the function A(x) = px 2 + 1 x 2 , 20 x p/(2 + ).The derivative A (x) = p (4 + )x is 0 only at x = p/(4 + ). Since A(0) = A[ p/(2 + )] = 0, and since A (x) > 0 for 0 < x < p/(4 + ) and A (x) < 0 for p/(4 + ) < x < p/(2 + ), the function A is maximized by setting x = p/(4 + ). For the window to have maximum area, the radius of the semicircular part must be p/(4 + ). Example 4 The highway department is asked to construct a road between point A and point B. Point A lies on an abandoned road that runs east-west. Point B is 3 miles north of the point of the old road that is 5 miles east of A. The engineering13:26 209. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.5 SOME MAX-MIN PROBLEMSdivision proposes that the road be constructed by restoring a section of the old road from A to some point P and constructing a new road from P to B. Given that the cost of restoring the old road is $2,000,000 per mile and the cost of a new road is $4,000,000 per mile, how much of the old road should be restored so as to minimize the cost of the project. SOLUTION Figure 4.5.5 shows the geometry of the problem. Notice that we have chosen a straight line joining P and B rather than some curved path. (The shortest connection between two points is provided by the straight-line path.) We let x be the amount of old road that will be restored. Then9 + (5 x)2 =185 B 9 + (5 x)2AxP35xFigure 4.5.534 10x + x 2is the length of the new part. The total cost of constructing the two sections of road is C(x) = 2 106 x + 4 106 [34 10x + x 2 ]1/2 ,0 x 5.We want to nd the value of x that minimizes this function. Differentiation gives C (x) = 2 106 + 4 106 = 2 106 +1 2[34 10x + x 2 ]1/2 (2x 10)4 106 (x 5) , [34 10x + x 2 ]1/20 < x < 5.Setting C (x) = 0, we nd that 2(x 5) =0 [34 10x + x 2 ]1/21+2(x 5) = [34 10x + x 2 ]1/2 4(x 2 10x + 25) = 34 10x + x 2 3x 2 30x + 66 = 0 x 2 10x + 22 = 0. By the general quadratic formula, we have x=10 100 4(22) = 5 3. 2 The value x = 5 + 3 is not in the domain of our function; the value we want is x = 5 3. We analyze the sign of C : sign of C': behavior of C: 00 + + + + + 5 3 5 decreases increasesx Since the function is continuous on [0, it decreases on [0, 5 3] and increases 5], on [5 3, 5]. The number x = 5 3 3.27 gives the minimum value of C. = The highway department will minimize its costs by restoring 3.27 miles of the old road. Example 5 (The angle of incidence equals the angle of reection.) Figure 4.5.6 depicts light from point A reected by a mirror to point B. Two angles have been marked: the angle of incidence, i , and the angle of reection, r . Experiment showsA B i rFigure 4.5.613:26 210. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006186 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES A (a1, a2) B (b1, b2)SOLUTION We write the length of the path as a function of x. In the setup of Figure 4.5.7, i r a1xthat i = r . Derive this result by postulating that the light travels from A to the mirror and then on to B by the shortest possible path.b1xL(x) =Figure 4.5.72 (x a1 )2 + a2 +2 (x b1 )2 + b2 ,x [a1 , b1 ].Differentiation gives L (x) =x a1 2 (x a1 )2 + a2+x b1 2 (x b1 )2 + b2.Therefore L (x) = 0iffx a1 2 (x a1 )2 + a2=b1 x 2 (x b1 )2 + b2iff sin i = sin r(see the gure)iff i = r .That L(x) is minimal when i = r can be seen by noting that L (x) is always positive, L (x) =2 2 a2 b2 + > 0, 2 2 [(x a1 )2 + a2 ]3/2 [(x b1 )2 + b2 ]3/2and applying the second-derivative test. (We must admit that there is a much simpler way to do Example 5, a way that requires no calculus at all. Can you nd it?) Now we will work out a simple problem in which the function to be maximized is dened not on an interval or on a union of intervals, but on a discrete set of points, in this case a nite collection of integers.Example 6 A small manufacturer of ne rugs has the capacity to produce 25 rugs per week. Assume (for the sake of this example) that the production of the rugs per week leads to an annual prot which, measured in thousands of dollars, is given by the function P = 100n 600 3n 2 . Find the level of weekly production that maximizes P. SOLUTION Since n is an integer, it makes no sense to differentiateP = 100n 600 3n 2 with respect to n. Table 4.5.1, compiled by direct calculation, shows the prot P that corresponds to each production level n from 8 to 25. (For n < 8, P is negative; 25 is full capacity.) The table shows that the largest prot comes from setting production at 17 units per week. We can avoid the arithmetic required to construct the table by considering the function f (x) = 100x 600 3x 2 , 8 x 25.This is a special case of Fermats principle of least time, which says that, of all (neighboring) paths, light chooses the one that requires the least time. If light passes from one medium to another, the geometrically shortest path is not necessarily the path of least time.13:26 211. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.5 SOME MAX-MIN PROBLEMS187 Table 4.5.1nPnPnP8 9 10 11 12 138 57 100 137 168 19314 15 16 17 18 19212 225 232 233 228 21720 21 22 23 24 25200 177 148 113 72 25For integral values of x, the function agrees with P. It is continuous on [8, 25] and differentiable on (8, 25) with derivative f (x) = 100 6x. Obviously, f (x) = 0 at x = 100 = 16 2 . Since f (x) > 0 on (8, 16 2 ) and is continuous 6 3 3 at the endpoints, f increases on [8, 16 2 ]. Since f (x) < 0 on (16 2 , 25) and is continuous 3 3 at the endpoints, f decreases on [16 2 , 25]. The largest value of f corresponding to an 3 integer value of x will therefore occur at x = 16 or at x = 17. Direct calculation of f (16) and f (17) shows that the choice x = 17 is correct. EXERCISES 4.5 1. Find the greatest possible value of xy given that x and y are both positive and x + y = 40. 2. Find the dimensions of the rectangle of perimeter 24 that has the largest area. 3. A rectangular garden 200 square feet in area is to be fenced off against rabbits. Find the dimensions that will require the least amount of fencing given that one side of the garden is already protected by a barn. 4. Find the largest possible area for a rectangle with base on the x-axis and upper vertices on the curve y = 4 x 2 . 5. Find the largest possible area for a rectangle inscribed in a circle of radius 4. 6. Find the dimensions of the rectangle of area A that has the smallest perimeter. 7. How much fencing is needed to dene two adjacent rectangular playgrounds of the same width and total area 15,000 square feet? 8. A rectangular warehouse will have 5000 square feet of oor space and will be separated into two rectangular rooms by an interior wall. The cost of the exterior walls is $150 per linear foot and the cost of the interior wall is $100 per linear foot. Find the dimensions that will minimize the cost of building the warehouse. 9. Rework Example 3; this time assume that the semicircular portion of the window admits only one-third as much light per square foot as does the rectangular portion. 10. A rectangular plot of land is to be dened on one side by a straight river and on three sides by post-and-rail fencing. Eight hundred feet of fencing are available. How should the fencing be deployed so as to maximize the area of the plot? 11. Find the coordinates of P that maximize the area of the rectangle shown in the gure.y(0, 3)P(4, 0)x12. A triangle is to be formed as follows: the base of the triangle is to lie on the x-axis, one side is to lie on the line y = 3x, and the third side is to pass through the point (1, 1). Assign a slope to the third side that maximizes the area of the triangle. 13. A triangle is to be formed as follows: two sides are to lie on the coordinate axes and the third side is to pass through the point (2, 5). Assign a slope to the third side that minimizes the area of the triangle. 14. Show that, for the triangle of Exercise 13, it is impossible to assign a slope to the third side that maximizes the area of the triangle. 15. What are the dimensions of the base of the rectangular box of greatest volume that can be constructed from 100 square inches of cardboard if the base is to be twice as long as it is wide? Assume that the box has a top. 16. Exercise 15 under the assumption that the box has no top. 17. Find the dimensions of the isosceles triangle of largest area with perimeter 12. 18. Find the point(s) on the parabola y = 1 x 2 closest to the point 8 (0, 6).13:26 212. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006188 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 19. Find the point(s) on the parabola x = y 2 closest to the point (0, 3). 20. Find A and B given that the function y = Ax 1/2 + Bx 1/2 has a minimum of 6 at x = 9. 21. Find the maximal possible area for a rectangle inscribed in the ellipse 16x 2 + 9y 2 = 144. 22. Find the maximal possible area for a rectangle inscribed in the ellipse b2 x 2 + a 2 y 2 = a 2 b2 . 23. A pentagon with a perimeter of 30 inches is to be constructed by adjoining an equilateral triangle to a rectangle. Find the dimensions of the rectangle and triangle that will maximize the area of the pentagon. 24. A 10-foot section of gutter is made from a 12-inch-wide strip of sheet metal by folding up 4-inch strips on each side so that they make the same angle with the bottom of the gutter. Determine the depth of the gutter that has the greatest carrying capacity. Caution: There are two ways to sketch the trapezoidal cross section. (See the gure.)4' 4'31. A rectangular banner is to have a red border and a rectangular white center. The width of the border at top and bottom is to be 8 inches, and along the sides 6 inches. The total area is to be 27 square feet. Find the dimensions of the banner that maximize the area of the white center. 32. Conical paper cups are usually made so that the depth is 2 times the radius of the rim. Show that this design requires the least amount of paper per unit volume. 33. A string 28 inches long is to be cut into two pieces, one piece to form a square and the other to form a circle. How should the string be cut so as to (a) maximize the sum of the two areas? (b) minimize the sum of the two areas? 34. What is the maximum volume for a rectangular box (square base, no top) made from 12 square feet of cardboard? 35. The gure shows a cylinder inscribed in a right circular cone of height 8 and base radius 5. Find the dimensions of the cylinder that maximize its volume.4'B4' 4'4'25. From a 15 8 rectangular piece of cardboard four congruent squares are to be cut out, one at each corner. (See the gure.) The remaining crosslike piece is then to be folded into an open box. What size squares should be cut out so as to maximize the volume of the resulting box?836. As a variant of Exercise 35, nd the dimensions of the cylinder that maximize the area of its curved surface. 37. A rectangular box with square base and top is to be made to contain 1250 cubic feet. The material for the base costs 35 cents per square foot, for the top 15 cents per square foot, and for the sides 20 cents per square foot. Find the dimensions that will minimize the cost of the box. 38. What is the largest possible area for a parallelogram inscribed in a triangle ABC in the manner of the gure? C1526. A page is to contain 81 square centimeters of print. The margins at the top and bottom are to be 3 centimeters each and, at the sides, 2 centimeters each. Find the most economical dimensions given that the cost of a page varies directly with the perimeter of the page. 27. Let ABC be a triangle with vertices A = (3, 0), B = (0, 6), C = (3, 0). Let P be a point on the line segment that joins B to the origin. Find the position of P that minimizes the sum of the distances between P and the vertices. 28. Solve Exercise 27 with A = (6, 0), B = (0, 3), C = (6, 0). 29. An 8-foot-high fence is located 1 foot from a building. Determine the length of the shortest ladder that can be leaned against the building and touch the top of the fence. 30. Two hallways, one 8 feet wide and the other 6 feet wide, meet at right angles. Determine the length of the longest ladder that can be carried horizontally from one hallway into the other.AB39. Find the dimensions of the isosceles triangle of least area that circumscribes a circle of radius r. 40. What is the maximum possible area for a triangle inscribed in a circle of radius r? 41. The gure shows a right circular cylinder inscribed in a sphere of radius r. Find the dimensions of the cylinder that maximize the volume of the cylinder.r13:26 213. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.5 SOME MAX-MIN PROBLEMS42. As a variant of Exercise 41, nd the dimensions of the right circular cylinder that maximize the lateral area of the cylinder. 43. A right circular cone is inscribed in a sphere of radius r as in the gure. Find the dimensions of the cone that maximize the volume of the cone.r44. What is the largest possible volume for a right circular cone of slant height a? 45. A power line is needed to connect a power station on the shore of a river to an island 4 kilometers downstream and 1 kilometer offshore. Find the minimum cost for such a line given that it costs $50,000 per kilometer to lay wire under water and $30,000 per kilometer to lay wire under ground. 46. A tapestry 7 feet high hangs on a wall. The lower edge is 9 feet above an observers eye. How far from the wall should the observer stand to obtain the most favorable view? Namely, what distance from the wall maximizes the visual angle of the observer? HINT: Use the formula for tan (A B). 47. An object of weight W is dragged along a horizontal plane by means of a force P whose line of action makes an angle with the plane. The magnitude of the force is given by the formula W P= sin + cos where denotes the coefcient of friction. Find the value of that minimizes P. 48. The range of a projectile red with elevation angle at an inclined plane is given by the formula R=2v 2 cos sin( ) g cos2 where is the inclination of the target plane, and v and g are constants. Calculate for maximum range. 49. Two sources of heat are placed s meters aparta source of intensity a at A and a source of intensity b at B. The intensity of heat at a point P on the line segment between A and B is given by the formula b a , I = 2 + x (s x)2 where x is the distance between P and A measured in meters. At what point between A and B will the temperature be lowest? 50. The distance from a point to a line is the distance from that point to the closest point of the line. What point of the line Ax + By + C = 0 is closest to the point (x1 , y1 )? What is the distance from (x1 , y1 ) to the line? 51. Let f be a differentiable function dened on an open interval I. Let P(a, b) be a point not on the graph of f. Show that if189P Q is the longest or shortest line segment that joins P to the graph of f, then P Q is perpendicular to the graph of f. 52. Draw the parabola y = x 2 . On the parabola mark a point P = O. Through P draw the normal line. The normal line intersects the parabola at another point Q. Show that the distance between P and Q is minimized by setting P = 22 , 1 . 2 53. For each integer n, set f (n) = 6n 4 16n 3 + 9n 2 . Find the integer n that minimizes f (n). 54. A local bus company offers charter trips to Blue Mountain Museum at a fare of $37 per person if 16 to 35 passengers sign up for the trip. The company does not charter trips for fewer than 16 passengers. The bus has 48 seats. If more than 35 passengers sign up, then the fare for every passenger is reduced by 50 cents for each passenger in excess of 35 that signs up. Determine the number of passengers that generates the greatest revenue for the bus company. 55. The Hotwheels Rent-A-Car Company derives an average net prot of $12 per customer if it services 50 customers or fewer. If it services more than 50 customers, then the average net prot is decreased by 6 cents for each customer over 50. What number of customers produces the greatest total net prot for the company? 56. A steel plant has the capacity to produce x tons per day of low-grade steel and y tons per day of high-grade steel where 40 5x . y= 10 x Given that the market price of low-grade steel is half that of high-grade steel, show that about 5 1 tons of low-grade steel 2 should be produced per day for maximum revenue. 1 57. The path of a ball is the curve y = mx 400 (m 2 + 1)x 2 . Here the origin is taken as the point from which the ball is thrown and m is the initial slope of the trajectory. At a distance which depends on m, the ball returns to the height from which it was thrown. What value of m maximizes this distance? 58. Given the trajectory of Exercise 57, what value of m maximizes the height at which the ball strikes a vertical wall 300 feet away? 59. A truck is to be driven 300 miles on a freeway at a constant speed of miles per hour. Speed laws require that 35 70. Assume that the fuel costs $2.60 per gallon and 1 is consumed at the rate of 1 + ( 400 ) 2 gallons per hour. Given that the drivers wages are $20 per hour, at what speed should the truck be driven to minimize the truck owners expenses? 60. A tour boat heads out on a 100-kilometer sight-seeing trip. Given that the xed costs of operating the boat total $2500 per hour, that the cost of fuel varies directly with the square of the speed of the boat, and at 10 kilometers per hour the cost of the fuel is $400 per hour, nd the speed that minimizes the boat owners expenses. Is the speed that minimizes the owners expenses dependent on the length of the trip? 61. An oil drum is to be made in the form of a right circular cylinder to contain 16 cubic feet. The upright drum is to be taller than it is wide, but not more than 6 feet tall. Determine the dimensions of the drum that minimize surface area.13:26 214. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006190 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVESCC62. The cost of erecting a small ofce building is $1,000,000 for the rst story, $1,100,000 for the second, $1,200,000 for the third, and so on. Other expenses (lot, basement, etc.) are $5,000,000. Assume that the annual rent is $200,000 per story. How many stories will provide the greatest return on investment? 63. Points A and B are opposite points on the shore of a circular lake of radius 1 mile. Maggie, now at point A, wants to reach point B. She can swim directly across the lake, she can walk along the shore, or she can swim part way and walk part way. Given that Maggie can swim at the rate of 2 miles per hour and walks at the rate of 5 miles per hour, what route should she take to reach point B as quickly as possible? (No running allowed.) 64. Our friend Maggie of Exercise 63 nds a row boat. Given that she can row at the rate of 3 miles per hour, what route should she take now? Row directly across, walk all the way, or row part way and walk part way? Here by return on investment we mean the ratio of income to cost.CC CC65. Set f (x) = x 2 x and let P be the point (4, 3). (a) Use a graphing utility to draw f and mark P. (b) Use a CAS to nd the point(s) on the graph of f that are closest to P. (c) Let Q be a point which satises part (b). Determine the equation for the line l P Q through P and Q; then display in one gure the graph of f, the point P, and the line l P Q . (d) Determine the equation of the line l N normal to the graph of f at (Q, f (Q)). (e) Compare l P Q and l N . 66. Exercise 65 with f (x) = x x 3 and P(1, 8). 67. Find the distance D(x) from a point (x, y) on the line y + 3x = 7 to the origin. Use a graphing utility to draw the graph of D and then use the trace function to estimate the point on the line closest to the origin. 68. Find the distance D(x) from a point (x, y) on the graph of f (x) = 4 x 2 to the point P(4, 3). Use a graphing utility to draw the graph of D and then use the trace function to estimate the point on the graph of f closest to P. PROJECT 4.5 Flight Paths of Birds Ornithologists studying the ight of birds have determined that certain species tend to avoid ying over large bodies of water during the daylight hours of summer. A possible explanation for this is that it takes more energy to y over water than land because on a summer day air typically rises over land and falls over water. Suppose that a bird with this tendency is released from an island that is 6 miles from the nearest point A of a straight shoreline. It ies to a point B on the shore and then ies along the shore to its nesting area C, which is 12 miles from A. (See the gure.)Problem 1. Show that the total energy E expended by the bird in ying from the island to its nesting area is given by E(x) = W 36 + x 2 + L(12 x),0 x 12where x is the distance from A to B measured in miles. Problem 2. Suppose that W = 1.5L; that is, suppose it takes 50% more energy to y over water than over land. (a) Use the methods of Section 4.5 to nd the point B to which the bird should y to minimize the total energy expended. (b) Use a graphing utility to graph E, and then nd the minimum value to conrm your result in part (a). Take L = 1. Problem 3. In general, suppose W = k L , k > 1.6AxBC 12Let W denote the energy per mile required to y over water, and let L denote the energy per mile required to y over land.(a) Find the point B (as a function of k) to which the bird should y to minimize the total energy expended. (b) Use a graphing utility to experiment with different values of k to nd out how the point B moves as k increases/decreases. Take L = 1. (c) Find the value(s) of k such that the bird will minimize the total energy expended by ying directly to its nest. (d) Are there any values of k such that the bird will minimize the total energy expended by ying directly to the point A and then along the shore to C? 4.6 CONCAVITY AND POINTS OF INFLECTION We begin with a sketch of the graph of a function f, Figure 4.6.1. To the left of c1 and between c2 and c3 , the graph curves up (we call it concave up); between c1 and c2 ,13:26 215. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.6 CONCAVITY AND POINTS OF INFLECTION191and to the right of c3 , the graph curves down (we call it concave down). These terms deserve a precise denition. y co nve down nca cocon ce upfe do avc avwnupc o ncav e c1c2c3xFigure 4.6.1DEFINITION 4.6.1 CONCAVITYLet f be a function differentiable on an open interval I. The graph of f is said to be concave up on I if f increases on I; it is said to be concave down on I if f decreases on I.Stated more geometrically, the graph is concave up on an open interval where the slope increases and concave down on an open interval where the slope decreases. One more observation: where concave up, the tangent line lies below the graph; where concave down, the tangent line lies above the graph. (Convince yourself of this by adding some tangent lines to the curve shown in Figure 4.6.1.) Points that join arcs of opposite concavity are called points of inection. The graph in Figure 4.6.1 has three of them: (c1 , f (c1 )), (c2 , f (c2 )), (c3 , f (c3 )). Here is the formal denition:ycave upyIf f is twice differentiable, we can determine the concavity of the graph from the sign of the second derivative.oncconca ve d owThus, the graph of f is concave down on (, 0) and concave up on (0, ). The origin, (0, 0) = (0, f (0)), is a point of inection, the only point of inection. (See Figure 4.6.3.) coreFigure 4.6.2Example 2 For the cubing function f (x) = x 3 , the derivative decreases on (, 0] and increases on [0, ).xf (x) = x2 4x + 3Example 1 The graph of the quadratic function f (x) = x 2 4x + 3 is concave up everywhere since the derivative f (x) = 2x 4 is everywhere increasing. (See Figure 4.6.2.) The graph has no point of inection. f (x) = 3x 2n cawheLet f be a function continuous at c and differentiable near c. The point (c, f (c)) is called a point of inection if there exists a > 0 such that the graph of f is concave in one sense on (c , c) and concave in the opposite sense on (c, c + ).ve upeveryDEFINITION 4.6.2 POINT OF INFLECTIONnxf (x) = x3Figure 4.6.313:26 216. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006192 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES THEOREM 4.6.3Suppose that f is twice differentiable on an open interval I. (i) If f (x) > 0 for all x in I, then f increases on I, and the graph of f is concave up. (ii) If f (x) < 0 for all x in I, then f decreases on I, and the graph of f is concave down.PROOFApply Theorem 4.2.2 to the function f .The following result gives us a way of identifying possible points of inection.THEOREM 4.6.4If the point (c, f (c)) is a point of inection, then f (c) = 0orf (c)does not exist.Suppose that (c, f (c)) is a point of inection. Lets assume that the graph of f is concave up to the left of c and concave down to the right of c. The other case can be handled in a similar manner. In this situation f increases on an interval (c , c) and decreases on an interval (c, c + ). Suppose now that f (c) exists. Then f is continuous at c. It follows that f increases on the half-open interval (c , c] and decreases on the half-open interval [c, c + ). This says that f has a local maximum at c. Since by assumption f (c) exists, f (c) = 0. (Theorem 4.3.2 applied to f .) We have shown that if f (c) exists, then f (c) = 0. The only other possibility is that f (c) does not exist. (Such is the case for the function examined in Example 4 below.) PROOFy5 f (2, 3) point of inflectionExample 3 For the function1f (x) = x 3 6x 2 + 9x + 1 123x(Figure 4.6.4)we have f (x) = 3x 2 12x + 9 = 3(x 2 4x + 3)f (x) = x3 6x2 + 9x + 1and f (x) = 6x 12 = 6(x 2).Figure 4.6.4Note that f (x) = 0 only at x = 2, and f keeps a constant sign on (, 2) and on (2, ). The sign of f on these intervals and the consequences for the graph of f are as follows: sign of f': graph of f : 0+ + + + + + + + + + + + + + + + + + + concave down2 point of inflectionThe point (2, f (2)) = (2, 3) is a point of inection.concave upx13:26 217. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006Example 4 For(1, 2)f (x) = 3x 5/3 5x(Figure 4.6.5)1934.6 CONCAVITY AND POINTS OF INFLECTION y 2point of inflection1we have f (x) = 5x 2/3 5andf (x) =210 1/3 x . 311 1The second derivative does not exist at x = 0. Since negative, f (x) is positive,x22for x < 0 for x > 0,(1, 2)f (x) = 3x 5/3 5xFigure 4.6.5the graph is concave down on (, 0) and concave up on (0, ). Since f is continuous at 0, the point (0, f (0)) = (0, 0) is a point of inection. CAUTION The fact that f (c) = 0 or f (c) does not exist does not guarantee that (c, f (c)) is a point of inection. (The statement that constitutes Theorem 4.6.4 is not an iff statement.) As you can verify, the function f (x) = x 4 satises the condition f (0) = 0, but the graph is always concave up and there are no points of inection. If f is discontinuous at c, then f (c) does not exist, but (c, f (c)) cannot be a point of inection. A point of inection occurs at c iff f is continuous at c and the point (c, f (c)) joins arcs of opposite concavity. Example 5 graph ofDetermine the concavity and nd the points of inection (if any) of the f (x) = x + cos x,x [0, 2 ].SOLUTION For x [0, 2 ], we havef (x) = 1 sin xandyf (x) = cos x.fOn the interval under consideration f (x) = 0 only at x = /2 and x = 3/2, and f keeps constant sign on (0, /2), on (/2, 3/2), and on (3/2, 2 ). The sign of f on these intervals and the consequences for the graph of f are as follows: sign of f': 0+++++++++++++++++++0 graph of f : 0 2concave downconcave up23 23 2 2xconcave down(3 , 3 2 2points of inflection)(, ) 2 2 223 2xf (x) = x + cos xThe points (/2, f (/2)) = (/2, /2), and (3/2, f (3/2)) = (3/2, 3/2) are points of inection. The graph of f is shown in Figure 4.6.6. Figure 4.6.6EXERCISES 4.6 1. The graph of a function f is given in the gure. (a) Determine the intervals on which f increases and the intervals on which f decreases; (b) determine the intervals on which the graph of f is concave up, the intervals on which the graph is concave down, and give the x-coordinate of each point of inection.y f c abdk lmnp13:26 218. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006194 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 9. f (x) = 1 x 4 1 x 2 . 4 2 x 11. f (x) = 2 . x 12. Exercise 1 applied to the function f graphed below. y13. f (x) = (1 x)2 (1 + x)2 . 1 x 15. f (x) = . 1+ xfabcdmnx3. The gure below gives the graph of a function f, the graph of its rst derivative f , and the graph of its second derivative f , but not in the correct order. Which curve is the graph of which function?ab(a)aC16. f (x) = (x 3)1/5 .f (x) = (x + 2)5/3 . 18. f (x) = x 4 x 2 . 2 f (x) = sin x, x [0, ]. f (x) = 2 cos2 x x 2 , x [0, ]. f (x) = x 2 + sin 2x, x [0, ]. f (x) = sin4 x, x [0, ].Exercises 2326. Find the points of inection of the graph of f by using a graphing utility. x 4 81 . 23. f (x) = x2 2 24. f (x) = sin x cos x, 2 x 2. 25. f (x) = x 5 + 9x 4 + 26x 3 + 18x 2 27x 27. x 26. f (x) = . 3x Exercises 2734. Find: (a) the intervals on which f increases and the intervals on which f decreases; (b) the local maxima and the local minima; (c) the intervals on which the graph is concave up and the intervals on which the graph is concave down; (d) the points of inection. Use this information to sketch the graph of f. 28. f (x) = 3x 4 + 4x 3 + 1. 27. f (x) = x 3 9x. 2x 29. f (x) = 2 . 30. f (x) = x 1/3 (x 6)2/3 . x +1 31. f (x) = x + sin x, x [, ]. 32. f (x) = sin x + cos x, x [0, 2 ]. x 3, x < 1 33. f (x) = 3x 2, x 1.b(b)a17. 19. 20. 21. 22.10. f (x) = x 3 (1 x). x +2 12. f (x) = . x 2 6x 14. f (x) = 2 . x +1b(c)4. A function f is continuous on [4, 4] and twice differentiable on (4, 4). Some information on f, f , and f is tabulated below:34. f (x) =x(4, 2)2(2, 0)0(0, 2)2positive negative0 negativenegative negativenegative 0negative positive0 0x 1 x > 1.(2, 4)f (x) f (x)2x + 4, 3 x 2,negative negative(a) Give the x-coordinates of the local maxima and minima of f . (b) Give the x-coordinates of the points of inection of the graph of f . (c) Given that f (0) = 0, sketch a possible graph for f . Exercises 522. Describe the concavity of the graph and nd the points of inection (if any). 1 1 6. f (x) = x + . 5. f (x) = . x x 7. f (x) = x 3 3x + 2. 8. f (x) = 2x 2 5x + 2.Exercises 3538. Sketch the graph of a continuous function f that satises the given conditions. 35. f (0) = 1, f (2) = 1; f (0) = f (2) = 0, f (x) > 0 for x 1 > 1, f (x) < 0 for x 1 < 1; f (x) < 0 for x < 1, f (x) > 0 for x > 1. 36. f (x) > 0 if x > 2, f (x) < 0 if x < 2; f (0) = 0, f (x) > 0 if x < 0, f (x) < 0 if x > 0; f (0) = 1, f (2) = f (2) = 1 , f (x) > 0 for all x, f is an even function. 2 37. f (x) < 0 if x < 0, f (x) > 0 if x > 0; f (1) = f (1) = 0, f (0) does not exist, f (x) > 0 if x > 1 f (x) < 013:26 219. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.7 VERTICAL AND HORIZONTAL ASYMPTOTES; VERTICAL TANGENTS AND CUSPSif x < 1 (x = 0); f (1) = 2, f (1) = 2; f is an odd function. 38. f (2) = 6, f (1) = 2, f (3) = 4; f (1) = f (3) = 0, f (x) < 0 if x 2 > 1, f (x) > 0 if x 2 < 1; f (x) < 0 if x + 1 < 1 or x > 2, f (x) > 0 if x 1 < 1 or x < 2. 39. Find d given that (d, f (d)) is a point of inection of the graph of f (x) = (x a)(x b)(x c). 40. Find c given that the graph of f (x) = cx 2 + x 2 has a point of inection at (1, f (1)). 41. Find a and b given that the graph of f (x) = ax 3 + bx 2 passes through the point (1, 1) and has a point of inection where x = 1 . 3 42. Determine A and B so that the curve y = Ax 1/2 + Bx 1/2Chas a point of inection at (1, 4). 43. Determine A and B so that the curve y = A cos 2x + B sin 3x 44.45.46.47.has a point of inection at (/6, 5). Find necessary and sufcient conditions on A and B for f (x) = Ax 2 + Bx + C (a) to decrease between A and B with graph concave up. (b) to increase between A and B with graph concave down. Find a function f with f (x) = 3x 2 6x + 3 for all real x and (1, 2) a point of inection. How many such functions are there? Set f (x) = sin x. Show that the graph of f is concave down above the x-axis and concave up below the x-axis. Does g(x) = cos x have the same property? Set p(x) = x 3 + ax 2 + bx + c.C195(a) Show that the graph of p has exactly one point of inection. What is x at that point? (b) Show that p has two local extreme values iff a 2 > 3b. (c) Show that p cannot have only one local extreme value. 48. Show that if a cubic polynomial p(x) = x 3 + ax 2 + bx + c has a local maximum and a local minimum, then the midpoint of the line segment that connects the local high point to the local low point is a point of inection. 49. (a) Sketch the graph of a function that satises the following conditions: for all real x, f (x) > 0, f (x) > 0, f (x) > 0; f (0) = 1. (b) Does there exist a function which satises the conditions: f (x) > 0, f (x) < 0, f (x) < 0 for all real x? Explain. 50. Prove that a polynomial of degree n can have at most n 2 points of inection. Exercises 5154. Use a graphing utility to graph the function on the indicated interval. (a) Estimate the intervals where the graph is concave up and the intervals where it is concave down. (b) Estimate the x-coordinate of each point of inection. Round off your estimates to three decimal places. 51. f (x) = x 4 5x 2 + 3; [4, 4]. 52. f (x) = x sin x; [2, 2]. 53. f (x) = 1 + x 2 2x cos x; [, ]. 54. f (x) = x 2/3 (x 2 4); [5, 5]. Exercises 5558. Use a CAS to determine where: (a) f (x) = 0, (b) f (x) > 0, (c) f (x) < 0. 55. f (x) = 2 cos2 x cos x, 0 x 2. x2 56. f (x) = 4 . x 1 57. f (x) = x 11 4x 9 + 6x 7 4x 5 + x 3 . 58. f (x) = x 16 x 2 . 4.7 VERTICAL AND HORIZONTAL ASYMPTOTES; VERTICAL TANGENTS AND CUSPS Vertical and Horizontal Asymptotes In Figure 4.7.1 you can see the graph of y1 f (x) = x c ffor x close to c.As x c, f (x) ; that is, given any positive number M, there exists a positive number such that if0 < x c < ,thenf (x) M.The line x = c is called a vertical asymptote. Figure 4.7.2 shows the graph of g(x) = 1 x c for x close to c.vertical asymptote x=c cFigure 4.7.1x13:26 220. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006196 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES yyy c x vertical asymptote x=cvertical asymptote x=cvertical asymptote x=cf cxfcgx ggFigure 4.7.2Figure 4.7.3Figure 4.7.4In this case, as x c, g(x) . Again, the line x = c is called a vertical asymptote. Vertical asymptotes can arise from one-sided behavior. With f and g as in Figure 4.7.3, we write as x c ,yf (x) andg(x) .andg(x) .With f and g as in Figure 4.7.4, we write as x c+ ,f (x) In each case the line x = c is a vertical asymptote for both functions.2 4 vertical asymptote x=4Figure 4.7.5xExample 1 The graph of f (x) =x23(x + 2) 3x + 6 = 2x 8 (x + 2)(x 4)has a vertical asymptote at x = 4: as x 4+ , f (x) and as x 4 , f (x) . The vertical line x = 2 is not a vertical asymptote since as x 2, f (x) tends to a nite limit: lim f (x) = lim 3/(x 4) = 1 . (Figure 4.7.5) 2 x2x2From your knowledge of trigonometry you know that as x /2 , tan x and as x /2+ , tan x . Therefore the line x = /2 is a vertical asymptote. In fact, the lines x = (2n + 1)/2, n = 0, 1, 2, . . ., are all vertical asymptotes for the tangent function. (Figure 4.7.6) y x=23 2x= 2x= 2x=Figure 4.7.63 22 x13:26 221. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.7 VERTICAL AND HORIZONTAL ASYMPTOTES; VERTICAL TANGENTS AND CUSPS197The graph of a function can have a horizontal asymptote. Such is the case (see Figures 4.7.7 and 4.7.8) if, as x or as x , f (x) tends to a nite limit. y yhorizontal asymptotey=Lhorizontal asymptotefy=Lf as x , f (x) Lx as x , f (x) LFigure 4.7.7x yFigure 4.7.8Example 2 Figure 4.7.9 shows the graph of the functionhorizontal asymptote y = 1x . f (x) = x 2 vertical asymptote x=2+As x 2 , f (x) ; as x 2 , f (x) . The line x = 2 is a vertical asymptote. As x , f (x) =f (x) =1 x = 1. x 2 1 2/xThe same holds true as x . The line y = 1 is a horizontal asymptote.cos x , xx x2Figure 4.7.9Example 3 Figure 4.7.10 shows the graph of the function f (x) =yx > 0. xAs x 0+ , cos x 1, 1/x , and f (x) =1 cos x = (cos x) x x .The line x = 0 (the y-axis) is a vertical asymptote. As x , f (x) =1 cos x 0 x x and therefore f (x) =xcos x 0. xThe line y = 0 (the x-axis) is a horizontal asymptote. In this case the graph does not stay to one side of the asymptote. Instead, it wiggles about it with oscillations of ever decreasing amplitude. Example 4 Here we examine the behavior of x +1 x x +1 x = . g(x) = 2 x 2x + 1 (x 1)2f (x) =cos x xFigure 4.7.1013:26 222. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006198 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES First, two observations: (a) Because of the presence of x, g is not dened for negative numbers. The domain of g is [0, 1) (1, ). (b) On its domain, g remains positive. As x 1, (x 1)2 0, and g(x) . x + 1 x 1, Thus, the line x = 1 is a vertical asymptote. As x , x +1 x 1 + 1/x 1/ x g(x) = 2 = 0. x 2x + 1 x 2 + 1/x (The numerator tends to 1 and the denominator tends to .) The line y = 0 (the x-axis) is a horizontal asymptote. The behavior of a rational function R(x) =an x n + + a1 x + a0 bk x k + + b1 x + b0(an = 0, bk = 0)as x and as x is readily understood after division of numerator and denominator by the highest power of x that appears in the conguration.Examples (a) For x = 0, set f (x) =x 4 4x 3 1 1/x 4/x 2 1/x 5 . = 2x 5 x 2 1/x 4Both as x and as x , 1/x 4/x 2 1/x 5 0,2 1/x 4 2,andf (x) 0.(b) For x = 0, set f (x) =1 3/x + 1/x 2 x 2 3x + 1 . = 4x 2 1 4 1/x 2Both as x and as x , 1 3/x + 1/x 2 1,4 1/x 2 4,andf (x) 1/4.(c) For x = 0, set f (x) =3x 3 7x 2 + 1 3 7/x + 1/x 3 . = x2 9 1/x 9/x 3Note that for large positive x, f (x) is positive, but for large negative x, f (x) is negative. As x , the numerator tends to 3, the denominator tends to 0, and the quotient, being positive, tends to ; as x , the numerator still tends to 3, the denominator still tends to 0, and the quotient, being negative this time, tends to . Vertical Tangents; Vertical Cusps Suppose that f is a function continuous at x = c. We say that the graph of f has a vertical tangent at the point (c, f (c)) if asx c,f (x) orf (x) .13:26 223. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.7 VERTICAL AND HORIZONTAL ASYMPTOTES; VERTICAL TANGENTS AND CUSPSExamples (Figure 4.7.11) (a) The graph of the cube-root function f (x) = x 1/3 has a vertical tangent at the point (0, 0): asf (x) = 1 x 2/3 . 3x 0,The vertical tangent is the line x = 0 (the y-axis). (b) The graph of the function f (x) = (2 x)1/5 has a vertical tangent at the point (2, 0): x 2,asf (x) = 1 (2 x)4/5 . 5The vertical tangent is the line x = 2. yy vertical tangent x=2vertical tangent x=0 xxf (x) = x1/3f (x) = (2 x)1/5(a)(b)Figure 4.7.11Occasionally you will see a graph tend to the vertical from one side, come to a sharp point, and then virtually double back on itself on the other side. Such a pattern signals the presence of a vertical cusp. Suppose that f is continuous at x = c. We say that the graph of f has a vertical cusp at the point (c, f (c)) if as x tends to c from one side, f (x) and as x tends to c from the other side, f (x) .Examples (Figure 4.7.12) (a) The function f (x) = x 2/3 is continuous at x = 0 and has derivative f (x) = 2 x 1/3 . 3 As x 0+ , f (x) ; as x 0 , f (x) . This tells us that the graph of f has a vertical cusp at the point (0, 0). (b) The function f (x) = 2 (x 1)2/5 is continuous at x = 1 and has derivative f (x) = 2 (x 1)3/5 . As x 1 , f (x) ; as x 1+ , f (x) . The 5 graph has a vertical cusp at the point (1, 2). yy(1, 2)xf (x) = x2/3xf (x) = 2 (x 1)2/5Figure 4.7.1219913:26 224. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006200 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES EXERCISES 4.7 1. The graph of a function f is given in the gure. y1 1f 1xyc f33. f (x) = abxp34. f (x) =dAs x , f (x) ? As x b+ , f (x) ? What are the vertical asymptotes? What are the horizontal asymptotes? Give the numbers c, if any, at which the graph of f has a vertical tangent. (f) Give the numbers c, if any, at which the graph of f has a vertical cusp. Exercises 320. Find the vertical and horizontal asymptotes x3 x . 4. f (x) = . 3. f (x) = 3x 1 x +2 x2 4x 5. f (x) = . 6. f (x) = 2 . x 2 x +1 2x x 7. f (x) = 2 . 8. f (x) = . x 9 4 xx 2x 1 4 + 3x11. f (x) =3x . (2x 5)22.10. f (x) =4x 2 . (3x 1)212. f (x) =x 1 2x3.x 1/3 + 2,x 01 x 1/5 , x > 0; 1 + x, x 0 (4x x 2 )1/3 ,c = 0.x > 0;c = 0.Exercises 3538. Sketch the graph of the function, showing all asymptotes. 1 x +1 . 36. f (x) = . 35. f (x) = x 2 (x + 1)2(a) (b) (c) (d) (e)9. f (x) =14. f (x) =Exercises 2134. Determine whether or not the graph of f has a vertical tangent or a vertical cusp at c. 21. f (x) = (x + 3)4/3 ; c = 3. 22. f (x) = 3 + x 2/5 ; c = 0. 23. f (x) = (2 x)4/5 ; c = 2. 24. f (x) = (x + 1)1/3 ; c = 1. 25. f (x) = 2x 3/5 x 6/5 ; c = 0. 26. f (x) = (x 5)7/5 ; c = 5. 27. f (x) = (x + 2)2/3 ; c = 2. 28. f (x) = 4 (2 x)3/7 ; c = 2. 29. f (x) = x 1 ; c = 1. 30. f (x) = x(x 1)1/3 ; c = 1. 31. f (x) = (x + 8)1/3 ; c = 8. 32. f (x) = 4 x 2 ; c = 2.(a) As x 1, f (x) ? (b) As x 1 , f (x) ? (c) As x 1+ , f (x) ? (d) As x , f (x) ? (e) As x , f (x) ? (f) What are the vertical asymptotes? (g) What are the horizontal asymptotes? 2. The graph of a function f is given in the gure.qx 1/3 . x 2/3 4 2x . 16. f (x) = x2 1 18. f (x) = x x 2. 1 20. f (x) = . sec x 13x . 13. f (x) = 4x 2 + 1 x 15. f (x) = . 2 x x 1 17. f (x) = x + 4 x. sin x . 19. f (x) = sin x 137. f (x) = Cx . 1 + x238. f (x) =x 2 . x 2 5x + 6Exercises 3942. Find (a) the intervals on which f increases and the intervals on which f decreases, and (b) the intervals on which the graph of f is concave up and the intervals on which it is concave down. Also, determine whether the graph of f has any vertical tangents or vertical cusps. Conrm your results with a graphing utility. 40. f (x) = x 2/3 x 1/3 . 39. f (x) = x 3x 1/3 . 42. f (x) = x . 41. f (x) = 3 x 5/3 3x 2/3 . 5 Exercises 4346. Sketch the graph of a function f that satises the given conditions. Indicate whether the graph of f has any horizontal or vertical asymptotes, and whether the graph has any vertical tangents or vertical cusps. If you nd that no function can satisfy all the conditions, explain your reasoning. 43. f (x) 1 for all x, f (0) = 1; f (x) < 0 for all x = 0; f (x) as x 0+ , f (x) as x 0 .13:26 225. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.8 SOME CURVE SKETCHING44. f (0) = 0, f (3) = f (3) = 0; f (x) as x 1, f (x) as x 1, f (x) 1 as x , f (x) 1 as x ; f (x) < 0 for all x = 1. 45. f (0) = 0; f (x) 1 as x , f (x) 1 as x ; f (x) as x 0; f (x) < 0 for x < 0, f (x) > 0 for x > 0; f is an odd function. 46. f (0) = 1; f (x) 4 as x , f (x) as x ; f (x) as x 0; f (x) > 0 for x < 0, f (x) < 0 for x > 0. 47. Let p and q be positive integers, q odd, p < q. Let f (x) = x p/q . Specify conditions on p and q so that (a) the graph of f has a vertical tangent at (0, 0). (b) the graph of f has a vertical cusp at (0, 0). 48. (Oblique asymptotes) Let r (x) = p(x)/q(x) be a rational function. If (degree of p) = (degree of q) + 1, then r can be written in the form Q(x) r (x) = ax + b + with (degree Q) < (degree q). q(x) Show that [r (x) (ax + b)] 0 both as x and as x . Thus the graph of f approaches the line y =201ax + b both as x and as x . The line y = ax + b is called an oblique asymptote. Exercises 4952. Sketch the graph of the function showing all vertical and oblique asymptotes. x2 4 2x 2 + 3x 2 49. f (x) = . 50. f (x) = . x x +1 1 + x 3x 2 x3 . . 52. f (x) = (x 1)2 x Exercises 5354. Use a CAS to nd the oblique asymptotes. Then use a graphing utility to draw the graph of f and its asymptotes, and thereby conrm your ndings. 3x 4 4x 3 2x 2 + 2x + 2 . 53. f (x) = x3 x 5x 3 3x 2 + 4x 4 54. f (x) = x2 + 1 Exercises 5556. Use a graphing utility to determine whether or not the graph of f has a horizontal asymptote. Conrm your ndings analytically. 56. f (x) = x 4 x 2 x 2 . 55. f (x) = x 2 + 2x x.51. f (x) = CC 4.8 SOME CURVE SKETCHING During the course of the last few sections you have seen how to nd the extreme values of a function, the intervals on which a function increases, and the intervals on which it decreases; how to determine the concavity of a graph and how to nd the points of inection; and, nally, how to determine the asymptotic properties of a graph. This information enables us to sketch a pretty accurate graph without having to plot point after point after point. Before attempting to sketch the graph of a function, we try to gather together the information available to us and record it in an organized form. Here is an outline of the procedure we will follow to sketch the graph of a function f. (1) Domain Determine the domain of f ; identify endpoints; nd the vertical asymptotes; determine the behavior of f as x and as x . (2) Intercepts Determine the x- and y-intercepts of the graph. [The y-intercept is the value f (0); the x-intercepts are the solutions of the equation f (x) = 0.] (3) Symmetry/periodicity If f is an even function [ f (x) = f (x)], then the graph of f is symmetric about the y-axis; if f is an odd function [ f (x) = f (x)], then the graph of f is symmetric about the origin. If f is periodic with period p, then the graph of f replicates itself on intervals of length p. (4) First derivative Calculate f . Determine the critical points; examine the sign of f to determine the intervals on which f increases and the intervals on which f decreases; determine the vertical tangents and cusps. (5) Second derivative Calculate f . Examine the sign of f to determine the intervals on which the graph is concave up and the intervals on which the graph is concave down; determine the points of inection. (6) Points of interest and preliminary sketch Plot the points of interest in a preliminary sketch: intercept points, extreme points (local extreme points, absolute extreme points, endpoint extreme points), and points of inection. (7) The graph Sketch the graph of f by connecting the points in a preliminary sketch, making sure that the curve rises, falls, and bends in the proper way. You may wish to verify your sketch by using a graphing utility.13:26 226. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006202 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES Figure 4.8.1 gives some examples of elements to be included in a preliminary sketch yyy(2, 4)(3, 2) (4, 1) xxlocal maximum: (2, 4)xpoint of inflection: (3, 2)endpoint minimum: (4, 1)Figure 4.8.1Example 1 Sketch the graph of f (x) = 1 x 4 2x 2 + 7 . 4 4 SOLUTION(1) Domain This is a polynomial function; so its domain is the set of all real numbers. Since the leading term is 1 x 4 , f (x) both as x and as x . There 4 are no asymptotes. (2) Intercepts The y-intercept is f (0) = 7 . To nd the x-intercepts, we solve the 4 equation f (x) = 0: 1 4 x 4 2x 2 +7 4= 0.x 4 8x 2 + 7 = 0, (x 2 1)(x 2 7) = 0, (x + 1)(x 1)(x + 7)(x 7) = 0, The x-intercepts are x = 1 and x = 7. (3) Symmetry/periodicity Since f (x) = 1 (x)4 2(x 2 ) + 47 4= 1 x 4 2x 2 + 47 4= f (x),f is an even function, and its graph is symmetric about the y-axis; f is not a periodic function. (4) First derivative f (x) = x 3 4x = x(x 2 4) = x(x + 2)(x 2). The critical points are x = 0, x = 2. The sign of f and behavior of f : sign of f ': 0 + + + + + + + + + + 0 0 + + + + + + + + + + behavior of f :decreases2 local minimumincreases0decreaseslocal maximum2 local minimum(5) Second derivative 2 f (x) = 3x 2 4 = 3 x 32 x+ . 3increasesx13:26 227. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.8 SOME CURVE SKETCHINGThe sign of f and the concavity of the graph of f : sign of f ': + + + + + + + + + + + + + + + 0 0 + + + + + + + + + + + + + concavity:2concave up3point of inflectionconcave down2xconcave up3point of inflection(6) Points of interest and preliminary sketch (Figure 4.8.2) (0, 7 ) : 4 (1, 0), (1, 0), ( 7, 0), ( 7, 0) : (0, 7 ) 4 (2, 9 ), (2, 9 ) 4 4y-intercept point. x-intercept points.:local maximum point.: (2/ 3, 17/36), (2/ 3, 17/36) : local and absolute minimum points. points of inection.(7) The graph Since the graph is symmetric about the y-axis, we can sketch the graph for x 0, and then obtain the graph for x 0 by a reection in the y-axis. See Figure 4.8.3. y2y2( ) 7 40,(2/3, 13 2 (7, 0)(2 3, )1)217 36(2, 2(1, 0) (1, 0)9 4(2 3, (3 (7, 0)17 362, x32117 ) 36)Figure 4.8.27 ) 4(2/3, (1, 0) (1, 0)17 ) 36231)9 4(0,9 42(2, ) f (x) =1 4x 4 2x 2 +7 49 4(2, )Figure 4.8.3Example 2 Sketch the graph of f (x) = x 4 4x 3 + 1, 1 x < 5. SOLUTION(1) Domain The domain is [1, 5); 1 is the left endpoint, and 5 is a missing right endpoint. There are no asymptotes. We do not consider the behavior of f as x since f is dened only on [1, 5). (2) Intercepts The y-intercept is f (0) = 1. To nd the x-intercepts, we must solve the equation x 4 4x 3 + 1 = 0. We cannot do this exactly, but we can verify that f (0) > 0 and f (1) < 0, and that f (3) < 0 and f (4) > 0. Thus there are x-intercepts in the interval (0, 1) and in the interval (3, 4). We could nd approximate values for these intercepts, but we wont stop to do this since our aim here is a sketch of the graph, not a detailed drawing. (3) Symmetry/periodicity The graph is not symmetric about the y-axis: f (x) = f (x). It is not symmetric about the origin: f (x) = f (x). The function is not periodic.x20313:26 228. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006204 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES (4) First derivative For x (1, 5) f (x) = 4x 3 12x 2 = 4x 2 (x 3). The critical points are x = 0 and x = 3. sign of f ':00+ + + + + + + + + + + + +behavior of f : 1 decreases 0 endpoint maximum3decreasesno extremelocal minimumx5increasesmissing endpoint(5) Second derivative. f (x) = 12x 2 24x = 12x(x 2). The sign of f and the concavity of the graph of f : sign of f ':y++++++0 0 + + + + + + + + + + + + + + + + + + +concavity: 1 concave 0 up point of inflection(5, 126)concave down2 point of inflectionconcave up5x(6) Points of interest and preliminary sketch (Figure 4.8.4) (1, 6)(0, 1) x (2, 15)(3, 26)(1, 6)y-intercept point; point of inection with horizontal tangent. endpoint maximum point. point of inection. local and absolute minimum point.As x approaches the missing endpoint 5 from the left, f (x) increases toward a value of 126. (7) The graph Since the range of f makes a scale drawing impractical, we must be content with a rough sketch as in Figure 4.8.5. In cases like this, it is particularly important to give the coordinates of the points of interest. Figure 4.8.4y(0, 1) : (1, 6) : (2, 15) : (3, 26) :(5, 126)Example 3 Sketch the graph of f (x) =x2 3 . x3SOLUTION (0, 1) x (2, 15) (3, 26)Figure 4.8.5(1) Domain The domain of f consists of all x = 0, the set (, 0) (0, ). The y-axis (the line x = 0) is a vertical asymptote: f (x) as x 0 and f (x) as x 0+ . The x-axis (the line y = 0) is a horizontal asymptote: f (x) 0 both as x and as x . (2) Intercepts. There is no y-intercept since f is not dened at x = 0. The x-intercepts are x = 3. (3) Symmetry Since f (x) =x2 3 (x)2 3 = = f (x), (x)3 x3the graph is symmetric about the origin; f is not periodic. (4) First derivative. It is easier to calculate f if we rst rewrite f (x) using negative exponents: f (x) =x2 3 = x 1 3x 3 x313:26 229. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.8 SOME CURVE SKETCHINGgives 9 x2 . x4 The critical points of f are x = 3. NOTE: x = 0 is not a critical point since 0 is not in the domain of f. The sign of f and the behavior of f : f (x) = x 2 + 9x 4 =sign of f ': behavior of f: 0 + + + + + + + + + + + + dne + + + + + + + + + + + + 0 decreases3 local minimumincreases0increases3 decreases local maximumx(5) Second derivative 2(x 2 18) 2(x 3 2)(x + 3 2) f (x) = 2x 36x = = . x5 x5 The sign of f and the concavity of the graph of f : 3sign of f ':5 0 + + + + + + + + + + + + dne 0 + + + + + + + +concavity: concave 32 down point of inflectionconcave upconcave down032 point of inflectionxconcave up(6) Points of interest and preliminary sketch (Figure 4.8.6) ( 3, 0), ( 3, 0) : x-intercept points. (3, 2/9) : local minimum point. (3, 2/9) : local maximum point. points of inection. (3 2, 5 2/36), (3 2, 5 2/36) : (7) The graphSee Figure 4.8.7.yy( ( ) 3,0.24 3 2 1(32,) (3, 2 932,52 36)( 3, ) 2 90.1352 362 91 2 3)34x3(32, Figure 4.8.652 36)(3, 2 93)Figure 4.8.7Example 4 Sketch the graph of f (x) = 3 x 5/3 3x 2/3 . 5 SOLUTION1. Domain The domain of f is the set of real numbers. Since we can express f (x) as 3 2/3 x (x 5), we see that, as x , f (x) , and as x , f (x) . 5 There are no asymptotes.(32,52 36) x20513:26 230. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006206 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 2. Intercepts Since f (0) = 0, the graph passes through the origin. Thus x = 0 is an x-intercept and y = 0 is the y-intercept; x = 5 is also an x-intercept. 3. Symmetry/periodicity There is no symmetry; f is not periodic. 4. First derivative x 2 f (x) = x 2/3 2x 1/3 = 1/3 . x The critical points are x = 0 and x = 2. The sign of f and the behavior of f : + + + + + dne 0 + + + + + + + +sign of f ':behavior of f : increases0 local maximumx2 increases local minimumdecreasesNote that, as x 0 , f (x) , and as x 0+ , f (x) . Since f is continuous at x = 0, and f (0) = 0, the graph of f has a vertical cusp at (0, 0). 5. Second derivative f (x) = 2 x 1/3 + 2 x 4/3 = 2 x 4/3 (x + 1). 3 3 3 The sign of f and the concavity of the graph of f : sign of f ': 0 + + + + + + + + + + + + + + + + + + + + + + + + + dne + + + + + + +concavity:concave down1 point of inflection0concave upxconcave up6. Points of interest and preliminary sketch (Figure 4.8.8) (0, 0): y-intercept point, local maximum point; vertical cusp. (0, 0), (5, 0): x-intercepts points. (2, 9 3 4/5): local minimum point, f (2) 2.9. = (1, 18/5) : point of inection. See Figure 4.8.9. 7. The graph yy1(1, 3.6)5(2, 2.9)Figure 4.8.8x5(1, 3.6)x(2, 2.9)Figure 4.8.9Example 5 Sketch the graph of f (x) = sin 2x 2 sin x. SOLUTION(1) Domain The domain of f is the set of all real numbers. There are no asymptotes and, as you can verify, the graph of f oscillates between 3 3 and 3 3 both as 2 2 x and as x .13:26 231. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.8 SOME CURVE SKETCHING(2) Intercepts2072 3The y-intercept is f (0) = 0. To nd the x-intercepts, we set f (x) = 0: sin 2x 2 sin x = 2 sin x cos x 2 sin x = 2 sin x(cos x 1) = 0.Since sin x = 0 at all integral multiples of and cos x = 1 at all integral multiples of 2 , the x-intercepts are the integral multiples of : all x = n . (3) Symmetry/periodicity Since the sine is an odd function, f (x) = sin(2x) 2 sin(x) = sin 2x + 2 sin x = f (x). Thus, f is an odd function and the graph is symmetric about the origin. Also, f is periodic with period 2 . On the basis of these two properties, it would be sufcient to sketch the graph of f on the interval [0, ]. The result could then be extended to the interval [, 0] using the symmetry, and then to (, ) using the periodicity. However, for purposes of illustration here, we will consider f on the interval [, ]. (4) First derivative f (x) = 2 cos 2x 2 cos x = 2(2 cos2 x 1) 2 cos x = 4 cos2 x 2 cos x 2 = 2(2 cos x + 1)(cos x 1). The critical points in [, ] are x = 2/3, x = 0, x = 2/3. sign of f ': behavior of f : +++++++++000+++++++++0 increasesdecreases2 30decreaseslocal maximum2 3increasesxlocal minimum(5) Second derivative f (x) = 4 sin 2x + 2 sin x = 8 sin x cos x + 2 sin x = 2 sin x(4 cos x + 1). f (x) = 0 at x = , 0, , and at the numbers x in [, ] where cos x = 1 , 4 which are approximately 1.3. The sign of f and the concavity of the graph on [, ]: sign of f ': 0 0 + + + + + 0 0 + + + + + + + + + + + + + 0 concavity: 1.3 concave down01.3concave concave up down point of point of point of inflection inflection inflectionconcave upx(6) Points of interest and preliminary sketch (Figure 4.8.10) (0, 0): y-intercept point. (, 0), (0, 0), (, 0): x-intercept points; (0, 0) is also a point of inection. ( 2 , 3 3): local and absolute maximum point; 3 3 2.6. = 3 2 2 2 3 3 ( 3 , 2 3): local and absolute minimum point; 2 3 2.6. = (1.3, 1.4), (1.3, 1.4): points of inection (approximately).(3 2 , 2 3)y3(1.3, 1.4) 2 1.3 31.3(1.3, 1.4)( , 3) 2 3Figure 4.8.103 2x13:26 232. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006208 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES(3 2 , 2 3)y3y (1.3, 1.4) 22 33x 22x(1.3, 1.4)( , 3) 2 33 2Figure 4.8.11Figure 4.8.12(7) The graph The graph of f on the interval [, ] is shown in Figure 4.8.11. An indication of the complete graph is given in Figure 4.8.12. EXERCISES 4.8 Exercises 154. Sketch the graph of the function using the approach presented in this section. 1. 3. 4. 5. 6. 7.2. f (x) = 1 (x 2)2 . f (x) = (x 2)2 . f (x) = x 3 2x 2 + x + 1. f (x) = x 3 9x 2 + 24x 7. f (x) = x 3 + 6x 2 , x [4, 4]. f (x) = x 4 8x 2 , x (0, ). f (x) = 2 x 3 1 x 2 10x 1. 3 28. f (x) = x(x 2 + 4)2 . 2 1 9. f (x) = x 2 + . 10. f (x) = x . x x x +2 x 4 . 12. f (x) = . 11. f (x) = x2 x3 13. f (x) = 2 x x, x [0, 4]. 14. f (x) = 1 x x, x [0, 9]. 4 15. f (x) = 2 + (x + 1)6/5 . 16. f (x) = 2 + (x + 1)7/5 . 18. f (x) = 3x 4 + 4x 3 . 17. f (x) = 3x 5 + 5x 3 . 5/3 20. f (x) = 1 + (x 2)4/3 . 19. f (x) = 1 + (x 2) . 2x 2 x2 . 22. f (x) = . 21. f (x) = 2 x +4 x +1 x x 23. f (x) = . 24. f (x) = 2 . (x + 3)2 x +1 1 x2 . 26. f (x) = 3 . 25. f (x) = 2 x 4 x x 27. f (x) = x 1 x. 28. f (x) = (x 1)4 2(x 1)2 . 29. f (x) = x + sin 2x, x [0, ]. 30. f (x) = cos3 x + 6 cos x, x [0, ]. 31. f (x) = cos4 x, x [0, ]. f (x) = 3x cos 2x, x [0, ]. f (x) = 2 sin3 x + 3 sin x, x [0, ]. f (x) = sin4 x, x [0, ]. f (x) = (x + 1)3 3(x + 1)2 + 3(x + 1). 37. f (x) = x 2 (5 x)3 . f (x) = x 3 (x + 5)2 . 39. f (x) = 3 x 2 1 . f (x) = 4 2x x 2 . 1/3 41. f (x) = x(x 1)1/5 . f (x) = x x . 43. f (x) = x 2 6x 1/3 . f (x) = x 2 (x 7)1/3 . x 2x . 45. f (x) = . 44. f (x) = x 2 x2 + 132. 33. 34. 35. 36. 38. 40. 42.x x2 . . 47. f (x) = x +4 x2 2 48. f (x) = 3 cos 4x, x [0, ]. 49. f (x) = 2 sin 3x, x [0, ]. 50. f (x) = 3 + 2 cot x + csc2 x, x (0, 1 ). 246. f (x) =51. f (x) = 2 tan x sec2 x,x (0, 1 ). 252. f (x) = 2 cos x + sin2 x. sin x 53. f (x) = , x (, ). 1 sin x 1 54. f (x) = , x (, ). 1 cos x, 55. Given: f is everywhere continuous, f is differentiable at all x = 0, f (0) = 0, and the graph of f is as indicated below. y f'531123x13:26 233. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.9 VELOCITY AND ACCELERATION; SPEED(a) Determine the intervals on which f increases and the intervals on which it decreases; nd the critical points of f. (b) Sketch the graph of f ; determine the intervals on which the graph of f is concave up and those on which it is concave down. (c) Sketch the graph of f. 56. Set F(x) =sin(1/x), 0,x =0 x = 0,G(x) =x sin(1/x), 0,H (x) =x 2 sin(1/x), 0,x =0 x = 0, x =0 x = 0.(a) Sketch a gure that shows the general nature of the graph of F. (b) Sketch a gure that shows the general nature of the graph of G. (c) Sketch a gure that shows the general nature of the graph of H. (d) Which of these functions is continuous at 0? (e) Which of these functions is differentiable at 0?f (x) =y2 x2 2 = 1. 2 a b (a) Draw a gure that illustrates this asymptotic behavior. (b) Show that the rst-quadrant arc of the hyperbola, the curve b y= x 2 a2, a is indeed asymptotic to the line y = (b/a)x by showing that b b x 2 a 2 x 0 as x . a a (c) Proceeding as in part (b), show that the second-quadrant arc of the hyperbola is asymptotic to the line y = (b/a)x by taking a suitable limit as x . (The asymptotic behavior in the other quadrants can be veried in an analogous manner, or by appealing to symmetry.)Suppose that an object (some solid object) moves along a straight line. On the line of motion we choose a point of reference, a positive direction, a negative direction, and a unit distance. This gives us a coordinate system by which we can indicate the position of the object at any given time. Using this coordinate system, we denote by x(t) the position of the object at time t. There is no loss in generality in taking the line of motion as the x-axis. We can arrange this by choosing a suitable frame of reference. You have seen that the derivative of a function gives the rate of change of that function at the point of evaluation. Thus, if x(t) gives the position of the object at time t and the position function is differentiable, then the derivative x (t) gives the rate of change of the position function at time t. We call this the velocity at time t and denote it by v(t). In symbols,v(t) = x (t).Velocity at a particular time t (some call it instantaneous velocity at time t) can be obtained as the limit of average velocities. At time t the object is at x(t) and at time t + h it is at x(t + h). If h > 0, then [t, t + h] is a time interval and the quotient x(t + h) x(t) x(t + h) x(t) = (t + h) t h 209x 3 x 1/3 . Show that f (x) x 2 0 as x x . This says that the graph of f is asymptotic to the parabola y = x 2 . Sketch the graph of f and feature this asymptotic behavior. 58. The lines y = (b/a)x and y = (b/a)x are called asymptotes of the hyperbola 57. Set 4.9 VELOCITY AND ACCELERATION; SPEED(4.9.1)If the object is larger than a point mass, we can choose a spot on the object and view the location of that spot as the position of the object. In a course in physics an object is usually located by the position of its center of mass. (Section 17.6.)13:26 234. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006210 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES gives the average velocity during this time interval. If h < 0, then [t + h, t] is a time interval and the quotient x(t) x(t + h) x(t) x(t + h) = , t (t + h) h which also can be written x(t + h) x(t) , h gives the average velocity during this time interval. Thus, whether h is positive or negative, the difference quotient x(t + h) x(t) h gives the average velocity of the object during the time interval of length h that begins or ends at t. The statement v(t) = x (t) = limh0x(t + h) x(t) hexpresses the velocity at time t as the limit as h 0 of these average velocities. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,a(t) = v (t) = x (t).(4.9.2)In the Leibniz notation,(4.9.3)v=dx dtanda=dv d2x = 2. dt dtThe magnitude of the velocity, by which we mean the absolute value of the velocity, is called the speed of the object:(4.9.4)speed at time t = (t) = v(t) .The four notions that we have just introduced position, velocity, acceleration, speed provide the framework for the description of all straight-line motion. The following observations exploit the connections that exist between these fundamental notions: (1) Positive velocity indicates motion in the positive direction (x is increasing). Negative velocity indicates motion in the negative direction (x is decreasing). (2) Positive acceleration indicates increasing velocity (increasing speed in the positive direction, decreasing speed in the negative direction). Negative acceleration indicates decreasing velocity (decreasing speed in the positive direction, increasing speed in the negative direction). Extended by vector methods (Chapter 14), these four notions provide the framework for the description of all motion.13:26 235. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.9 VELOCITY AND ACCELERATION; SPEED(3) If the velocity and acceleration have the same sign, the object is speeding up, but if the velocity and acceleration have opposite signs, the object is slowing down. Note that v = x . If v > 0, then x > 0 and x increases. If v < 0, then x < 0 and x decreases. PROOF OF (1)PROOF OF (2)Note that =v, v,in the positive direction in the negative direction.Suppose that a > 0. Then v increases. In the positive direction, = v and therefore increases; in the negative direction, = v and therefore decreases. Suppose that a < 0. Then v decreases. In the positive direction, = v and therefore decreases; in the negative direction, = v and therefore increases. PROOF OF (3)Note that v2 = 2andd 2 (v ) = 2vv = 2va. dtIf v and a have the same sign, then va > 0 and v 2 = 2 increases. Therefore increases, which means the object is speeding up. If v and a have opposite sign, then va < 0 and v 2 = 2 decreases. Therefore decreases, which means the object is slowing down. Example 1 An object moves along the x-axis; its position at each time t given by the function x(t) = t 3 12t 2 + 36t 27. Lets study the motion from time t = 0 to time t = 9. The object starts out at 27 units to the left of the origin: x(0) = 03 12(0)2 + 36(0) 27 = 27 and ends up 54 units to the right of the origin: x(9) = 93 12(9)2 + 36(9) 27 = 54. We nd the velocity function by differentiating the position function: v(t) = x (t) = 3t 2 24t + 36 = 3(t 2)(t 6). We leave it to you to verify that positive for 0 t < 2 0, at t = 2 v(t) is negative, for 2 < t < 6 0, at t = 6 positive, for 6 < t 9. We can interpret all this as follows: the object begins by moving to the right [v(t) is positive for 0 t < 2]; it comes to a stop at time t = 2[v(2) = 0]; it then moves left [v(t) is negative for 2 < t < 6]; it stops at time t = 6[v(6) = 0]; it then moves right and keeps going right [v(t) > 0 for 6 < t 9].21113:26 236. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006212 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES We nd the acceleration by differentiating the velocity: a(t) = v (t) = 6t 24 = 6(t 4). We note that negative, for 0 t < 4 0, at t = 4 a(t) is positive, for 4 < t 9.At the beginning the velocity decreases, reaching a minimum at time t = 4. Then the velocity starts to increase and continues to increase. Figure 4.9.1 shows a diagram for the sign of the velocity and a corresponding diagram for the sign of the acceleration. Combining the two diagrams, we have a brief description of the motion in convenient form. The direction of the motion at each time t [0, 9] is represented schematically in Figure 4.9.2. sign of v(t): + + + + + 0 0 + + + + + + + + + + 206object moving rightobject moving left9tobject moving rightsign of a(t): 0 + + + + + + + + + + + + + + + 9t940tvelocity increasingvelocity decreasingsign of v(t):++sign of a(t):++0 motion of object2 to the right, slowing down4 to the left, speeding up6 to the left, slowing downto the right, speeding upFigure 4.9.1 t=9t=6 t=2t=0x(t) 54270554x(t)Figure 4.9.25 224627Figure 4.9.39 tAnother way to represent the motion is to graph x as a function of t, as we do in Figure 4.9.3. The velocity v(t) = x (t) then appears as the slope of the curve. From the gure, we see that we have positive velocity from t = 0 up to t = 2, zero velocity at time t = 2, then negative velocity up to t = 6, zero velocity at t = 6, then positive velocity to t = 9. The acceleration a(t) = v (t) can be read from the concavity of the curve. Where the graph is concave down (from t = 0 to t = 4), the velocity decreases; where the graph is concave up (from t = 4 to t = 9), the velocity increases. The speed is reected by the steepness of the curve. The speed decreases from t = 0 to t = 2, increases from t = 2 to t = 4, decreases from t = 4 to t = 6, increases from t = 6 to t = 9. 13:26 237. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.9 VELOCITY AND ACCELERATION; SPEEDA few words about units. The units of velocity, speed, and acceleration depend on the units used to measure distance and the units used to measure time. The units of velocity are units of distance per unit time: feet per second,meters per second,miles per hour,and so forth.The units of acceleration are units of distance per unit time per unit time: feet per second per second, miles per hour per hour,meters per second per second, and so forth.Free Fall Near the Surface of the Earth (In what follows, the line of motion is clearly vertical. So, instead of writing x(t) to indicate position, well follow custom and write y(t). Velocity is then y (t), acceleration is y (t), and speed is y (t) .) Imagine an object (for example, a rock or an apple) falling to the ground. (Figure 4.9.4.) We will assume that the object is in free fall; namely, that the gravitational pull on the object is constant throughout the fall and that there is no air resistance. Galileos formula for the free fall gives the height of the object at each time t of the fall:(4.9.5)y(t) = 1 gt 2 + v0 t + y0 2where g is a positive constant the value of which depends on the units used to measure time and the units used to measure distance. Lets examine this formula. First, the point of reference is at ground level and the positive y direction is up. Next, since y(0) = y0 , the constant y0 represents the height of the object at time t = 0. This is called the initial position. Differentiation gives y (t) = gt + v0 . Since y (0) = v0 , the constant v0 gives the velocity of the object at time t = 0. This is called the initial velocity. A second differentiation gives y (t) = g. This indicates that the object falls with constant negative acceleration g. The constant g is a gravitational constant. If time is measured in seconds and distance in feet, then g is approximately 32 feet per second per second ; if time is measured in seconds and distance in meters, then g is approximately 9.8 meters per In practice, neither of these conditions is ever fully met. Gravitational attraction near the surface of the earth does vary somewhat with altitude, and there is always some air resistance. Nevertheless, in the setting in which we will be working, the results that we obtain are good approximations of the actual motion. Galileo Galilei (15641642), a great Italian astronomer and mathematician, is popularly known today for his early experiments with falling objects. His astronomical observations led him to support the Copernican view of the solar system. For this he was brought before the Inquisition. The value of this constant varies slightly with latitude and elevation. It is approximately 32 feet per second per second at the equator, elevation zero. In Greenland it is about 32.23.yFigure 4.9.421313:26 238. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006214 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES second per second. In making numerical calculations, we will take g as 32 feet per second per second or as 9.8 meters per second per second. Equation 4.9.5 then reads y(t) = 16t 2 + v0 t + y0(distance in feet)or y(t) = 4.9t 2 + v0 t + y0 .(distance in meters)Example 2 A stone is dropped from a height of 98 meters. In how many seconds does it hit the ground? What is the speed at impact? SOLUTION Here y0 = 98 and v0 = 0. Consequently, we havey(t) = 4.9t 2 + 98. To nd the time t at impact, we set y(t) = 0. This gives t = 20 = 2 5. We disregard the negative value and conclude that it takes 2 5 4.47 seconds for the = stone to hit the ground. The velocity at impact is the velocity at time t = 2 5. Since 4.9t 2 + 98 = 0,t 2 = 20,v(t) = y (t) = 9.8t, we have v(2 5) = (19.6) 5 43.83. =The speed at impact is about 43.83 meters per second. Example 3 An explosion causes some debris to rise vertically with an initial velocity of 72 feet per second. (a) In how many seconds does this debris attain maximum height? (b) What is this maximum height? (c) What is the speed of the debris as it reaches a height of 32 feet (i) going up? (ii) coming back down? SOLUTION Since we are measuring distances in feet, the basic equation readsy(t) = 16t 2 + v0 t + y0 . Here y0 = 0 (it starts at ground level) and v0 = 72 (the initial velocity is 72 feet per second). The equation of motion is therefore y(t) = 16t 2 + 72t. Differentiation gives v(t) = y (t) = 32t + 72. The maximum height is attained when the velocity is 0. This occurs at time t = Since y( 9 ) = 81, the maximum height attained is 81 feet. 4 To answer part (c), we must nd those times t for which y(t) = 32. Since y(t) = 16t 2 + 72t,72 32= 9. 413:26 239. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.9 VELOCITY AND ACCELERATION; SPEED215the condition y(t) = 32 yields 16t 2 + 72t = 32, which simplies to 16t 2 72t + 32 = 0. This quadratic has two solutions, t = 1 and t = 4. Since v( 1 ) = 56 and v(4) = 56, 2 2 the velocity going up is 56 feet per second and the velocity coming down is 56 feet per second. In each case the speed is 56 feet per second. EXERCISES 4.9 Exercises 16. An object moves along a coordinate line, its position at each time t 0 given by x(t). Find the position, velocity, and acceleration at time t0 . What is the speed at time t0 ?Exercises 2128. An object moves along the x-axis, its position at each time t 0 given by x(t). Determine the time interval(s), if any, during which the object satises the given condition. 21. x(t) = t 4 12t 3 + 28t 2 ; moves right. 22. x(t) = t 3 12t 2 + 21t; moves left. 23. x(t) = 5t 4 t 5 ; speeds up. 24. x(t) = 6t 2 t 4 ; slows down. 25. x(t) = t 3 6t 2 15t; moves left slowing down. 26. x(t) = t 3 6t 2 15t; moves right slowing down. 27. x(t) = t 4 8t 3 16t 2 ; moves right speeding up. 28. x(t) = t 4 8t 3 16t 2 ; moves left speeding up.1. x(t) = 4 + 3t t 2 ; t0 = 5 2. x(t) = 5t t 3 ; t0 = 3. 2t 18 4. x(t) = ; t0 = 1. ; t0 = 3. 3. x(t) = t +2 t +3 2 2 5. x(t) = (t + 5t)(t + t 2); t0 = 1. 6. x(t) = (t 2 3t)(t 2 + 3t); t0 = 2. Exercises 710. An object moves along the x-axis, its position at each time t 0 given by x(t). Determine the times, if any, at which (a) the velocity is zero, (b) the acceleration is zero. 7. x(t) = 5t + 1. 8. x(t) = 4t 2 t + 3. 9. x(t) = t 3 6t 2 + 9t 1. 10. x(t) = t 4 4t 3 + 4t 2 + 2. Exercises 1120. Objects A, B, C move along the x-axis. Their positions x(t) from time t = 0 to time t = t3 have been graphed in the gure as functions of t.Exercises 2932. An object moves along a coordinate line, its position at each time t 0 being given by x(t). Find the times t at which the object changes direction. 30. x(t) = t(t 8)3 . 29. x(t) = (t + 1)2 (t 9)3 . 32. x(t) = (t 2 8t + 15)3 . 31. x(t) = (t 3 12t)4 .x(t) object Cobject At111. 12. 13. 14. 15. 16. 17. 18. 19. 20.t2t3 object BWhich object begins farthest to the right? Which object nishes farthest to the right? Which object has the greatest speed at time t1 ? Which object maintains the same direction during the time interval [t1 , t3 ]? Which object begins moving left? Which object nishes moving left? Which object changes direction at time t2 ? Which object speeds up throughout the time interval [0, t1 ]? Which objects slow down during the time interval [t1 , t2 ]? Which object changes direction during the time interval [t2 , t3 ].tExercises 3338. An object moves along the x-axis, its position at each time t given by x(t). Determine those times from t = 0 to t = 2 at which the object is moving to the right with increasing speed. 33. x(t) = sin 3t. 34. x(t) = cos 2t. 35. x(t) = sin t cos t. 36. x(t) = sin t + cos t. 37. x(t) = t + 2 cos t. 38. x(t) = t 2 sin t. In Exercises 3952, neglect air resistance. For the numerical calculations, take g as 32 feet per second per second or as 9.8 meters per second per second. 39. An object is dropped and hits the ground 6 seconds later. From what height, in feet, was it dropped? 40. Supplies are dropped from a stationary helicopter and seconds later hit the ground at 98 meters per second. How high was the helicopter? 41. An object is projected vertically upward from ground level with velocity v. Find the height in meters attained by the object. 42. An object projected vertically upward from ground level returns to earth in 8 seconds. Give the initial velocity in feet per second. 43. An object projected vertically upward passes every height less than the maximum twice, once on the way up and once on the way down. Show that the speed is the same in each direction. Measure height in feet.13:26 240. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006216 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 44. An object is projected vertically upward from the ground. Show that it takes the object the same amount of time to reach its maximum height as it takes for it to drop from that height back to the ground. Measure height in meters. 45. A rubber ball is thrown straight down from a height of 224 feet at a speed of 80 feet per second. If the ball always rebounds with one-fourth of its impact speed, what will be the speed of the ball the third time it hits the ground? 46. A ball is thrown straight up from ground level. How high will the ball go if it reaches a height of 64 feet in 2 seconds? 47. A stone is thrown upward from ground level. The initial speed is 32 feet per second. (a) In how many seconds will the stone hit the ground? (b) How high will it go? (c) With what minimum speed should the stone be thrown so as to reach a height of at least 36 feet? 48. To estimate the height of a bridge, a man drops a stone into the water below. How high is the bridge (a) if the stone hits the water 3 seconds later? (b) if the man hears the splash 3 seconds later? (Use 1080 feet per second as the speed of sound.) 49. A falling stone is at a certain instant 100 feet above the ground. Two seconds later it is only 16 feet above the ground. (a) From what height was it dropped? (b) If it was thrown down with an initial speed of 5 feet per second, from what height was it thrown? (c) If it was thrown upward with an initial speed of 10 feet per second, from what height was it thrown? 50. A rubber ball is thrown straight down from a height of 4 feet. If the ball rebounds with one-half of its impact speed and returns exactly to its original height before falling again, how fast was it thrown originally? 51. Ballast dropped from a balloon that was rising at the rate of 5 feet per second reached the ground in 8 seconds. How high was the balloon when the ballast was dropped? 52. Had the balloon of Exercise 51 been falling at the rate of 5 feet per second, how long would it have taken for the ballast to reach the ground? 53. Two race horses start a race at the same time and nish in a tie. Prove that there must have been at least one time t during the race at which the two horses had exactly the same speed. 54. Suppose that the two horses of Exercise 53 cross the nish line together at the same speed. Show that they had the same acceleration at some instant during the race. 55. A certain tollroad is 120 miles long and the speed limit is 65 miles per hour. If a drivers entry ticket at one end of the tollroad is stamped 12 noon and she exits at the other end at 1:40 p.m., should she be given a speeding ticket? Explain. 56. At 1:00 p.m. a cars speedometer reads 30 miles per hour and at 1:15 p.m. it reads 60 miles per hour. Prove that the cars acceleration was exactly 120 miles per hour per hour at least once between 1:00 and 1:15. 57. A car is stationary at a toll booth. Twenty minutes later, at a point 20 miles down the road, the car is clocked at 60 mph. Explain how you know that the car must have exceeded the60-mph speed limit some time before being clocked at 60 mph. 58. The results of an investigation of a car accident showed that the driver applied his brakes and skidded 280 feet in 6 seconds. If the speed limit on the street where the accident occurred was 30 miles per hour, was the driver exceeding the speed limit at the instant he applied his brakes? Explain. HINT: 30 miles per hour = 44 feet per second. 59. (Simple harmonic motion) A bob suspended from a spring oscillates up and down about an equilibrium point, its vertical position at time t given by y(t) = A sin (t + 0 ) where A, , 0 are positive constants. (This is an idealization in which we are disregarding friction.)A0A(a) Show that at all times t the acceleration of the bob y (t) is related to the position of the bob by the equation y (t) + 2 y(t) = 0.C(b) It is clear that the bob oscillates from A to A, and the speed of the bob is zero at these points. At what position does the bob attain maximum speed? What is this maximum speed? (c) What are the extreme values of the acceleration function? Where does the bob attain these extreme values? 60. An object moves along the x-axis, its position from t = 0 to t = 5 given by x(t) = t 3 7t 2 + 10t + 5. (a) Determine the velocity function v. Use a graphing utility to graph v as a function of t. (b) Use the graph to estimate the times when the object is moving right and the times when it is moving left. (c) Use the graphing utility to graph the speed v of the object as a function of t. Estimate the time(s) when the object stops. Estimate the maximum speed from t = 1 to t = 4. (d) Determine the acceleration function a and use the graphing utility to graph it as a function of t. Estimate the times when the object is speeding up and the times when it is slowing down. (e) Graph the velocity and acceleration functions on the same set of axes and use the graphs to estimate the times when the object is speeding up and the times when it is slowing down.13:26 241. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.9B VELOCITY AND ACCELERATION; SPEED217 PROJECT 4.9A Angular Velocity; Uniform Circular Motion As a particle moves along a circle of radius r, it effects a change in the central angle, marked in Figure A. We measure in radians. The angular velocity, , of the particle is the time rate of change of ; that is, = d/dt . Circular motion with constant, positive angular velocity is called uniform circular motion. Problem 1. A particle in uniform circular motion traces out a circular arc. The time rate of change of the length of that arc is called the speed of the particle. What is the speed of a particle that moves around a circle of radius r with constant, positive angular velocity ? Problem 2. The kinetic energy, KE, of a particle of mass m is given by the formulaProblem 3. A point P moves uniformly along the circle x 2 + y 2 = r 2 with constant angular velocity . Find the x- and y-coordinates of P at time t given that the motion starts at time t = 0 with = 0 . Then nd the velocity and acceleration of the projection of P onto the x-axis and onto the y-axis. [The projection of P onto the x-axis is the point (x, 0); the projection of P onto the y-axis is the point (0, y).] Problem 4. Figure B shows a sector in a circle of radius r. The sector is the union of the triangle T and the segment S. Suppose that the radius vector rotates counterclockwise with a constant angular velocity of radians per second. Show that the area of the sector changes at a constant rate but that the area of T and the area of S do not change at a constant rate.KE = 1 m 2 2 where is the speed of the particle. Suppose the particle in Problem 1 has mass m. What is the kinetic energy of the particle?rS T r rFigure BFigure A The symbol is the lowercase Greek letter omega.Problem 5. Take S and T as in Problem 4. While the area of S and the area of T change at different rates, there is one value of between 0 and at which both areas change at the same rate. Find this value of . PROJECT 4.9B Energy of a Falling Body (Near the Surface of the Earth) If we lift an object, we counteract the force of gravity. In so doing, we increase what physicists call the gravitational potential energy of the object. The gravitational potential energy of an object is dened by the formula GPE = weight height. Since the weight of an object of mass m is mg where g is the gravitational constant (we take this from physics), we can write GPE = mgy where y is the height of the object. If we lift an object and release it, the object drops. As it drops, it loses height and therefore loses gravitational potential energy, but its speed increases. The speed with which the objectfalls gives the object a form of energy called kinetic energy, the energy of motion. The kinetic energy of an object in motion is given by the formula KE = 1 m 2 2 where is the speed of the object. For straight-line motion with velocity v we have v 2 = 2 and therefore KE = 1 mv 2 . 2 Problem 1. Prove the law of conservation of energy: GPE + KE = C, constant. HINT: Differentiate the expression GPE + KE and use the fact that dv/dt = g.13:26 242. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006218 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES Problem 2. An object initially at rest falls freely from height y0 . Show that the speed of the object at height y is given by =2g(y0 y).Problem 3. According to the results in Section 4.9, the position of an object that falls from rest from a height y0 isgiven by y(t) = 1 gt 2 + yo . 2 Calculate the speed of the object from this equation and show that the result obtained is equivalent to the result obtained in Problem 2. 4.10 RELATED RATES OF CHANGE PER UNIT TIME In Section 4.9 we studied straight-line motion and dened velocity as the rate of change of position with respect to time and acceleration as the rate of change of velocity with respect to time. In this section we work with other quantities that vary with time. The fundamental point is this: if Q is any quantity that varies with time, then the derivative dQ/dt gives the rate of change of that quantity with respect to time.Example 1 A spherical balloon is expanding. Given that the radius is increasing at the rate of 2 inches per minute, at what rate is the volume increasing when the radius is 5 inches? SOLUTION Find d V /dt when r = 5 inches, given that dr/dt = 2 in./min andV = 4 r 3 . 3r(volume of a sphere of radius r )Both r and V are functions of t. Differentiating V = 4 r 3 with respect to t, we have 3 dV dr = 4r 2 . dt dt Setting r = 5 and dr/dt = 2, we nd that dV = 4 (52 )2 = 200. dt When the radius is 5 inches, the volume is increasing at the rate of 200 cubic inches per minute. Example 2 A particle moves clockwise along the unit circle x 2 + y 2 = 1. As it passes through the point (1/2, 3/2), its y-coordinate decreases at the rate of 3 units per second. At what rate does the x-coordinate change at this point? SOLUTION Find d x/dt when x = 1/2 and y = 3/2, given that dy/dt = 3 units/sec andy (x, y)x 2 + y 2 = 1.1 x(equation of circle)Differentiating x 2 + y 2 = 1 with respect to t, we have dx dy dx dy + 2y =0 and thus x +y = 0. dt dt dt dt Setting x = 1/2, y = 3/2, and dy/dt = 3, we nd that dx 3 1 dx + (3) = 0 and therefore = 3 3. 2 dt 2 dt As the object passes through the point (1/2, 3/2), the x-coordinate increases at the rate 3 3 units per second. 2x13:26 243. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.10 RELATED RATES OF CHANGE PER UNIT TIME219Example 3 A 13-foot ladder leans against the side of a building, forming an angle with the ground. Given that the foot of the ladder is being pulled away from the building at the rate of 0.1 feet per second, what is the rate of change of when the top of the ladder is 12 feet above the ground? 13 ftSOLUTION Find d/dt when y = 12 feet, given that d x/dt = 0.1 ft/sec andcos =x . 13Differentiation with respect to t givesx1 dx d = . sin dt 13 dt When y = 12, sin = 12 . 13Setting sin =d 1 = (0.1) dt 1312 1312 13and d x/dt = 0.1, we haveand thusd 1 = . dt 120When the top of the ladder is 12 feet above the ground, decreases at the rate of radians per second (about half a degree per second). 1 120Example 4 Two ships, one heading west and the other east, approach each other on parallel courses 8 nautical miles apart. Given that each ship is cruising at 20 nautical miles per hour (knots), at what rate is the distance between them diminishing when the ships are 10 nautical miles apart? SOLUTION Let y be the distance between the ships measured in nautical miles. Since the ships are moving in opposite directions at the rate of 20 knots each, their horizontal separation (see the gure) is decreasing at the rate of 40 knots. Thus, we want to nd dy/dt when y = 10, given that d x/dt = 40 knots. (We take d x/dt as negative since x is decreasing.) The variables x and y are related by the equationx 2 + 82 = y 2 .(Pythagorean theorem)Differentiating x + 8 = y with respect to t, we nd that 222dx dy dx dy + 0 = 2y and consequently x =y . dt dt dt dt When y = 10, x = 6. (Explain.) Setting x = 6, y = 10, and d x/dt = 40, we have 2xdy dy so that = 24. dt dt (Note that dy/dt is negative since y is decreasing.) When the two ships are 10 miles apart, the distance between them is diminishing at the rate of 24 knots. 6(40) = 10The preceding examples were solved by the same general method, a method that we recommend to you for solving problems of this type. Step 1. Draw a suitable diagram, and indicate the quantities that vary. Step 2. Specify in mathematical form the rate of change you are looking for, and record all relevant information. Step 3. Find an equation that relates the relevant variables. Step 4. Differentiate with respect to time t the equation found in Step 3. Step 5. State the nal answer in coherent form, specifying the units that you are using. The international nautical mile measures 6080 feet.yship8 milesyshipx13:26 244. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006220 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES Example 5 A conical paper cup 8 inches across the top and 6 inches deep is full of water. The cup springs a leak at the bottom and loses water at the rate of 2 cubic inches per minute. How fast is the water level dropping when the water is exactly 3 inches deep?4'r6'hSOLUTION We begin with a diagram that represents the situation after the cup has been leaking for a while. (Figure 4.10.1.) We label the radius and height of the remaining cone of water r and h. We can relate r and h by similar triangles. (Figure 4.10.2.) We measure r and h in inches. Now we seek dh/dt when h = 3, given that d V /dt = 2 in3 / min,Figure 4.10.14 2 r 1 2 r h = = . (volume of cone) (similar triangles) and 3 h 6 3 Using the second equation to eliminate r from the rst equation, we have V =4 r1 V = 36 h2h 32h=4 h3. 27Differentiation with respect to t gives r 4 = 6 hdV dh 4 = h2 . dt 9 dtFigure 4.10.2Setting h = 3 and d V /dt = 2, we have 2 =4 dh (3)2 9 dtand thusdh 1 = . dt 2When the water is exactly 3 inches deep, the water level is dropping at the rate of 1/2 inches per minute (about 0.16 inches per minute). Example 6 A balloon leaves the ground 500 feet away from an observer and rises vertically at the rate of 140 feet per minute. At what rate is the inclination of the observers line of sight increasing when the balloon is exactly 500 feet above the ground?balloonx observer500 ftSOLUTION Let x be the altitude of the balloon and the inclination of the observers line of sight. Find d/dt when x = 500, given that d x/dt = 140 ft/min and x . tan = 500 Differentiation with respect to t givessec2 1 dx d = . dt 500 dt When x = 500, triangle is isosceles. This implies that = /4 and sec = 2. the Setting sec = 2 and d x/dt = 140, we have d 1 d ( 2)2 = (140) and therefore = 0.14. dt 500 dt When the balloon is exactly 500 feet above the ground, the inclination of the observers line of sight is increasing at the rate of 0.14 radians per minute (about 8 degrees per minute). Example 7 A water trough with vertical cross section in the form of an equilateral triangle is being lled at a rate of 4 cubic feet per minute. Given that the trough is 12 feet long, how fast is the level of the water rising when the water reaches a depth of 1 1 feet? 213:26 245. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.10 RELATED RATES OF CHANGE PER UNIT TIME221SOLUTION Let x be the depth of the water measured in feet and V the volume ofwater measured in cubic feet. Find d x/dt when x = 3/2, given that d V /dt = 4 ft3 /min. 3 2 1 2x x . area of cross section = x= 2 3 3 3 2 x = 4 3x 2 . volume of water = 12 3 2 Differentiation of V = 4 3x with respect to t gives dx dV = 8 3x . dt dt Setting x = 3/2 and d V /dt = 4, we have 3 dx dx 1 1 3. and thus = = 4=8 3 2 dt dt 9 3 312'water levelx 3 3 3 3xcross section of troughWhen the water reaches a depth of 1 1 feet, the water level is rising at the rate of 2 1 3 feet per minute (about 0.19 feet per minute). 9 EXERCISES 4.10 1. A point moves along the line x + 2y = 2. Find (a) the rate of change of the y-coordinate, given that the x-coordinate is increasing at the rate of 4 units per second; (b) the rate of change of the x-coordinate, given that the y-coordinate is decreasing at the rate of 2 units per second. 2. A particle is moving in the circular orbit x 2 + y 2 = 25. As it passes through the point (3, 4), its y-coordinate is decreasing at the rate of 2 units per second. At what rate is the x-coordinate changing? 3. A particle is moving along the parabola y 2 = 4(x + 2). As it passes through the point (7, 6), its y-coordinate is increasing at the rate of 3 units per second. How fast is the x-coordinate changing at this instant? 4. A particle is moving along the parabola 4y = (x + 2)2 in such a way that its x-coordinate is increasing at the constant rate of 2 units per second. How fast is the particles distance from the point (2, 0) changing as it passes through the point (2, 4)? 5. A particle is moving along the ellipse x 2 /16 + y 2 /4 = 1. At each time t its x- and y-coordinates are given by x = 4 cos t, y = 2 sin t. At what rate is the particles distance from the origin changing at time t? At what rate is this distance from the origin changing when t = /4? 6. A particle is moving along the curve y = x x, x 0. Find the points on the curve, if any, at which both coordinates are changing at the same rate. 7. A heap of rubbish in the shape of a cube is being compacted into a smaller cube. Given that the volume decreases at the rate of 2 cubic meters per minute, nd the rate of change of an edge of the cube when the volume is exactly 27 cubic meters. What is the rate of change of the surface area of the cube at that instant?8. The volume of a spherical balloon is increasing at the constant rate of 8 cubic feet per minute. How fast is the radius increasing when the radius is exactly 10 feet? How fast is the surface area increasing at that time? 9. At a certain instant the side of an equilateral triangle is centimeters long and increasing at the rate of k centimeters per minute. How fast is the area increasing? 10. The dimensions of a rectangle are changing in such a way that the perimeter remains 24 inches. Show that when the area is 32 square inches, the area is either increasing or decreasing 4 times as fast as the length is increasing. 11. A rectangle is inscribed in a circle of radius 5 inches. If the length of the rectangle is decreasing at the rate of 2 inches per second, how fast is the area changing when the length is 6 inches? 12. A boat is held by a bow line that is wound about a bollard 6 feet higher than the bow of the boat. If the boat is drifting away at the rate of 8 feet per minute, how fast is the line unwinding when the bow is 30 feet from the bollard? 13. Two boats are racing with constant speed toward a nish marker, boat A sailing from the south at 13 mph and boat B approaching from the east. When equidistant from the marker, the boats are 16 miles apart and the distance between them is decreasing at the rate of 17 mph. Which boat will win the race? 14. A spherical snowball is melting in such a manner that its radius is changing at a constant rate, decreasing from 16 cm to 10 cm in 30 minutes. How fast is the volume of the snowball changing when the radius is 12 cm? 15. A 13-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 2 feet per second, how fast is the area of the13:26 246. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006222 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES16.17.18.19.20.21.22.triangle formed by the wall, the ground, and the ladder changing when the bottom of the ladder is 12 feet from the wall? A ladder 13 feet long is leaning against a wall. If the foot of the ladder is pulled away from the wall at the rate of 0.5 feet per second, how fast will the top of the ladder be dropping when the base is 5 feet from the wall? A tank contains 1000 cubic feet of natural gas at a pressure of 5 pounds per square inch. Find the rate of change of the volume if the pressure decreases at a rate of 0.05 pounds per square inch per hour. (Assume Boyles law: pressure volume = constant.) The adiabatic law for the expansion of air is P V 1.4 = C. At a given instant the volume is 10 cubic feet and the pressure is 50 pounds per square inch. At what rate is the pressure changing if the volume is decreasing at a rate of 1 cubic foot per second? A man standing 3 feet from the base of a lamppost casts a shadow 4 feet long. If the man is 6 feet tall and walks away from the lamppost at a speed of 400 feet per minute, at what rate will his shadow lengthen? How fast is the tip of his shadow moving? A light is attached to the wall of a building 64 feet above the ground. A ball is dropped from that height, but 20 feet away from the side of the building. The height y of the ball at time t is given by y(t) = 64 16t 2 . Here we are measuring y in feet and t in seconds. How fast is the shadow of the ball moving along the ground 1 second after the ball is dropped? An object that weighs 150 pounds on the surface of the earth 1 will weigh 150(1 + 4000 r )2 pounds when it is r miles above the earth. Given that the altitude of the object is increasing at the rate of 10 miles per minute, how fast is the weight decreasing when the object is 400 miles above the surface? In the special theory of relativity the mass of a particle moving at speed is given by the expression27.28.29.30.31.32.33.m 1 2 /c223.24.25.26.where m is the mass at rest and c is the speed of light. At what rate is the mass of the particle changing when the speed of the particle is 1 c and is increasing at the rate of 0.01c per 2 second? Water is dripping through the bottom of a conical cup 4 inches across and 6 inches deep. Given that the cup loses half a cubic inch of water per minute, how fast is the water level dropping when the water is 3 inches deep? Water is poured into a conical container, vertex down, at the rate of 2 cubic feet per minute. The container is 6 feet deep and the open end is 8 feet across. How fast is the level of the water rising when the container is half full? At what rate is the volume of a sphere changing at the instant when the surface area is increasing at the rate of 4 square centimeters per minute and the radius is increasing at the rate of 0.1 centimeter per minute? Water ows from a faucet into a hemispherical basin 14 inches in diameter at the rate of 2 cubic inches per second.34.35.36.37.How fast does the water rise (a) when the water is exactly halfway to the top? (b) just as it runs over? (The volume of a spherical segment is given by r h 2 1 h 3 where r is the 3 radius of the sphere and h is the depth of the segment.) The base of an isosceles triangle is 6 feet. Given that the altitude is 4 feet and increasing at the rate of 2 inches per minute, at what rate is the vertex angle changing? As a boy winds up the cord, his kite is moving horizontally at a height of 60 feet with a speed of 10 feet per minute. How fast is the inclination of the cord changing when the cord is 100 feet long? A revolving searchlight 1 mile from a straight shoreline 2 makes 1 revolution per minute. How fast is the light moving along the shore as it passes over a shore point 1 mile from the shore point nearest to the searchlight? A revolving searchlight 1 mile from a straight shoreline turns at the rate of 2 revolutions per minute in the counterclockwise direction. (a) How fast is the light moving along the shore when it makes an angle of 45 with the shore? (b) How fast is the light moving when the angle is 90 ? A man starts at a point A and walks 40 feet north. He then turns and walks due east at 4 feet per second. A searchlight placed at A follows him. At what rate is the light turning 15 seconds after the man started walking east? The diameter and height of a right circular cylinder are found at a certain instant to be 10 centimeters and 20 centimeters, respectively. If the diameter is increasing at the rate of 1 centimeter per second, what change in height will keep the volume constant? A horizontal trough 12 feet long has a vertical cross section in the form of a trapezoid. The bottom is 3 feet wide, and the sides are inclined to the vertical at an angle with sine 4 . 5 Given that water is poured into the trough at the rate of 10 cubic feet per minute, how fast is the water level rising when the water is exactly 2 feet deep? Two cars, car A traveling east at 30 mph and car B traveling north at 22.5 mph, are heading toward an intersection I. At what rate is the angle IAB changing when cars A and B are 300 feet and 400 feet, respectively, from the intersection? A rope 32 feet long is attached to a weight and passed over a pulley 16 feet above the ground. The other end of the rope is pulled away along the ground at the rate of 3 feet per second. At what rate is the angle between the rope and the ground changing when the weight is exactly 4 feet off the ground? A slingshot is made by fastening the two ends of a 10-inch rubber strip 6 inches apart. If the midpoint of the strip is drawn back at the rate of 1 inch per second, at what rate is the angle between the segments of the strip changing 8 seconds later? A balloon is released 500 feet away from an observer. If the balloon rises vertically at the rate of 100 feet per minute and at the same time the wind is carrying it away horizontally at the rate of 75 feet per minute, at what rate is the inclination of the observers line of sight changing 6 minutes after the balloon has been released?13:26 247. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.11 DIFFERENTIALS38. A searchlight is continually trained on a plane that ies directly above it at an altitude of 2 miles at a speed of 400 miles per hour. How fast does the light turn 2 seconds after the plane passes directly overhead? 39. A baseball diamond is a square 90 feet on a side. A player is running from second base to third base at the rate of 15 feet per second. Find the rate of change of the distance from the player to home plate at the instant the player is 10 feet from third base. (If you are not familiar with baseball, skip this problem.) 40. An airplane is ying at constant speed and altitude on a line that will take it directly over a radar station on the ground. At the instant the plane is 12 miles from the station, it is noted that the planes angle of elevation is 30 and is increasing at the rate of 0.5 per second. Give the speed of the plane in miles per hour. 41. An athlete is running around a circular track of radius 50 meters at the rate of 5 meters per second. A spectator isIn Figure 4.11.1 we have sketched the graph of a differentiable function f and below it the tangent line at the point (x, f (x)). y(x + h, f (x + h)) f (x + h) f (x) f'(x)h (x, f (x)) hx+hxFigure 4.11.1As the gure suggests, for small h = 0, f (x + h) f (x), the change in f from x to x + h can be approximated by the product f (x)h:(4.11.1)f (x + h) f (x) f (x)h. =How good is this approximation? It is good in the sense that, for small h the difference between the two quantities, [ f (x + h) f (x)] f (x)h, is small compared to h. How small compared to h? Small enough compared to h that its ratio to h, the quotient [ f (x + h) f (x)] f (x)h , h223200 meters from the center of the track. How fast is the distance between the two changing when the runner is approaching the spectator and the distance between them is 200 meters? Exercises 4244. Here x and y are functions of t and are related as indicated. Obtain the desired derivative from the information given. dy = 2 when x = 3 and y = 42. 2x y 2 y = 22. Given that dt dx . 2, nd dt dy = 3 when x = 8 and y = 2, 43. x x y = 4. Given that dt dx . nd dt dx = 1 when x = and 44. sin x = 4 cos y 1. Given that dt dy . y = , nd 3 dt 4.11 DIFFERENTIALSx13:26 248. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006224 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES tends to 0 as h tends to 0: [ f (x + h) f (x)] f (x)h f (x + h) f (x) f (x)h = lim lim lim h0 h0 h0 h h h = f (x) f (x) = 0. The quantities f (x + h) f (x) and f (x)h have names: DEFINITION 4.11.2For h = 0 the difference f (x + h) f (x) is called the increment of f from x to x + h and is denoted by f : f = f (x + h) f (x). The product f (x)h is called the differential of f at x with increment h and is denoted by df : d f = f (x)h.Display 4.11.1 says that, for small h,f and d f are approximately equal: d f. f =How close is the approximation? Close enough (as we just showed) that the quotient f df h tends to 0 as h tends to 0. Lets see what all this amounts to in a very simple case. The area of a square of side x is given by the function f (x) = x 2 ,hxhh2x > 0.If the length of each side increases from x to x + h, the area increases from f (x) to f (x + h). The change in area is the increment f : f = f (x + h) f (x) = (x + h)2 x 2xx2= (x 2 + 2xh + h 2 ) x 2xh= 2xh + h 2 . xhAs an estimate for this change, we can use the differential d f = f (x)h = 2xh.Figure 4.11.2The error of this estimate, the difference between the actual change change df, is the difference f d f = h2. As promised, the error is small compared to h in the sense that f df h2 = =h h h tends to 0 as h tends to 0. is a Greek letter, the capital of .f is read delta f.(Figure 4.11.2)f and the estimated13:26 249. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.11 DIFFERENTIALSExample 1 Use a differential to estimate the change in f (x) = x 2/5 9 (a) as x increases from 32 to 34, (b) as x decreases from 1 to 10 . SOLUTION Since f (x) = 2 x 3/5 = 2/(5x 3/5 ), we have 52 h. 5x 3/5 (a) We set x = 32 and h = 2. The differential then becomes d f = f (x)h =df =2 4 (2) = = 0.1. 3/5 5(32) 40A change in x from 32 to 34 increases the value of f by approximately 0.1. For comparison, our hand calculator gives f = f (34) f (32) 4.0982 4 = 0.0982. = 1 (b) We set x = 1 and h = 10 . In this case, the differential isdf =2 5(1)3/51 10=2 = 0.04. 509 A change in x from 1 to 10 decreases the value of f by approximately 0.04. For comparison, our hand calculator givesf = f (0.9) f (1) = (0.9)2/5 (1)2/5 0.9587 1 = 0.0413. =Example 2 Use a differential to estimate: (a) SOLUTION(a) We know that 104,(b) cos 40 . 100 = 10. We need an estimate for the increase of f (x) = xas x increases from 100 to 104. Here 1 f (x) = and 2 xh d f = f (x)h = . 2 xWith x = 100 and h = 4, d f becomes 4 1 = = 0.2. 5 2 100 A change in x from 100 to 104 increases the value of the square root by approximately 0.2. It follows that 104 100 + 0.2 = 10 + 0.2 = 10.2. = As you can check, (10.2)2 = 104.04. Our estimate is not far off. (b) Let f (x) cos x, where as usual x is given in radians. We know that cos 45 = = cos (/4) = 2/2. Converting 40 to radians, we have 40 = 45 5 = 5= radians. 4 180 4 36 We use a differential to estimate the change in cos x as x decreases from /4 to (/4) (/36): f (x) = sin xandd f = f (x)h = h sin x.22513:26 250. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006226 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES With x = /4 and h = /36, d f is given by 2 2 df = sin = = = 0.0617. 36 4 36 2 72 A decrease in x from /4 to (/4) (/36) increases the value of the cosine by approximately 0.0617. Therefore, cos 40 cos 45 + 0.0617 0.7071 + 0.0616 = 0.7688. = = Our hand calculator gives cos 40 = 0.7660. Example 3 A metal sphere with a radius of 10 cm is to be covered by a 0.02 cm coating of silver. Approximately how much silver will be required? SOLUTION We will use a differential to estimate the increase in the volume of a sphere if the radius is increased from 10 cm to 10.02 cm. The volume of a sphere of radius r is given by the formula V = 4 r 3 . Therefore 3d V = 4r 2 h. Taking r = 10 and h = 0.02, we have d V = 4 (10)2 (0.02) = 8 25.133. = It will take approximately 25.133 cubic cm of silver to coat the sphere.Example 4 A metal cube is heated and the length of each edge is thereby increased by 0.1%. Use a differential to show that the surface area of the cube is then increased by about 0.2%. SOLUTION Let x be the initial length of an edge. The initial surface area is thenS(x) = 6x 2 . As the length increases from x to x + h, the surface area increases from S(x) to S(x + h). We will estimate the ratio S(x + h) S(x) S = S S(x)by dS Stakingh = 0.001x.Here S(x) = 6x 2 ,dS = 12xh = 12x(0.001x),and therefore dS 12x(0.001x) = 0.002. = S 6x 2 If the length of each edge is increased by 0.1%, the surface area is increased by about 0.2%. EXERCISES 4.11 1. Use a differential to estimate the change in the volume of a cube caused by an increase h in the length of each side. Interpret geometrically the error of your estimate V d V . 2. Use a differential to estimate the area of a ring of inner radius r and width h. What is the exact area?CExercises 38. Use a differential to estimate the value of the indicated expression. Then compare your estimate with the result given by a calculator. 3. 3 1002. 4. 1/ 24.5.13:26 251. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.11 DIFFERENTIALSC 5. 4 15.5. 6. (26)2/3 . 8. (33)1/5 . 7. (33)3/5 Exercises 912. Use a differential to estimate the value of the expression. (Remember to convert to radian measure.) Then compare your estimate with the result given by a calculator. 10. cos 62 . 9. sin 46 . 12. sin 43 . 11. tan 28 13. Estimate f (2.8) given that f (3) = 2 and f (x) = (x 3 + 5)1/5 . 14. Estimate f (5.4) given that f (5) = 1 and f (x) = 3 x 2 + 2. 15. Find the approximate volume of a thin cylindrical shell with open ends given that the inner radius is r, the height is h, and the thickness is t. 16. The diameter of a steel ball is measured to be 16 centimeters, with a maximum error of 0.3 centimeters. Estimate by differentials the maximum error (a) in the surface area as calculated from the formula S = 4r 2 ; (b) in the volume as calculated from the formula V = 4 r 3 . 3 17. A box is to be constructed in the form of a cube to hold 1000 cubic feet. Use a differential to estimate how accurately the inner edge must be made so that the volume will be correct to within 3 cubic feet. 18. Use differentials to estimate the values of x for which (a) x + 1 x < 0.01. (b) 4 x + 1 4 x < 0.002. 19. A hemispherical dome with a 50-foot radius will be given a coat of paint 0.01 inch thick. The contractor for the job wants to estimate the number of gallons of paint that will be needed. Use a differential to obtain an estimate. (There are 231 cubic inches in a gallon.) 20. View the earth as a sphere of radius 4000 miles. The volume of ice that covers the north and south poles is estimated to be 8 million cubic miles. Suppose that this ice melts and the water produced distributes itself uniformly over the surface of the earth. Estimate the depth of this water. 21. The period P of the small oscillations of a simple pendulum is related to the length L of the pendulum by the equation P = 2L gwhere g is the (constant) acceleration of gravity. Show that a small change dL in the length of a pendulum produces a change dP in the period that satises the equation dP 1 dL = . P 2 L 22. Suppose that the pendulum of a clock is 90 centimeters long. Use the result in Exercise 21 to determine how the length of the pendulum should be adjusted if the clock is losing 15 seconds per hour. 23. A pendulum of length 3.26 feet goes through one complete oscillation in 2 seconds. Use Exercise 21 to nd the approximate change in P if the pendulum is lengthened by 0.01 feet.22724. A metal cube is heated and the length of each edge is thereby increased by 0.1%. Use a differential to show that the volume of the cube is then increased by about 0.3%. 25. We want to determine the area of a circle by measuring the diameter x and then applying the formula A = 1 x 2 . 4 Use a differential to estimate how accurately we must measure the diameter for our area formula to yield a result that is accurate within 1%. 26. Estimate by differentials how precisely x must be determined (a) for our calculation of x n to be accurate within 1%; (b) for our estimate of x 1/n to be accurate within 1%. (Here n is a positive integer.) Little-o(h) Let g be a function dened at least on some open interval that contains the number 0. We say that g(h) is little-o(h) and write g(h) = o(h) to indicate that, for small h, g(h) is so small compared to h that limh0g(h) = 0. h27. Determine whether the statement is true. (a) h 3 = o(h) h2 = o(h). (b) h1 (c) h 1/3 = o(h). 28. Show that if g(h) = o(h), then lim g(h) = 0. h029. Show that if g1 (h) = o(h) and g2 (h) = o(h), then g1 (h) + g2 (h) = o(h)g1 (h)g2 (h) = o(h).and30. The gure shows the graph of a differentiable function f and a line with slope m that passes through the point (x, f (x)). The vertical separation at x + h between the line with slope m and the graph of f has been labeled g(h).y(x + h, f (x + h))g(h)slope mmh(x, f (x)) hxx+hx(a) Calculate g(h). (b) Show that, of all lines that pass through (x, f (x)), the tangent line is the line that best approximates the graph of f near the point (x, f (x)) by showing that g(h) = o(h)iffm = f (x).13:26 252. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006228 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES Here we dispense with g(h) and call o(h) any expression in h which, for small h, is so small compared to h that o(h) = 0. h0 h 31. (A differentiable function is locally almost linear.) If f is a linear function, f (x) = mx + b, then f (x + h) f (x) = mh. limShow that a function f is differentiable at x iff there exists a number m such that(4.11.3)f (x + h) f (x) = mh + o(h). What is m here? PROJECT 4.11 Marginal Cost, Marginal Revenue, Marginal Profit In business and economics we recognize that costs, revenues, and prots depend on many factors. Of special interest to us here is the study of how costs, revenues, and prots are affected by changes in production and sales. In this brief section we make the simplifying assumption that all production is sold and therefore units sold = units produced. Suppose that C(x) represents the cost of producing x units. Although x is usually a nonnegative integer, in theory and practice it is convenient to assume that C(x) is dened for x in some interval and that the function C is differentiable. In this context, the derivative C (x) is called the marginal cost at x. This terminology deserves some explanation. The difference C(x + 1) C(x) represents the cost of increasing production from x units to x + 1 units and, as such, gives the cost of producing the (x + 1)-st unit. The derivative C (x) is called the marginal cost at x because it provides an estimate for the cost of the (x + 1)-st unit: in general, C(x + h) C(x) C (x)h. =(differential estimate)At h = 1 this estimate reads C(x + 1) C(x) C (x). = Thus, as asserted, C (x) cost of (x + 1)-st unit. = By studying the marginal cost function C , we can obtain an overall view of the changing cost patterns. Similarly, if R(x) represents the revenue obtained from the sale of x units, then the derivative R (x), called the marginal revenue at x, provides an estimate for the revenue obtained from the sale of the (x + 1)-st unit. If C = C(x) and R = R(x) are the cost and revenue functions associated with the production and sale of x units, then the functionare called break-even points. The derivative P (x) is called the marginal prot at x. By Theorem 4.3.2, maximum prot occurs at a point where P (x) = 0, a point where the marginal prot is zero, which, since P (x) = R (x) C (x), is a point where the marginal revenue R (x) equals the marginal cost C (x). A word about revenue functions. The revenue obtained from the sale of x units at an average price p(x) is the product of x and p(x): R(x) = x p(x). In classical supply-demand theory, if too many units are sold, the price at which they can be sold comes down. It may come down so much that the product x p(x) starts to decrease. If the market is ooded with units, p(x) tends to zero and revenues are severely impaired. Thus it is that the revenue R(x) increases with x up to a point and then decreases. The gure gives a graphical representation of a pair of cost and revenue functions, shows the break-even points, and indicates the regions of prot and loss. RC C R profitlosslossbreak-even pointsP(x) = R(x) C(x) is called the prot function. The points x (if any) at which C(x) = R(x) that is, the values at which cost = revenuex13:26 253. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.12 NEWTON-RAPHSON APPROXIMATIONSProblem 1. A manufacturer determines that the total cost of producing x units per hour is given by the function C(x) = 2000 + 50x x2 . 20(dollars)What is the marginal cost at production level 20 units per hour? What is the exact cost of producing the 21st component? Problem 2. A manufacturer determines that the costs and revenues that result from the production and sale of x units per day are given by the functions C(x) = 12,000 + 30xR(x) = 650x 5x 2 .andFind the prot function and determine the break-even points. Find the marginal prot function and determine the production/sales level for maximum prot. Problem 3. The cost and revenue functions for the production and sale of x units are C(x) = 4x + 1400R(x) = 20x andx2 . 50(a) Find the prot function and determine the break-even points. (b) Find the marginal prot function and determine the production level at which the marginal prot is zero. (c) Sketch the cost and revenue functions in the same coordinate system and indicate the regions of prot and loss. Estimate the production level that produces maximum prot.C(x) = 3000 + 5xfcFigure 4.12.1 R(x) = 60x 2x x,with x measured in hundreds of units. (a) Use a graphing utility to graph the cost function together with the revenue function. Estimate the break-even points. (b) Graph the prot function and estimate the production level that yields the maximum prot. Exactly how many units should be produced to maximize prot? Problem 6. Let C(x) be the cost of producing x units. The average cost per unit is A(x) = C(x)/x. Show that if C (x) > 0, then the average cost per unit is a minimum at the production levels x where the marginal cost equals the average cost per unit. Problem 7. Let R(x) be the revenue that results from the sale of x units. The average revenue per unit is B(x) = R(x)/x. Show that if R (x) < 0, then the average revenue per unit is a maximum at the values x where the marginal revenue equals the average revenue per unit.yx2andwith x measured in thousands of units. (a) Use a graphing utility to graph the cost function together with the revenue function. Estimate the break-even points. (b) Graph the prot function and estimate the production level that yields the maximum prot. Problem 5. The cost and revenue functions are 10x , C(x) = 4 + 0.75x and R(x) = 1 + 0.25x 2Figure 4.12.1 shows the graph of a function f. Since the graph of f crosses the x-axis at x = c, the number c is a solution (root) of the equation f (x) = 0. In the setup of Figure 4.12.1, we can approximate c as follows: Start at x1 (see the gure). The tangent line at (x1 , f (x1 )) intersects the x-axis at a point x2 which is closer to c than x1 . The tangent line at (x2 , f (x2 )) intersects the x-axis at a point x3 , which in turn is closer to c than x2 . In this manner, we obtain numbers x1 , x2 , x3 , . . . , xn , xn+1 , which more and more closely approximate c.x3229Problem 4. The cost and revenue functions are 4.12 NEWTON-RAPHSON APPROXIMATIONSx1x13:26 254. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006230 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES There is an algebraic connection between xn and xn+1 that we now develop. The tangent line at (xn , f (xn )) has the equation y f (xn ) = f (xn )(x xn ). The x-intercept of this line, xn+1 , can be found by setting y = 0: 0 f (xn ) = f (xn )(xn+1 xn ). Solving this equation for xn+1 , we havexn+1 = xn (4.12.1)yf (xn ) . f (xn )y = f (xn )y = f (x) xnxFigure 4.12.2This method of locating a root of an equation f (x) = 0 is called the NewtonRaphson method. The method does not work in all cases. First, there are some conditions that must be placed on the function f. Clearly, f must be differentiable at points near the root c. Also, if f (xn ) = 0 for some n, then the tangent line at (xn , f (xn )) is horizontal and the next approximation xn+1 cannot be calculated. See Figure 4.12.2. Thus, we will assume that f (x) = 0 at points near c. The method can also fail for other reasons. For example, it can happen that the rst approximation x1 produces a second approximation x2 , which in turn takes us back to x1 . In this case the approximations simply alternate between x1 and x2 . See Figure 4.12.3. Another type of difculty can arise if f (x1 ) is close to zero. In this case the second approximation x2 can be worse than x1 , the third approximation x3 can be worse than x2 , and so forth. See Figure 4.12.4. yy yy = f (x) c x2c x1xx1y = f (x)y = f (x) (x1, f (x1))x2 x (x2, f (x2)) c x3Figure 4.12.3Figure 4.12.4y x1x2x3 c(x2, f (x2))(x1, f (x1)) y = f (x)Figure 4.12.6xx2x1xFigure 4.12.5There is a condition that guarantees that the Newton-Raphson method will work. Suppose that f is twice differentiable and that f (x) f (x) > 0 for all x between c and x1 . If f (x) > 0 for such x, then f (x) > 0 for such x and (as shown in Section 4.6) the graph bends up, and we have the situation pictured in Figure 4.12.5. On the other hand, if f (x) < 0 for such x, then f (x) < 0 for such x and (as shown in Section 4.6) the graph bends down, and we have the situation pictured in Figure 4.12.6. In each of these cases the approximations x1 , x2 , x3 , . . . tend to the root c. Example 1 The number 3 is a root of the equation x 2 3 = 0. We will estimate 3 by applying the Newton-Raphson method to the function f (x) = x 2 3 starting13:26 255. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.12 NEWTON-RAPHSON APPROXIMATIONS231 at x1 = 2. [As you can check, f (x) f (x) > 0 on ( 3, 2) and therefore we can be sure that the method applies.] Since f (x) = 2x, the Newton-Raphson formula gives xn+1 = xn 2 xn 3 2xn=2 xn + 3 . 2xnSuccessive calculations with this formula (using a calculator) are given in the following table: nxn1 2 32 1.75000 1.73214xn+1 =2 xn +3 2xn1.75000 1.73214 1.73205Since (1.73205)2 2.999997, the method has generated a very accurate estimate of = 3 in only three steps. EXERCISES 4.12 CExercises 18. Use the Newton-Raphson method to estimate a root of the equation f (x) = 0 starting at the indicated value of x: (a) Express xn+1 in terms of xn . (b) Give x4 rounded off to ve decimal places and evaluate f at that approximation. f (x) = x 2 24; x1 = 5. f (x) = x 3 4x + 1; x1 = 2. 4. f (x) = x 5 30; x1 = 2. f (x) = x 3 25; x1 = 3. f (x) = cos x x; x1 = 1. f (x) = sin x x 2 ; x1 = 1. f (x) = x + 3 x; x1 = 1. f (x) = x + tan x; x1 = 2. The function f (x) = x 1/3 is 0 at x = 0. Verify that the condition f (x) f (x) > 0 fails everywhere. Show that the Newton-Raphson method starting at any number x1 = 0 fails to generate numbers that approach the solution x = 0. Describe the numbers x1 , x2 , x3 , . . . that the method generates. 10. What results from the application of the Newton-Raphson method to a function f if the starting approximation x1 is precisely the desired zero of f ? 11. Set f (x) = 2x 3 3x 2 1. (a) Show that the equation f (x) = 0 has a root between 1 and 2. (b) Show that the Newton-Raphson method process started at x1 = 1 fails to generate numbers that approach the root that lies between 1 and 2. (c) Estimate this root by starting at x1 = 2. Determine x4 rounded off to four decimal places and evaluate f (x4 ). 12. The function f (x) = x 4 2x 2 17 has two zeros, one at a 16 point a between 0 and 2, and the other at a. ( f is an even function.) (a) Show that the Newton-Raphson method fails in the search for a if we start at x = 1 . What are the outputs 2 x1 , x2 , x3 , . . . in this case? 1. 2. 3. 5. 6. 7. 8. 9.CC(b) Estimate a by starting at x1 = 2. Determine x4 rounded off to ve decimal places and evaluate f (x4 ). 13. Set f (x) = x 2 a, a > 0. The roots of the equation f (x) = 0 are a. (a) Show that if x1 is any initial estimate for a, then the Newton-Raphson method gives the iteration formula xn+1 =1 a xn + 2 xn,n 1.(b) Take a = 5. Starting at x1 = 2, use the formula in part (a) to calculate x4 to ve decimal places and evaluate f (x4 ). 14. Set f (x) = x k a, k a positive integer, a > 0. The number a 1/k is a root of the equation f (x) = 0. (a) Show that if x1 is any initial estimate for a 1/k , then the Newton-Raphson method gives the iteration formula xn+1 =1 a (k 1)xn + k1 . k xnNote that for k = 2 this formula reduces to the formula given in Exercise 13. (b) Use the formula in part (a) to approximate 3 23. Begin at x1 = 3 and calculate x4 rounded off to ve decimal places. Evaluate f (x4 ). 1 15. Set f (x) = a, a = 0. x (a) Apply the Newton-Raphson method to derive the iteration formula 2 xn+1 = 2xn axn ,n 1.Note that this formula provides a method for calculating reciprocals without recourse to division. (b) Use the formula in part (a) to calculate 1/2.7153 rounded off to ve decimal places.13:26 256. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006232 CHAPTER 4 THE MEAN-VALUE THEOREM; APPLICATIONS OF THE FIRST AND SECOND DERIVATIVES 16. Set f (x) = x 4 7x 2 8x 3. (a) Show that f has exactly one critical point c in the interval (2, 3). (b) Use the Newton-Raphson method to estimate c by calculating x3 . Round off your answer to four decimal places. Does f have a local maximum at c, a local minimum, or neither? 17. Set f (x) = sin x + 1 x 2 2x. 2 (a) Show that f has exactly one critical point c in the interval (2, 3). (b) Use the Newton-Raphson method to estimate c by calculating x3 . Round off your answer to four decimal places.Does f have a local maximum at c, a local minimum, or neither? 18. Approximations to can be obtained by applying the Newton-Raphson method to f (x) = sin x starting at x1 = 3. (a) Find x4 rounded off to four decimal places. (b) What are the approximations if we start at x1 = 6? 19. The equation x + tan x = 0 has an innite number of positive roots r1 , r2 , r3 , . . . , rn slightly larger than (n 1 ). 2 Use the Newton-Raphson method to nd r1 and r2 to three decimal place accuracy. CHAPTER 4. REVIEW EXERCISES Exercises 12. Show that f satises the conditions of Rolles theorem on the indicated interval and nd all the numbers c on the interval for which f (c) = 0. 1. f (x) = x 3 x; [1, 1]. 2. f (x) = sin x + cos x 1; [0, 2]. Exercises 36. Verify that f satises the conditions of the meanvalue theorem on the indicated interval and nd all the numbers c that satisfy the conclusion of the theorem. 3. f (x) = x 3 2x + 1; [2, 3]. 4. f (x) = x 1; [2, 5]. x +1 5. f (x) = ; [2, 4]. 6. f (x) = x 3/4 ; [0, 16]. x 1 7. Set f (x) = x 1/3 x. Note that f (1) = f (1) = 0. Verify that there does not exist a number c in (1, 1) for which f (c) = 0. Explain how this does not violate Rolles theorem. 8. Set f (x) = (x + 1)/(x 2). Show that there does not exist a number c in (1, 4) for which f (4) f (1) = f (c)(4 1). Explain how this does not violate the mean-value theorem. 9. Does there exist a differentiable function f with f (1) = 5, f (4) = 1, and f (x) 1 for all x in (1, 4)? If not, how do you know? 10. Let f (x) = x 3 3x + k, k constant. a. Show that f (x) = 0 for at most one x in [1, 1]. b. For what values of k does f (x) = 0 for some x in [1, 1]? Exercises 1116. Find the intervals on which f increases and the intervals on which f decreases; nd the critical points and the local extreme values. 12. f (x) = x 4 4x + 3. 11. f (x) = 2x 3 + 3x 2 + 1. 4 13. f (x) = (x + 2)2 (x 1)3 . 14. f (x) = x + 2 . x x 15. f (x) = . 1 + x2 16. f (x) = sin x cos x, 0 x 2 . Exercises 1722. Find the critical points. Then nd and classify all the extreme values. 17. f (x) = x 3 + 2x 2 + x + 1; x [2, 1].18. f (x) = x 4 8x 2 + 2; x [1, 3]. 4 19. f (x) = x 2 + 2 ; x [1, 4]. x 20. f (x) = cos2 x + sin x; x [0, 2]. 21. f (x) = x 1 x; x (, 1]. x2 22. f (x) = ; x (2, ). x 2 Exercises 2325. Find all vertical, horizontal, and oblique (see Exercises 4.7) asymptotes. x2 4 3x 2 9x . 24. f (x) = 2 . 23. f (x) = 2 x x 12 x 5x + 6 4 x 25. f (x) = 3 . x 1 Exercises 2628. Determine whether or not the graph of f has a vertical tangent or a vertical cusp at c. 26. f (x) = (x 1)3/5 ; c = 1. 27. f (x) = 5 x 7/5 5x 2/5 ; c = 0. 7 28. f (x) = 3x 1/3 (2 + x);c = 0.Exercises 2936. Sketch the graph of the function using the approach outlined in Section 4.8. 30. f (x) = 3x 5 5x 3 + 1. 29. f (x) = 6 + 4x 3 3x 4 . 2x 31. f (x) = 2 . 32. f (x) = x 2/3 (x 10). x +4 34. f (x) = x 4 2x 2 + 3. 33. f (x) = x 4 x. 35. f (x) = sin x + 3 cos x, x [0, 2]. 36. f (x) = sin2 x cos x, x [0, 2]. 37. Sketch the graph of a function f that satises the following conditions: f (1) = 3, f (0) = 0, f (2) = 4; f (1) = f (2) = 0; f (x) > 0 for x < 1 and for x > 2, f (x) < 0 if 1 < x < 2; f (x) < 0 for x < 1 , f (x) > 0 for x > 1 . 2 213:26 257. P1: PBU/OVYP2: PBU/OVYJWDD023-04JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20064.12 NEWTON-RAPHSON APPROXIMATIONS38. Given that the surface area of a sphere plus the surface area of a cube is constant, show that the sum of the volumes is minimized by letting the diameter of the sphere equal the length of a side of the cube. What dimensions maximize the sum of the volumes? 39. A closed rectangular box with a square base is to be built subject to the following conditions: the volume is to be 27 cubic feet, the area of the base may not exceed 18 square feet, the height of the box may not exceed 4 feet. Determine the dimensions of the box (a) for minimal surface area; (b) for maximal surface area. 40. The line through P(1, 2) intersects the positive x-axis at A(a, 0) and the positive y-axis at B(0, b). Determine the values of a and b that minimize the area of the triangle OAB. 41. A right circular cylinder is generated by revolving a rectangle of given perimeter P about one of its sides. What dimensions of the rectangle will generate the cylinder of maximum volume? 42. A printed page is to have a total area of 80 square inches. The margins at the top and on the sides are to be 1 inch each; the bottom margin is to be 1.5 inches. Determine the dimensions of the page that maximize the area available for print. 43. An object moves along a coordinate line, its position at time t given by the function x(t) = t + 2 cos t. Find those times from t = 0 to t = 2 when the object is slowing down. 44. An object moves along a coordinate line, its position at time t given by the function x(t) = (4t 1)(t 1)2 , t 0. (a) When is the object moving to the right? When to the left? When does it change direction? (b) What is the maximum speed of the object when moving left? 45. An object moves along a coordinate line, its position at time t given by the function x(t) = t + 1, t 0. (a) Show that the acceleration is negative and proportional to the cube of the velocity. (b) Use differentials to obtain numerical estimates for the position, velocity, and acceleration at time t = 17. Base your estimate on t = 15. 46. A rocket is red from the ground straight up with an initial velocity of 128 feet per second. (a) When does the rocket reach maximum height? What is maximum height? (b) When does the rocket hit the ground and at what speed? 47. Ballast dropped from a balloon that was rising at the rate of 8 feet per second reached the ground in 10 seconds. How high was the balloon when the ballast was released?C23348. A ball thrown straight up from the ground reaches a height of 24 feet in 1 second. How high will the ball go? 49. A boy walks on a straight, horizontal path away from a light that hangs 12 feet above the path. How fast does his shadow lengthen if he is 5 feet tall and walks at the rate of 168 feet per minute? 50. The radius of a cone increases at the rate of 0.3 inches per minute, but the volume remains constant. At what rate does the height of the cone change when the radius is 4 inches and the height is 15 inches? 51. A railroad track crosses a highway at an angle of 60 . A locomotive is 500 feet from the intersection and moving away from it at the rate of 60 miles per hour. A car is 500 feet from the intersection and moving toward it at the rate of 30 miles per hour. What is the rate of change of the distance between them? 52. A square is inscribed in a circle. Given that the radius of the circle is increasing at the rate of 5 centimeters per minute, at what rate is the area of the square changing when the radius is 10 centimeters? 53. A horizontal water trough 12 feet long has a vertical cross section in the form of an isosceles triangle (vertex down). The base and height of the triangle are each 2 feet. Given that water is being drained out of the trough at the rate of 3 cubic feet per minute, how fast is the water level falling when the water is 1.5 feet deep? 54. Use a differential to estimate f (3.8) given that f (4) = 2 and f (x) = 3 3x 4. 55. Use a differential to estimate f (4.2) given that f (x) = x + 1/ x. Exercises 5657. Use a differential to estimate the value of the expression. 57. tan 43 . 56. 4 83. 58. A spherical tank with a diameter of 20 feet will be given a coat of paint 0.05 inches thick. Estimate by a differential the amount of paint needed. (Assume that there are 231 cubic inches in a gallon.) Exercises 5960. Use the Newton-Raphson method to estimate a root of f (x) = 0 starting at the indicated value: (a) Express xn+1 in terms of xn . (b) Give x4 rounded off to ve decimal places and evaluate f at that approximation. 59. f (x) = x 3 10; x1 = 2. 60. f (x) = x sin x cos x; x1 = 1.13:26 258. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006CHAPTER5 INTEGRATION 5.1 AN AREA PROBLEM; A SPEED-DISTANCE PROBLEM An Area Problem In Figure 5.1.1 you can see a region bounded above by the graph of a continuous function f, bounded below by the x-axis, bounded on the left by the line x = a, and bounded on the right by the line x = b. The question before us is this: What number, if any, should be called the area of ? y y = f (x)abxFigure 5.1.1To begin to answer this question, we split up the interval [a, b] into a nite number of subintervals [x0 , x1 ], [x1 , x2 ], . . . , [xn1 , xn ] This breaks up the regionwitha = x0 < x1 < < xn = b.into n subregions: 1,2, . . . ,n.(Figure 5.1.2)We can estimate the total area of by estimating the area of each subregion i and adding up the results. Lets denote by Mi the maximum value of f on [xi1 , xi ] and by23417:59 259. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.1 AN AREA PROBLEM; A SPEED-DISTANCE PROBLEM y1 2 3a = x0 x1 x2nxn = bx3xFigure 5.1.2m i the minimum value. (We know that there are such numbers because f is continuous.) Consider now the rectangles ri and Ri of Figure 5.1.3. Since ri i Ri ,Mi miaxi 1Ri1rixib axi 1xib axi 1xibFigure 5.1.3we must have area of ri area of area of Ri .iSince the area of a rectangle is the length times the width, m i (xi xi1 ) area of Settingi Mi (xi xi1 ).i Mi x i .xi = xi xi1 , we have m i xi area ofThis inequality holds for i = 1, i = 2, . . . , i = n. Adding up these inequalities, we get on the one hand(5.1.1)m 1 x1 + m 2 x2 + + m n xn area of,and on the other hand(5.1.2)area of M1 x 1 + M2 x 2 + + Mn x n .A sum of the form m 1 x1 + m 2 x2 + + m n xn(Figure 5.1.4)23517:59 260. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006236 CHAPTER 5 INTEGRATION is called a lower sum for f. A sum of the form M1 x 1 + M2 x 2 + + Mn x n(Figure 5.1.5)is called an upper sum for f. yyxx area of shaded region is a lower sum for farea of shaded region is an upper sum for fFigure 5.1.4Figure 5.1.5Inequalities 5.1.1 and 5.1.2 together tell us that for a number to be a candidate for the title area of , it must be greater than or equal to every lower sum for f and it must be less than or equal to every upper sum. It can be proven that with f continuous on [a, b] there is one and only one such number. This number we call the area of .A Speed-Distance Problem If an object moves at a constant speed for a given period of time, then the total distance traveled is given by the familiar formula distance = speed time. Suppose now that during the course of the motion the speed does not remain constant; suppose that it varies continuously. How can we calculate the distance traveled in that case? To answer this question, we suppose that the motion begins at time a, ends at time b, and during the time interval [a, b] the speed varies continuously. As in the case of the area problem, we begin by breaking up the interval [a, b] into a nite number of subintervals: [t0 , t1 ], [t1 , t2 ], . . . , [tn1 , tn ]witha = t0 < t1 < < tn = b.On each subinterval [ti1 , ti ] the object attains a certain maximum speed Mi and a certain minimum speed m i . (How do we know this?) If throughout the time interval [ti1 , ti ] the object were to move constantly at its minimum speed, m i , then it would cover a distance of m i ti units. If instead it were to move constantly at its maximum speed, Mi , then it would cover a distance of Mi ti units. As it is, the actual distance traveled, call it si , must lie somewhere in between; namely, we must have m i ti si Mi ti .17:59 261. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTIONThe total distance traveled during full the time interval [a, b], call it s, must be the sum of the distances traveled during the subintervals [ti1 , ti ]; thus we must have s = s1 + s2 + + sn . Since m 1 t1 s1 M1 t1 m 2 t2 s2 M2 t2 , . . .m n tn sn Mn tn , it follows by the addition of these inequalities that m 1 t1 + m 2 t2 + + m n tn s M1 t1 + M2 t2 + + Mn tn . A sum of the form m 1 t1 + m 2 t2 + + m n tn is called a lower sum for the speed function. A sum of the form M1 t1 + M2 t2 + + Mn tn is called an upper sum for the speed function. The inequality we just obtained for s tells us that s must be greater than or equal to every lower sum for the speed function, and it must be less than or equal to every upper sum. As in the case of the area problem, it turns out that there is one and only one such number, and this is the total distance traveled. 5.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTION The process we used to solve the two problems in Section 5.1 is called integration, and the end results of this process are called denite integrals. Our purpose here is to establish these notions in a more general way. First, some auxiliary notions.(5.2.1)By a partition of the closed interval [a, b], we mean a nite subset of [a, b] which contains the points a and b.We index the elements of a partition according to their natural order. Thus, if we say that P = {x0 , x1 , x2 , . . . , xn1 , xn }is a partition of [a, b],you can conclude that a = x0 < x1 < < xn = b.Example 1 The sets {0, 1}, {0, 1 , 1}, 2{0, 1 , 1 , 1}, 4 2{0, 1 , 1 , 1 , 5 , 1} 4 3 2 8are all partitions of the interval [0, 1]. If P = {x0 , x1 , x2 , . . . , xn1 , xn } is a partition of [a, b], then P breaks up [a, b] into n subintervals [x0 , x1 ], [x1 , x2 ], . . . , [xn1 , xn ]of lengthsx1 , x2 , . . . , xn .23717:59 262. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006238 CHAPTER 5 INTEGRATION Suppose now that f is continuous on [a, b]. Then on each subinterval [xi1 , xi ] the function f takes on a maximum value, Mi , and a minimum value, m i .The number U f (P) = M1 x1 + M2 x2 + + Mn xn (5.2.2)is called the P upper sum for f, and the number L f (P) = m 1 x1 + m 2 x2 + + m n xn is called the P lower sum for f.Example 2 The function f (x) = 1 + x 2 is continuous on [0, 1]. The partition P = {0, 1 , 3 , 1} breaks up [0, 1] into three subintervals 2 4 [x0 , x1 ] = 0, 1 , 2[x1 , x2 ] =1 3 , 2 4,[x2 , x3 ] =3 ,1 4of lengths x1 =1 2 0 = 1, 2x2 =3 4= 1, 41 2x3 = 1 3 4= 1. 4Since f increases on [0, 1], it takes on its maximum value at the right endpoint of each subinterval: M1 = f1 2= 5, 4M2 = f=3 425 , 16M3 = f (1) = 2.The minimum values are taken on at the left endpoints: m 1 = f (0) = 1,m2 = f1 2= 5, 4m3 = f3 4=25 . 16Thus U f (P) = M1 x1 + M2 x2 + M2 x3 =5 41 2+25 16+21 41 4=97 64 1.52 =and L f (P) = m 1 x1 + m 2 x2 + m 3 x3 = 1+1 25 4+1 425 161 4=77 64For a geometric interpretation of these sums, see Figure 5.2.1. yy111 23 41x1 23 4upper sumlower sum(a)(b)Figure 5.2.11x 1.20. =17:59 263. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTIONExample 3 The function f (x) = cos x is continuous on 0, 3 . The partition 4 P = 0, 1 , 1 , 1 , 2 , 3 breaks up 0, 3 into ve subintervals 6 4 2 3 4 4 [x0 , x1 ] = 0, 1 , 6[x1 , x2 ] =[x3 , x4 ] =1 2 , 2 3,1 1 , 6 4,[x2 , x3 ] =[x4 , x5 ] =1 1 , 4 2239y 1,2 3 , 3 4of lengths x1 = 1 , 6x2 =1 , 12x3 = 1 , 4x4 = 1 , 6x5 =See Figure 5.2.2. The maximum values of f on these subintervals are as follows: M1 = f (0) = cos 0 = 1,M2 = f 2,M3 = f1 4= cos 1 = 4M5 = f2 3= cos 1 = 61 2M4 = fm3 = f1 2m5 = f3 4= cos 1 = 0, 2U f (P) = 1 and 1 231 61 6+1 2 3 21 12+ 1 21 12m2 = f m4 = f++01 21 4= cos 1 = 0, 2 21 42 3+0+ 1 21 61 4= cos 1 = 41 22,= cos 2 = 1 , 3 21 6+ 1 2+ 1 2 0.37 =1 1221 12 0.06. =Both in Example 2 and in Example 3 the separation between U f (P) and L f (P) was quite large. Had we added more points to the partitions we chose, the upper sums would have been smaller, the lower sums would have been greater, and the separation between them would have been lessened. By an argument that we omit here (it appears in Appendix B.4), it can be proved that, with f continuous on [a, b], there is one and only one number I that satises the inequality L f (P) I U f (P)for all partitions P of [a, b].This is the number we want.DEFINITION 5.2.3 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTIONLet f be continuous on [a, b]. The unique number I that satises the inequality L f (P) I U f (P)for all partitions P of [a, b]is called the denite integral (or more simply the integral) of f from a to b and is denoted by bf (x) dx. a1 2Figure 5.2.2 = cos 3 = 1 2. 4 2Therefore1 21 1 6 4 3,= cos 2 = 1 . 3 21 2and the minimum values are as follows: m 1 = f 1 = cos 1 = 1 3, 6 6 2L f (P) =1 61 . 122 3 3 4x17:59 264. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006240 CHAPTER 5 INTEGRATION The symbol dates back to Leibniz and is called an integral sign. It is really an elongated Sas in Sum. The numbers a and b are called the limits of integration (a is the lower limit and b is the upper limit), and we will speak of integrating a function f from a to b. The function f being integrated is called the integrand. This is not the b only notation. Some mathematicians omit the dx and simply write a f . We will keep the dx. As we go on, you will see that it does serve a useful purpose. In the expression bf (x) dx athe letter x is a dummy variable; in other words, it can be replaced by any letter not already in use. Thus, for example, bbf (x) d(x),bf (t) dt,af (z) dzaaall denote exactly the same quantity, the denite integral of f from a to b. From the introduction to this chapter, you know that if f is nonnegative and continuous on [a, b], then the integral of f from x = a to x = b gives the area below the graph of f from x = a to x = b: bA=f (x) dx. aYou also know that if an object moves with continuous speed (t) = v(t) from time t = a to time t = b, then the integral of the speed function gives the distance traveled by the object during that time period: bs=b(t) dt =a v(t) dt.aWell come back to these applications and introduce others as we go on. Right now we carry out some computations.Example 4 (The integral of a constant function)b(5.2.4)k dx = k(b a).aIn this case the integrand is the constant function f (x) = k. To verify the formula, we take P = {x0 , x1 , . . . , xn } as an arbitrary partition of [a, b]. Since f is constantly k on [a, b], f is constantly k on each subinterval [xi1 , xi ]. Thus both m i and Mi are k, and both L f (P) and U f (P) are k x1 + k x2 + + k xn = k( x1 +x2 + +xn ) = k(b a). explain Therefore it is certainly true that L f (P) k(b a) U f (P). There is no connection between the term limit as used here and the limits introduced in Chapter 2.17:59 265. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTION241Since this inequality holds for all partitions P of [a, b], we can conclude that bf (x) dx = k(b a). ayFor example, 1 13 dx = 3[1 (1)] = 3(2) = 610andf (x) = kk2 dx = 2(10 4) = 2(6) = 12.4aIf k > 0, the region between the graph and the x-axis is a rectangle of height k erected on the interval [a, b]. (Figure 5.2.3.) The integral gives the area of this rectangle.bxFigure 5.2.3Example 5 (The integral of the identity function)b(5.2.5) ax dx = 1 (b2 a 2 ). 2Here the integrand is the identity function f (x) = x. (Figure 5.2.4.) To verify the formula we take P = {x0 , x1 , . . . , xn } as an arbitrary partition of [a, b]. On each subinterval [xi1 , xi ], the function f (x) = x has a maximum value Mi and a minimum value m i . Since f is an increasing function, the maximum value occurs at the right endpoint of the subinterval and the minimum value occurs at the left endpoint. Thus Mi = xi and m i = xi1 . It follows thaty f (x) = xaU f (P) = x1 x1 + x2 x2 + + xn xnband L f (P) = x0 x1 + x1 x2 + + xn1 xn .Figure 5.2.4For each index i xi1 1 (xi + xi1 ) xi . 2() Multiplication by(explain)xi = xi xi1 gives xi1 xi 1 (xi + xi1 )(xi xi1 ) xi xi , 2which we write as xi1 xi 1 22 xi2 xi1 xi xi .Summing from i = 1 to i = n, we nd that ()L f (P) 1 22 2 x1 x0 +1 22 2 x2 x1 + +1 2The sum in the middle collapses to 1 22 2 xn x0 = 1 (b2 a 2 ). 2Consequently L f (P) 1 (b2 a 2 ) U f (P). 22 2 xn xn1 U f (P).x17:59 266. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006242 CHAPTER 5 INTEGRATION ySince P was chosen arbitrarily, we can conclude that this inequality holds for all partitions P of [a, b]. It follows that f (x) = xb ax dx = 1 (b2 a 2 ). 2For example, a3xbb area of shaded region: a x dx = 1 2 a2) (b122x dx = 1 [32 (1)2 ] = 1 (8) = 4 2 2and 2x dx = 1 [22 (2)2 ] = 0. 2If the interval [a, b] lies to the right of the origin, then the region below the graph ofFigure 5.2.5f (x) = x,x [a, b]is the trapezoid shown in Figure 5.2.5. The integral byx dx a9gives the area of this trapezoid: A = (b a)[ 1 (a + b)] = 1 (b2 a 2 ). 2 2f (x) = x2Example 6 3x 2 dx =11 12area of shaded region:x3 3 1 x 2dx =26 326 . 3(Figure 5.2.6)Let P = {x0 , x1 , . . . , xn } be an arbitrary partition of [1, 3]. On each subinterval 2 [xi1 , xi ] the function f (x) = x 2 has a maximum Mi = xi2 and a minimum m i = xi1 . It follows that 2 2 U f (P) = x1 x1 + + xn xnandFigure 5.2.62 2 L f (P) = x0 x1 + + xn1 xn .For each index i, 1 i n, 2 2 3xi1 xi1 + xi1 xi + xi2 3xi2 .(Verify this)Division by 3 gives 2 xi1 1 32 xi1 + xi1 xi + xi2 xi2 .We now multiply this inequality by 1 3xi = xi xi1 . The middle term then becomes2 xi1 + xi1 xi + xi2 (xi xi1 ) =1 33 xi3 xi1 ,and shows that 2 xi1 xi 1 33 xi3 xi1 xi2 xi .The sum of the terms on the left is L f (P). The sum of all the middle terms collapses to 26 : 3 1 33 3 3 3 3 3 x1 x0 + x2 x1 + + xn xn1 =1 33 3 xn x0 = 1 (33 13 ) = 3The sum of the terms on the right is U f (P). Clearly, then, L f (P) 26 3 U f (P).26 . 317:59 267. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.2 THE DEFINITE INTEGRAL OF A CONTINUOUS FUNCTION243Since P was chosen arbitrarily, we can conclude that this inequality holds for all partitions P of [1, 3]. It follows that 3x 2 dx =26 . 31The Integral as the Limit of Riemann Sums For a function f continuous on [a, b], we have dened the denite integral bf (x) dx aas the unique number that satises the inequality bL f (P) f (x) dx U f (P)for all partitions P of [a, b].aThis method of obtaining the denite integral (squeezing toward it with upper and lower sums) is called the Darboux method. There is another way to obtain the integral that, in some respects, has distinct advantages. Take a partition P = {x0 , x1 , . . . , xn } of [a, b]. P breaks up [a, b] into n subintervals [x0 , x1 ], [x1 , x2 ], . . . , [xn1 , xn ] of lengths x1 , x2 , . . . , xn . Now pick a point x1 from [x0 , x1 ] and form the product f (x1 ) x1 ; pick a point x2 from [x1 , x2 ] and form the product f (x2 ) x2 ; go on in this manner until you have formed the products f (x1 ) x1 , f (x2 ) x2 , . . . , f (xn ) xn .The sum of these products S (P) = f (x1 ) x1 + f (x2 ) x2 + + f (xn ) xn is called a Riemann sum.Since m i f (xi )y Mi for each index i, its clear that9L f (P) S (P) U f (P).(5.2.6)This inequality holds for all partitions P of [a, b].Example 7 Let f (x) = x 2 , x [1, 3]. Take P = 1, 3 , 2, 3 and set 2 x1Here= x37 , 41(Figure 5.2.7)5 41 + f 27 41 + f 25 21 =25 1611 2+49 161 2+25 (1) 4=137 16 8.5625. =23 2 5 4After the French mathematician J. G. Darboux (18421917). After the German mathematician G. F. B. Riemann (18261866).=5 . 2x1 = 1 , x2 = 1 , x3 = 1. Therefore 2 2S (P) = f = x25 , 47 43 5 2Figure 5.2.7x17:59 268. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006244 CHAPTER 5 INTEGRATION In Example 6 we showed that 3x 2 dx =11 3 3 x 1 3=27 31 3=26 3 8.667. =Our Riemann approximation is pretty good. For each partition P of [a, b], we dene P , the norm of P, by setting P = max xi ,i = 1, 2, . . . , n.The denite integral of f is the limit of Riemann sums in the following sense: given any > 0, there exists a > 0 such that P < ,ifb S (P) thenf (x) dx a g(x) dx. Which of the statements necessarily holds for all partitions P of [a, b]? Justify your answer. 26. L g (P) < L f (P). 25. L g (P) < U f (P).t2 + ti [ti1 , ti ],b29. U f (P) > abf (x) dx. a31. A partition P = {x0 , x1 , x2 , . . . , xn1 , xn } of [a, b] is said to be regular if the subintervals [xi1 , xi ] all have the same length x = (b a)/n. Let P = {x0 , x1 , . . . , xn1 , xn } be17:59 270. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006246 CHAPTER 5 INTEGRATION (b) Show thata regular partition of [a, b]. Show that if f is continuous and increasing on [a, b], thenb2 [1 + 2 + 3 + + n]. n2 (c) Use Exercise 35 to show that U f (P) =U f (P) L f (P) = [ f (b) f (a)] x. 32. Let P = {x0 , x1 , x2 , . . . , xn1 , xn } be a regular partition of the interval [a, b]. (See Exercise 31.) Show that if f is continuous and decreasing on [a, b], thenCU f (P) L f (P) = [ f (a) f (b)] x. 33. Set f (x) = 1 + x 2 . (a) Verify that f increases on [0, 2]. (b) Let P = {x0 , x1 , . . . , xn1 , xn } be a regular partition of [0, 2]. Determine a value of n such that 20L f (P) = 1 b2 (1 P ) and U f (P) = 1 b2 (1 + P ). 2 2 (d) Show that for all choices of xi -pointsf (x) dx L f (P) 0.1.L f (P) =b3 2 [0 + 12 + 22 + + (n 1)2 ]. n3b3 2 [1 + 22 + 32 + + n 2 ]. n3 (c) Use Exercise 36 to show that U f (P) =L f (P) = 1 b3 (2 3 P + P 2 ) 6 U f (P) =f (x) dx L f (P) 0.1.1 + 2 + 3 + + k = 1 k(k + 1). 2 36. Show by induction that for each positive integer k, 12 + 22 + 32 + + k 2 = 1 k(k + 1)(2k + 1). 6 37. Let P = {x0 , x1 , x2 , . . . , xn1 , xn } be a regular partition of the interval [0, b], and set f (x) = x. (a) Show that1 3 b (2 6lim S(P) = 1 b3 3+ 3 P + P 2 ). P 0band therefore 0x 2 dx = 1 b3 . 339. Let f be a function continuous on [a, b]. Show that if P is a partition of [a, b], then L f (P), U f (P), and 1 [L f (P) + U f (P)] are all Riemann sums. 2 Exercises 4043. Using a regular partition P with 10 subintervals, estimate the integral (a) by L f (P) and by U f (P), (b) by 1 [L f (P) + U f (P)], 2 (c) by S (P) using the midpoints of the subintervals. How does this result compare with your result in part (b)? 2 1 40. (x 3 + 2) dx. 41. x dx. 0b2 L f (P) = 2 [0 + 1 + 2 + 3 + + (n 1)]. nand(d) show that for all choices of xi -points0(c) Use a programmable calculator or computer to calculate 1 0 f (x) dx with an error of less than 0.05. NOTE: You will see in Chapter 7 that the exact value of this integral is /4. 35. Show by induction that for each positive integer k,x dx = 1 b2 . 2(b) Show that34. Set f (x) = 1/(1 + x 2 ). (a) Verify that f decreases on [0, 1]. (b) Let P = {x0 , x1 , . . . , xn1 , xn } be a regular partition of [0, 2]. Determine a value of n such that 0038. Let P = {x0 , x1 , x2 , . . . , xn1 , xn } be a regular partition of [0, b], and let f (x) = x 2 . (a) Show that(c) Use a programmable calculator or computer to calculate 2 0 f (x) dx with an error of less than 0.1.2and thereforeP 00Cblim S (P) = 1 b2 20 242. 011 dx. 1 + x2x a 5.3 THE FUNCTION F(x) =43.sin x dx.0f (t) dtThe evaluation of the denite integral bf (x) dx adirectly from upper and lower sums or from Riemann sums is usually a laborious and difcult process. Try, for example, to evaluate 5 2x3 + x2 2x 1 x21/4dxor 1/2x dx 1 x217:59 271. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006x a5.3 THE FUNCTION F(x) =f (t) dt247from such sums. Theorem 5.4.2, called the fundamental theorem of integral calculus, gives us another way to evaluate such integrals. This other way depends on a connection between integration and differentiation described in Theorem 5.3.5. Along the way we will pick up some information that is of interest in itself. THEOREM 5.3.1Suppose that f is continuous on [a, b], and P and Q are partitions of [a, b]. If Q P, then L f (P) L f (Q)andU f (Q) U f (P).This result can be justied as follows: By adding points to a partition, we make the subintervals [xi1 , xi ] smaller. This tends to make the minima, m i , larger and the maxima, Mi , smaller. Thus the lower sums are made bigger, and the upper sums are made smaller. The idea is illustrated (for a positive function) in Figures 5.3.1 and 5.3.2.as points are added to a partition, the lower sums tend to get biggerFigure 5.3.1as points are added to a partition, the upper sums tend to get smallerFigure 5.3.2The next theorem says that the integral is additive on intervals.THEOREM 5.3.2If f is continuous on [a, b] and a < c < b, then c abf (t) dt +bf (t) dt =cf (t) dt. afFor nonnegative functions f , this theorem is easily understood in terms of area. The area of part I in Figure 5.3.3 is given by cf (t) dt; aI aII cFigure 5.3.3b17:59 272. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006248 CHAPTER 5 INTEGRATION the area of part II by bf (t) dt; cand the area of the entire region by bf (t) dt. aThe theorem says that the area of part I + the area of part II = the area of the entire region.The fact that the additivity theorem is so easy to understand does not relieve us of the necessity to prove it. Here is a proof. PROOF OF THEOREM 5.3.2To prove the theorem, we need only show that for each partitionP of [a, b] cL f (P) bf (t) dt +af (t) dt U f (P).(Why?)cWe begin with an arbitrary partition of [a, b]: P = {x0 , x1 , . . . , xn }. Since the partition Q = P {c} contains P, we know from Theorem 5.3.1 that L f (P) L f (Q)andU f (Q) U f (P).Q 1 = Q [a, c](1)andQ 2 = Q [c, b]The setsare partitions of [a, c] and [c, b], respectively. Moreover L f (Q 1 ) + L f (Q 2 ) = L f (Q)(2)andU f (Q 1 ) + U f (Q 2 ) = U f (Q).andL f (Q 2 ) Since cL f (Q 1 ) f (t) dt U f (Q 1 )abf (t) dt U f (Q 2 ),cwe have cL f (Q 1 ) + L f (Q 2 ) bf (t) dt +af (t) dt U f (Q 1 ) + U f (Q 2 ),cand thus by (2), cL f (Q) bf (t) dt +af (t) dt U f (Q).cTherefore, by (1), cL f (P) abf (t) dt + cf (t) dt U f (P).17:59 273. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.3 THE FUNCTION F(x) =Until now we have integrated only from left to right: from a number a to a number b greater than a. We integrate in the other direction by dening a(5.3.3)bf (t) dt = bf (t) dt. aThe integral from any number to itself is dened to be zero: c(5.3.4)f (t) dt = 0.cWith these additional conventions, the additivity condition cbf (t) dt +abf (t) dt =cf (t) dt aholds for all choices of a, b, c from an interval on which f is continuous, no matter what the order of a, b, c happens to be. We have left the proof of this to you as an exercise. (Exercise 16) We are now ready to state the all-important connection that exists between integration and differentiation. Our rst step is to point out that if f is continuous on [a, b] and c is any number in [a, b], then for each x in [a, b], the integral xf (t) dt cis a number, and consequently we can dene a function F on [a, b] by setting xF(x) =f (t) dt. cTHEOREM 5.3.5Let f be continuous on [a, b] and let c be any number in [a, b]. The function F dened on [a, b] by setting xF(x) =f (t) dt cis continuous on [a, b], differentiable on (a, b), and has derivative F (x) = f (x)for all x in (a, b).We will prove the theorem for the special case where the integration that denes F is begun at the left endpoint a; namely, we will prove the theorem for the following function: PROOFxF(x) =f (t) dt. a(The more general case is left to you as Exercise 34.)We begin with x in the half-open interval [a, b) and show that lim+h0F(x + h) F(x) = f (x). hx af (t) dt24917:59 274. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006250 CHAPTER 5 INTEGRATION A pictorial argument that applies to the case where f > 0 is roughed out in Figure 5.3.4. ff (x)af (x + h)xbx+hhF(x) = area from a to x and F(x + h) = area from a to x + h. Therefore F (x + h) F(x) = area from x to x + h. For small h this is approximately f (x) h. Thus F(x + h) F(x) f (x) h = f (x). is approximately h hFigure 5.3.4Now to a proof. For a x < x + h < b, xx+hf (t) dt +ax+hf (t) dt =xf (t) dt. aTherefore x+hxf (t) dt ax+hf (t) dt =af (t) dt, xwhich, by the denition of F, gives x+hF(x + h) F(x) =(1)f (t) dt. xOn the interval [x, x + h], an interval of length h, f takes on a maximum value Mh and a minimum value m h . On [x, x + h], the product Mh h is an upper sum for f and m h h is a lower sum for f . (Use the partition {x, x + h}.) Therefore x+hmh h f (t) dt Mh h.xIt follows from (1) and the fact that h is positive that F(x + h) F(x) Mh . h Since f is continuous on [x, x + h], mh lim m h = f (x) = lim+ Mhh0+h0and thus F(x + h) F(x) = f (x). h0 h This last statement follows from the pinching theorem, Theorem 2.5.1, which, as we remarked in Section 2.5, applies also to one-sided limits. In a similar manner we can prove that, for x in the half-open interval (a, b], (2)(3)lim+F(x + h) F(x) = f (x). h For x in the open interval (a, b), both (2) and (3) hold, and we have limh0F (x) = limh0F(x + h) F(x) = f (x). h17:59 275. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.3 THE FUNCTION F(x) =This proves that F is differentiable on (a, b) and has derivative F (x) = f (x). All that remains to be shown is that F is continuous from the right at a and continuous from the left at b. Limit (2) at x = a gives F(a + h) F(a) = f (a). hlim+h0Now, for h > 0, F(a + h) F(a) =F(a + h) F(a) h, hand so F(a + h) F(a) h hlim+ [F(a + h) F(a)] = lim+h0h0= f (a) lim+ h = 0. h0Therefore lim F(a + h) = F(a).h0+This shows that F is continuous from the right at x = a. The continuity of F from the left at x = b can be shown in a similar manner by applying limit (3) at x = b. Example 1 The function F(x) = F (x) = 2x + xx 1(2t + t 2 ) dt for all x [1, 5] has derivative for all x (1, 5). 2Example 2 For all real x, dene xF(x) =sin t dt.0Find F ( 3 ) and F ( 1 ). 4 2 SOLUTION By Theorem 5.3.5,Thus, F3 4= sin3 4F (x) = sin x = 1 2 and F 2for all real x. 1 2= sin 1 = 1. 2Example 3 Set xF(x) = 01 dt 1 + t2for all real numbers x.(a) Find the critical points of F and determine the intervals on which F increases and the intervals on which F decreases. (b) Determine the concavity of the graph of F and nd the points of inection (if any). (c) Sketch the graph of F. SOLUTION(a) To nd the intervals on which F increases and the intervals on which F decreases, we examine the rst derivative of F. By Theorem 5.3.5, F (x) =1 1 + x2for all real x.x af (t) dt25117:59 276. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006252 CHAPTER 5 INTEGRATION Since F (x) > 0 for all real x, F increases on (, ); there are no critical points. (b) To determine the concavity of the graph and to nd the points of inection, we use the second derivative 2x . F (x) = (1 + x 2 )2yThe sign of F and the behavior of the graph of F are as follows: sign of F': behavior of graph: x+ + + + + +0 0concave upconcave downpoint of inflection(c) Since F(0) = 0 and F (0) = 1, the graph passes through the origin with slope 1. A sketch of the graph is shown in Figure 5.3.5. As youll see in Chapter 7, the graph has two horizontal asymptotes: y = 1 and y = 1 . 2 2Figure 5.3.5EXERCISES 5.3 1. Given that 12f (x) dx = 6,05f (x) dx = 4,0f (x) dx = 1,2nd the following: 5(a)2f (x) dx.(b)0(c)f (x) dx.1 0(d)1 0f (x) dx.(e)(f)0f (x) dx.2 418f (x) dx = 7,3f (x) dx = 11,183(b)f (x) dx.4(c)4 8(d)3f (x) dx.f (x) dx.t t 2 + 1 dt.(e)4f (x) dx.3(f)8f (x) dx. 43. Use upper and lower sums to show that 2dx < 1. 1 x 4. Use upper and lower sums to show that 0.5 1, set F(x) =1dx0 1+x x 0 t t +2cos t dt.< 1.1 dt.(b) Find F (x). (c) Find F (2). (d) Express F(2) as an integral of t t + 1. (e) Express F(x) as an integral of t t + 1. x 6. Let F(x) = t sin t dt. (b) Find F (x).sin t dt.x12. F(x) =(t + 1)3 dt.213. Show that statements (a) and (b) are false. (a) U f (P1 ) = 4 for the partition P1 = 0, 1, 3 , 2 , and 2 for the partitionP2 = 0, 1 , 1, 3 , 2 . 4 2(b) L f (P1 ) = 5for the partitionP1 = 0, 1, 3 , 2 , and 2L f (P2 ) = 4 for the partition P2 = 0, 1 , 1, 3 , 2 . 4 2 14. (a) Which continuous functions f dened on [a, b], have the property that L f (P) = U f (P) for some partition P? (b) Which continuous functions f dened on [a, b] have the property that L f (P) = U f (Q) for some partitions P and Q? 15. Which continuous functions f dened on [a, b] have the property that all lower sums L f (P) are equal? 16. Show that if f is continuous on an interval I, then(a) Find F(0).(a) Find F( ).x10. F(x) =U f (P2 ) = 51 4t 2 + 1 dt.x1 x11. F(x) =(d) F (x). 08. F(x) =1nd the following: f (x) dx.dt . t2 + 9xf (x) dx = 5,(a)19. F(x) =52. Given that 4x7. F(x) =1f (x) dx.0(c) F ( 1 ). 2(a) F (1). (b) F (0).5f (x) dx.(d) Express F(2) as an integral of t sin t. (e) Express F(x) as an integral of t sin t. Exercises 712. Calculate the following for each F given below:(c) Find F ( 1 ). 2c abf (t) dt + cbf (t) dt =f (t) dt afor every choice of a, b, c from I. HINT: Assume a < b and consider the four cases: c = a, c = b, c < a, b < c. Then consider what happens if a > b or a = b.17:59 277. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.3 THE FUNCTION F(x) =Exercises 17 and 18. Find the critical points for F and, at each critical point, determine whether F has a local maximum, a local minimum, or neither. x x t 1 t 4 17. F(x) = dt. 18. F(x) = dt. 2 2 0 1+t 0 1+t x19. For x > 0, set F(x) =(1/t) dt. 1(a) Find the critical points for F, if any, and determine the intervals on which F increases and the intervals on which F decreases. (b) Determine the concavity of the graph of F and nd the points of inection, if any. (c) Sketch the graph of F. x20. Let F(x) =(a) Find the critical points for F and determine the intervals on which F increases and the intervals on which F decreases. (b) Determine the concavity of the graph of F and nd the points of inection, if any. (c) Sketch the graph of F. 21. Suppose that f is differentiable with f (x) > 0 for all x, and suppose that f (1) = 0. Set xF(x) =f (t) dt. 0Justify each statement. (a) F is continuous. (b) F is twice differentiable. (c) x = 1 is a critical point for F. (d) F takes on a local minimum at x = 1. (e) F(1) < 0. Make a rough sketch of the graph of F. 22. Suppose that g is differentiable with g (x) < 0 for all x < 1, g (1) = 0, and g (x) > 0 for all x > 1, and suppose that g(1) = 0. Set xG(x) =g(t) dt. 0Justify each statement. (a) G is continuous. (b) G is twice differentiable. (c) x = 1 is a critical point for G. (d) The graph of G is concave down for x < 1 and concave up for x > 1. (e) G is an increasing function. Make a rough sketch of the graph of G. 23. (a) Sketch the graph of the function 2 x, 2 + x, x 125324. (a) Sketch the graph of the function x 2 + x, 0 x 1 2x, 1 < x 3.f (x) = x(b) Calculate F(x) =f (t) dt, 0 x 3, and sketch the0graph of F. (c) What can you conclude about f and F at x = 1? Exercises 2528. Calculate F (x). x3HINT: Set u = x 3 and use the chain rule.t cos t dt.25. F(x) 0cos x26. F(x) =t(t 3)2 dt.(b) Calculate F(x) =f (t) dt1 t 2 dt.10f (x) =x a1 x 0 0 < x 3.127. F(x) =(t sin2 t) dt.x2 xt2 dt. 1 + t4 0 x sin 2t 29. Set F(x) = 2x + dt. Determine 1 + t2 0 (a) F(0). (b) F (0). (c) F (0). 28. F(x) =x2sin 2t dt. Determine 1 + t2 0 (a) F(0). (b) F (x). 31. Assume that f is continuous and 30. Set F(x) = 2x +x2x . 4 + x2f (t) dt =0(a) Determine f (0). (b) Find the zeros of f , if any. 32. Assume that f is continuous and xf (t) dt = sin x x cos x.0(a) Determine f 1 . (b) Find f (x). 2 33. (A mean-value theorem for integrals) Show that if f is continuous on [a, b], then there is a least one number c in (a, b) for which bf (x) dx = f (c)(b a).a34. We proved Theorem 5.3.5 only in the case that the integration which denes F is begun at the left endpoint a. Show that the result still holds if the integration is begun at an arbitrary point c (a, b). 35. Let f be continuous on [a, b]. For each x [a, b] set xF(x) =f (t) dt,andcf (t) dt, 1 x 3, and sketchthe graph of F. (c) What can you conclude about f and F at x = 0?f (t) dt dtaking c and d from [a, b]. (a) Show that F and G differ by a constant. (b) Show that F(x) G(x) =xG(x) =d cf (t) dt.17:59 278. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006254 CHAPTER 5 INTEGRATION 36. Let f be everywhere continuous and set xF(x) =t 0Con which the graph of F is concave down. Produce a gure that displays both the graph of F and the graph of F .tf (u) du dt. 1Find (a) F (x). (b) F (1). (c) F (x). (d) F (1). Exercises 3740. Use a CAS to carry out the following steps: (a) Solve the equation F (x) = 0. Determine the intervals on which F increases and the intervals on which F decreases. Produce a gure that displays both the graph of F and the graph of F . (b) Solve the equation F (x) = 0. Determine the intervals on which the graph of F is concave up and the intervalsx37. F(x) =(t 2 3t 4) dt.0 x38. F(x) =(2 3 cos t) dt,x [0, 2]0 039. F(x) =sin 2t dt,x [0, 2]x 040. F(x) =(2 t)2 dt.x 5.4 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS The natural setting for differentiation is an open interval. For functions f dened on an open interval, the antiderivatives of f are simply the functions with derivative f . For continuous functions dened on a closed interval [a, b], the term antiderivative takes into account the endpoints a and b.DEFINITION 5.4.1ANTIDERIVATIVE ON [a, b]Let f be continuous on [a, b]. A function G is called an antiderivative for f on [a, b] if G is continuous on [a, b]G (x) = f (x)andfor all x (a, b).Theorem 5.3.5 tells us that if f is continuous on [a, b], then xF(x) =f (t) dt ais an antiderivative for f on [a, b]. This gives us a prescription for constructing antiderivatives. It tells us that we can construct an antiderivative for f by integrating f . The theorem below, called the fundamental theorem, goes the other way. It gives us a prescription, not for nding antiderivatives, but for evaluating integrals. It tells us that we can evaluate the integral bf (t) dt afrom any antiderivative of f by evaluating the antiderivative at b and at a.THEOREM 5.4.2THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUSLet f be continuous on [a, b]. If G is any antiderivative for f on [a, b], then b af (t) dt = G(b) G(a).17:59 279. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.4 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS PROOFFrom Theorem 5.3.5 we know that the function xF(x) =f (t) dt ais an antiderivative for f on [a, b]. If G is also an antiderivative for f on [a, b], then both F and G are continuous on [a, b] and satisfy F (x) = G (x) for all x in (a, b). From Theorem 4.2.4 we know that there exists a constant C such that F(x) = G(x) + Cfor all x in [a, b].Since F(a) = 0, G(a) + C = 0and thusC = G(a).It follows that F(x) = G(x) G(a)for all x in [a, b].In particular, bf (t) dt = F(b) = G(b) G(a). aWe now evaluate some integrals by applying the fundamental theorem. In each case we use the simplest antiderivative we can think of. 4x 2 d x.Example 1 Evaluate 1SOLUTION As an antiderivative for f (x) = x 2 , we can use the functionG(x) = 1 x 3 . 3(Verify this.)By the fundamental theorem, 4 1x 2 d x = G(4) G(1) = 1 (4)3 1 (1)3 = 3 364 31 3= 21.NOTE: Any other antiderivative of f (x) = x 2 has the form H (x) = 1 x 3 + C for 3 some constant C. Had we chosen such an H instead of G, then we would have had 4x 2 d x = H (4) H (1) =11 (4)3 3the Cs would have canceled out.+C 1 (1)3 3+C =64 3+C 1 3 C = 21;/2Example 2 Evaluatesin x dx. 0SOLUTION Here we use the antiderivative G(x) = cos x: /2sin x dx = G(/2) G(0)0= cos(/2) [ cos(0)] = 0 (1) = 1. Expressions of the form G(b) G(a) are conveniently writtenNotationbG(x) . aIn this notation 4x 2 dx =14 1 3 x 3 1= 1 (4)3 1 (1)3 = 21 3 3and /2 0sin x dx = cos x/2 0= cos(/2) [ cos(0)] = 1. 25517:59 280. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006256 CHAPTER 5 INTEGRATION To calculate bf (x) d x aby the fundamental theorem, we need to nd an antiderivative for f . We do this by working back from the results of differentiation. For rational r, d r +1 (x ) = (r + 1)x r . dx Thus, if r = 1, x r +1 r +1d dx= xr .This tells us that G(x) =x r +1 r +1is an antiderivative for f (x) = x r .Some common trigonometric antiderivatives are listed in Table 5.4.1. Note that in each case the function on the left is the derivative of the function on the right. Table 5.4.1FunctionAntiderivativeFunctionAntiderivativesin x sec2 x sec x tan x cos x tan x sec xcos x csc2 x csc x cot xsin x cot x csc xWe continue with computations. 2 1 1dx = x3t 5/3 dt =0 /3 /4 /2 /62x 3 d x =1 13 8/3 t 8 0sec2 t dt = tan t1Example 3 Evaluate2= 11 2x 22= 1 1 = 3, 8 2 8 1= 3 (1)8/3 3 (0)8/3 = 3 . 8 8 8/3 /4csc x cot x dx = csc xx 2 2= tan/2 /6 tan = 3 (1) = 3 + 1. 3 4= csc csc = 1 (2) = 1. 2 6(2x 6x 4 + 5)d x.0SOLUTION As an antiderivative we use G(x) = x 2 6 x 5 + 5x: 5 1 0(2x 6x 4 + 5)d x = x 2 6 x 5 + 5x 5 1Example 4 Evaluate11 0=1(x 1)(x + 2) d x.SOLUTION First we carry out the indicated multiplication:(x 1)(x + 2) = x 2 + x 2.6 5+5=24 . 517:59 281. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.4 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUSAs an antiderivative we use G(x) = 1 x 3 + 1 x 2 2x: 3 2 1 1(x 1)(x + 2) d x =1 3 x 31+ 1 x 2 2x 2= 10 . 31We now give some slightly more complicated examples. The essential step in each case is the determination of an antiderivative. Check each computation in detail. 2 1 5x4 + 1 dx = x2(x 2 + x 2 ) d x =5(x 1)1/2 d x =1x)2 d x =0 x 1 1)3/22 (x 31(4 1 3 x 31x 1 dx =1 1221 (16 8 x + x) d x = 16x 1dt = (t + 2)22(t + 2)2 dt = (t + 2)112 117 . 6=516 . 316 3/2 x 30 2=1+ 1 x2 21 0=67 . 61 = 12 .The Linearity of the Integral The preceding examples suggest some simple properties of the integral that are used regularly in computations. Throughout, take f and g as continuous functions and and as constants. I. Constants may be factored through the integral sign: b(5.4.3)b f (x) dx = f (x) dx.aaFor example, 4 1 /43 x 7dx =4 3 71 /42 cos x dx = 20x 1/2 d x =3 x 3/2 7 3/24= 1cos x dx = 2 sin x0/4 02 7(4)3/2 (1)3/2 = 2 [8 1] = 2. 7= 2 sin sin 0 42 = 2. 2 II. The integral of a sum is the sum of the integrals: =2bb[ f (x) + g(x)] dx =(5.4.4) abf (x) dx +ag(x) dx. aFor example, /2 0/2(sin x + cos x) dx =/2sin x dx +0= cos xcos x dx 0/2 0+ sin x/2 0= ( cos /2) ( cos 0) + sin /2 sin 0 = 1 + 1 = 2.25717:59 282. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006258 CHAPTER 5 INTEGRATION III. The integral of a linear combination is the linear combination of the integrals: bb[ f (x) + g(x)] dx = (5.4.5) abf (x) dx + g(x) dx.aaThis applies to the linear combination of more than two functions. For example, 11(2x 6x 4 + 5) dx = 201x dx 60=20 1x2 26 0x5 51x 4d x +5 dx 011+ 5x 0=16 5+5=24 . 50This is the result obtained in Example 3. Properties I and II are particular instances of Property III. To prove III, let F be an antiderivative for f and let G be an antiderivative for g. Then, since [ F(x) + G(x)] = F (x) + G (x) = f (x) + g(x), it follows that F + G is an antiderivative for f + g. Therefore, b[ f (x) + g(x)] dx = F(x) + G(x)ab a= [ F(b) + G(b)] [ F(a) + G(a)] = [F(b) F(a)] + [G(b) G(a)] b= a /4Example 5 Evaluatebf (x) dx + g(x) dx. asec x[2 tan x 5 sec x]d x.0SOLUTION /4/4sec x[2 tan x 5 sec x] dx =0[2 sec x tan x 5 sec2 x] dx0 /4=2/4sec x tan x dx 50= 2 sec xsec2 x dx0 /4 0 5 tan x/4 0 = 2 sec sec 0 5 tan tan 0 4 4 = 2[ 2 1] 5[1 0] = 2 2 7. EXERCISES 5.4 Exercises 134. Evaluate the integral. 11.(2x 3) dx.12.04(3x + 2) dx.5.025x 4 d x. 14. 146.103. 2 x d x.7. 1x d x.0 5(2x + x 2 ) dx. 3 2 x 1d x.28. 13 + 5x x3d x.17:59 283. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.4 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS 09. 2(x + 1)(x 2) dx. 4 3t + 2 t11. 113.1dt.(x 3/2 x 1/2 ) dx.12.545. (a)14.07x 6 d x.1(x + 1)17 d x.2(x 3/4 2x 1/2 ) dx.a16.0(a 2 x x 3 ) dx.0 a17. ( a x)2 d x.18.6t dt. t310 219. 1 22x(x + 1)d x. 221.x2 1 x223x (x + 1) dx.50.2 324.03 sin x dx.51. 20 /422 sec xd x.25.2/326.sec x tan x dx. /60 /427./3csc u cot u du.28./6/4 229.sin x dx.30.00 /30 /232. /4 333. 01 2csc2 u du. cos x dx.2 x 2 sec2 x d x. 31.x/2 0(t + 2)2 dt.x36.111 1 dx = x3 2x 22 2 22 00= 1 1 = 0.= 0 0 = 0.1 1 = 8 8= 0. x d x =v(t) = 10t t 2 ,0 t 10.(a) What is the position of the object at any time t, 0 t 10? (b) When is the objects velocity a maximum, and what is its position at that time? 54. The velocity of a bob suspended on a spring is given: t 0.At time t = 0, the bob is one unit below the equilibrium position. (See the gure.) (a) Determine the position of the bob at each time t 0. (b) What is the bobs maximum displacement from the equilibrium position?d (sin3 x) d x. dx(cos t sin t) dt.0 2x+137.sec2 x dx = tan x2v(t) = 3 sin t + 4 cos t,34. cos x dx./2x cos x dx = x sin x + cos xcsc x(cot x 3 csc x) dx. d ( 4 + x 2 ) d x. dx x 2 1 dx.1 = 2 (2) = 0. x x 2 2 2 53. An object starts at the origin and moves along the x-axis with velocity 52.Exercises 3538. Calculate the derivative with respect to x (a) without integrating; that is, using the results of Section 5.3; (b) by integrating and then differentiating the result. 35.(b)0 2cos x dx.23. 2x + 3 dx.2 00 /2cos x dx.48. (a)49.222.(b)2 d x.11(x 2 1) dx.4 2Exercises 4952. Determine whether the calculation is valid. If it is not valid, explain why it is not valid.(x 2)2 d x.31(b)4 247. (a) x 3 dx.2(2x + 3) dx./220.1(b) 246. (a)015.5(x 3) dx.21 12 1 2sec u tan u du.38.259Exercises 4548. Evaluate.(t 3 + t 2 ) dt.121010.t(t 1) dt.x239. Dene a function F on [1, 8] such that F (x) = 1/x and (a) F(2) = 0; (b) F(2) = 3. 40. Dene a function F on [0, 4] such that F (x) = 1 + x 2 and (a) F(3) = 0; (b) F(3) = 1. Exercises 4144. Verify that the function is nonnegative on the given interval, and then calculate the area below the graph on that interval. 41. f (x) = 4x x 2 ; [0, 4]. 42. f (x) = x x + 1; [1, 9]. 43. f (x) = 2 cos x; [/2, /4]. 44. f (x) = sec x tan x; [0, /3].0equilibrium position1Exercises 5558. Evaluate the integral. 4f (x)d x;f (x) =f (x)d x;55.f (x) =0 456. 22x + 1, 0 x 1 4 x, 1 < x 4. 2 + x 2 , 2 x < 0 + 2, 0 x 4.1 x 217:59 284. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006260 CHAPTER 5 INTEGRATION f (x)d x;f (x) =1 + 2 cos x, /2 x /3 (3/ )x + 1, /3 < x .f (x)d x;57.f (x) =2 sin x, 0 x /2 2 + cos x, /2 < x 3/2./2 3/258. 0(a) Carry out the integration. (b) Sketch the graphs of f and g. (c) Where is f continuous? Where is f differentiable? Where is g differentiable? 61. (Important) If f is a function and its derivative f is continuous on [a, b], then x + 2, 2 x 0 2, 0 0bfor all x [a, b],thenf (x) dx > 0.aReasoning: (5.8.1) holds because in this case all of the lower sums L f (P) are nonnegative; (5.8.2) holds because in this case all the lower sums are positive. II. The integral is order-preserving: for continuous functions f and g,(5.8.3)iff (x) g(x) for all x [a, b],bthenbf (x) d x ag(x) d x aand (5.8.4)if f (x) < g(x)for all x [a, b],bthen(5.8.3) Thus by (5.8.1)bf (x) d x< aPROOF OF r 2 x 2, x r 2 x 2 dx.Set x = r sin u, d x = r cos u du. 86. Find the area enclosed by the ellipse b2 x 2 + a 2 y 2 = a 2 b2 . 5.8 ADDITIONAL PROPERTIES OF THE DEFINITE INTEGRALif281085. (The area of a circular region) The circle x 2 + y 2 = r 2 encloses a circular disc of radius r. Justify the familiar formula A = r 2 by integration. HINT: The quarter-disk in the rst(5.8.1)rA=4g(x) d x. aIf f (x) g(x) on [a, b], then f (x) f (x) 0 on [a, b]. b[g(x) f (x)] dx 0.aThis gives bbg(x) dx af (x) dx 0aand shows that bbf (x) dx aThe proof of (5.8.4) is similarly simple.g(x) dx. a17:59 306. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006282 CHAPTER 5 INTEGRATION III. Just as the absolute value of a sum of numbers is less than or equal to the sum of the absolute values of those numbers, x1 + x2 + + xn x1 + x2 + + xn , the absolute value of an integral of a continuous function is less than or equal to the integral of the absolute value of that function: b(5.8.5)bf (x) dx aPROOF OF (5.8.5) f (x) dx.aSince f (x) f (x) f (x) , it follows from (5.8.3) that bb f (x) dx abf (x) dx a f (x) dx.aThis pair of inequalities is equivalent to (5.8.5).IV. If f is continuous on [a, b], then bm(b a) (5.8.6)f (x) dx M(b a)awhere m is the minimum value of f on [a, b] and M is the maximum. Reasoning: m(b a) is a lower sum for f and M(b a) is an upper sum.You know from Theorem 5.3.5 that, if f is continuous on [a, b], then for all x (a, b) x d f (t) dt = f (x). dx a Below we give an extension of this result that plays a large role in Chapter 7. V. If f is continuous on [a, b] and u is a differentiable function of x with values in [a, b], then for all u(x) (a, b) u(x)d dx(5.8.7)PROOF OF (5.8.7)f (t) dt = f (u(x))u (x).aSince f is continuous on [a, b], the function uF(u) =f (t) dt ais differentiable on (a, b) and F (u) = f (u). This we know from Theorem 5.3.5. The result that we are trying to prove follows from noting that u(x)f (t) dt = F(u(x))aand applying the chain rule: d dxu(x) af (t) dt =d [F(u(x))] = F (u(x))u (x) = f (u(x))u (x). dx17:59 307. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.8 ADDITIONAL PROPERTIES OF THE DEFINITE INTEGRAL x3d Example 1 Find dx2831 dt . 1+t0SOLUTION At this stage you probably cannot carry out the integration: it requires the natural logarithm function. (Not introduced in this text until Chapter 7.) But for our purposes, that doesnt matter. By (5.8.7), x3d dxExample 2 Find01 1 3x 2 dt = 3x 2 = . 1+t 1 + x3 1 + x3 2xd dx1 dt . 1 + t2xSOLUTION The idea is to express the integral in terms of integrals that have constant lower limits of integration. Once we have done that, we can apply (5.8.7). In this case, we choose 0 as a convenient lower limit. Then, by the additivity of the integral, x 02x1 dt + 1 + t2x2x1 dt = 1 + t21 dt. 1 + t2x1 dt. 1 + t20Thus 2x x1 dt = 1 + t22x1 dt 1 + t200Differentiation gives 2xd dxxd 1 dt = 2 1+t dx2x 0xd 1 dt 2 1+t dx01 dt 1 + t21 2 1 1 (2) (1) = . 2 2 2 1 + (2x) 1+x 1 + 4x 1 + x2= by (5.8.7) VI. Now a few words about the role of symmetry in integration. Suppose that f is continuous on an interval of the form [a, a], a closed interval symmetric about the origin. a(5.8.8)(a) iff is odd on [a, a],then a a(b) iff is even on [a, a],then ay1f (x) dx = 0.aax2af (x) dx = 2f (x) dx.f odd0Figure 5.8.1These assertions can be veried by a simple change of variables. (Exercise 34.) Here we look at these assertions from the standpoint of area. For convenience we refer to Figures 5.8.1 and 5.8.2.yFor the odd function, a af (x) dx =0 aaf (x) dx +f (x) dx = area of1 area of2= 0.021 aaFor the even function, a af (x) dx = area of1+ area of2= 2(area of2)f evena=2f (x) dx. 0Figure 5.8.2x17:59 308. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006284 CHAPTER 5 INTEGRATION Suppose we were asked to evaluate (sin x x cos x)3 d x.A laborious calculation would show that this integral is zero. We dont have to carry out that calculation. The integrand is an odd function, and the interval of integration is symmetric about the origin. Thus we can tell immediately that the integral is zero: (sin x x cos x)3 d x = 0.EXERCISES 5.8 Exercises 1723. Calculate. 1+x 2 d dt 17. . dx 2t + 5 0Assume that f and g are continuous on [a, b] and bbf (x) dx >ag(x) dx. aAnswer questions 16, giving supporting reasons.19.b a[f (x) g(x)] dx > 0? 1. Does it necessarily follow that 2. Does it necessarily follow that f (x) > g(x) for all x [a, b]? 3. Does it necessarily follow that f (x) > g(x) for at least some x [a, b]? 4. Does it necessarily follow that bbf (x) dx >a21.23.f (x) dx >6. Does it necessarily follow that f (x) dx > Assume that f is continuous on [a, b] and bb a g(x) dx? b a g(x) dx?f (x) dx = 0.Answer questions 715, giving supporting reasons. 7. Does it necessarily follow that f (x) = 0 for all x [a, b]? 8. Does it necessarily follow that f (x) = 0 for at least some x [a, b]? 10. 11. 12. 13.b a f (x) dx = 0? b a f (x) dx = 0?Does it necessarily follow that Must all upper sums U f (P) be nonnegative? Must all upper sums U f (P) be positive? Can a lower sum L f (P) be positive?14. Does it necessarily follow that 15. Does it necessarily follow that 16. Derive a formula for d dxb a[ b a[20.x 3d dxx2 d dxsin t dt . t x122.d dx d dxx2 1 x3dt . t 0dt 1 + t2.4sin t 2 dt . tan xt2 dt . 1 + t2f (x)]2 dx = 0? f (x) + 1] dx = b a?v(x)d dxf (t) dt = f (v(x))v (x) f (u(x))u (x)u(x)given that u and v are differentiable and f is continuous. Exercises 2528. Calculate. HINT: Exercise 24. 25.a9. Does it necessarily follow thatf (t) dt .d dx24. Show thatg(x) dx ? b a b a ad dxa5. Does it necessarily follow that18.27.d dx d dxx2 x 2xdt . t26.t 1 + t 2 dt .28.tan xd dx d dxx 2 +x x1/xcos 2t dt . 3x29. Prove (5.8.4). 30. (Important) Prove that, if f is continuous on [a, b] and b f (x) dx = 0,athen f (x) = 0 for all x in [a, b]. HINT: Exercise 50, Section 2.4. 31. Find H (2) given that x 3 4H (x) =2xx dt. 1+ t32. Find H (3) given thatbf (t) dt u(x)given that u is differentiable and f is continuous.dt . 2+ tH (x) =1 xx 3[2t 3H (t)] dt.17:59 309. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.9 MEAN-VALUE THEOREMS FOR INTEGRALS; AVERAGE VALUE OF A FUNCTION a33. (a) Let f be continuous on [a, 0]. Use a change of variable to show that 0 aaf (x) dx = 0aaf (x) dx =/3[ f (x) + f (x)] dx.a a37. /3 /4034. Let f be a function continuous on [a, a]. Prove the statement basing your argument on Exercise 33. (a)af (x) dx = 0285af (x) dx = 2f (x) dxif f is even.0Exercises 3538. Evaluate using symmetry considerations. /4 3 t3 35. (x + sin 2x) dx. 36. dt. 2 /4 3 1 + tf (x) dx.(b) Let f be continuous on [a, a]. Show that a(b)38. /4(1 + x 2 cos x) dx. (x 2 2x + sin x + cos 2x) dx.if f is odd. 5.9 MEAN-VALUE THEOREMS FOR INTEGRALS; AVERAGE VALUE OF A FUNCTION We begin with a result that we asked you to prove earlier. (Exercise 33, Section 5.3.)THEOREM 5.9.1 THE FIRST MEAN-VALUE THEOREM FOR INTEGRALSIf f is continuous on [a, b], then there is at least one number c in (a, b) for which bf (x) dx = f (c)(b a).aThis number f (c) is called the average value (or mean value) of f on [a, b].We now have the following identity:b(5.9.2)ff (x) dx = (the average value of f on [a, b]) (b a).aThis identity provides a powerful, intuitive way of viewing the denite integral. Think for a moment about area. If f is constant and positive on [a, b], then , the region below the graph, is a rectangle. Its area is given by the formula area of= (the constant value of f on [a, b]) (b a).b=f (x) dx, aand the area formula readsarea of= (the average value of f on [a, b]) (b a).Figure 5.9.1 (Figure 5.9.1)If f is now allowed to vary continuously on [a, b], then we have area ofa b area = (the constant value of f ) (b a)(Figure 5.9.2)17:59 310. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006286 CHAPTER 5 INTEGRATION fyf (c)acbxFigure 5.9.2Think now about motion. If an object moves along a line with constant speed v during the time interval [a, b], then distance traveled = (the constant value of on [a, b]) (b a). If the speed varies, then we have bdistance traveled =(t) dt,aand the formula reads distance traveled = (the average speed on [a, b]) (b a). Lets calculate some simple averages. Writing f avg for the average value of f on [a, b], we have f avg =1 babf (x) dx. aThe average value of a constant function f (x) = k is, of course, k: f avg =b1 bak dx =ak x bab a=k (b a) = k. baThe average value of a constant multiple of the identity function f (x) = x is the arithmetical average of the values taken on by the function at the endpoints of the interval: b 1 1 2 b x dx = x f avg = ba a ba 2 a =1 b + a 2 f (b) + f (a) (b a 2 ) = = . ba 2 2 2What is the average value of the squaring function f (x) = x 2 ? f avg = =1 babx 2d x =a1 bax3 3b= a1 bab3 a 3 3(b + ab + a )(b a) 1 = 1 (b2 + ab + a 2 ). 3 ba 3 22The average value of the squaring function on [a, b] is not 1 (b2 + a 2 ); it is 2 1 2 (b + ab + a 2 ). On [1, 3] the values of the squaring function range from 1 to 9. 3 While the arithmetic average of these two values is 5, the average value of the squaring function on the entire interval [1, 3] is not 5; it is 13 . 317:59 311. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.9 MEAN-VALUE THEOREMS FOR INTEGRALS; AVERAGE VALUE OF A FUNCTIONThere is an extension of Theorem 5.9.1 which, as youll see, is useful in applications. THEOREM 5.9.3 THE SECOND MEAN-VALUE THEOREM FOR INTEGRALSIf f and g are continuous on [a, b] and g is nonnegative, then there is a number c in (a, b) for which bbf (x)g(x) dx = f (c)ag(x) dx. aThis number f (c) is called the g-weighted average of f on [a, b].We will prove this theorem (and thereby obtain a proof of Theorem 5.9.1) at the end of this section. First, some physical considerations. The Mass of a Rod Imagine a thin rod (a straight material wire of negligible thickness) lying on the x-axis from x = a to x = b. If the mass density of the rod (the mass per unit length) is constant, then the total mass M of the rod is simply the density times the length of the rod: M = (b a). If the density varies continuously from point to point, say = (x), then the mass of the rod is the average density of the rod times the length of the rod:M = (average density) (length). This is an integral: bM=(5.9.4)(x) dx.aThe Center of Mass of a Rod Continue with that same rod. If the rod is homogeneous (constant density), then the center of mass of the rod (we denote this point by x M ) is simply the midpoint of the rod:x M = 1 (a + b). 2(the average of x from a to b)If the rod is not homogeneous, the center of mass is still an average, but now a weighted average, the density-weighted average of x from a to b; namely, x M is the point for which bxMb(x) dx =ax(x) dx. aSince the integral on the left is M, we have(5.9.5)bxM M =x(x) dx. aExample 1 A rod of length L is placed on the x-axis from x = 0 to x = L. Find the mass of the rod and the center of mass given that the density of the rod varies directly as the distance from the x = 0 endpoint of the rod. The symbol is the Greek letter lambda.28717:59 312. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006288 CHAPTER 5 INTEGRATION SOLUTION Here (x) = kx where k is some positive constant. Therefore LM=kx dx =0L 1 kx 2 2 0= 1 k L2 2and LxM M =Lx(kx) d x =0Division by M gives x M =L 1 kx 3 3 0kx 2 dx =0= 1 k L 3. 32 L. 302 xM = L 3LIn this instance the center of mass is to the right of the midpoint. This makes sense. After all, the density increases from left to right. Thus mass accumulates near the right tip of the rod. We know from physics that, close to the surface of the earth, where the force of gravity is given by the familiar formula W = mg, the center of mass is the center of gravity. This is the balance point. For the rod of Example 1, the balance point is at x = 2 L. Supported at that point, the rod will be in balance. 3 Later (in Project 10.6) you will see that a projectile red at an angle follows a parabolic path. (Here we are disregarding air resistance.) Suppose that a rod is hurled into the air end over end. Certainly not every point of the rod can follow a parabolic path. What moves in a parabolic path is the center of mass of the rod. We go back now to Theorem 5.9.3 and prove it. [There is no reason to construct a separate proof for Theorem 5.9.1. It is Theorem 5.9.3 with g(x) identically 1.] PROOF OF THEOREM 5.9.3Since f is continuous on [a, b], f takes on a minimum value m on [a, b] and a maximum value M. Since g is nonnegative on [a, b], mg(x) f (x)g(x) Mg(x)for all x in [a, b].Therefore bbmg(x) dx abf (x)g(x) dx aM g(x) dx aand bmbg(x) dx aabbf (x)g(x) dx Mg(x) dx. abWe know that a g(x) dx 0. If a g(x) dx = 0, then, by the inequality we just deb rived, a f (x)g(x) dx = 0 and the theorem holds for all choices of c in (a, b). If b a g(x) dx > 0, then bf (x)g(x) dx mMa bg(x) dx a17:59 313. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.9 MEAN-VALUE THEOREMS FOR INTEGRALS; AVERAGE VALUE OF A FUNCTION289and by the intermediate-value theorem (Theorem 2.6.1) there exists a number c in (a, b) for which bf (x)g(x) dx f (c) =.a bg(x) dx aObviously, then, bf (c)bg(x) dx =af (x)g(x) dx. aEXERCISES 5.9 Exercises 112. Determine the average value of the function on the indicated interval and nd an interior point of this interval at which the function takes on its average value. f (x) = mx + b, x [0, c]. f (x) = x 2 , x [1, 1]. f (x) = x 3 , x [1, 1]. f (x) = x 2 , x [1, 4]. f (x) = x , x [2, 2]. f (x) = x 1/3 , x [8, 8]. f (x) = 2x x 2 , x [0, 2]. f (x) = 3 2x, x [0, 3]. f (x) = x, x [0, 9]. f (x) = 4 x 2 , x [2, 2]. f (x) = sin x, x [0, 2]. f (x) = cos x, x [0, ]. Let f (x) = x n , n a positive integer. Determine the average value of f on the interval [a, b]. 14. Given that f is continuous on [a, b], compare 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.f (b)(b a)bf (x)d x.and a15. 16.17.18.(a) if f is constant on [a, b]; (b) if f increases on [a, b]; (c) if f decreases on [a, b]. Suppose that f has a continuous derivative on [a, b]. What is the average value of f on [a, b]? Determine whether the assertion is true or false on an arbitrary interval [a, b] on which f and g are continuous. (a) ( f + g)avg = f avg + gavg . (b) ( f )avg = f avg . (c) ( f g)avg = ( f avg )(gavg ). (d) ( f g)avg = ( f avg )/gavg ). Let P(x, y) be an arbitrary point on the curve y = x 2 . Express as a function of x the distance from P to the origin and calculate the average of this distance as x ranges from 0 to 3. Let P(x, y) be an arbitrary point on the line y = mx. Express as a function of x the distance from P to the originand calculate the average of this distance as x ranges from 0 to 1. 19. A stone falls from rest in a vacuum for t seconds. (Section 4.9). (a) Compare its terminal velocity to its average velocity; (b) compare its average velocity during the rst 1 t 2 seconds to its average velocity during the next 1 t seconds. 2 20. Let f be continuous. Show that, if f is an odd function, then its average value on every interval of the form [a, a] is zero. b 21. Suppose that f is continuous on [a, b] and a f (x) d x = 0. Prove that there is at least one number c in (a, b) for which f (c) = 0. 22. Show that the average value of the functions f (x) = sin x and g(x) = cos x is zero on every interval of length 2n, n a positive integer. 23. An object starts from rest at the point x0 and moves along the x-axis with constant acceleration a. (a) Derive formulas for the velocity and position of the object at each time t 0. (b) Show that the average velocity over any time interval [t1 , t2 ] is the arithmetic average of the initial and nal velocities on that interval. 24. Find the point on the rod of Example 1 that breaks up that rod into two pieces of equal mass. (Observe that this point is not the center of mass.) 25. A rod 6 meters long is placed the x-axis from x = 0 to on x = 6. The mass density is 12/ x + 1 kilograms per meter. (a) Find the mass of the rod and the center of mass. (b) What is the average mass density of the rod? 26. For a rod that extends from x = a to x = b and has mass density = (x), the integral b(x c)(x)d xagives what is called the mass moment of the rod about the point x = c. Show that the mass moment about the center of mass is zero. (The center of mass can be dened as the point about which the mass moment is zero.)17:59 314. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 2006290 CHAPTER 5 INTEGRATION 27. A rod of length L is placed on the x-axis from x = 0 to x = L. Find the mass of the rod and the center of mass if the mass density of the rod varies directly: (a) as the square root of the distance from x = 0; (b) as the square of the distance from x = L. 28. A rod of varying mass density, mass M, and center of mass x M , extends from x = a to x = b. A partition P = {x0 , x1 , . . . , xn } of [a, b] decomposes the rod into n pieces in the obvious way. Show that, if the n pieces have masses M1 , M2 , . . . , Mn and centers of mass x M1 , x M2 , . . . , x Mn , then36. The arithmetic average of n numbers is the sum of the numbers divided by n. Let f be a function continuous on [a, b]. Show that the average value of f on [a, b] is the limit of arithmetic averages of values taken on by f on [a, b] in the following sense: Partition [a, b] into n subintervals of equal length (b a)/n and let S (P) be a corresponding Riemann sum. Show that S (P)/(b a) is an arithmetic average of n values taken on by f and the limit of these arithmetic averages as P 0 is the average value of f on [a, b]. 37. A partition P = {x0 , x1 , x2 , . . . , xn } of [a, b] breaks up [a, b] into n subintervals [x0 , x1 ], [x1 , x2 , . . . , [xn1 , xn ].x M M = x M1 M 1 + x m 2 M 2 + + x Mn M n . 29. A rod that has mass M and extends from x = 0 to x = L consists of two pieces with masses M1 , M2 . Given that the center of mass of the entire rod is at x = 1 L and the center 4 of mass of the rst piece is at x = 1 L, determine the center 8 of mass of the second piece. 30. A rod that has mass M and extends from x = 0 to x = L consists of two pieces. Find the mass of each piece given that the center of mass of the entire rod is at x = 2 L, the 3 center of mass of the rst piece is at x = 1 L, and the center 4 of mass of the second piece is at x = 7 L. 8 31. A rod of mass M and length L is to be cut from a long piece that extends to the right from x = 0. Where should the cuts be made if the density of the long piece varies directly as the distance from x = 0? (Assume that M 1 k L 2 where k is 2 the constant of proportionality in the density function.) 32. Is the conclusion of Theorem 5.9.3 valid if g is negative throughout [a, b]? If so, prove it. 33. Prove Theorem 5.9.1 without invoking Theorem 5.9.3. 34. Let f be continuous on [a, b]. Let a < c < b. Prove that f (c) = lim+ (average value of f on [c h, c + h]).Show that if f is continuous on [a, b], then there are n numbers xi [xi1 , xi ] such that b aC38.C39. 40. 41.C C f (x) dx = f (x1 ) x1 + f (x2 ) x2 + + f (xn ) xn .(Thus each partition P of [a, b] gives rise to a Riemann sum which is exactly equal to the denite integral.) Let f (x) = x 3 x + 1 for x [1, 2]. (a) Find the average value of f on this interval. (b) Estimate with three decimal place accuracy a number c in the interval at which f takes on its average value. (c) Use a graphing utility to illustrate your results with a gure similar to Figure 5.9.2. Exercise 38 taking f (x) = sin x with x [0, ]. Exercise 38 taking f (x) = 2 cos 2x with x [/4, /6]. Set f (x) = x 4 + 10x 2 + 25. (a) Estimate the numbers a and b with a < b for which f (a) = f (b) = 0. (b) Use a graphing utility to draw the graph of f on [a, b]. (c) Estimate the numbers c in (a, b) for which bh035. Prove that two distinct continuous functions cannot have the same average on every interval.f (x)d x = f (c)(b a).aC42. Exercise 41 taking f (x) = 8 + x 2 x 4 . CHAPTER 5. REVIEW EXERCISES Exercises 122. Calculate.3. 5. 7. 9.(sec tan )2 d .(tan 3 cot 3)2 d.14.x sin3 x 2 cos x 2 dx.15.1 dx. 1 + cos 2x16.1 dx. 1 sin 2x13.1 + sin xdx.x 2x + 1 dx. x2.t 2 (1 + t 3 )10 dt.4. (1 + 2 x)2 dx.6.x x 2 2 dx.17.sec3 x tan x dx.18.ax 1 + bx 2 dx.8.x 2 (2 + 2x 3 )4 dx.19. ax 1 + bx dx.20. ax 2 1 + bx dx.sin(1/x) dx. x221.22.g (x) dx. g 3 (x)31.cos x12.11.(t 2/3 1)2 dt. t 1/3 x 2 x dx. (1 + x)5 dx. x10.(x 3/5 3x 5/3 ) dx.g(x)g (x) 1 + g 2 (x)dx.17:59 315. P1: PBU/OVYP2: PBU/OVYJWDD023-05JWDD023-Salas-v13QC: PBU/OVYT1: PBU October 11, 20065.9 MEAN-VALUE THEOREMS FOR INTEGRALS; AVERAGE VALUE OF A FUNCTIONExercises 2328. Evaluate. 223.(x 2 2x + 3) dx.1 /425.124. 0x dx. (x 2 + 1)3/8sin3 2x cos 2x dx.26.0(tan2 2x + sec2 2x) dx.0 2(x + 1)(x + 3x 6) 227.31/3dx.0(1 + x 1/3 )2 dx. x 2/3 1 29. Assume that f is a continuous function and that 828.23f (x) dx = 3,05f (x) dx = 1,041.d dxcos x 0291dt . 1 t2 42. At each point (x, y) of a curve the slope is x x 2 + 1. Find an equation y = f (x) for given that passes through the point (0, 1). x 1 dt, x real 43. Let F(x) = 2 0 t + 2t + 2 (a) Does F take on the value 0? If so, where? (b) Show that F increases (, ). (c) Determine the concavity of the graph of F. (d) Sketch the graph of F. 44. Assume that f is a continuous function and thatf (x) dx = 8.x3t f (t) dt = x sin x + cos x 1.0(a) Find (b) Find3 2 5 2f (x) dx. f (x) dx.(c) Explain how we know that f (x) 4 for at least one x in [3, 5]. (d) Explain how we know that f (x) < 0 for at least one x in [2, 3]. 30. Let f be a function continuous on [2, 8] and let g(x) = f (x) + 3. If8 2f (x)d x = 4, what is8g(x)d x? 2Exercises 3136. Sketch the region bounded by the curves and nd its area. 31. y = 4 x 2 , y = x + 2. 32. y = 4 x 2 , x + y + 2 = 0. 33. y 2 = x, x = 3y. 34. y = x, the x-axis, y = 6 x. 35. y = x 3 , the x-axis, x + y = 2. 36. 4y = x 2 x 4 , x + y + 1 = 0. Exercises 3741. Carry out the differentiation. 37.39.d dx d dxx 0 x2 xdt . 1 + t2 dt . 1 + t238.40.d dx d dxx2 0dt . 1 + t2sin x 0dt . 1 t2(a) Find f (). (b) Calculate f (x). Exercises 4446. Find the average value of f on the indicated interval. x ; [0, 4]. 44. f (x) = 2+9 x 45. f (x) = x + 2 sin x; [0, ]. 46. Find the average value of f (x) = cos x on every closed interval of length 2. Exercises 4750. Let f be a function continuous on [, ] and let be the region between the graph of f and the x-axis from x = to x = . Draw a gure. Do not assume that f keeps constant sign. 47. Write an integral over [, ] that gives the area of the portion of that lies above the x-axis minus the area of the portion of that lies below the x-axis. 48. Write an integral over [, ] that gives the area of . 49. Write an integral over [, ] that gives the area of the portion of that lies above the x-axis. 50. Write an integral over [, ] that gives the area of the portion of that lies below the x-axis. 51. A rod extends from x = 0 to x = a, a > 0. Find the center of mass if the density of the rod varies directly as the distance from x = 2a. 52. A rod extends from x = 0 to x = a, a > 0. Find the center of mass if the density of the rod varies directly as the distance from x = 1 a. 417:59 316. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER6SOME APPLICATIONS OF THE INTEGRAL 6.1 MORE ON AREA Representative Rectangles You have seen that the denite integral can be viewed as the limit of Riemann sums: b(1)f (x)d x = limP 0a f x1 x1 + f x2 x2 + + f xnxn .With xi chosen arbitrarily from [xi1 , xi ], you can think of f (xi ) as a representative value of f for that interval. If f is positive, then the product f xixigives the area of the representative rectangle shown in Figure 6.1.1. Formula (1) tells us that we can approximate the area under the curve as closely as we wish by adding up the areas of representative rectangles. (Figure 6.1.2) y y y = f (x)y = f (x)f (xi*) axi 1xi*xi xiFigure 6.1.1bf (xi*) x a xibxFigure 6.1.2Figure 6.1.3 shows a region bounded above by the graph of a function f and bounded below by the graph of a function g. As you know, we can obtain the area of29214:26 317. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.1 MORE ON AREA293by integrating the vertical separation f (x) g(x) from x = a to x = b: bA=[ f (x) g(x)] d x.aIn this case the approximating Riemann sums are of the form f x1 g x1 x1 + f x2 g x2 x2 + + f xn g xnyxn .y y = f (x)y = f (x)f (xi*) g (xi*)y = g (x)y = g(x)axbaxb xiFigure 6.1.3Here a representative rectangle has f xi g xi ,heightxi ,widthand area f xi g xixi . yExample 1 Find the area A of the set shaded in Figure 6.1.4.(2, 8)SOLUTION From x = 1 to x = 2 the vertical separation is the differencey = 2x 22x 2 (x 4 2x 2 ). Therefore A= =2 1[2x 2 (x 4 2x 2 )] d x =4 3 x 32 1 5 x 5 1=32 332 52 1(4x 2 x 4 ) d x (1, 2) 4 3+1 5=27 5y = x 4 2x 2 11Areas Obtained by Integration with Respect to y We can interchange the roles played by x and y. In Figure 6.1.5 you see a region , the boundaries of which are given not in terms of x but in terms of y. Here we set the representative rectangles horizontally and calculate the area of the region as the limit of sums of the form F y1 G y1 y1 + F y2 G y2 y2 + + F yn G ynyn .These are Riemann sums for the integral of F G. The area formula now reads dA=[F(y) G(y)] dy.oIn this case we are integrating with respect to y the horizontal separation F(y) G( y) from y = c to y = d.Figure 6.1.42x14:26 318. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006294 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL yydd F( yi*) G( yi*) yi x = G( y)x = G( y) x = F ( y)cx = F( y)cxxFigure 6.1.5Example 2 Find the area of the region bounded on the left by the curve x = y 2 and bounded on the right by the curve x = 3 2y 2 .y(1, 1)SOLUTION The region is sketched in Figure 6.1.6. The points of intersection can be found by solving the two equations simultaneously:13 x = 3 2y 21x = y2xandx = 3 2y 2together imply that y = 1.(1, 1)x = y2Figure 6.1.6The points of intersection are (1, 1) and (1, 1). The easiest way to calculate the area is to set our representative rectangles horizontally and integrate with respect to y. We then nd the area of the region by integrating the horizontal separation (3 2y 2 ) y 2 = 3 3y 2 from y = 1 to y = 1: 1A=1(3 3y 2 ) dy = 3y y 31 1= 4.NOTE: Our solution did not take advantage of the symmetry of the region. The region is symmetric about the x-axis (the integrand is an even function of y), and so 1A=2 0(3 3y 2 ) dy = 2 3y y 31 0= 4. Example 3 Calculate the area of the region bounded by the curves x = y 2 and x y = 2 rst (a) by integrating with respect to x and then (b) by integrating with respect to y. SOLUTION Simple algebra shows that the two curves intersect at the points (1, 1) and (4, 2).(a) To obtain the area of the region by integration with respect to x, we set the representative rectangles vertically and express the bounding curves as functions of x. Solving x = y 2 for y we get y = x; y = x is the upper half of the parabola and y = x is the lower half. The equation of the line can be written y = x 2. (See Figure 6.1.7.)14:26 319. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.1 MORE ON AREA y295y2(4, 2)y = xx = y2(4, 2)y=x21x=y+2 x4x(1, 1)(1, 1)1y = xFigure 6.1.7Figure 6.1.8 The upper boundary of the region is the curve y = x. However, the lower boundary consists of two parts: y = x from x = 0 to x = 1, and y = x 2 from x = 1 to x = 4. Thus, we use two integrals: 1A= [ x ( x)] d x +04x (x 2) d x1 1=2 ( x x + 2) d x =4x dx +011 4 3/2 x 3 0+2 3/2 x 34 1 x 2 + 2x = 9 . 2 2 1(b) To obtain the area by integration with respect to y, we set the representative rectangles horizontally. (See Figure 6.1.8.) The right boundary is the line x = y + 2 and the left boundary is the curve x = y 2 . Since y ranges from 1 to 2, A=2 1[(y + 2) y 2 ] dy =1 2 y 2+ 2y 1 y 3 32 1= 9. 2In this instance integration with respect to y was the more efcient route to take. EXERCISES 6.1 Exercises 114. Sketch the region bounded by the curves. Represent the area of the region by one or more integrals (a) in terms of x; (b) in terms of y. Evaluation not required. 2. y = x 2 , y = 4x. 1. y = x 2 , y = x + 2. 4. y = x, y = x 3 . 3. y = x 3 , y = 2x 2 . 5. y = x, y = x 6, y = 0. 7. y = x , 3y x = 8. 6. x = y 3 , x = 3y + 2. 8. y = x, y = 2x, y = 3. 10. x = y , x = 2. 9. x + 4 = y 2 , x = 5. 11. y = 2x, x + y = 9, y = x 1. 12. y = x 3 , y = x 2 + x 1. 13. y = x 1/3 , y = x 2 + x 1. 14. y = x + 1, y + 3x = 13, 3y + x + 1 = 0. Exercises 1526. Sketch the region bounded by the curves and calculate the area of the region. 15. 4x = 4y y 2 , 4x y = 0. 16. x + y 2 4 = 0, x + y = 2. 17. x = y 2 , x = 12 2y 2 .18. 19. 20. 21. 22. 23. 24. 25. 26.x + y = 2y 2 , y = x 3 . x + y y 3 = 0, x y + y 2 = 0. 8x = y 3 , 8x = 2y 3 + y 2 2y. y = cos x, y = sec2 x, x [/4, /4]. y = sin2 x, y = tan2 x, x [/4, /4]. HINT: sin2 x = 1 (1 cos 2x). 2 y = 2 cos x, y = sin 2x, x [, ]. y = sin x, y = sin 2x, x [0, /2]. y = sin4 x cos x, x [0, /2]. y = sin 2x, y = cos 2x, x [0, /4].Exercises 2728. Use integration to nd the area of the triangle with the given vertices. 27. (0, 0), (1, 3), (3, 1). 28. (0, 1), (2, 0), (3, 4). 29. Use integration to nd the area of the trapezoid with vertices (2, 2), (1, 1), (5, 1), (7, 2). 30. Sketch the region bounded by y = x 3 , y = x, and y = 1. Find the area of the region.14:26 320. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006296 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 31. Sketch the region bounded by y = 6 x 2 , y = x (x 0), and y = x (x 0). Find the area of the region. 32. Find the area of the region bounded by the parabolas x 2 = 4 py and y 2 = 4 px, p a positive constant. 33. Sketch the region bounded by y = x 2 and y = 4. This region is divided into two subregions of equal area by a line y = c. Find c. 34. The region between y = cos x and the x-axis for x [0, /2] is divided into two subregions of equal area by a line x = c. Find c. Exercises 3538. Represent the area of the given region by one or more integrals. 35. The region in the rst quadrant bounded by the x-axis, the line y = 3x, and the circle x 2 + y 2 = 4. 36. The region in the rst quadrant bounded by the y-axis, the line y = 3x, and the circle x 2 + y 2 = 4. 37. The region determined by the intersection of the circles x 2 + y 2 = 4 and (x 2)2 + (y 2)2 = 4. 38. The region in the rst quadrant bounded by the x-axis, the parabola y = x 2 /3, and the circle x 2 + y 2 = 4. 39. Take a > 0, b > 0, n a positive integer. A rectangle with sides parallel to the coordinate axes has one vertex at the origin and opposite vertex on the curve y = bx n at a point where x = a. Calculate the area of the part of the rectangle that lies below the curve. Show that the ratio of this area to the area of the entire rectangle is independent of a and b, and depends solely on n.CC40. (a) Calculate the area of the region in the rst quadrant bounded by the coordinate axes and the parabola y = 1 + a ax 2 , a > 0. (b) Determine the value of a that minimizes this area. 41. Use a graphing utility to draw the region bounded by the curves y = x 4 2x 2 and y = x + 2. Then nd (approximately) the area of the region. 42. Use a graphing utility to sketch the region bounded by the curves y = sin x and y = x 1 . Then nd (approximately) the area of the region. 43. A section of rain gutter is 8 feet long. Vertical cross sections of the gutter are in the shape of the parabolic region bounded by y = 4 x 2 and y = 4, with x and y measured in 9 inches. What is the volume of the rain gutter? HINT: V = (cross-sectional area) length. 44. (a) Calculate the area A of the region bounded by the graph of f (x) = 1/x 2 and the x-axis with x [1, b]. (b) What happens to A as b ? 45. (a) Calculate the A of the region bounded by the graph area of f (x) = 1/ x and the x-axis with x [1, b]. (b) What happens to A as b ? 46. (a) Let r > 1, r rational. Calculate the area A of the region bounded by the graph of f (x) = 1/x r and the x-axis with x [1, b]. What happens to A as b ? (b) Let 0 < r < 1, r rational. Calculate the area A of the region bounded by the graph of f (x) = 1/x r and the x-axis with x [1, b]. What happens to A as b ? 6.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERS Figure 6.2.1 shows a plane region and a solid formed by translating along a line perpendicular to the plane of . Such a solid is called a right cylinder with cross section .Figure 6.2.1If has area A and the solid has height h, then the volume of the solid is a simple product: (cross-sectional area height) V = A h.14:26 321. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERS297Two elementary examples are given in Figure 6.2.2.h rhw l V = ( r 2)h = (cross-sectional area) heightV = (l w) h = (cross-sectional area) heightFigure 6.2.2To calculate the volume of a more general solid, we introduce a coordinate axis and then examine the cross sections of the solid that are perpendicular to that axis. In Figure 6.2.3 we depict a solid and a coordinate axis that we label the x-axis. As in the gure, we suppose that the solid lies entirely between x = a and x = b. The gure shows an arbitrary cross section perpendicular to the x-axis. By A(x) we mean the area of the cross section at coordinate x. If the cross-sectional area A(x) varies continuously with x, then we can nd the volume V of the solid by integrating A(x) from x = a to x = b:x axbFigure 6.2.3bV =(6.2.1)area A (x)A(x) d x. aDERIVATION OF THE FORMULA Let P = {x 0 , x 1 , x 2 , . . . , x n } be a partition of [a, b]. On each subinterval [xi1 , xi ] choose a point xi . The solid from xi1 to xi can be approximated by a slab of cross-sectional area A(xi ) and thickness xi . The volume of this slab is the productA xixi .x0 = aThe sum of these products, A x1 x1 + A x2 x2 + + A xnxn ,is a Riemann sum which approximates the volume of the entire solid. As P 0, such Riemann sums converge to bA(x) d x. aRemark(6.2.2)x(Figure 6.2.4)(Average-value point of view) Formula (6.2.1) can be writtenV = (average cross-sectional area) (b a).xi 1 xixn = bFigure 6.2.414:26 322. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006298 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL Example 1 Find the volume of the pyramid of height h given that the base of the pyramid is a square with sides of length r and the apex of the pyramid lies directly above the center of the base SOLUTION Set the x-axis as in Figure 6.2.5. The cross section at coordinate x is asquare. Let s denote the length of the side of that square. By similar triangles 1 s 2hx=1 r 2and thereforehs=r (h x). hxQQhxx 1 s 2h s xr OP1 r 2OPrFigure 6.2.5The area A(x) of the square at coordinate x is s 2 = (r 2 / h 2 )(h x)2 . Thus hV = 0A(x)d x =r2 h2h 0(h x)2 d x =(h x)3 r2 h2 3h 0= 1 r 2 h. 3Example 2 The base of a solid is the region enclosed by the ellipse y2 x2 + 2 = 1. 2 a b Find the volume of the solid given that each cross section perpendicular to the x-axis is an isosceles triangle with base in the region and altitude equal to one-half the base. SOLUTION Set the x-axis as in Figure 6.2.6. The cross section at coordinate x is anisosceles triangle with base P Q and altitude 1 P Q. The equation of the ellipse can be 2 written b2 y 2 = 2 (a 2 x 2 ). a Since 2b length of P Q = 2y = a2 x 2, a14:26 323. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERSyQx = a2y xPxx=aFigure 6.2.6the isosceles triangle has area A(x) = 1 bh = 22b a2 x 2 a1 2b a2 x 2 a=b2 2 (a x 2 ). a2We can nd the volume of the solid by integrating A(x) from x = a to x = a: V =a aaA(x) d x = 2 A(x) d x 0 by symmetry 2b2 = 2 aa2b2 x3 (a x ) d x = 2 a 2 x a 3 20a20= 4 ab2 . 3Example 3 The base of a solid is the region between the parabolas x = y2x = 3 2y 2 .andFind the volume of the solid given that the cross sections perpendicular to the x-axis are squares. SOLUTION The solid is pictured in Figure 6.2.7. The two parabolas intersect at (1, 1) and (1, 1). From x = 0 to x = 1 the cross section at coordinate x has areaA(x) = (2y)2 = 4y 2 = 4x. yx = y2x x = 3 2y 2x=0x=1x=3Figure 6.2.729914:26 324. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006300 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL (Here we are measuring the span across the rst parabola x = y 2 .) The volume of the solid from x = 0 to x = 1 is 1V1 =4x d x = 2x 201 0= 2.From x = 1 to x = 3, the cross section at coordinate x has area A(x) = (2y)2 = 4y 2 = 2(3 x) = 6 2x. (Here we are measuring the span across the second parabola x = 3 2y 2 .) The volume of the solid from x = 1 to x = 3 is 3V2 =(6 2x) d x = 6x x 213 1= 4.The total volume is V1 + V2 = 6. Solids of Revolution: Disk Method Suppose that f is nonnegative and continuous on [a, b]. (See Figure 6.2.8.) If we revolve about the x-axis the region bounded by the graph of f and the x-axis, we obtain a solid. yyf f (x) axbxxFigure 6.2.8The volume of this solid is given by the formula(6.2.3)bV = [ f (x)]2 d x.aVERIFICATION The cross section at coordinate x is a circular disk of radius f (x). The area of this disk is [ f (x)]2 . We can get the volume of the solid by integrating this function from x = a to x = b. Among the simplest solids of revolution are the circular cone and sphere.Example 4 We can generate a circular cone of base radius r and height h by revolving about the x-axis the region below the graph of r (Figure 6.2.9) f (x) = x, 0 x h. h14:26 325. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERS y(h, r)ry = hx r xxhxhFigure 6.2.9By (6.2.3), hvolume of cone = 02r x hr 2 dx = 2 hhr 2 x dx = 2 h 20x3 3h 0= 1 r 2 h. 3Example 5 A sphere of radius r can be obtained by revolving about the x-axis the region below the graph of f (x) =r 2 x 2,r x r.(Draw a gure.)Therefore volume of sphere =r r (r 2 x 2 ) d x = r 2 x 1 x 3 3r r= 4 r 3 . 3NOTE: Archimedes derived this formula (by somewhat different methods) in the third century b.c. We can interchange the roles played by x and y. By revolving about the y-axis the region of Figure 6.2.10, we obtain a solid of cross-sectional area A(y) = [g(y)]2 and volume dV =(6.2.4) [g(y)]2 dy.cyyd g y g(y) cxFigure 6.2.10x30114:26 326. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006302 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL Example 6 Let be the region bounded below by the curve y = x 2/3 + 1, bounded to the left by the y-axis, and bounded above by the line y = 5. Find the volume of the solid generated by revolving about the y-axis. (See Figure 6.2.11.) y y(8, 5)5y = x 2/3 + 11 8xxFigure 6.2.11SOLUTION To apply (6.2.4) we need to express the right boundary of function of y:y = x 2/3 + 1givesx 2/3 = y 1The volume of the solid obtained by revolving 5V =x = (y 1)3/2 .about the y-axis is given by the integral 5 [g(y)]2 dy = 1and thusas a[(y 1)3/2 ]2 dy1 5=(y 1)3 dy = 15(y 1)4 4= 641Solids of Revolution: Washer Method The washer method is a slight generalization of the disk method. Suppose that f and g are nonnegative continuous functions with g(x) f (x) for all x in [a, b]. (See Figure 6.2.12.) If we revolve the region about the x-axis, we obtain a solid. The volume of yy y = f (x)f (x)g(x)y = g(x) axbxFigure 6.2.12axbx14:26 327. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERSthis solid is given by the formulabV =(6.2.5) ([ f (x)]2 [g(x)]2 ) d x.(washer method about the x-axis)aVERIFICATION The cross section at coordinate x is a circular ring (in this setting we call it a washer) of outer radius f (x), inner radius g(x), and areaA(x) = [ f (x)]2 [g(x)]2 = ([ f (x)]2 [g(x)]2 ). We can get the volume of this solid by integrating A(x) from x = a to x = b. As before, we can interchange the roles played by x and y. By revolving the region depicted in Figure 6.2.13 about the y-axis, we obtain a solid of volumedV =(6.2.6) ([F(y)]2 [G(y)]2 ) dy.(washer method about the y-axis)cyydF (y) G(y)x = F(y)x = G (y)yy cxxFigure 6.2.13Example 7 Find the volume of the solid generated by revolving the region between and y = 2x y = x2 (a) about the x-axis.(b) about the y-axis.SOLUTION The curves intersect at the points (0, 0) and (2, 4).(a) We refer to Figure 6.2.14. For each x from 0 to 2, the x cross section is a washer of outer radius 2x and inner radius x 2 . By (6.2.5), 2V = 02 [(2x)2 (x 2 )2 ] d x = 0(4x 2 x 4 ) d x = 4 3 x 3 1 x5 52 0=64 . 1530314:26 328. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006304 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL y (2, 4)y (2, 4)2x x2 xx 1 y 2yFigure 6.2.14Figure 6.2.15(b) The solid is depicted in Figure 6.2.15. For each y from 0 to 4, the y cross section is a washer of outer radius y and inner radius 1 y. By (6.2.6), 2 4V = ( y)2 0=1 2 y 24 1 3 y 12 02 1 y 2=8 . 34dy = 0y 1 y 2 dy 4EXERCISES 6.2 Exercises 116. Sketch the region bounded by the curves and nd the volume of the solid generated by revolving this region about the x-axis. 1. y = x, y = 0, x = 1. 2. x + y = 3, y = 0, x = 0 4. y = x 3 , y = 8, x = 0. 3. y = x 2 , y = 9 6. y = x 2 , y = x 1/3 . 5. y = x, y = x 3 . 3 7. y = x , x + y = 10, y = 1. 8. y = x, x + y = 6, y = 1. 10. y = x 2 , y = 2 x 9. y = x 2 , y = x + 2. 12. y = 1 x , y = 0. 11. y = 4 x 2 , y = 0. 1 13. y = sec x, x = 0, x = 4 , y = 0. 14. y = csc x,x = 1 , 415. y = cos x,y = x + 1,x = 3, 4 x=y = 0.1 . 216. y = sin x, x = x = 1 , y = 0. 2 Exercises 1726. Sketch the region bounded by the curves and nd the volume of the solid generated by revolving this region about the y-axis. 17. y = 2x, y = 4, x = 0. 18. x + 3y = 6, x = 0, y = 0. 19. x = y 3 , x = 8, y = 0. 1 , 4 21. y = x, y = x 3 . 20. x = y 2 , x = 4. 23. y = x, y = 2x, x = 4. 22. y = x 2 , y = x 1/3 . 24. x + y = 3, 2x + y = 6, x = 0. 26. x = 9 y 2 , x = 0. 25. x = y 2 , x = 2 y 2 . 27. The base of a solid is the disk bounded by the circle x 2 + y 2 = r 2 . Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) squares; (b) equilateral triangles. 28. The base of a solid is the region bounded by the ellipse 4x 2 + 9y 2 = 36. Find the volume of the solid given that cross sections perpendicular to the x-axis are: (a) equilateral triangles; (b) squares. 29. The base of a solid is the region bounded by y = x 2 and y = 4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) squares; (b) semicircles; (c) equilateral triangles. 30. The base of a solid is the region between the parabolas x = y 2 and x = 3 2y 2 . Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) rectangles of height h; (b) equilateral triangles; (c) isosceles right triangles, hypotenuse on the xy-plane. 31. Carry out Exercise 29 with the cross sections perpendicular to the y-axis.14:26 329. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.2 VOLUME BY PARALLEL CROSS SECTIONS; DISKS AND WASHERS30545. Let f (x) = x 2/3 for x > 0. (a) Sketch the graph of f. (b) Calculate the area of the region bounded by the graph of f and the x-axis from x = 1 to x = b. Take b > 1. (c) The region in part (b) is rotated about the x-axis. Find the volume of the resulting solid. (d) What happens to the area of the region as b ? What happens to the volume of the solid? 46. This is a continuation of Exercise 45. (a) Calculate the area of the region bounded by the graph of f and the x-axis from x = c to x = 1. Take 0 < c < 1. (b) The region in part (a) is rotated about the x-axis. Find the volume of the resulting solid. (c) What happens to the area of the region as c 0+ ? What happens to the volume of the solid? 47. With x and y measured in feet, the conguration shown in the gure is revolved about the y-axis to form a parabolic container, no top. Given that a liquid is poured into the container at the rate of two cubic feet per minute, how fast is the level of the liquid rising when the depth of the liquid is 1 foot? 2 feet?32. Carry out Exercise 30 with the cross sections perpendicular to the y-axis. 33. The base of a solid is the triangular region bounded by the y-axis and the lines x + 2y = 4, x 2y = 4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) squares; (b) isosceles right triangles with hypotenuse on the xy-plane. 34. The base of a solid is the region bounded by the ellipse b2 x 2 + a 2 y 2 = a 2 b2 . Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) isosceles right triangles, hypotenuse on the xy-plane; (b) squares; (c) isosceles triangles of height 2. 35. The base of a solid is the region bounded by y = 2 sin x and the x-axis with x [0, /2]. Find the volume of the solid given that cross sections perpendicular to the x-axis are: (a) equilateral triangles; (b) squares. 36. The base of a solid is the region bounded by y = sec x and y = tan x with x [0, /4]. Find the volume of the solid given that cross sections perpendicular to the x-axis are: (a) semicircles; (b) squares. 37. Find the volume enclosed by the surface obtained by revolving the ellipse b2 x 2 + a 2 y 2 = a 2 b2 about the x-axis. 38. Find the volume enclosed by the surface obtained by revolving the ellipse b2 x 2 + a 2 y 2 = a 2 b2 about the y-axis. 39. Derive a formula for the volume of the frustum of a right circular cone in terms of the height h, the lower base radius R, and the upper base radius r. (See the gure.)y 32y = x2 11r1hR48. Let x = f (y) be continuous and positive on the interval [0, b]. The conguration in the gure is revolved about the y-axis to form a container, no top. Suppose that the container is lled with water which then evaporates at a rate proportional to the area of the surface of the water. Show that the water level drops at a constant rate.40. Find the volume enclosed by the surface obtained by revolving the equilateral triangle with vertices (0, 0), (a, 0), ( 1 a, 1 3a) about the x-axis. 2 2 41. A hemispherical basin of radius r feet is being used to store water. To what percent of capacity is it lled when the water is: (b) 1 r feet deep? (a) 1 r feet deep? 2 3 42. A sphere of radius r is cut by two parallel planes: one, a units above the equator; the other, b units above the equator. Find the volume of the portion of the sphere that lies between the two planes. Assume that a < b. 43. A sphere of radius r is cut by a plane h units above the equator. Take 0 < h < r . The top portion is called a cap. Derive the formula for the volume of a cap. 44. A hemispherical punch bowl 2 feet in diameter is lled to within 1 inch of the top. Thirty minutes after the party starts, there are only 2 inches of punch left at the bottom of the bowl. (a) How much punch was there at the beginning? (b) How much punch was consumed?x2y bx = f (y)xC49. Set f (x) = x 5 and g(x) = 2x, x 0. (a) Use a graphing utility to display the graphs of f and g in one gure. (b) Use a CAS to nd the points of intersection of the two graphs. (c) Use a CAS to nd the area of the region bounded by the two graphs.14:26 330. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006306 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL50. 51.52.53.54.55.56.57.(d) The region in part (c) is revolved about the x-axis. Use a CAS to nd the volume of the resulting solid. Carry out Exercise 49 taking f (x) = 2x 1 and g(x) = x 2 4x + 4. The region between the graph of f (x) = x and the x-axis, 0 x 4, is revolved about the line y = 2. Find the volume of the resulting solid. The region bounded by the curves y = (x 1)2 and y = x + 1 is revolved about the line y = 1. Find the volume of the resulting solid. The region between the graph of y = sin x and the x-axis, 0 x , is revolved about the line y = 1. Find the volume of the resulting solid. The region bounded by y = sin x and y = cos x, with /4 x , is revolved about the line y = 1. Find the volume of the resulting solid. The region bounded by the curves y = x 2 2x and y = 3x is revolved about the line y = 1. Find the volume of the resulting solid. Find the volume of the solid generated by revolving the region bounded by y = x 2 and x = y 2 : (a) about the line x = 2; (b) about the line x = 3. Find the volume of the solid generated by revolving the region bounded by y 2 = 4x and y = x: (a) about the x-axis; (b) about the line x = 4.58. Find the volume of the solid generated by revolving the region bounded by y = x 2 and y = 4x: (a) about the line x = 5; (b) about the line x = 1. 59. Find the volume of the solid generated by revolving the region OAB in the gure about: (a) the x-axis; (b) the line AB; (c) the line CA; (d) the y-axis. y A(4, 8)C(0, 8)y = x 3/2OB(4, 0)x60. Find the volume of the solid generated by revolving the region OAC in the gure accompanying Exercise 59 about: (a) the y-axis; (b) the line CA; (c) the line AB; (d) the x-axis. 6.3 VOLUME BY THE SHELL METHOD Figure 6.3.1 shows the region below the curve y = 5x x 5 from x = 0 to x = 1. By revolving about the y-axis we obtain a solid of revolution. This solid has a certain volume. Call it V. To calculate V by the washer method we would have to express the curved boundary of in the form x = (y), and this we cant do: given that y = 5x x 5 , we have no way of expressing x in terms of y. Thus, in this instance, the washer method fails. Below we introduce another method of calculating volume, a method by which we can avoid the difculty just cited. It is called the shell method. To describe the shell method of calculating volumes, we begin with a solid cylinder of radius R and height h, and from it we cut out a cylindrical core of radius r. (Figure 6.3.2) Since the original cylinder has volume R 2 h and the piece removed has volume 2 r h, the cylindrical shell that remains has volumey 4 1xFigure 6.3.1r(6.3.1) Rh R 2 h r 2 h = h(R + r )(R r ).We will use this shortly. Now let [a, b] be an interval with a 0 and let f be a nonnegative function continuous on [a, b]. If the region bounded by the graph of f and the x-axis is revolved about the y-axis, then a solid is generated. (Figure 6.3.3) The volume of this solid can be obtained from the formulaFigure 6.3.2 (6.3.2)bV =2 x f (x) d x. a14:26 331. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.3 VOLUME BY THE SHELL METHODThis is called the shell-method formula. yyff (xi*) axi1xi*xxbxiFigure 6.3.3 DERIVATION OF THE FORMULA We take a partition P = {x 0 , x 1 , . . . , x n } of [a, b] and concentrate on what happens on the ith subinterval [xi1 , xi ]. Recall that when we form a Riemann sum we are free to choose xi as any point from [xi1 , xi ]. For convenience we take xi as the midpoint 1 (xi1 + xi ). The representative rectangle of height f (xi ) and 2 base xi (see Figure 6.3.2) generates a cylindrical shell of height f (xi ), inner radius r = xi1 , and outer radius R = xi . We can calculate the volume of this shell by (6.3.1). Sinceh = f xiR + r = xi + xi1 = 2xiandR r =andxi ,the volume of this shell is h(R + r )(R r ) = 2 xi f xixi .The volume of the entire solid can be approximated by adding up the volumes of these shells: V 2 x f x x1 + 2 x f x x2 + + 2 x f x xn . = 1122nnThe sum on the right is a Riemann sum. As P 0, such Riemann sums converge to b2 x f (x) d x. aRemark (Average-value point of view) To give some geometric insight into the shellmethod formula, we refer to Figure 6.3.4. As the region below the graph of f is revolved about the y-axis, the vertical line segment at x generates a cylindrical surface of radius x, height f (x), and lateral area 2 x f (x). As x ranges from x = a to x = b, the cylindrical surfaces form a solid. The shell-method formula bV =2 x f (x) d x ayyfxf (x)f (x) xxFigure 6.3.430714:26 332. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006308 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL states that the volume of this solid can be expressed by writing the average lateral area of the component cylindrical surfacesV =(6.3.3) (b a)This is the point of view well take.Example 1 The region bounded by the graph of f (x) = 4x x 2 and the x-axis from x = 1 to x = 4 is revolved about the y-axis. Find the volume of the resulting solid. SOLUTION See Figure 6.3.5. The line segment x units from the y-axis, 1 x 4,ygenerates a cylinder of radius x, height f (x), and lateral area 2 x f (x). Thus 44V = 131x 4x42 x(4x x 2 ) d x = 2(4x 2 x 3 ) d x = 214 3 x 34 1 x4 41=81 . 2The shell-method formula can be generalized. With the region from x = a to x = b shown in Figure 6.3.6, the volume generated by revolving about the y-axis is given by the formulaFigure 6.3.5bV =(6.3.4)2 x[ f (x) g(x)] d x.(shell method about the y-axis)aThe integrand 2 x[ f (x) g(x)] is the lateral area of the cylindrical surface, which is at a distance x from the axis of rotation. As usual, we can interchange the roles played by x and y. With the region from y = c to y = d shown in Figure 6.3.7, the volume generated by revolving about the x-axis is given by the formula dV =(6.3.5)2 y[F(y) G(y)] dy.(shell method about the x-axis)eyx = G(y)x = F( y)dyF ( y) G ( y) y = f (x) cyx x f (x) g(x) y = g(x) aFigure 6.3.6bxFigure 6.3.714:26 333. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.3 VOLUME BY THE SHELL METHOD309The integrand 2 y [F(y) G(y)] is the lateral area of the cylindrical surface, which is at a distance y from the axis of rotation.Example 2 Find the volume of the solid generated by revolving the region between y = x2 (a) about the y-axis,y = 2xand(b) about the x-axis.SOLUTION The curves intersect at the points (0, 0) and (2, 4).(a) We refer to Figure 6.3.8. For each x from 0 to 2 the line segment at a distance x from the y-axis generates a cylindrical surface of radius x, height (2x x 2 ), and lateral area 2 x(2x x 2 ). By (6.3.4), 2V =22 x(2x x 2 ) d x = 20(2x 2 x 3 ) d x = 202 3 x 3 1 x4 42 0= 8 . 3 (b) We begin by expressing the bounding curves as functions of y. We write x = y for the right boundary and x = 1 y for the left boundary. (See Figure 6.3.9.) For 2 each y from 0 to 4 the line segment at a distance y from the x-axis generates a cylindrical surface of radius y, height ( y 1 y), and lateral area 2 y( y 1 y). 2 2 By (6.3.5), 4V = 0= 2 y( y 1 y) dy = 24 5/2 y 5 1 y3 34 0=64 . 154(2y 3/2 y 2 ) dy0 y (2, 4) y 1 2y1 2x= y x = yyy (2, 4)x y = 2xxy = x2 2x x2xFigure 6.3.8yFigure 6.3.9Example 3 A round hole of radius r is drilled through the center of a half-ball of radius a (r < a). Find the volume of the remaining solid.x 2 + y 2 = a2SOLUTION A half-ball of radius a can be formed by revolving about the y-axis therst quadrant region bounded by x 2 + y 2 = a 2 . What remains after the hole is drilled is the solid formed by revolving about the y-axis only that part of the region which is shaded in Figure 6.3.10.raFigure 6.3.10x14:26 334. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006310 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL y(a) By the washer method. We refer to Figure 6.3.11. (r,a2r 2) a 2 r 2V =a2 y20x = a2 y2= (a r )y 2yrax22 r 2 dy = a 2 r 21 3 y 3 0 a 2 r 2(a 2 r 2 y 2 ) dy0= 2 (a 2 r 2 )3/2 . 3(b) By the shell method. We refer to Figure 6.3.12. aV =Figure 6.3.112 x a 2 x 2 d x.rSet u = a 2 x 2 , du = 2x d x. At x = r, u = a 2 r 2 ; at x = a, u = 0. Therefore,y(r,a2aV =r 2)2 x a 2 x 2 d x = ry = a2 x2rxaFigure 6.3.12= xa 2 r 22 3/2 u 3 00 a 2 r 2a 2 r 2u 1/2 du = u 1/2 du0= 2 (a 2 r 2 )3/2 . 3If r = 0, no hole is drilled and V = 2 a 3 , the volume of the entire half-ball. 3 In our last example we revolve a region about a line parallel to the y-axis. Example 4 The region between y = x and y = x 2 , 0 x 1, is revolved about the line x = 2. (See Figure 6.3.13.) Find the volume of the solid which is generated. y1y = x y = x21x = 21xxFigure 6.3.13SOLUTION We use the shell method. For each x in [0, 1] the line segment at x generates a cylindrical surface of radius x + 2, height x x 2 , and lateral area 2 (x + 2)( x x 2 ). Therefore 1 V = 2 (x + 2)( x x 2 ) d x 0 1= 2(x 3/2 + 2x 1/2 x 3 2x 2 ) d x0= 22 5/2 x 5+ 4 x 3/2 1 x 4 2 x 3 3 4 31 0=49 . 3014:26 335. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.3 VOLUME BY THE SHELL METHOD311Remark We began this section by explaining that the washer method does not provide a way for us to calculate the volume generated by revolving about the y-axis the region shown in Figure 6.3.1. By the shell method we can easily calculate this volume: 1V =12 x(5x x 5 ) d x = 20(5x 2 x 6 ) d x = 205 3 x 3 1 x7 71 0=64 . 21EXERCISES 6.3 Exercises 112. Sketch the region bounded by the curves and use the shell method to nd the volume of the solid generated by revolving about the y-axis. 1. 2. 3. 4. 5. 7. 8. 9.y = x, y = 0, x = 1. x + y = 3, y = 0, x = 0. y = x, x = 4, y = 0. y = x 3 , x = 2, y = 0. 6. y = x 2 , y = x, y = x 3 . y = x, y = 2x, y = 4. y = x, y = 1, x + y = 6. 10. x = y 2 , x = y 2 , x = y + 2.y = x 1/3 .x = 2 y.11. x = 9 y 2 , x = 0. 12. x = y , x = 2 y 2 . Exercises 1324. Sketch the region bounded by the curves and use the shell method to nd the volume of the solid generated by revolving about the x-axis. 13. x + 3y = 6, y = 0, x = 0. 14. y = x, y = 5, x = 0. 16. y = x 3 , y = 8, x = 0. 15. y = x 2 , y = 9. 18. y = x 2 , y = x 1/3 . 17. y = x, y = x 3 . 2 20. y = x 2 , y = 2 x. 19. y = x , y = x + 2. 21. y = x, y = 2x, x = 4. 22. y = x, x + y = 8, x = 1. 23. y = 1 x 2 , x + y = 1. 24. y = x 2 , y = 2 x . Exercises 2530. The gure shows three regions within the unit square. Express the volume obtained by revolving the indicated region about the indicated line: (a) by an integral with respect to x; (b) by an integral with respect to y. Calculate each volume by evaluating one of these integrals y25. 1 , the y-axis. 26. 1 , the line y = 2. 28. 2 , the line x = 3. 27. 2 , the x-axis. 30. 3 , the line y = 1. 29. 3 , the y-axis. 31. Use the shell method to nd the volume enclosed by the surface obtained by revolving the ellipse b2 x 2 + a 2 y 2 = a 2 b2 about the y-axis. 32. Carry out Exercise 31 with the ellipse revolved about the x-axis. 33. Find the volume enclosed by the surface generated by revolving the equilateral triangle with vertices (0, 0), (a, 0), ( 1 a, 1 3a) about the y-axis. 2 2 34. A ball of radius r is cut into two pieces by a horizontal plane a units above the center of the ball. Determine the volume of the upper piece by using the shell method. 35. Carry out Exercise 59 of Section 6.2, this time using the shell method. 36. Carry out Exercise 60 of Section 6.2, this time using the shell method. 37. (a) Verify that F(x) = x sin x + cos x is an antiderivative of f (x) = x cos x. (b) Find the volume generated by revolving about the yaxis the region between y = cos x and the x-axis, 0 x /2. 38. (a) Sketch the region in the right half-plane that is outside the parabola y = x 2 and is between the lines y = x + 2 and y = 2x 2. (b) The region in part (a) is revolved about the y-axis. Use the method that you nd most practical to calculate the volume of the solid generated. For Exercises 3942, set 3x, 0 x < 1 f (x) = 4 x 2 , 1 x 2, and let be the region between the graph of f and the x-axis. (See the gure.) y(1, 1)1(1, 3)y = xy = 3 x1y = 4 x22 3 y = x2 1x12x14:26 336. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006312 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 39. Revolve about the y-axis. (a) Express the volume of the resulting solid as an integral in x. (b) Express the volume of the resulting solid as an integral in y. (c) Calculate the volume of the solid by evaluating one of these integrals. 40. Carry out Exercise 39 for revolved about the x-axis. 41. Carry out parts (a) and (b) of Exercise 39 for revolved about the line x = 2. 42. Carry out parts (a) and (b) of Exercise 39 for revolved about the line y = 1. 43. Let be the circular disk (x b)2 + y 2 a 2 , 0 < a < b. The doughnut-shaped region generated by revolving about the y-axis is called a torus. Express the volume of the torus as: (a) a denite integral in x. (b) a denite integral in y. 44. The circular disk x 2 + y 2 a 2 , a > 0, is revolved about the line x = a. Find the volume of the resulting solid. 45. Let r and h be positive numbers. The region in the rst quadrant bounded by the line x/r + y/ h = 1 and the coordinate axes is rotated about the y-axis. Use the shell method to derive the formula for the volume of a cone of radius r and height h. 46. A hole is drilled through the center of a ball of radius r, leaving a solid with a hollow cylindrical core of height h. Show that the volume of this solid is independent of the radius of the ball. 47. The region in the rst quadrant bounded by the parabola y = r 2 x 2 and the coordinate axes is revolved about the y-axis. The resulting solid is called a paraboloid. A vertical hole of radius a, a < r , centered along the y-axis, is drilledCCCthrough the paraboloid. Find the volume of the solid that remains: (a) by integrating with respect to x; (b) by integrating with respect to y. 48. (a) Draw the graph of f (x) = sin x 2 , x [3, 3]. (b) Let be the region bounded by the graph of f and the x-axis with x [0, 1]. If is revolved about the x-axis and the disk method is used to calculate the volume, then the resulting integral cannot be readily evaluated by the fundamental theorem of calculus. Use a CAS to estimate this volume. (c) If is revolved about the y-axis and the shell method is used to calculate the volume, then the resulting integral can be evaluated by the fundamental theorem of calculus. Calculate this volume. 49. Set f (x) = sin x and g(x) = 1 x. 2 (a) Use a graphing utility to display the graphs of f and g in one gure. (b) Use a CAS to nd the points of intersection of the two graphs. (c) Use a CAS to nd the area of the region bounded by the two graphs. (d) The region in part (c) is revolved about the y-axis. Use a CAS to nd the volume of the resulting solid. 2 and g(x) = 3 1 x. 50. Set f (x) = 2 2 (x + 1)2 (a) Use a graphing utility to display the graphs of f and g in one gure. (b) Use a CAS to nd the points of intersection of the two graphs. (c) Use a CAS to nd the area of the region in the rst quadrant bounded by the graphs. (d) The region in part (c) is revolved about the y-axis. Use a CAS to nd the volume of the resulting solid. 6.4 THE CENTROID OF A REGION; PAPPUS'S THEOREM ON VOLUMES The Centroid of a Region yxFigure 6.4.1In Section 5.9 you saw how to locate the center of mass of a thin rod. Suppose now that we have a thin distribution of matter, a plate, laid out in the xy-plane in the shape of some region . (Figure 6.4.1) If the mass density of the plate varies from point to point, then the determination of the center of mass of the plate requires the evaluation of a double integral. (Chapter 17) If, however, the mass density of the plate is constant throughout , then the center of mass depends only on the shape of and falls on a point (x, y) that we call the centroid. Unless has a very complicated shape, we can locate the centroid by ordinary one-variable integration. We will use two guiding principles to locate the centroid of a plane region. The rst is obvious. The second we take from physics; the result conforms to physical intuition and is easily justied by double integration Principle 1: Symmetry If the region has an axis of symmetry, then the centroid (x, y) lies somewhere along that axis. In particular, if the region has a center, then the center is the centroid.14:26 337. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006Principle 2: AdditivityIf the region, having area A, consists of a nite number of pieces with areas A1 , . . . , An and centroids (x 1 , y 1 ), . . . , (x n , y n ), then x A = x 1 A1 + + x n An(6.4.1)yfy A = y 1 A1 + + y n An .and3136.4 THE CENTROID OF A REGION; PAPPUSS THEOREM ON VOLUMESWe are now ready to bring the techniques of calculus into play. Figure 6.4.2 shows the region under the graph of a continuous function f. Denote the area of by A. The centroid (x, y) of can be obtained from the following formulas:bxbaxFigure 6.4.2 bxA =(6.4.2)byA =x f (x) d x, a1 [ 2af (x)]2 d x.To derive these formulas we choose a partition P = {x0 , x1 , . . . , xn } of [a, b]. This breaks up [a, b] into n subintervals [xi1 , xi ]. Choosing xi as the midpoint of [xi1 , xi ], we form the midpoint rectangles Ri shown in Figure 6.4.3. The area of Ri is f (xi ) xi , and the centroid of Ri is its center (xi , 1 f (xi )). By (6.4.1), the centroid 2 (x p , y p ) of the union of all these rectangles satises the following equations: DERIVATION x p A p = x1 f x1y p Ap =1 2 x1 + + xn f xn 2 f x1x1 + +1 2 f xnyxn , 2f (xi*)xn .a(Here A P represents the area of the union of the n rectangles.) As P 0, the union of rectangles tends to the shape of and the equations we just derived tend to the formulas given in (6.4.2). xi 1xi*xiFigure 6.4.3Before we start looking for centroids, we should explain what we are looking for. We learn from physics that, in our world of W = mg, the centroid of a plane region is the balance point of the plate , at least in the following sense: If has centroid (x, y), then the plate can be balanced on the line x = x and it can be balanced on the line y = y. If (x, y) is actually in , which is not necessarily the case, then the plate can be balanced at this point.Example 1 Locate the centroid of the quarter-disk shown in Figure 6.4.4.y y=xSOLUTION The quarter-disk is symmetric about the line y = x. Therefore we knowthat x = y. Herer ryA = 0b(x) =1 [ 2rf (x)]2 d x = r2 x2 01 2 (r 2 x 2) d x =1 2r 2x 1 x3 3r 0= 1 r 3. 3(x, y )y = r 2 x 2rFigure 6.4.4Since A = 1 r 2 , 4 y=1 3 r 3 1 r 2 4=4r . 3The centroid of the quarter-disk is the point 4r 4r , 3 3.x14:26 338. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006314 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL NOTE: It is almost as easy to calculate x A: rxA = 0x r2 x2 dx0 0= 1 2[u = (r 2 x 2 ), du = 2x d x]u 1/2 du r2 0 2 3/2 u 3 r2= 1 2= 1 r 3, 3and proceed from there. yrx f (x) d x =Example 2 Locate the centroid of the triangular region shown in Figure 6.4.5.(0, h)y=h bx+hSOLUTION The hypotenuse lies on the linehh y = x +h . b(x, y ) (b, 0) xbHenceFigure 6.4.5bxA =bx f (x) d x =01 h x 2 + hx d x = b2 h b 60and byA = 01 [ 2bf (x)]2 d x =1 2h 2 2 2h 2 x x + h 2 d x = 1 bh 2 . 6 b2 b0Since A = 1 bh, we have 2 x=1 2 b h 6 1 bh 2= 1b 3y=and1 bh 2 6 1 bh 2= 1h . 3The centroid is the point ( 1 b, 1 h). 3 3 yFigure 6.4.6 shows the region between the graphs of two continuous functions f and g. In this case, if has area A and centroid (x, y), thenf(6.4.3)bFigure 6.4.6x[ f (x) g(x)] d x,xbyA =ag abxA =a1 ([ 2f (x)]2 [g(x)]2 ) d x.VERIFICATION Let A f be the area below the graph of f and let A g be the area below the graph of g. Then, in obvious notation,x A + x g Ag = x f A fy A + y g Ag = y f A fandTherefore bx A = x f A f x g Ag =bx f (x) d x abxg(x) d x =ax[ f (x) g(x)] d xaand by A = y f A f y g Ag =1 [ 2a b= abf (x)]2 d x 1 ([ 2a1 [g(x)]2 2dxf (x)]2 [g(x)]2 ) d x. 14:26 339. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006yExample 3 Locate the centroid of the region shown in Figure 6.4.7. SOLUTION Here there is no symmetry we can appeal to. We must carry out the calculations. 2A=2[ f (x) g(x)] d x =02x[ f (x) g(x)] d x =30 20(2x x ) d x = 20yA =20 2x=(2x x ) d x = x 21 ([ 2Therefore x =f (x)] [g(x)] ) d x = 24 4 / 3 32= 1 and y =32 4 / 15 32 1 2=2 1 3 x 3 02 3 x 3(4x x ) d x = 240 8 . 5=1 4 3 x 2 3g (x) = x 2= 2 1 5 x 5 0The centroid is the point 1,8 5x=32 . 15.All the formulas that we have derived for volumes of solids of revolution are simple corollaries to an observation made by a brilliant ancient Greek, Pappus of Alexandria (circa 300 a.d.). THEOREM 6.4.4 PAPPUS'S THOREM ON VOLUMESA plane region is revolved about an axis that lies in its plane. If the region does not cross the axis, then the volume of the resulting solid of revolution is the area of the region multiplied by the circumference of the circle described by the centroid of the region: V = 2 R A where A is the area of the region and R is the distance from the axis of revolution to the centroid of the region. (See Figure 6.4.8.) axis axisRcentroidFigure 6.4.8Basically we have derived only two formulas for the volumes of solids of revolution: If the region x-axis, the resulting solid has volume bVx =of Figure 6.4.6 is revolved about the ([ f (x)]2 [g(x)]2 ) d x.a f (x) = 2x4 , 3Pappuss Theorem on VolumesThe Washer-Method Formula(2, 4)(x, y )4 , 32 1 4 x 4 03156.4 THE CENTROID OF A REGION; PAPPUSS THEOREM ON VOLUMESThis theorem is found in Book VII of Pappuss Mathematical Collection, largely a survey of ancient geometry to which Pappus made many original contributions (among them this theorem). Much of what we know today of Greek geometry we owe to Pappus.Figure 6.4.714:26 340. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006316 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL The Shell-Method Formula If the region the resulting solid has volume bVy =of Figure 6.4.6 is revolved about the y-axis,2 x[ f (x) g(x)] d x.aNote that bVx = ([ f (x)]2 [g(x)]2 ) d xa b= 2[ f (x)]2 [g(x)]2 d x = 2 y A = 2 R A1 2aand bVy =2 x[ f (x) g(x)] d x = 2 x A = 2 R A,aas asserted by Pappus.Remark In stating Pappuss theorem, we assumed a complete revolution. If is only partially revolved about a given axis, then the volume of the resulting solid is simply the area of multiplied by the length of the circular arc described by the centroid. Applying Pappuss Theorem Example 4 Earlier we saw that the region in Figure 6.4.7 has area 4 and centroid 3 1, 8 . Find the volumes of the solids formed by revolving this region (a) about the 5 y-axis, (b) about the line y = 5. SOLUTION(a) We have already calculated this volume by two methods: by the washer method and by the shell method. The result was V = 8 . Now we calculate the volume by 3 Pappuss theorem. Here we have R = 1 and A = 4 . Therefore 3 V = 2 (1) (b) In this case R = 5 8 5=17 54 3= 8 . 3and A = 4 . Therefore 3V = 217 54 3=136 . 15Example 5 Find the volume of the torus generated by revolving the circular disk (x h)2 + (y k)2 r 2 , (a) about the x-axis,h, k r(Figure 6.4.9)(b) about the y-axis.SOLUTION The centroid of the disk is the center (h, k). This lies k units from thex-axis and h units from the y-axis. The area of the disk is r 2 . Therefore (a) Vx = 2 (k)(r 2 ) = 2 2 kr 2 .(b) Vy = 2 (h)(r 2 ) = 2 2 hr 2 .14:26 341. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.4 THE CENTROID OF A REGION; PAPPUSS THEOREM ON VOLUMES y317yr khxxxFigure 6.4.9Example 6 Locate the centroid of the half-disk x 2 + y2 r 2,y0by appealing to Pappuss theorem. SOLUTION Since the half-disk is symmetric about the y-axis, we know that x = 0.All we need is y. If we revolve the half-disk about the x-axis, we obtain a solid ball of volume 4 r 3 . 3 The area of the half-disk is 1 r 2 . By Pappuss theorem 2 4 r 3 3= 2 y1 r 2 2.Simple division gives y = 4r/3 . RemarkCentroids of solids of revolution are introduced in Project 6.4.EXERCISES 6.4 Exercises 114. Sketch the region bounded by the curves. Locate the centroid of the region and nd the volume generated by revolving the region about each of the coordinate axes. 2. y = x 3 , y = 0, x = 2. 1. y = x, y = 0, x = 4. 4. y = x 3 , y = x. 3. y = x 2 , y = x 1/3 . 5. y = 2x, y = 2, x = 3. 6. y = 3x, y = 6, x = 1. 7. y = x 2 + 2, y = 6, x = 0. 8. y = x 2 + 1, y = 1, x = 3. 9. x + y = 1, x + y = 1. 10. y = 1 x 2 , x + y = 1. 11. y = x 2 , y = 0, x = 1, x = 2. 12. y = x 1/3 , y = 1, x = 8. 13. y = x, x + y = 6, y = 1. 14. y = x, y = 2x, x = 3. Exercises 1524 Locate the centroid of the bounded region determined by the curves.15. 17. 18. 19. 21. 22. 23. 24. 25.16. y = 4x x 2 , y = 2x 3. y = 6x x 2 , y = x. x 2 = 4y, x 2y + 4 = 0. y = x 2 , 2x y + 3 = 0. 20. y 2 = 2x, y = x x 2 . y 3 = x 2 , 2y = x. 2 2 y = x 2x, y + 6x x . y = 6x x 2 , x + y = 6. x + 1 = 0, x + y 2 = 0. x + y = a, x = 0, y = 0. Let be the annular region (ring) formed by the circles x 2 + y2 =1 4x 2 + y 2 = 4.(a) Locate the centroid of . (b) Locate the centroid of the rst-quadrant part of . (c) Locate the centroid of the upper half of . 26. The ellipse b2 x 2 + a 2 y 2 = a 2 b2 encloses a region of area ab. Locate the centroid of the upper half of the region.14:26 342. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006318 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 27. The rectangle in the accompanying gure is revolved about the line marked l. Find the volume of the resulting solid.32. The region in the accompanying gure consists of a square S of side 2r and four semidisks of radius r. Locate the centroid of each of the following. (a) . (b) 1 . (c) S 1 . (d) S 3 . (e) S 1 3 . (f) S 1 2 . (g) S 1 2 3 .lyc b1a 428. In Example 2 of this section you saw that the centroid of the triangle in Figure 6.4.5 is at the point 1 b, 1 h . 3 3 (a) Verify that the line segments that join the centroid to the vertices divide the triangle into three triangles of equal area. (b) Find the distance d from the centroid of the triangle to the hypotenuse. (c) Find the volume generated by revolving the triangle about the hypotenuse. 29. The triangular region in the gure is the union of two right triangles 1 , 2 . Locate the centroid: (a) of 1 , (b) of 2 , (c) of the entire region.S2x333. Give an example of a region that does not contain its centroid. 34. The centroid of a triangular region can be located without integration. Find the centroid of the region shown in the accompanying gure by applying Principles 1 and 2. Then verify that this point x, y lies on each median of the triangle, two-thirds of the distance from the vertex to the opposite side.y a(b, h) h 12 xb30. Find the volume of the solid generated by revolving the entire triangular region of Exercise 29. (a) about the x-axis; (b) about the y-axis. 31. (a) Find the volume of the ice-cream cone of Figure A. (A right circular cone topped by a solid hemisphere.) (b) Find x for the region in Figure B.(0, 0)CyCCRRR CFigure AFigure Bx(a, 0) 35. Use a graphing utility to draw the graphs of y = 3 x and y = x 3 for x 0. Let be the region bounded by the two curves. Use a CAS to nd: (a) the area of . (b) the centroid of ; plot the centroid. (c) the volume of the solid generated by revolving about the x-axis. (d) the volume of the solid generated by revolving about the y-axis. 36. Exercise 35 with y = x 2 2x + 4 and y = 2x + 1. 37. Use a graphing utility to draw the graphs of y = 16 8x and y = x 4 5x 2 + 4. Let be the region bounded by the two curves. Use a CAS to nd: (a) the area of . (b) the centroid of . 38. Exercise 37 with y = 2 + x + 2 and y = 1 (5x 2 + 6 3x 2).14:26 343. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.5 THE NOTION OF WORK319 PROJECT 6.4 Centroid of a Solid of Revolution If a solid is homogeneous (constant mass density), then the center of mass depends only on the shape of the solid and is called the centroid. In general, determination of the centroid of a solid requires triple integration. (Chapter 17.) However, if the solid is a solid of revolution, then the centroid can be found by onevariable integration. Let be the region shown in the gure and let T be the solid generated by revolving around the x-axis. By symmetry, the centroid of T is on the x-axis. Thus the centroid of T is determined solely by its x-coordinate x.HINT: Use the following principle: if a solid of volume V consists of a nite number of pieces with volumes V1 , V2 , . . . , Vn and the pieces have centroids x 1 , x 2 , . . . , x n , then x V = x 1 V1 + x 2 V2 + + x n Vn . Now revolve around the y-axis and let S be the resulting solid. By symmetry, the centroid of S lies on the y-axis and is determined solely by its y-coordinate y. Problem 2. Show that yV = volume of S. x[ f (x)]2 d x where V is theProblem 3. Use the results in Problems 1 and 2 to locate the centroid of each of the following solids:yfWaProblem 1. Show that x V = volume of T.b b ax x[ f (x)]2 d x where V is the(a) A solid cone of base radius r and height h. (b) A ball of radius r. (c) The solid generated by revolving about the x-axis the rstquadrant region bounded by the ellipse b2 x 2 + a 2 y 2 = a 2 b2 and the coordinate axes. (d) The solid generated by revolving the region below the graph of f (x) = x, x [0, 1], (i) about the x-axis; (ii) about the y-axis. (e) The solid generated by revolving the region below the graph of f (x) = 4 x 2 , x [0, 2], (i) about the x-axis; (ii) about the y-axis. 6.5 THE NOTION OF WORK We begin with a constant force F directed along some line that we call the x-axis. By convention we view F as positive if it acts in the direction of increasing x and negative if it acts in the direction of decreasing x. (Figure 6.5.1) F>0F 0, but if F acts against the motion, then W < 0. Thus, for example, if an object slides off a table and falls to the oor, then the work done by gravity is positive (earths gravity points down). But if an object is lifted from the oor and raised to tabletop level, then the work done by gravity is negative. However, the work done by the hand that lifts the object is positive.14:26 344. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006320 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL To repeat, if an object moves from x = a to x = b subject to a constant force F, then the work done by F is the constant value of F times b a. What is the work done by F if F does not remain constant but instead varies continuously as a function of x? As you would expect, we then dene the work done by F as the average value of F times b a: bW =(6.5.2)f (x) d x.(Figure 6.5.2)aa constant forceabvariable forceW = F (b a)bbW = aF(x) dxFigure 6.5.2Hookes Law You can sense a variable force in the action of a steel spring. Stretch a spring within its elastic limit and you feel a pull in the opposite direction. The greater the stretching, the harder the pull of the spring. Compress a spring within its elastic limit and you feel a push against you. The greater the compression, the harder the push. According to Hookes law (Robert Hooke, 16351703), the force exerted by the spring can be written F(x) = kx where k is a positive number, called the spring constant, and x is the displacement from the equilibrium position. The minus sign indicates that the spring force always acts in the direction opposite to the direction in which the spring has been deformed (the force always acts so as to restore the spring to its equilibrium state). Remark Hookes law is only an approximation, but it is a good approximation for small displacements. In the problems that follow, we assume that the restoring force of the spring is given by Hookes law. Example 1 A spring of natural length L, compressed to length force F0 .7 L, 8exerts a(a) Find the work done by the spring in restoring itself to natural length. (b) What work must be done to stretch the spring to length 11 L? 10 SOLUTION Place the spring on the x-axis so that the equilibrium point falls at the origin. View compression as a move to the left. (See Figure 6.5.3.) Our rst step is to determine the spring constant. Compressed 1 L units to the left, 8 the spring exerts a force F0 . Thus, by Hookes lawF0 = F 1 L = k 1 L = 1 k L . 8 8 8 Therefore k = 8F0 /L. The force law for this spring reads F(x) = 8F0 Lx.14:26 345. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.5 THE NOTION OF WORK0x0x0xLLx011 L 107 L 8Figure 6.5.3Figure 6.5.4(a) To nd the work done by this spring in restoring itself to equilibrium, we integrate F(x) from x = 1 L to x = 0: 8 0W =L/8F(x) d x =0 L/88F0 Lx dx = x2 28F0 L0 L/8=LF0 . 16(b) We refer to Figure 6.5.4. To stretch the spring, we must counteract the force of the spring. The force exerted by the spring when stretched x units is F(x) = 8F0 Lx.To counter this force, we must apply the opposite force F(x) =8F0 Lx.The work we must do to stretch the spring to length 11 L can be found by integrating 10 1 F(x) from x = 0 to x = 10 L: L/10W = 0L/10F(x) d x = 08F0 L8F0 x dx = Lx2 2L/10= 0LF0 . 25UnitsThe unit of work is the work done by a unit force which displaces an object a unit distance in the direction of the force. If force is measured in pounds and distance is measured in feet, then the work is given in foot-pounds. In the SI system force is measured in newtons, distance is measured in meters, and work is given in newtonmeters. These are called joules. There are other units used to quantify work, but for our purposes foot-pounds and joules are sufcient.Example 2 Stretched 1 meter beyond its natural length, a certain spring exerts a 3 restoring force with a magnitude of 10 newtons. What work must be done to stretch the spring an additional 1 meter? 3 SOLUTION Place the spring on the x-axis so that the equilibrium point falls at theorigin. View stretching as a move to the right and assume Hookes law: F(x) = kx. When the spring is stretched 1 meter, it exerts a force of 10 newtons (10 newtons 3 to the left). Therefore, 10 = k( 1 ) and k = 30. 3 The term newton deserves denition. In general, force is measured by the acceleration that it imparts. The denition of a newton of force is made on that basis; namely, a force is said to measure 1 newton if it acts in the positive direction and imparts an acceleration of 1 meter per second per second to a mass of 1 kilogram.32114:26 346. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006322 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL To nd the work necessary to stretch the spring an additional the opposite force F(x) = 30x from x = 1 to x = 2 : 3 3 W =2/32/3 1 2 x 2 1/330x d x = 301/31 3meter, we integrate= 5 joules. Counteracting the Force of Gravity To lift an object we must counteract the force of gravity. Therefore, the work required to lift an object is given by the equation (6.5.3)work = (weight of the object) (distance lifted).If an object is lifted from level x = a to level x = b and the weight of the object varies continuously with xsay the weight is w(x)then the work done by the lifting force is given by the integral bW =(6.5.4)w(x) d x. aThis is just a special case of (6.5.2).Example 3 A 150-pound bag of sand is hoisted from the ground to the top of a 50-foot building by a cable of negligible weight. Given that sand leaks out of the bag at the rate of 0.75 pounds for each foot that the bag is raised, nd the work required to hoist the bag to the top of the building. SOLUTION Once the bag has been raised x feet, the weight of the bag has been reduced to 150 0.75x pounds. Therefore xn5050W =xn 10(150 0.75x) d x = 150x 1 (0.75)x 2 250 0= 150(50) 1 (0.75)(50)2 = 6562.5 foot-pounds 250 xi x2 x1xixi* xi 1x2 x1 0Example 4 What is the work required to hoist the sandbag of Example 3 given that the cable weighs 1.5 pounds per foot? SOLUTION To the work required to hoist the sandbag of Example 3, which we found to be 6562.5 foot-pounds, we must add the work required to hoist the cable. Instead of trying to apply (6.5.4), we go back to fundamentals. We partition the interval [0,50] as in Figure 6.5.5 and note that the ith piece of cable weighs 1.5 xi pounds and is approximately 50 xi feet from the top of the building. Thus the work required to lift this piece to the top is approximately(1.5) xi 50 xi = 1.5 50 xiFigure 6.5.5xi foot-pounds.The weight of an object of mass m is the product mg where g is the magnitude of the acceleration due to gravity. The value of g is approximately 32 feet per second per second; in the metric system, approximately 9.8 meters per second per second.14:26 347. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.5 THE NOTION OF WORKIt follows that the work required to hoist the entire cable is approximately 1.5(50 x1 ) x1 + 1.5(50 x2 ) x2 + + 1.5(50 xn ) xn foot-pounds.This sum is a Riemann sum which, as max 50 0xi 0, converges to the denite integral1.5(50 x) d x = 1.5 50x 1 x 2 250 0= 1875.The work required to hoist the cable is 1875 foot-pounds. The work required to hoist the sandbag by this cable is therefore 6562.5 foot-pounds + 1875 foot-pounds = 8437.5 foot-pounds. Remark We just found that a hanging 50-foot cable that weighs 1.5 pounds per foot can be lifted to the point from which it hangs by doing 1875 foot-pounds of work. This result can be obtained by viewing the weight of the entire cable as concentrated at the center of mass of the cable: The cable weighs 1.5 50 = 75 pounds. Since the cable is homogeneous, the center of mass is at the midpoint of the cable, 25 feet below the suspension point. The work required to lift 75 pounds a distance of 25 feet is 75 pounds 25 feet = 1875 foot-pounds. We have found that this simplication works. But how come? To understand why this simplication works, we reason as follows: Initially the cable hangs from a suspension point 50 feet high. The work required to lift the bottom half of the cable to the 25-foot level can be offset exactly by the work done in lowering the top half of the cable to the 25-foot level. Thus, without doing any work (on a net basis), we can place the entire cable at the 25-foot level and proceed from there. (NOTE: In Exercise 31 you are asked to extend the center-of-mass argument to the nonhomogeneous case.) Pumping Out a TankFigure 6.5.6 depicts a storage tank lled to within a feet of the top with some liquid. Assume that the liquid is homogeneous and weighs pounds per cubic foot. Suppose now that this storage tank is pumped out from above so that the level of the liquid drops to b feet below the top of the tank. How much work has been done? For each x [a, b], we let A(x) = cross-sectional area x feet below the top of the tank,0 abs(x) = distance that the x-level must be lifted. We let P = {x0 , x1 , . . . , xn } be an arbitrary partition of [a, b] and focus our attention on the ith subinterval [xi1 , xi ]. (Figure 6.5.7) Taking xi as an arbitrary point in the ith subinterval, we have A(xi ) xi = approximate volume of the ith layer of liquid, A(xi ) xi = approximate weight of this volume, s(xi ) = approximate distance this weight is to be lifted. The symbol is the lowercase Greek letter sigma.Figure 6.5.632314:26 348. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006324 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 0 a xi 1 xixi bFigure 6.5.7Therefore s(xi )A(xi ) xi = approximate work (weight distance) required to pump out this layer of liquid. The work required to pump out all the liquid can be approximated by adding up all these terms: W s(x1 )A(x1 ) x1 + s(x2 )A(x2 ) x2 + + s(xn )A(xn ) xn . =The sum on the right is a Riemann sum. As P 0, such Riemann sums converge to givebW =(6.5.5) s(x) A(x) d x.aWe use this result in Example 5.xFigure 6.5.810Example 5 A hemispherical water tank of radius 10 feet is being pumped out. (See Figure 6.5.8.) Find the work done in lowering the water level from 2 feet below the top of the tank to 4 feet below the top of the tank given that the pump is placed (a) at the top of the tank, (b) 3 feet above the top of the tank. SOLUTION Take 62.5 pounds per cubic foot as the weight of water. From the gure you can see that the cross section x feet below the top of the tank is a disk of radius 100 x 2 . The area of this disk isA(x) = (100 x 2 ). In case (a) we have s(x) = x. Therefore 4W = 262.5 x(100 x 2 ) d x = 33,750 106, 029 foot-pounds. =14:26 349. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20066.5 THE NOTION OF WORK325In case (b) we have s(x) = x + 3. Therefore 4W =62.5(x + 3)(100 x 2 )d x = 67,750 212, 843 foot-pounds. =2Suggestion: Work out Example 5 without invoking Formula (6.5.5). Simply construct the pertinent Riemann sums. EXERCISES 6.5 Exercises 12. An object moves along the x-axis coordinatized in feet under the action of a force of F(x) pounds. Find the work done by F in moving the object from x = a to x = b. 1. F(x) = x(x + 1) ; 2. F(x) = 2x x + 1; 2(a) Verify that the work done in pumping out the oil to the top of the tank is given by the integral 3960a = 1, b = 4.2a = 3, b = 8.Evaluate this integral by evaluating the integralsExercises 36. An object moves along the x-axis coordinatized in meters under the action of a force of F(x) newtons. Find the work done by F in moving the object from x = a to x = b. a = 0, b = 3. 3. F(x) = x x 2 + 7; 4. F(x) = x 2 + cos 2x;a = 0, b = 1 . 45. F(x) = x + sin 2x; 6. F(x) = 7.8.9.10.11.12.13.cos 2xa= ;1 , b 63 0C14.= .a = 0, b =(x + 3) 9 x 2 d x.01 . 22 + sin 2x A 600-pound force compresses a 10-inch automobile coil exactly 1 inch. How much work must be done to compress that coil to 5 inches? Five foot-pounds of work are needed to stretch a certain spring from 1 foot beyond natural length to 3 feet beyond natural length. How much stretching beyond natural length is achieved by a 6-pound force? Stretched 4 feet beyond natural length, a certain spring exerts a restoring force of 200 pounds. How much work is required to stretch the spring: (a) 1 foot beyond natural length? (b) 1 1 feet beyond natural length? 2 A certain spring has natural length L. Given that W is the work required to stretch the spring from L feet to L + a feet, nd the work required to stretch the spring: (a) from L feet to L + 2a feet; (b) from L feet to L + na feet; (c) from L + a feet to L + 2a feet; (d) from L + a feet to L + na feet. Find the natural length of a spring given that the work required to stretch it from 2 feet to 2.1 feet is one-half of the work required to stretch it from 2.1 feet to 2.2 feet. A cylindrical tank of height 6 feet standing on a base of radius 2 feet is full of water. Find the work required to pump the water: (a) to an outlet at the top of the tank; (b) to a level of 5 feet above the top of the tank. (Take the weight of water as 62.5 pounds per cubic foot.) A cylindrical tank of radius 3 feet and length 8 feet is laid out horizontally. The tank is half full of oil that weighs 60 pounds per cubic foot.15. 16.17.18. 19.20.x 9 x2 dx3and9 x2 dx0separately. (b) What is the work required to pump out the oil to a level 4 feet above the top of the tank? Exercise 12 with the same tank laid out horizontally. Use a CAS for the integration. Calculate the work required to hoist the cable of Example 4 by applying (6.5.4). In the coordinate system used to derive (6.5.5) the liquid moves in the negative direction. How come W is positive? A conical container (vertex down) of radius r feet and height h feet is full of liquid that weighs pounds per cubic foot. Find the work required to pump out the top 1 h feet of liquid: 2 (a) to the top of the tank; (b) to a level k feet above the top of the tank. What is the work done by gravity if the tank of Exercise 17 is completely drained through an opening at the bottom? A container of the form obtained by revolving the parabola y = 3 x 2 , 0 x 4, about the y-axis is full of water. Here 4 x and y are given in meters. Find the work done in pumping the water: (a) to an outlet at the top of the tank; (b) to an outlet 1 meter above the top of the tank. Take = 9800. The force of gravity exerted by the earth on a mass m at a distance r from the center of the earth is given by Newtons formula, mM F = G 2 , rwhere M is the mass of the earth and G is the universal gravitational constant. Find the work done by gravity in pulling a mass m from r = r1 to r = r2 . 21. A chain that weighs 15 pounds per foot hangs to the ground from the top of an 80-foot building. How much work is required to pull the chain to the top of the building? 22. A box that weighs w pounds is dropped to the oor from a height of d feet. (a) What is the work done by gravity?14:26 350. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006326 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL23.24.25.26.27.28.29.30.31.32.(b) Show that the work is the same if the box slides to the oor along a smooth inclined plane. (By saying smooth, we are saying disregard friction.) A 200-pound bag of sand is hoisted at a constant rate by a chain from ground level to the top of a building 100 feet high. (a) How much work is required to hoist the bag if the weight of the chain is negligible? (b) How much work is required to hoist the bag if the chain weighs 2 pounds per foot? Suppose that the bag in Exercise 23 has a tear in the bottom and sand leaks out at a constant rate so that only 150 pounds of sand are left when the bag reaches the top. (a) How much work is required to hoist the bag if the weight of the chain is negligible? (b) How much work is required to hoist the bag if the chain weighs 2 pounds per foot? A 100-pound bag of sand is lifted for 2 seconds at the rate of 4 feet per second. Find the work done in lifting the bag if the sand leaks out at the rate of half a pound per second. A rope is used to pull up a bucket of water from the bottom of a 40-foot well. When the bucket is full of water, it weighs 40 pounds; however, there is a hole in the bottom, and the water leaks out at the constant rate of 1 gallon for each 2 10 feet that the bucket is raised. Given that the weight of the rope is negligible, how much work is done in lifting the bucket to the top of the well? (Assume that water weighs 8.3 pounds per gallon.) A rope of length l feet that weighs pounds per foot is lying on the ground. What is the work done in lifting the rope so that it hangs from a beam: (a) l feet high; (b) 2l feet high? A load of weight w is lifted from the bottom of a shaft h feet deep. Find the work done given that the rope used to hoist the load weighs pounds per foot. An 800-pound steel beam hangs from a 50-foot cable which weighs 6 pounds per foot. Find the work done in winding 20 feet of the cable about a steel drum. A water container initially weighing w pounds is hoisted by a crane at the rate of n feet per second. What is the work done if the container is raised m feet and the water leaks out constantly at the rate of p gallons per second? (Assume that the water weighs 8.3 pounds per gallon.) A chain of variable mass density hangs to the ground from the top of a building of height h. Show that the work required to pull the chain to the top of the building can be obtained by assuming that the weight of the entire chain is concentrated at the center of mass of the chain. An object moves along the x-axis. At x = a it has velocity va , and at x = b it has velocity vb . Use Newtons second law of motion, F = ma = m(dv/dt), to show that bW = aF(x)d x =1 2 1 2 mv mv . 2 b 2 aThe term 1 mv 2 is called the kinetic energy of the object. 2 What you have been asked to show is that the work done onan object equals the change in kinetic energy of that object. This is an important result. 33. An object of mass m is dropped from a height h. Express the impact velocity in terms of the gravitational constant g and the height h. In Exercises 3437 use the relation between work and kinetic energy given in Exercise 32. 34. The same amount of work on two objects results in the speed of one being three times that of the other. How are the masses of the two objects related? 35. A major league baseball weighs 5 oz. How much work is required to throw a baseball at a speed of 95 mph? (The balls mass is its weight in pounds divided by 32 ft /sec2 , the acceleration due to gravity.) 36. How much work is required to increase the speed of a 2000pound vehicle from 30 mph to 55 mph? 37. The speed of an earth satellite at an altitude of 100 miles is approximately 17,000 mph. How much work is required to launch a 1000-lb satellite into a 100-mile orbit? (Power) Power is work per unit time. Suppose an object moves along the x-axis under the action of a force F. The work done by F in moving the object from x = a to arbitrary x is given by the integral xW =F(u)du. aViewing position as a function of time, setting x = x(t), we have P=dx dW = F(x(t)) = F[x(t)] v(t). dt dtThis is called the power expended by the force F. If force is measured in pounds, distance in feet, and time in seconds, then power is given in foot-pounds per second. If force is measured in newtons, distance in meters, and time in seconds, then power is given in joules per second. These are called watts. Commonly used in engineering is the term horsepower: 1 horsepower = 550 foot-pounds per second = 746 watts. 38. (a) Assume constant acceleration. What horsepower must an engine produce to accelerate a 3000-pound truck from 0 to 60 miles per hour (88 feet per second) in 15 seconds along a level road? (b) What horsepower must the engine produce if the road rises 4 feet for every 100 feet of road? HINT: Integration is not required to answer these questions. 39. A cylindrical tank set vertically with height 10 feet and radius 5 feet is half-lled with water. Given that a 1-horsepower pump can do 550 foot-pounds of work per second, how long will it take a 1 -horsepower pump: 2 (a) to pump the water to an outlet at the top of the tank?14:26 351. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006*6.6 FLUID FORCE(b) to pump the water to a point 5 feet above the top of the tank? 40. A storage tank in the form of a hemisphere topped by a cylinder is lled with oil that weighs 60 pounds per cubic foot. The hemisphere has a 4-foot radius; the height of the cylinder is 8 feet.327(a) How much work is required to pump the oil to the top of the tank? (b) How long would it take a 1 -horsepower motor to empty 2 out the tank? 41. Show that the rate of change of the kinetic energy of an object is the power of the force expended on it. 6.6 FLUID FORCE If you pour oil into a container of water, youll see that the oil soon rises to the top. Oil weighs less than water. For any uid, the weight per unit volume is called the weight density of the uid. Well denote this by the Greek letter . An object submerged in a uid experiences a compressive force that acts at right angles to the surface of the body exposed to the uid. (It is to counter these compressive forces that submarines have to be built so structurally strong.) Fluid in a container exerts a downward force on the base of the container. What is the magnitude of this force? It is the weight of the column of uid directly above it. (Figure 6.6.1.) If a container with base area A is lled to a depth h by a uid of weight density , the downward force on the base of the container is given by the product F = h A.(6.6.1)Fluid force acts not only on the base of the container but also on the walls of the container. In Figure 6.6.2, we have depicted a vertical wall standing against a body of liquid. (Think of it as the wall of a container or as a dam at the end of a lake.) We want to calculate the force exerted by the liquid on this wall. level of the liquidw(x)0 axbFigure 6.6.2As in the gure, we assume that the liquid extends from depth a to depth b, and we let w(x) denote the width of the wall at depth x. A partition P = {x0 , x1 , . . . , xn } of [a, b] of small norm subdivides the wall into n narrow horizontal strips. (See Figure 6.6.3.) We can estimate the force on the ith strip by taking xi as the midpoint of [xi1 , xi ]. Then w(xi ) = the approximate width of the ith strip and w(xi ) xi = the approximate area of the ith strip.liquid levelhFigure 6.6.114:26 352. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006328 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 0 axi 1 xibFigure 6.6.3Since this strip is narrow, all the points of the strip are approximately at depth xi . Thus, using (6.6.1), we can estimate the force on the ith strip by the product xi w(xi ) xi . Adding up all these estimates, we have an estimate for the force on the entire wall: F x w(x ) x1 + x w(x ) x2 + + x w(x ) xn . = ii22nnThe sum on the right is a Riemann sum for the integral b xw(x) d x.aAs P 0, such Riemann sums converge to that integral. Thus we have(6.6.2)buid force against the wall = xw(x) d x.aThe Weight Density of Water The weight density of water is approximately 62.5 pounds per cubic foot; equivalently, about 9800 newtons per cubic meter. Well use these values in carrying out computations.Example 1 A cylindrical water main 6 feet in diameter is laid out horizontally. (Figure 6.6.4) Given that the main is capped half-full, calculate the uid force on the cap.033x 29 x 2 3Figure 6.6.414:26 353. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006*6.6 FLUID FORCESOLUTION Here = 62.5 pounds per cubic foot. From the gure we see that w(x) = 2 9 x 2 . The uid force on the cap can be calculated as follows: 3F=3(62.5)x(2 9 x 2 ) d x = 62.502x 9 x 2 d x = 1125 pounds. 0Example 2 A metal plate in the form of a trapezoid is afxed to a vertical dam as in Figure 6.6.5. The dimensions shown are in meters. Calculate the uid force on the plate taking the weight density of water as 9800 newtons per cubic meter.water level 4 22448Figure 6.6.5SOLUTION First we nd the width of the plate x meters below the water level. By similar triangles (see Figure 6.6.6),t = 1 (8 x) 2w(x) = 8 + 2t = 16 x.so thatwater level0x 24 xt 8x88Figure 6.6.6The uid force against the plate is 8 489800x(16 x) d x = 9800(16x x 2 ) d x4= 9800 8x 2 1 x 3 38 4 2,300,000 newtons. =32914:26 354. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006330 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL EXERCISES 6.6 1. A rectangular plate 8 feet by 6 feet is submerged vertically in a tank of water, an 8-foot edge at the surface of the water. Find the force of the water on each side of the plate. 2. A square plate 6 feet by 6 feet is submerged vertically in a tank of water, one edge parallel to the surface of the water. Calculate the uid force on each side of the plate given that the center of the plate is 4 feet below the surface of the water. 3. A vertical dam at the end of a reservoir is in the form of an isosceles trapezoid: 100 meters across at the surface of the water, 60 feet across at the bottom. Given that the reservoir is 20 meters deep, calculate the force of the water on the dam. 4. A square metal plate 5 meters by 5 meters is afxed to the lowermost portion of the dam of Exercise 3. What is the force of the water on the plate? 5. A plate in the form of an isosceles trapezoid 4 meters at the top, 6 meters at the bottom, and 3 meters high has its upper edge 10 meters below the top of the dam of Exercise 3. Calculate the force of the water on this plate. 6. A vertical dam in the shape of a rectangle is 1000 feet wide and 100 feet high. Calculate the force on the dam given that (a) the water at the dam is 75 feet deep; (b) the water at the dam is 50 feet deep. 7. Each end of a horizontal oil tank is elliptical, with horizontal axis 12 feet long, vertical axis 6 feet long. Calculate the force on an end when the tank is half full of oil that weighs 60 pounds per cubic foot. 8. Each vertical end of a vat is a segment of a parabola (vertex down) 8 feet across the top and 16 feet deep. Calculate the force on an end when the vat is full of liquid that weighs 70 pounds per cubic foot. 9. The vertical ends of a water trough are isosceles right triangles with the 90 angle at the bottom. Calculate the force on an end of the trough when the trough is full of water given that the legs of the triangle are 8 feet long. 10. The vertical ends of a water trough are isosceles triangles 5 feet across the top and 5 feet deep. Calculate the force on an end when the trough is full of water. 11. The ends of a water trough are semicircular disks with radius 2 feet. Calculate the force of the water on an end given that the trough is full of water. 12. The ends of a water trough have the shape of the parabolic segment bounded by y = x 2 4 and y = 0; the measure-13.14. 15.16.17.18. 19.20.ments are in feet. Assume that the trough is full of water and set up an integral that gives the force of the water on an end. A horizontal cylindrical tank of diameter 8 feet is half full of oil that weighs 60 pounds per cubic foot. Calculate the force on an end. Calculate the force on an end of the tank of Exercise 13 when the tank is full of oil. A rectangular metal plate 10 feet by 6 feet is afxed to a vertical dam, the center of the plate 11 feet below water level. Calculate the force on the plate given that (a) the 10-foot sides are horizontal, (b) the 6-foot sides are horizontal. A vertical cylindrical tank of diameter 30 feet and height 50 feet is full of oil that weighs 60 pounds per cubic foot. Calculate the force on the curved surface. A swimming pool is 8 meters wide and 14 meters long. The pool is 1 meter deep at the shallow end and 3 meters deep at the deep end; the depth increases linearly from the shallow end to the deep end. Given that the pool is full of water, calculate (a) the force of the water on each of the sides, (b) the force of the water on each of the ends. Relate the force on a vertical dam to the centroid of the submerged surface of the dam. Two identical metal plates are afxed to a vertical dam. The centroid of the rst plate is at depth h 1 , and the centroid of the second plate is at depth h 2 . Compare the forces on the two plates given that the two plates are completely submerged. Show that if a plate submerged in a liquid makes an angle with the vertical, then the force on the plate is given by the formula bF= x w(x) sec dxawhere is the weight density of the liquid and w(x) is the width of the plate at depth x, a x b. 21. Find the force of the water on the bottom of the swimming pool of Exercise 17. 22. The face of a rectangular dam at the end of a reservoir is 1000 feet wide, 100 feet tall, and makes an angle of 30 with the vertical. Find the force of the water on the dam given that (a) the water level is at the top of the dam; (b) the water at the dam is 75 feet deep. CHAPTER 6. REVIEW EXERCISES Exercises 14. Sketch the region bounded by the curves. Represent the area of the region by one or more denite integrals (a) in terms of x; (b) in terms of y. Find the area of the region using the more convenient representation. 1. y = 2 x 2 , y = x2. 3. 4. 5.y = x 3 , y = x, y = 1 y 2 = 2(x 1), x y = 5 y 3 = x 2 , x 3y + 4 = 0 Find the area of the region bounded by y = sin x and y = cos x between consecutive intersections of the two graphs.14:26 355. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006*6.6 FLUID FORCE6. Find the area of the region bounded by y = tan2 x and the x-axis from x = 0 to x = /4. 7. The curve y 2 = x(1 x)2 is shown in the gure. Find the area of the loop. y 11x18. The curve x 1/2 + y 1/2 = a 1/2 is shown in the gure. Find the area of the region bounded by the curve and the coordinate axes.14. x 2 = 4y,y = 1 x; 2331y-axis.15. y = x , y = 1, x = 0; x-axis. 16. y = x 3 , y = 1, x = 0; y-axis. 17. y = sec x, y = 0, 0 x /4; x-axis. 18. y = cos x, /2 x /2; x-axis. 19. y = sin x 2 , 0 x ; y-axis. 20. y = cos x 2 , 0 x /2; y-axis. 21. y = 3x x 2 , y = x 2 3x, y-axis. 22. y = 3x x 2 , y = x 2 3x, x = 4. 23. y = (x 1)2 , y = x + 1; x-axis. 24. y = x 2 2x, y = 3x; y-axis. Exercises 2530. The gure shows three regions within the rectangle bounded by the coordinate axes and the lines x = 4 and y = 2. Express the volume obtained by revolving the indicated region about the indicated line: (a) by an integral with respect to x; (b) by an integral with respect to y. Calculate each volume by evaluating one of these integrals 3yy(0, a)21 2 3 (a, 0)x9. The base of a solid is the disk bounded by the circle x 2 + y 2 = r 2 . Find the volume of the solid given that the cross sections perpendicular to the x-axis are: (a) semicircles; (b) isosceles right triangles with hypotenuse on the xy-plane. 10. The base of a solid is the region bounded by the equilateral triangle of side length a with one vertex at the origin and altitude along the positive x-axis. Find the volume of the solid given that cross-sections perpendicular to the x-axis are squares with one side on the base of the solid. 11. The base of a solid is the region in the rst quadrant bounded by the coordinate axes and the line 2x + 3y = 6. Find the volume of the solid given that the cross sections perpendicular to the x-axis are semicircles. 12. A solid in the shape of a right circular cylinder of radius 3 has its base on the xy-plane. A wedge is cut from the cylinder by a plane that passes through a diameter of the base and is inclined to the xy-plane at an angle of 30 . Find the volume of the wedge. Exercises 1324. Sketch the region bounded by the curves and nd the volume of the solid generated by revolving about the axis indicated. 13. x 2 = 4y, y = 1 x; x-axis. 24x25. 1 ; the x-axis. 26. 1 ; the line y = 2. 28. 2 ; the y-axis. 27. 2 ; the line x = 1. 30. 3 ; the line y = 2. 29. 3 ; the y-axis. Exercises 3134. Find the centroid of the bounded region determined by the curves. 32. y = x 3 , y = 4x. 31. y = 4 x 2 , y = 0. 33. y = x 2 4, y = 2x x 2 . 34. y = cos x, y = 0 from x = /2 to x = /2. Exercises 3536. Sketch the region bounded by the curves. Determine the centroid of the region and the volume of the solid generated by revolving the region about each of the coordinate axes. 35. y = x, y = 2 x 2 , 0 x 1 36. y = x 3 , x = y 3 , 0 x 1. 37. An object moves along the x-axis from x = 0 to x = 3 sub ject to a force F(x) = x 7 + x 2 . Given that x is measured in feet and F in pounds, determine the work done by F. 38. One of the springs that supports a truck has a natural length of 12 inches. Given that a force of 8000 pounds compresses this spring 1 inch, nd the work required to compress the 2 spring from 12 inches to 9 inches.14:26 356. P1: PBU/OVYP2: PBU/OVYJWDD023-06JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006332 CHAPTER 6 SOME APPLICATIONS OF THE INTEGRAL 39. The work required to stretch a spring from 9 inches to 10 inches is 1.5 times the work needed to stretch the spring from 8 inches to 9 inches. What is the natural length of the spring? 40. A conical tank 10 feet deep and 8 feet across the top is lled with water to a depth of 5 feet. Find the work done in pumping the water (a) to an outlet at the top of the tank; (b) to an outlet 1 foot below the top of the tank. Take = 62.5 pounds per cubic foot as the weight density of water. 41. A 25-foot chain that weighs 4 pounds per foot hangs from the top of a tall building. How much work is required to pull the chain to the top of the building? 42. A bucket that weighs 5 pounds when empty rests on the ground lled with 60 pounds of sand. The bucket is lifted to the top of a 20 foot building at a constant rate. The sand leaks out of the bucket at a constant rate and only two-thirds of the sand remains when the bucket reaches the top. Find the work done in lifting the bucket of sand to the top of the building.43. A spherical oil tank of radius 10 feet is half full of oil that weighs 60 pounds per cubic foot. Find the work required to pump the oil to an outlet at the top of the tank. 44. A rectangular sh tank has length 1 meter, width 1 meter, 2 depth 1 meter. Given that the tank is full of water, nd 2 (a) the force of the water on each of the sides of the tank; (b) the force of the water on the bottom of the tank. Take the weight density of water as 9800 newtons per cubic meter. 45. A vertical dam is in the form of an isosceles trapezoid 300 meters across the top, 200 meters across the bottom, 50 meters high. (a) What is the force of the water on the face of the dam when the water level is even with the top of the dam? (b) What is the force of the water on the dam when the water level is 10 meters below the the top of the dam? Take the weight density of water as 9800 newtons per cubic meter.14:26 357. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006CHAPTER7THE TRANSCENDENTAL FUNCTIONSSome real numbers satisfy polynomial equations with integer coefcients: satises the equation5x 3 = 0;2 satises the equationx 2 2 = 0.3 5Such numbers are called algebraic. There are, however, numbers that are not algebraic, among them . Such numbers are called transcendental. Some functions satisfy polynomial equations with polynomial coefcients: x f (x) = satises the equation ( x + 2) f (x) x = 0; x + 2 f (x) = 2 x 3x 2 satises the equation [ f (x)]2 + 6x 2 f (x) + (9x 4 4x) = 0. Such functions are called algebraic. There are, however, functions that are not algebraic. Such functions are called transcendental. You are already familiar with some transcendental functionsthe trigonometric functions. In this chapter we introduce other transcendental functions: the logarithm function, the exponential function, and the trigonometric inverses. But rst, a little more on functions in general. 7.1 ONE-TO-ONE FUNCTIONS; INVERSES One-to-One Functions A function can take on the same value at different points of its domain. Constant functions, for example, take on the same value at all points of their domains. The quadratic function f (x) = x 2 takes on the same value at c as it does at c; so does the absolute-value function g(x) = x . The function f (x) = 1 + (x 3)(x 5) takes on the same value at x = 5 as it does at x = 3: f (3) = 1,f (5) = 1.33313:36 358. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006334 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Functions for which this kind of repetition does not occur are called one-to-one functions. yDEFINITION 7.1.1A function f is said to be one-to-one if there are no two distinct numbers in the domain of f at which f takes on the same value. f (x1 ) = f (x2 ) x1x2f is not one-to-one:x1 = x2 .impliesxf (x1) = f (x2)Thus, if f is one-to-one and x1 , x2 are different points of the domain, then f (x1 ) = f (x2 ).Figure 7.1.1The functions f (x) = x 3yxf is one-to-oneFigure 7.1.2andf (x) =xare both one-to-one. The cubing function is one-to-one because no two distinct numbers have the same cube. The square-root function is one-to-one because no two distinct nonnegative numbers have the same square root. There is a simple geometric test, called the horizontal line test, which can be used to determine whether a function is one-to-one. Look at the graph of the function. If some horizontal line intersects the graph more than once, then the function is not one-to-one. (Figure 7.1.1) If, on the other hand, no horizontal line intersects the graph more than once, then the function is one-to-one (Figure 7.1.2).Inverses We begin with a theorem about one-to-one functions.THEOREM 7.1.2If f is a one-to-one function, then there is one and only one function g dened on the range of f that satises the equation f (g(x)) = xfor all x in the range of f.PROOF The proof is straightforward. If x is in the range of f, then f must take on the value x at some number. Since f is one-to-one, there can be only one such number. We have called that number g(x). The function that we have named g in the theorem is called the inverse of f and is usually denoted by the symbol f 1 . fDEFINITION 7.1.3 INVERSE FUNCTIONf 1(x)x = f (f 1(x))f 1Figure 7.1.3Let f be a one-to-one function. The inverse of f, denoted by f 1 , is the unique function dened on the range of f that satises the equation f ( f 1 (x)) = xfor all x in the range of f.(Figure 7.1.3)13:36 359. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.1 ONE-TO-ONE FUNCTIONS; INVERSESRemark The notation f 1 for the inverse function is standard, at least in the United States. Unfortunately, there is the danger of confusing f 1 with the reciprocal of f, that is, with 1/ f (x). The 1 in the notation for the inverse of f is not an exponent; f 1 (x) does not mean 1/ f (x). On those occasions when we want to express 1/ f (x) using the exponent 1, we will write [ f (x)]1 . 335yfExample 1 You have seen that the cubing function f (x) = x 3f11is one-to-one. Find the inverse. SOLUTION We set y = f 1 (x) and apply f to both sides:1f (y) = xfy =x 3y=x1 11( f is the cubing function) 1/3.Recalling that y = f 1 (x), we haveff 1 (x) = x 1/3 .Figure 7.1.4The inverse of the cubing function is the cube-root function. The graphs of f (x) = x and f 1 (x) = x 1/3 are shown in Figure 7.1.4. 3Remark We set y = f 1 (x) to avoid clutter. It is easier to work with a single letter y than with the expression f 1 (x). Example 2 Show that the linear function y = 3x 5 is one-to-one. Then nd the inverse. SOLUTION To show that f is one-to-one, lets suppose thatf (x1 ) = f (x2 ). Then 3x1 5 = 3x2 5 3x1 = 3x2 x1 = x2 . The function is one-to-one since f (x1 ) = f (x2 )impliesx1 = x2 .(Viewed geometrically, the result is obvious. The graph is a line with slope 3 and as such cannot be intersected by any horizontal line more than once.) Now lets nd the inverse. To do this, we set y = f 1 (x) and apply f to both sides: f (y) = x 3y 5 = x 3y = x + 5 y = 1x + 5. 3 3x13:36 360. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006336 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Recalling that y = f 1 (x), we haveyf 1 (x) = 1 x + 5 . 3 3fThe graphs of f and f 1 are shown in Figure 7.1.5. f1Example 3 Find the inverse of the function5/3f (x) = (1 x 3 )1/5 + 2. 5/35xSOLUTION We set y = f 1 (x) and apply f to both sides:f (y) = x (1 y 3 )1/5 + 2 = x 5(1 y 3 )1/5 = x 2 1 y 3 = (x 2)5 y 3 = 1 (x 2)5Figure 7.1.5y = [1 (x 2)]5 ]1/3 . Recalling that y = f 1 (x), we have f 1 (x) = [1 (x 2)5 ]1/3 . Example 4 Show that the function F(x) = x 5 + 2x 3 + 3x 4 is one-to-one. SOLUTION Setting F(x 1 ) = F(x2 ), we have 5 3 5 3 x1 + 2x1 + 3x1 4 = x2 + 2x2 + 3x2 4 5 3 5 3 x1 + 2x1 + 3x1 = x2 + 2x2 + 3x2 .How to go on from here is far from clear. The algebra becomes complicated. Here is another approach. Differentiating F, we get F (x) = 5x 4 + 6x 2 + 3. Note that F (x) > 0 for all x and therefore F is an increasing function. Increasing functions are clearly one-to-one: x1 < x2 implies F(x1 ) < F(x2 ), and so F(x1 ) cannot possibly equal F(x2 ). Remark In Example 4 we used the sign of the derivative to test for one-to-oneness. For functions dened on an interval, the sign of the derivative and one-to-oneness can be summarized as follows: functions with positive derivative are increasing functions and therefore one-to-one; functions with negative derivative are decreasing functions and therefore one-to-one. Suppose that the function f has an inverse. Then, by denition, f 1 satises the equation (7.1.4)f ( f 1 (x)) = xfor all x in the range of f.13:36 361. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.1 ONE-TO-ONE FUNCTIONS; INVERSESf (x)It is also true that f(7.1.5)337f1( f (x)) = x1for all x in the domain of f. fPROOFTake x in the domain of f and set y = f (x). Since y is the range of f,xf 1( f (x)) = xf ( f 1 (y)) = y.Figure 7.1.6 x = f ( f 1(x))This means that f ( f 1 ( f (x))) = f (x)f 1and tells us that f takes on the same value at f 1 ( f (x)) as it does at x. With f one-to-one, this can only happen ifff 1 ( f (x)) = x. f 1(x)Equation (7.1.5) tells us that f 1 undoes what is done by f : f takes x to f (x);f 1 takes f (x) back to x.Equation (7.1.4) tells us that f undoes what is done by f f 1 takes x to f 1 (x);1Figure 7.1.7 (Figure 7.1.6)yf 1 y=x:f takes f 1 (x) back to x.Q ( f (x), x)(Figure 7.1.7)fIt is evident from this that domain of f 1 = range of fandP (x, f (x))range of f 1 = domain of f.xThe Graphs of f and f 1Figure 7.1.8 yThe graph of f consists of points (x, f (x)). Since f 1 takes on the value x at f (x), the graph of f 1 consists of points ( f (x), x). If, as usual, we use the same scale on the y-axis as we do on the x-axis, then the points (x, f (x)) and ( f (x), x) are symmetric with respect to the line y = x. (Figure 7.1.8.) Thus we see thatfxthe graph of f 1 is the graph of f reected in the line y = x. This idea pervades all that follows. Figure 7.1.9Example 5 Sketch the graph of f 1 for the function f graphed in Figure 7.1.9.yy=xSOLUTION First we draw the line y = x. Then we reect the graph of f in that line. The result is shown in Figure 7.1.10. fContinuity and Differentiability of Inverses Let f be a one-to-one function. Then f has an inverse, f 1 . Suppose, in addition, that f is continuous. Since the graph of f has no holes or gaps, and since the graph of f 1 is simply the reection of the graph of f in the line y = x, we can conclude that the graph of f 1 also has no holes or gaps; namely, we can conclude that f 1 is also continuous. We state this result formally; a proof is given in Appendix B.3.xf1Figure 7.1.1013:36 362. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006338 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS THEOREM 7.1.6Let f be a one-to-one function dened on an open interval I. If f is continuous, then its inverse f 1 is also continuous. Now suppose that f is differentiable. Is f 1 necessarily differentiable? Lets assume so for the moment. From the denition of inverse, we know that f ( f 1 (x)) = xfor all x in the range of f.Differentiation gives d [ f ( f 1 (x))] = 1. dx However, by the chain rule, d [ f ( f 1 (x))] = f ( f 1 (x))( f 1 ) (x). dx Therefore f ( f 1 (x))( f 1 ) (x) = 1, and, if f ( f 1 (x)) = 0, ( f 1 ) (x) =(7.1.7)1 f (f 1 (x)).For a geometric understanding of this relation, we refer you to Figure 7.1.11. yy=xl2(f 1(x), x)l1 (x, f 1(x)) f(b, b) f 1(x)xxf 1Figure 7.1.11The graphs of f and f 1 are reections of each other in the line y = x. The tangent lines l1 and l2 are also reections of each other. From the gure, ( f 1 ) (x) = slope of l1 =f 1 (x) b , x bf ( f 1 (x)) = slope of l2 =so that ( f 1 ) (x) and f ( f 1 (x)) are indeed reciprocals.x b , bf 1 (x)13:36 363. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.1 ONE-TO-ONE FUNCTIONS; INVERSESThe gure shows two tangents intersecting the line y = x at a common point. If the tangents have slope 1, they do not intersect that line at all. However, in that case, both graphs have slope 1, the derivatives are 1, and the relation holds. One more observation. We have assumed that f ( f 1 (x)) = 0. If f ( f 1 (x)) = 0, then the tangent to the graph of f at ( f 1 (x), x) is horizontal, and the tangent to the graph of f 1 at (x, f 1 (x)) is vertical. In this case f 1 is not differentiable at x. Formula (7.1.7) has an unwieldy look about it; too many fussy little symbols. The following characterization of ( f 1 ) may be easier to understand.THEOREM 7.1.8Let f be a one-to-one function differentiable on an open interval I. Let a be a point of I and let f (a) = b. If f (a) = 0, then f 1 is differentiable at b and ( f 1 ) (b) =1 . f (a)This theorem, proven in Appendix B.3, places our discussion on a rm footing. RemarkNote that a = f 1 (b), and therefore ( f 1 ) (b) =1 1 = . f (a) f ( f 1 (b))This is simply (7.1.7) at x = b. We rely on Theorem 7.1.8 when we cannot solve for f 1 explicitly and yet we want to evaluate ( f 1 ) at a particular number.Example 6 The function f (x) = x 3 + 1 x is differentiable and has range 2 (, ). (a) Show that f is one-to-one. (b) Calculate ( f 1 ) (9). SOLUTION(a) To show that f is one-to-one, we note that f (x) = 3x 2 +1 2>0for all real x.Thus f is an increasing function and therefore one-to-one. (b) To calculate ( f 1 ) (9), we want to nd a number a for which f (a) = 9. Then ( f 1 ) (9) is simply 1/ f (a). The assumption f (a) = 9 gives a3 + 1 a = 9 2 and tells us a = 2. (We must admit that this example was contrived so that the algebra would be easy to carry out.) Since f (2) = 3(2)2 + 1 = 25 , we conclude that 2 2 ( f 1 ) (9) =1 1 2 = 25 = . f (2) 25 233913:36 364. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006340 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Finally, a few words about differentiating inverses in the Leibniz notation. Suppose that y is a one-to-one function of x: y = y(x). Then x is a one-to-one function of y: x = x(y). Moreover, y(x(y)) = yfor all y in the domain of x.Assuming that y is a differentiable function of x and x is a differentiable function of y, we have y (x(y))x (y) = 1, which, if y (x(y)) = 0, gives x (y) =1 . y (x(y))In the Leibniz notation, we have(7.1.9)dx 1 = . dy dy/d xThe rate of change of x with respect to y is the reciprocal of the rate of change of y with respect to x. Where are these rates of change to be evaluated? Given that y(a) = b, the right side is to be evaluated at x = a and the left side at y = b.EXERCISES 7.1 Exercises 126. Determine whether or not the function is oneto-one and, if so, nd the inverse. If the function has an inverse, give the domain of the inverse. 1. 3. 5. 7. 9. 11. 13. 15. 17. 18. 19. 21.f (x) = 5x + 3. f (x) = 1 x 2 . f (x) = x 5 + 1. f (x) = 1 + 3x 3 . f (x) = (1 x)3 . f (x) = (x + 1)3 + 2. f (x) = x 3/5 . f (x) = (2 3x)3 . f (x) = sin x, x , 2 2 f (x) = cos x, x , 2 2 1 f (x) = . x 1 f (x) = x + . x2. 4. 6. 8. 10. 12. 14. 16.f (x) = 3x + 5. f (x) = x 5 . f (x) = x 2 3x + 2. f (x) = x 3 1. f (x) = (1 x)4 . f (x) = (4x 1)3 . f (x) = 1 (x 2)1/3 . f (x) = (2 3x 2 )3 .1 1 . 24. f (x) = 1. +1 1x x +2 1 25. f (x) = . . 26. f (x) = x +1 (x + 1)2/3 27. What is the relation between a one-to-one function f and the function ( f 1 )1 ? 23. f (x) =x3Exercises 2831. Sketch the graph of the inverse of the function graphed below. 29. 28. y1 11 20. f (x) = . 1x x 22. f (x) = . x y1 1x1x13:36 365. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.1 ONE-TO-ONE FUNCTIONS; INVERSES30.341ax + b . cx + d 49. (a) Show that f is one-to-one iff ad bc = 0. (b) Suppose that ad bc = 0. Find f 1 . 50. Determine the constants a, b, c, d for which f = f 1 . 51. Set31.For Exercises 49 and 50, let f (x) =yy11 11xxxf (x) =1 + t 2 dt.2(a) Show that f has an inverse. (b) Find ( f 1 ) (0). 52. Set32. (a) Show that the composition of two one-to-one functions, f and g, is one-to-one. (b) Express ( f g)1 in terms of f 1 and g 1 . 33. (a) Let f (x) = 1 x 3 + x 2 + kx, k a constant. For what val3 ues of k is f one-to-one? (b) Let g(x) = x 3 + kx 2 + x, k a constant. For what values of k is g one-to-one? 34. (a) Suppose that f has an inverse, f (2) = 5, and f (2) = 3 . What is ( f 1 ) (5)? 4 (b) Suppose that f has an inverse, f (2) = 3, and f (2) = 2 . If g = 1/ f 1 , what is g (3)? 3 Exercises 3544. Verify that f has an inverse and nd ( f 35. f (x) = x 3 + 1; c = 9.139. f (x) = 2x + cos x; x +3 , 40. f (x) = x 1(a) Show that f has an inverse. (b) Find ( f 1 ) (0). 53. Let f be a twice differentiable one-to-one function and set g = f 1 . (a) Show that g (x) = ) (c).c = 1. 2c = 3.41. f (x) = tan x, 1 < x < 1 ; 2 2 42. f (x) = x 5 + 2x 3 + 2x;c= 3.c = 5.1 43. f (x) = 3x 3 , x > 0; c = 2. x 44. f (x) = x + cos x, 0 < x < 2 ;Exercises 5760. Find f 1 . 57. f (x) = 4 + 3 x 1, x 1. 3x 58. f (x) = , x = 5/2. 2x + 5 59. f (x) = 3 8 x + 2. 1x 60. f (x) = . 1+xc = 1.Exercises 4547. Find a formula for ( f 1 ) (x) given that f is one-to-one and its derivative satises the equation given. 46. f (x) = 1 + [ f (x)]2 . 45. f (x) = f (x). 47. f (x) = 48. Set1 [ f (x)]2 . Cf (x) =x 3 1, x 2,x 1;16 + t 4 dt.136. f (x) = 1 2x x 3 ; c = 4. 37. f (x) = x + 2 x, x > 0; c = 8. 38. f (x) = sin x, 1 < x < 1 ; 2 22xf (x) =Exercises 6164. Use a graphing utility to draw the graph of f. Show that f is one-to-one by consideration of f . Draw a gure that displays both the graph of f and the graph of f 1 . 62. f (x) = x 3/5 1. 61. f (x) = x 3 + 3x + 2. 63. f (x) = 4 sin 2x, /4 x /4. 64. f (x) = 2 cos 3x, 0 x /3.13:36 366. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006342 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 7.2 THE LOGARITHM FUNCTION, PART I You have seen that if n is an integer different from 1, then the function f (x) = x n is a derivative: xn =d dxx n+1 . n+1This formula breaks down if n = 1, for then n + 1 = 0 and the right side of the formula is meaningless. No function that we have studied so far has derivative x 1 = 1/x. However, we can easily construct one: set xL(x) = 11 dt. tFrom Theorem 5.3.5 we know that L is differentiable and 1 for all x > 0. L (x) = xyy=1 tThis function has a remarkable property that well get to in a moment. First some preliminary observations: Make sure you understand them. (1) L is dened for all x > 0. (2) Since L (x) =1xarea of shaded region = L(x) =Figure 7.2.1t1dt t x1 xfor all x > 0,L increases on (0, ). (3) L(x) is negative if 0 < x < 1,L(1) = 0,L(x) is positive for x > 1.(4) For x > 1, L(x) gives the area of the region shaded in Figure 7.2.1. Now to the remarkable property.THEOREM 7.2.1For all positive numbers a and b, L(ab) = L(a) + L(b).PROOFSet b > 0. For all x > 0, L(xb) and L(x) have the same derivative: 1 1 d d [L(xb)] = b = = [L(x)]. dx x dx xb chain rule Therefore the two functions differ by some constant C: L(xb) = L(x) + C. We can evaluate C by setting x = 1: L(b) = L(1 b) = L(1) + C = C. L(1) = 0 (Theorem 4.2.4)13:36 367. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.2 THE LOGARITHM FUNCTION, PART IIt follows that, for all x > 0, L(x b) = L(x) + L(b). We get the statement made in the theorem by setting x = a. From Theorem 7.2.1 and the fact that L(1) = 0, it readily follows that(1) for all positive numbers b, (7.2.2)L(1/b) = L(b)and (2) for all positive numbers a and b,L(a/b) = L(a) L(b).PROOF(1) 0 = L(1) = L(b 1/b) = L(b) + L(1/b)and therefore(2) L(a/b) = L(a 1/b) = L(a) + L(1/b) = L(a) L(b).L(1/b) = L(b);We now prove thatfor all positive numbers a and all rational numbers p/q, (7.2.3)L(a p/q ) =PROOFp L(a). qYou have seen that d[L(x)]/d x = 1/x. By the chain rule,d d p 1 d p/q 1 p 1 p [L(x p/q )] = p/q (x ) = p/q x ( p/q)1 = = L(x) . dx x dx x q q x dx q (3.7.1) p Since L(x p/q ) and L(x) have the same derivative, they differ by a constant: q L(x p/q ) =p L(x) + C. qSince both functions are zero at x = 1, C = 0. Therefore L(x p/q ) = x > 0. We get the theorem as stated by setting x = a. p L(x) for all qThe domain of L is (0, ). What is the range of L? (7.2.4)The range of L is (, ).Since L is continuous on (0, ), we know from the intermediate-value theorem that it skips no values. Thus, the range of L is an interval. To show that the interval is (, ), we need only show that the interval is unbounded above and unbounded below. We can do this by taking M as an arbitrary positive number and showing that L takes on values greater than M and values less than M.PROOF34313:36 368. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006344 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Let M be an arbitrary positive number. Since 21 dt 1 t is positive (explain), we know that some positive multiple of L(2) must be greater than M; namely, we know that there exists a positive integer n such that L(2) =n L(2) > M. Multiplying this equation by 1, we have n L(2) < M. Since n L(2) = L(2n )and n L(2) = L(2n ),L(2n ) > MandL(2n ) < M.(7.2.3)we haveyy=1 tThis proves that the range of L is unbounded in both directions. Since the range of L is an interval, it must be (, ), the set of all real numbers. The Number e A=1 12eFigure 7.2.23tSince the range of L is (, ) and L is an increasing function (and therefore one-toone), we know that L takes on as a value every real number and it does so only once. In particular, there is one and only one real number at which L takes on the value 1. This unique number is denoted throughout the world by the letter e . Figure 7.2.2 locates e on the number line: the area under the curve y = 1/t from t = 1 to t = e is exactly 1.The Logarithm Function Since eL(e) = 11 dt = 1, twe see from (7.2.3) that(7.2.5)for all rational numbers p/q p L(e p/q ) = . qThe function that we have labeled L is known as the natural logarithm function, or more simply as the logarithm function, and from now on L(x) will be written ln x. Here are the arithmetic properties of the logarithm function that we have already established. Both a and b represent arbitrary positive real numbers.(7.2.6)ln(1) = 0, ln(ab) = ln a + ln b, ln(a/b) = ln a ln b,ln(e) = 1, ln(1/b) = ln b, p ln a p/q = ln a. q After the celebrated Swiss mathematician Leonhard Euler (17071783), considered by many the greatest mathematician of the eighteenth century.13:36 369. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.2 THE LOGARITHM FUNCTION, PART I345The Graph of the Logarithm Function You know that the logarithm function xln x =1 dt t1has domain (0, ), range (, ), and derivative d 1 (ln x) = . dx x For small x the derivative is large (near 0, the curve is steep); for large x the derivative is small (far out, the curve attens out). At x = 1 the logarithm is 0 and its derivative 1/x is 1. [The graph crosses the x-axis at the point (1, 0), and the tangent line at that point is parallel to the line y = x.] The second derivative,yln x .12 12345x1 2 3Figure 7.2.3 y y=Example 1 We use upper and lower sums to estimate ln 2 =y = In x11 d2 (ln x) = 2 , 2 dx x is negative on (0, ). (The graph is concave down throughout.) We have sketched the graph in Figure 7.2.3. The y-axis is a vertical asymptote: as x 0+ ,y=x2dt t1 t(Figure 7.2.4) area = In 2from the partition P = {1 =10 11 12 13 14 15 16 17 18 19 20 , , , , , , , , , , 10 10 10 10 10 10 10 10 10 10 10= 2}.1Using a calculator, we nd that L f (P) =1 10=1 11U f (P) =1 10=1 1010 11++10 12+1 122tFigure 7.2.4+10 13+1 13+10 14+1 14+10 15+1 15+10 16+1 16+10 17+1 17+10 18+1 18+10 19+1 19+10 20> 0.6681 20and 10 10++1 1110 11++1 1210 12++1 1310 13++1 1410 14++1 1510 15++1 1610 16++1 1710 17++1 1810 18++1 1910 19< 0.719.We know then that 0.668 < L f (P) ln 2 U f (P) < 0.719. The average of these two estimates, 1 (0.668 2+ 0.719) = 0.6935,is not far off. Rounded off to four decimal places, our calculator gives ln 2 0.6931. = Table 7.2.1Table 7.2.1 gives the natural logarithms of the integers 1 through 10 rounded off to the nearest hundredth.Example 2 Use the properties of logarithms and Table 7.2.1 to estimate the following: (a) ln 0.2.(b) ln 0.25.(c) ln 2.4.(d) ln 90.n 1 2 3 4 5ln n 0.00 0.69 1.10 1.39 1.61n 6 7 8 9 10ln n 1.79 1.95 2.08 2.20 2.3013:36 370. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006346 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS SOLUTION(a) ln 0.2 = ln 1 = ln 5 1.61. = 5 (c) ln 2.4 = ln 12 = ln (3)(4) 5 5(b) ln 0.25 = ln 1 = ln 4 1.39. = 4 0.88. = ln 3 + ln 4 ln 5 =(d) ln 90 = ln [(9)(10)] = ln 9 + ln 10 4.50. =Example 3 Estimate e on the basis of Table 7.2.1. SOLUTION We know that ln e = 1. From the table you can see that3 ln 3 ln 10 1. = The expression on the left can be written ln 33 ln 10 = ln 27 ln 10 = ln 27 = ln 2.7. 10 1 and therefore e 2.7. This tells us that ln 2.7 = = Remark It can be shown that e is an irrational number, in fact a transcendental number. The decimal expansion of e to twelve decimal places reads e 2.718281828459. = Exercise 66 in Section 12.6 guides you through a proof of the irrationality of e. A proof that e is transcendental is beyond the reach of this text.EXERCISES 7.2 Exercises 110. Estimate the logarithm on the basis of Table 7.2.1; check your results on a calculator. 1. 3. 5. 7. 9. 11.ln 20. 2. ln 16. ln 1.6. 4. ln 34 . ln 0.1. 6. ln 2.5. ln 7.2. 8. ln 630. 10. ln 0.4. ln 2 Verify that the area under the curve y = 1/x from x = 1 to x = 2 equals the area from x = 2 to x = 4, the area from x = 3 to x = 6, the area from x = 4 to x = 8, and, more generally, the area from x = k to x = 2k. Draw some gures. 12. Verify that the area under the curve y = 1/x from x = 1 to x = m equals the area from x = 2 to x = 2m, the area from x = 3 to x = 3m, and, more generally, the area from x = k to x = km. 13. Estimate 1.5ln 1.5 = 1dt tby using the approximation 1 [L f (P) + U f (P)] with 214. Estimate1P = {1 1.5}.dt tby using the approximation 1 [L f (P) + U f (P)] with 2 P = {1 = 4 , 5 , 6 , 7 , 8 , 9 , 10 = 2.5}. 4 4 4 4 4 4 4 15. Taking ln 5 1.61, use differentials to estimate = (a) ln 5.2, (b) ln 4.8, (c) ln 5.5. 16. Taking ln 10 2.30, use differentials to estimate = (a) ln 10.3, (b) ln 9.6, (c) ln 11. Exercises 1722. Solve the equation for x. 17. ln x = 2. 18. ln x = 1. 19. (2 ln x) ln x = 0. 20. 1 ln x = ln (2x 1). 2 21. ln [(2x + 1)(x + 2)] = 2 ln (x + 2). 22. 2 ln (x + 2) 1 ln x 4 = 1. 2 23. Show that ln x = 1. lim x1 x 1 HINT: Note that8 9 10 11 12 , , , 8, 8 8 8 82.5ln 2.5 =as a derivative.ln x ln 1 ln x = and interpret the limit x 1 x 113:36 371. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.3 THE LOGARITHM FUNCTION, PART IIExercises 2425. Let n be a positive integer greater than 2. Draw relevant gures. 24. Find the greatest integer k for which 1 1 1 + + + < ln n. 2 3 k C25. Find the least integer k for which 1 1 1 + + + . 2 3 k Exercises 2628. A function g is given. (i) Use the intermediatevalue theorem to conclude that there is a number r in the indicated interval at which g(r ) = ln r . (ii) Use a graphing utility to draw a gure that displays both the graph of the logarithm and the graph ln n < 1 +C347of g on the indicated interval. Find r accurate to four decimal places. 26. g(x) = 2x 3; [1, 2]. 27. g(x) = sin x; [2, 3]. 1 28. g(x) = 2 ; [1, 2]. x Exercises 2930. Estimate the limit numerically by evaluating the function at the indicated values of x. Then use a graphing utility to zoom in on the graph and justify your estimate. ln x ; x = 1 0.5, 1 0.1, 1 0.01, 1 0.001, 29. lim x1 x 1 1 0.0001. 30. lim+ x ln x; x = 0.5, 0.1, 0.01, 0.001, 0.0001. x0 7.3 THE LOGARITHM FUNCTION, PART II Differentiating and Graphing We know that for x > 0 1 d (ln x) = . dx x As usual, we differentiate composite functions by the chain rule. Thus d 1 2x d [ln (1 + x 2 )] = (1 + x 2 ) = dx 1 + x2 dx 1 + x2for all real xand 1 d 3 d [ln (1 + 3x)] = (1 + 3x) = dx 1 + 3x d x 1 + 3xfor all x > 1 . 3Example 1 Determine the domain and nd f (x) for f (x) = ln (x 4 + x 2 ). SOLUTION For x to be in the domain of f, we must have x 4 + x 2 > 0, and thus wemust have x > 0. The domain of f is the set of positive numbers. Before differentiating f, we make use of the special properties of the logarithm: f (x) = ln (x 4 + x 2 ) = ln x + ln [(4 + x 2 )1/2 ] = ln x + 1 ln (4 + x 2 ). 2From this we see that 1 x 1 1 1 4 + 2x 2 . 2x = + = f (x) = + x 2 4 + x2 x 4 + x2 x(4 + x 2 )y y = In (x)Example 2 Sketch the graph ofy = In xf (x) = ln x .xSOLUTION The function, dened at all x = 0, is an even function: f (x) = f (x) forall x = 0. The graph has two branches: y = ln (x),x 0. y = In xFigure 7.3.113:36 372. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006348 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Example 3 (Important) Show that d 1 (ln x ) = dx x(7.3.1)for allx = 0.SOLUTION For x > 0,d d 1 (ln x ) = (ln x) = . dx dx x For x < 0, we have x = x > 0, and therefore d 1 d d (ln x ) = [ln (x)] = (x) = dx dx x d x1 x(1) =1 . xApplying the chain rule, we have d d 1 3x 2 3x 2 (ln 1 x 3 ) = (1 x 3 ) = = 3 dx 1 x3 dx 1 x3 x 1 and d dxlnx 1 x 2=d d 1 1 (ln x 1 ) (ln x 2 ) = . dx dx x 1 x 2Example 4 Set f (x) = x ln x. (a) Give the domain of f and indicate where f takes on the value 0. (b) On what intervals does f increase? decrease? (c) Find the extreme values of f. (d) Determine the concavity of the graph and give the points of inection. (e) Sketch the graph of f. SOLUTION Since the logarithm function is dened only for positive numbers, the domain of f is (0, ). The function takes on the value 0 at x = 1 : f (1) = 1 ln 1 = 0. Differentiating f, we have1 + ln x = 1 + ln x. x To nd the critical points of f, we set f (x) = 0: f (x) = x 1 (verify this) . e Since the logarithm is an increasing function, the sign chart for f looks like this: 1 + ln x = 0,ln x = 1,x=sign of f ': 0 + + + + + + + + + + + + + + + + + + + + + behavior of f :1decreases eincreasesxyf decreases on (0, 1/e] and increases on [1/e, ). Therefore 1f (1/e) =1 1 ee1f (x) = x ln xFigure 7.3.22x1 1 ln e e=1 1 1 (ln 1 ln e) = = = 0.368 e e 2.72is a local minimum for f and the absolute minimum. Since f (x) = 1/x > 0 for x > 0, the graph of f is concave up throughout. There are no points of inection. You can verify numerically that lim+ x ln x = 0. Finally note that as x , x0 x ln x . A sketch of the graph of f is shown in Figure 7.3.2. 13:36 373. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.3 THE LOGARITHM FUNCTION, PART II349x4 . x 1 (a) Specify the domain of f. (b) On what intervals does f increase? decrease? (c) Find the extreme values of f. (d) Determine the concavity of the graph and nd the points of inection. (e) Sketch the graph, specifying the asymptotes if any.Example 5 Set f (x) = lnSOLUTION Since the logarithm function is dened only for positive numbers, the domain of f is the open interval (1, ). Making use of the special properties of the logarithm, we writef (x) = ln x 4 ln (x 1) = 4 ln x ln (x 1). Differentiation gives f (x) =1 3x 4 4 = x x 1 x(x 1)4 1 (x 2)(3x 2) + = . 2 2 x (x 1) x 2 (x 1)2 Since f is dened only for x > 1, we disregard all x 1. Note that f (x) = 0 at x = 4/3 (critical point) and we have: f (x) = sign of f ': 0 + + + + + + + + + + + + + + + behavior of f : 1 decreasesx4 3increasesThus f decreases on (1, 4 ] and increases on [ 4 , ). The number 3 3 f 4 = 4 ln 4 3 ln 3 2.25 = 3is a local minimum and the absolute minimum. There are no other extreme values. Testing for concavity: observe that f (x) = 0 at x = 2. (We ignore x = 2/3 since 2/3 is not part of the domain of f.) The sign chart for f looks like this: y sign of f ': + + + + + + + + + + + + + + + + + + + + 0 concave up2concave downxThe graph is concave up on (1, 2) and concave down on (2, ). The point (2, f (2)) = (2, 4 ln 2) (2, 2.77) = is a point of inection, the only point of inection. Before sketching the graph, we note that the derivative 1 4 f (x) = x x 1 is very large negative for x close to 1 and very close to 0 for x large. This tells us that the graph is very steep for x close to 1 and very at for x large. See Figure 7.3.3. The line x = 1 is a vertical asymptote: as x 1+ , f (x) . Integration The integral counterpart of (7.3.1) takes the form(7.3.2)1 d x = ln x + C. xThe relation is valid on every interval that does not include 0.vertical asymptote x = 1concavity: 11point of inflectionminimum value 2.254 32Figure 7.3.3x13:36 374. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006350 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Integrals of the form u (x) dx u(x)with u(x) = 01 du ucan be writtenby setting u = u(x),du = u (x) d x.x2 d x. 1 4x 3Example 6 CalculateSOLUTION Up to a constant factor, x 2 is the derivative of 1 4x 3 . Therefore, we setu = 1 4x 3 , x2 1 d x = 12 1 4x 3du 1 1 = 12 ln u + C = 12 ln 1 4x 3 + C. u 2Example 7 Evaluate 16x 2 + 2 d x. x3 + x + 1u = x 3 + x + 1,SOLUTION Setdu = 12x 2 d x.du = (3x 2 + 1) d x.At x = 1, u = 3; at x = 2, u = 11. 2 16x 2 + 2 dx = 2 x3 + x + 111 3du = 2 ln u u= 2(ln 11 ln 3) = 2 ln11 3 11 3. Here is an example of a different sort.Example 8 Calculateln x d x. xSOLUTION Since 1/x is the derivative of ln x, we setu = ln x,du =1 d x. xThis gives ln x dx = xu du = 1 u 2 + C = 1 (ln x)2 + C. 2 2Integration of the Trigonometric Functions We repeat Table 5.6.1:sin x d x = cos x + Ccos x d x = sin x + Csec2 x d x = tan x + Ccsc2 x d x = cot x + Csec x tan x d x = sec x + Ccsc x cot x d x = csc x + C13:36 375. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.3 THE LOGARITHM FUNCTION, PART IINow that you are familiar with the logarithm function, we can add four more basic formulas to the table:tan x d x = ln cos x + C = ln sec x + C cot x d x = ln sin x + C(7.3.3)sec x d x = ln sec x + tan x + C csc x d x = ln csc x cot x + CThe derivation of these formulas runs as follows: sin x tan x d x = dx cos x du = ln u + C = u = ln cos x + C = ln(set u = cos x,du = sin x d x)1 +C cos x= ln sec x + C. cot x d x = = sec x d x = =cos x (set u = sin x, dx sin x du = ln u + C = ln sin x + C. u sec xsec x + tan x dx sec x + tan xsec x tan x + sec2 x dx sec x + tan x [set u = sec x + tan x,=du = cos x d x)du = (sec x tan x + sec2 x) d x]du = ln u + C = ln sec x + tan x + C. uThe derivation of the formula forcsc x d x is left to you.Example 9 Calculatecot x d x.SOLUTION Set u = x,du = d x.cot x d x =1 cot u du =1 1 ln sin u + C = ln sin x + C. Only experience prompts us to multiply numerator and denominator by sec x + tan x.35113:36 376. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006352 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Remark The u-substitution simplies many calculations, but you will nd with experience that you can carry out many of these integrations without it. /8Example 10 Evaluatesec 2x d x. 0 1 2SOLUTION As you can check,Therefore /8sec 2x d x =0=ln sec 2x + tan 2x 1 21 [ln ( 2/8 0 2 + 1) ln 1] =1 2 ln ( 2 + 1) 0.44 =sec2 3x d x. 1 + tan 3xExample 11 CalculateSOLUTION Set u = 1 + tan 3x,sec2 3x dx = 1 + tan 3xln sec 2x + tan 2x is an antiderivative for sec 2x.du = 3 sec2 3x d x.du = u1 31 3ln u + C =1 3ln 1 + tan 3x + C.Logarithmic Differentiation We can differentiate a lengthy product g(x) = g1 (x)g2 (x) gn (x) by rst writing ln g(x) = ln ( g1 (x) g2 (x) gn (x) ) = ln g1 (x) + ln g2 (x) + + ln gn (x) and then differentiating: g (x) g (x) g2 (x) g (x) = 1 + + + n . g(x) g1 (x) g2 (x) gn (x) Multiplication by g(x) then gives(7.3.4)g (x) = g(x)g1 (x) g2 (x) g (x) + + + n . g1 (x) g2 (x) gn (x)The process by which g (x) was obtained is called logarithmic differentiation. Logarithmic differentiation is valid at all points x where g(x) = 0. At points x where g(x) = 0, the process fails. A product of n factors, g(x) = g1 (x)g2 (x) gn (x) can, of course, also be differentiated by repeated applications of the product rule, Theorem 3.2.6. The great advantage of logarithmic differentiation is that it readily gives us an explicit formula for the derivative, a formula thats easy to remember and easy to work with.Example 12 Calculate the derivative of g(x) = x(x 1)(x 2)(x 3) by logarithmic differentiation.13:36 377. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.3 THE LOGARITHM FUNCTION, PART IISOLUTION We can write down g (x) directly from Formula (7.3.4):1 1 1 1 + + + ; x x 1 x 2 x 3g (x) = x(x 1)(x 2)(x 3)or we can go through the process by which we derived Formula (7.3.4): ln g(x) = ln x + ln x 1 + ln x 2 + ln x 3 , g (x) 1 1 1 1 = + + + g(x) x x 1 x 2 x 3 g (x) = x(x 1)(x 2)(x 3)1 1 1 1 + + + . x x 1 x 2 x 3The result is valid at all numbers x other than 0, 1, 2, 3. These are the numbers where g(x) = 0. Logarithmic differentiation can be applied to quotients.Example 13 Calculate the derivative of g(x) =(x 2 + 1)3 (2x 5)2 (x 2 + 5)2by logarithmic differentiation. SOLUTION Our rst step is to writeg(x) = (x 2 + 1)3 (2x 5)2 (x 2 + 5)2 . Then, according to (7.3.4), g (x) = =(x 2 + 1)3 (2x 5)2 3(x 2 + 1)2 (2x) 2(2x 5)(2) (2)(x 2 + 5)3 (2x) + + (x 2 + 5)2 (x 2 + 1)3 (2x 5)2 (x 2 + 5)2 (x 2 + 1)3 (2x 5)2 (x 2 + 5)24 4x 6x + 2 . + 1 2x 5 x + 5x2We dont have to rely on (7.3.4). We can simply write ln g(x) = ln (x 2 + 3)3 + ln (2x 5)2 ln (x 2 + 5)2 = 3 ln x 2 + 1 + 2 ln 2x 5 2 ln x 2 + 5 and go on from there: 3(2x) 2(2) 2(2x) g (x) = 2 + 2 g(x) x + 1 2x 5 x + 5 g (x) = g(x)6x 4 4x + 2 . + 1 2x 5 x + 5x2The result is valid at all numbers x other than 5 . At this number g(x) = 0. 235313:36 378. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006354 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS That logarithmic differentiation fails at the points where a product g(x) is 0 is not a serious deciency because at these points we can easily apply the product rule. For example, suppose that g(a) = 0. Then one of the factors of g(x) is 0 at x = a. We write that factor in front and call it g1 (x). We then have g(x) = g1 (x)[g2 (x) gn (x)]g1 (a) = 0.withBy the product rule, g (x) = g1 (x)d [g2 (x) gn (x)] + g1 (x) [g2 (x) gn (x)]. dxSince g1 (a) = 0, g (a) = g1 (a)[g2 (a) gn (a)]. We go back to the function of Example 12 and calculate the derivative of g(x) = x(x 1)(x 2)(x 3) at x = 3 by the method just described. Since it is the factor x 3 that is 0 at x = 3, we write g(x) = (x 3) [x(x 1)(x 2)]. By the product rule, g (x) = (x 3)d [x(x 1)(x 2)] + 1[x(x 1)(x 2)]. dxTherefore g (3) = 3(3 1)(3 2) = 6.EXERCISES 7.3 Exercises 114. Determine the domain and nd the derivative. f (x) = ln 4x. 2. 3 4. f (x) = ln (x + 1). 6. f (x) = ln 1 + x 2 . 4 8. f (x) = ln x 1 . f (x) = (2x + 1)2 ln (2x + 1). x +2 10. f (x) = ln 3 . 11. x 1 12. f (x) = ln 4 x 2 + 1. 13. 14. f (x) = cos (ln x). Exercises 1536. Calculate. dx . 16. 15. x +1 x 17. d x. 18. 3 x2 1. 3. 5. 7. 9.19.tan 3x d x.20.f (x) = ln(2x + 1). f (x) = ln [(x + 1)3 ]. f (x) = (ln x)3 . f (x) = ln (ln x). f (x) =1 . ln xf (x) = sin (ln x).x sec x 2 d x.22.23.x d x. (3 x 2 )224.27. 29. 31. 32. 33.dx . 3x x +1 d x. x2 sec1 x 2 221.sin x d x. 2 + cos x 1 d x. x ln x dx . x (ln x)2 sin x cos x d x. sin x + cos x 1 d x. x(1 + x) x d x. 1+x x25.d x.csc x d x. 2 + cot x ln (x + a) d x. x +a28. 30.HINT: Set u = 1 +(1 + sec x)2 d x.35.Exercises 3746. Evaluate. e dx 37. . x 1 e239. edx . x541. 4x2x d x. 1sec2 2x d x. 4 tan 2x x2 d x. 31 2x sec 2x tan 2x d x. 1 + sec 2x26.x.34.tan (ln x) d x. x36.(3 csc x)2 d x. e238. 1 1dx . x40. 0 1/342. 1/41 1 d x. x +1 x +2 tan x d x.13:36 379. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.3 THE LOGARITHM FUNCTION, PART II /2/2/6 /245.cos x d x. 1 + sin x44.cot x d x.43.46./4 e/4135567. Show that csc x d x = ln csc x cot x + C using the methods of this section. 68. (a) Show that for n = 2, (7.3.4) reduces to the product rule (3.2.6) except at those points where g(x) = 0. (b) Show that (7.3.4) applied to(1 + csc x)2 d x.ln x d x. x47. Pinpoint the error in the following: 5 11 d x = ln x 2 x 25 1g(x) ln 3.reduces to the quotient rule (3.2.10) except at those points where g(x) = 0.ln (1 + x) = 1 from the denition of derivax0 x48. Show that lim tive.Exercises 6974. (i) Find the domain of f, (ii) nd the intervals on which the function increases and the intervals on which it decreases, (iii) nd the extreme values, (iv) determine the concavity of the graph and nd the points of inection, and, nally, (v) sketch the graph, indicating asymptotes. 69. f (x) = ln (4 x). 70. f (x) = x ln x. 72. f (x) = ln (4 x)2 . 71. f (x) = x 2 ln x. x3 x . . 74. f (x) = ln 73. f (x) = ln 1 + x2 x 1 75. Show that the average slope of the logarithm curve from x = a to x = b isExercises 4952. Calculate the derivative by logarithmic differentiation and then evaluate g at the indicated value of x. 49. g(x) = (x 2 + 1)2 (x 1)5 x 3 ; x = 1. 50. g(x) = x(x + a)(x + b)(x + c); x = b. x 4 (x 1) 51. g(x) = ; x = 0. (x + 2)(x 2 + 1) 52. g(x) =(x 1)(x 2) (x 3)(x 4)2;x = 2.Exercises 5356. Sketch the region bounded by the curves and nd its area. 53. y = sec x, y = 2, x = 0, x = /6. 54. y = csc 1 x, y = x, x = 1 . 2 2 55. y = tan x, y = 1, x = 0. 56. y = sec x, y = cos x, x = 0, x = . 4 Exercises 5758. Find the area of the part of the rst quadrant that lies between the curves. 57. x + 4y 5 = 0 and x y = 1. 58. x + y 3 = 0 and x y = 2. 59. The region bounded by the graph of f (x) = 1/ 1 + x and the x-axis for 0 x 8 is revolved about the x-axis. Find the volume of the resulting solid. 60. The region bounded by the graph of f (x) = 3/(1 + x 2 ) and the x-axis for 0 x 3 is revolved about the y-axis. Find the volume of the resulting solid. 61. The region bounded by the graph of f (x) = sec x and the x-axis for /3 x /3 is revolved about the x-axis. Find the volume of the resulting solid. 62. The region bounded by the graph of f (x) = tan x and the x-axis for 0 x /4 is revolved about the x-axis. Find the volume of the resulting solid. 63. A particle moves along a coordinate line with acceleration a(t) = (t + 1)2 feet per second per second. Find the distance traveled by the particle during the time interval [0, 4] given that the initial velocity v(0) is 1 foot per second. 64. Exercise 63 taking v(0) as 2 feet per second. Exercises 6566. Find a formula for the nth derivative. dn dn (ln x). 66. [ln (1 x)]. 65. dxn dxng1 (x) g2 (x)1 b ln ba a.76. (a) Show that f (x) = ln 2x and g(x) = ln 3x have the same derivative. (b) Calculate the derivative of F(x) = ln kx, where k is any positive number. (c) Explain these results in terms of the properties of logarithms. CCCCExercises 7780. Use a graphing utility to graph f on the indicated interval. Estimate the x-intercepts of the graph of f and the values of x where f has either a local or absolute extreme value. Use four decimal place accuracy in your answers. (0, 10]. 77. f (x) = x ln x; 3 (0, 2]. 78. f (x) = x ln x; 79. f (x) = sin (ln x); (1, 100]. 2 (0, 2]. 80. f (x) = x ln (sin x); 81. A particle moves along a coordinate line with acceleration a(t) = 4 2(t + 1) + 3/(t + 1) feet per second per second from t = 0 to t = 3. (a) Find the velocity v of the particle at each time t during the motion given that v(0) = 2. (b) Use a graphing utility to graph v and a together. (c) Estimate the time t at which the particle has maximum velocity and the time at which it has minimum velocity. Use four decimal place accuracy. 82. Exercise 81 with a(t) = 2 cos 2(t + 1) + 2/(t + 1) feet per second per second from t = 0 to t = 7. 83. Set f (x) = 1/x and g(x) = x 2 + 4x 2. (a) Use a graphing utility to graph f and g together.13:36 380. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006356 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONSC(b) Use a CAS to nd the points where the two graphs intersect. (c) Use a CAS to nd the area of the region bounded by the two graphs. x 1 and g(x) = x 2 . 84. Exercise 83 taking f (x) = xCExercises 8586. Use a CAS to nd (i) f (x) and f (x); (ii) the points where f, f and f are zero; (iii) the intervals on which f, f and f are positive, negative; (iv) the extreme values of f. ln x 1 + 2 ln x . 85. f (x) = 2 . 86. f (x) = x 2 ln x 7.4 THE EXPONENTIAL FUNCTION Rational powers of e already have an established meaning: by e p/q we mean the qth root of e raised to the pth power. But what is meant by e 2 or e ? Earlier we proved that each rational power e p/q has logarithm p/q:(7.4.1)ln e p/q =p . qThe denition of e z for z irrational is patterned after this relation.DEFINITION 7.4.2If z is irrational, then by e z we mean the unique number that has logarithm z: ln e z = z. What is e 2 ? It is the unique number that has logarithm 2. What is e ? It is the unique number that has logarithm . Note that e x now has meaning for every real value of x: it is the unique number that has logarithm x.DEFINITION 7.4.3The function E(x) = e xfor all real xis called the exponential function.Some properties of the exponential function are listed below. (1) In the rst place,yexy=x(7.4.4)ln e x = xfor all real xIn xWriting L(x) = ln x and E(x) = e x , we have(0, 1) (1, 0)xL(E(x)) = xfor all real x.This says that the exponential function is the inverse of the logarithm function. Figure 7.4.1(2) The graph of the exponential function appears in Figure 7.4.1. It can be obtained from the graph of the logarithm by reection in the line y = x.13:36 381. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.4 THE EXPONENTIAL FUNCTION(3) Since the graph of the logarithm lies to the right of the y-axis, the graph of the exponential function lies above the x-axis: ex > 0(7.4.5)for all real x.(4) Since the graph of the logarithm crosses the x-axis at (1, 0), the graph of the exponential function crosses the y-axis at (0, 1): ln 1 = 0e0 = 1.gives(5) Since the y-axis is a vertical asymptote for the graph of the logarithm function, the x-axis is a horizontal asymptote for the graph of the exponential function: as x ,e x 0.(6) Since the exponential function is the inverse of the logarithm function, the logarithm function is the inverse of the exponential function; thus eln x = x(7.4.6)for all x > 0.You can verify this equation directly by observing that both sides have the same logarithm: ln (eln x ) = ln x since, for all real t, ln et = t. You know that for rational exponents e( p/q+r/s) = e p/q er/s . This property holds for all exponents, including irrational exponentsTHEOREM 7.4.7ea+b = ea ebfor all real a and b.PROOFln ea+b = a + b = ln ea + ln eb = ln (ea eb ). The one-to-oneness of the logarithm function gives ea+b = ea eb . We leave it to you to verify that(7.4.8)eb =1 ebandeab =ea . ebWe come now to one of the most important results in calculus. It is marvelously simple.35713:36 382. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006358 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS THEOREM 7.4.9The exponential function is its own derivative: for all real x, d x (e ) = e x . dxThe logarithm function is differentiable, and its derivative is never 0. It follows (Section 7.1) that its inverse, the exponential function, is also differentiable. Knowing this, we can show thatPROOFd x (e ) = e x dx by differentiating both sides of the identity ln e x = x. On the left-hand side, the chain rule gives 1 d x d (ln e x ) = x (e ). dx e dx On the right-hand side, the derivative is 1: d (x) = 1. dx Equating these derivatives, we have 1 d x (e ) = 1 ex d xand thusd x (e ) = e x . dxCompositions are differentiated by the chain rule.Example 1 d d kx (a) (e ) = ekx (kx) = ekx k = kekx . dx dx d d x 1 1 x (b) (e ) = e ( x) = e x = e x. dx dx 2 x 2 x d x 2 2 x 2 d 2 x 2 (e ) = e (x ) = e (2x) = 2x ex . (c) dx dxThe relation d kx d x and its corollary (e ) = e x (e ) = k ekx dx dx have important applications to engineering, physics, chemistry, biology, and economics. We take up some of these applications in Section 7.6.Example 2 Let f (x) = xex for all real x. (a) (b) (c) (d)On what intervals does f increase? decrease? Find the extreme values of f. Determine the concavity of the graph and nd the points of inection. Sketch the graph indicating the asymptotes if any.13:36 383. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.4 THE EXPONENTIAL FUNCTIONSOLUTIONf (x) = xex , f (x) = xex (1) + ex = (1 x) ex , f (x) = (1 x) ex (1) ex = (x 2) ex . Since ex > 0 for all x, we have f (x) = 0 only at x = 1. (critical point) The sign of f and the behavior of f are as follows: sign of f ': + + + + + + + + + + + + + + + + + + + + 0 behavior of f :1increasesxdecreasesThe function f increases on (, 1] and decreases on [1, ). The number 1 1 = = 0.368 e 2.72 is a local maximum and the absolute maximum. The function has no other extreme values. The sign of f and the concavity of the graph of f are as follows: f (1) =sign of f ': 0 + + + + + + + concavity:2 concave upconcave downxThe graph is concave down on (, 2) and concave up on (2, ). The point (2, f (2)) = (2, 2e2 ) 2, =2 (2.72)2 (2, 0.27) =is a point of inection, the only point of inection. In Section 11.6 we show that as x , f (x) = x/e x 0. Accepting this result for now, we conclude that the x-axis is a horizontal asymptote. The graph is given in Figure 7.4.2. y12horizontal asymptote y=0xf (x) = x e xFigure 7.4.2Example 3 Let f (x) = ex (a) (b) (c) (d) (e)2/2for all real x.Determine the symmetry of the graph and nd the asymptotes. On what intervals does f increase? decrease? Find the extreme values. Determine the concavity of the graph and nd the points of inection. Sketch the graph.35913:36 384. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006360 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS SOLUTION Since f (x) = e(x)= ex /2 = f (x), f is an even function. Thus 2 the graph is symmetric about the y-axis. As x , ex /2 0. Therefore, the x-axis is a horizontal asymptote. There are no vertical asymptotes. Differentiating f, we have f (x) = ex2/22/22(x) = xexf (x) = x(xex2/2) ex22/2/2= (x 2 1)ex2/2.Since ex /2 > 0 for all x, we have f (x) = 0 only at x = 0 (critical point). The sign of f and the behavior of f are as follows: 2sign of f ': + + + + + + + + + + + + + + + + + + + + 0 behavior of f :increases0xdecreasesThe function increases on (, 0] and decreases [0, ). The number f (0) = e0 = 1 is a local maximum and the absolute maximum. The function has no other extreme values. 2 Now consider f (x) = (x 2 1)ex /2 . The sign of f and the concavity of the graph of f are as follows:y(0, 1)sign of f ': + + + + + + + 0 0 + + + + + + +(1, e 1/2)(1, e 1/2)11 f (x) = e x 2/2Figure 7.4.3concavity:10concave point up of inflectionxconcave downx1 point concave up of inflectionThe graph of f is concave up on (, 1) and on (1, ); the graph is concave down on (1, 1). The points (1, e1/2 ) and (1, e1/2 ) are points of inection. The graph of f is the bell-shaped curve sketched in Figure 7.4.3. The integral counterpart of Theorem 7.4.9 takes the form e x d x = e x + C.(7.4.10)In practice eu(x) u (x) d xis reduced toeu duby setting u = u(x),Example 4 Find9 e3x d x.SOLUTION Set u = 3x,du = 3 d x.9 e3x d x = 3 du = u (x) d x.eu du = 3eu + C = 3 e3x + C.Bell-shaped curves play a big role in probability and statistics.13:36 385. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.4 THE EXPONENTIAL FUNCTIONIf you recognize at the very beginning that 3 e3x =d 3x (e ), dxthen you can dispense with the u-substitution and simply write 9 e3x d x = 33 e3x d x = 3e3x + C.e x d x. xExample 5 Find SOLUTION Set u =1 du = d x. 2 xx,e x dx = 2 xeu du = 2eu + C = 2ex+ C.If you recognize from the start that 1 2e x x= d e x , dxthen you can dispense with the u-substitution and integrate directly: e x dx = 2 xExample 6 Finde x x1 2d x = 2ex+ C.e3x d x. e3x + 1SOLUTION We can put this integral in the form1 du u by setting u = e3x + 1,du = 3e3x d x.Then e3x dx = e3x + 11 du = u1 31 3ln u + C =1 3ln (e3x + 1) + C.Example 7 Evaluate 2 ln 3xex2/2d x.0SOLUTION Set u = 1 x 2 , 2x = 0, u = 0;At 2 ln 3 0xex2/2du = x d x. at x = 2 ln 3, u = ln 3. Thus ln 3dx = 0eu du = eu ln 3 0= 1 e ln 3 = 1 1 3= 2. 336113:36 386. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006362 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 1e x (e x + 1)1/5 d x.Example 8 Evaluate 0SOLUTION Set u = e x + 1,x = 0, u = 2;At1at x = 1, e+1e x (e x + 1)1/5 d x =0du = e x d x. u = e + 1. Thus u 1/5 du =2e+1 5 6/5 u 6 2= 5 [(e + 1)6/5 26/5 ]. 6EXERCISES 7.4 ln Exercises 124. Differentiate. 2x=e . 2 = e x 1 . = e x ln x. = x 1 ex . = 1 (e x + ex ). 2 11. y = e x ln x. 1. 3. 5. 7. 9.y y y y y2. 4. 6. 8. 10.= 3e . = 2e4x . = x 2 ex . = e x+1 . = 1 (e x ex ). 2 2x+114. y = (e2x e2x )2 .215. y = (x 2x + 2)e . ex 1 17. y = x . e +1 19. y = e4 ln x . 21. f (x) = sin (e2x ). 23. f (x) = e2x cos x. xx216. y = x e xe . e2x 1 18. y = 2x . e +1 20. y = ln e3x . 22. f (x) = esin 2x . 24. f (x) = ln (cos e2x ). 2 xExercises 2542. Calculate. 25.e2x d x.26.e2x d x.27.ekx d x.28.eax+b d x.29.xe x d x.30.xex d x.31.e1/x d x. x232.e2 33.ln e x d x.34.eln x d x.35.4 d x. ex36.37.2excos xesin x d x.41.exxxd x.ex d x. +1 2d x.ex + 1 e2x d x. 2x + 3 2e39.2xeax d x. ax 2 + 1 e sin (e2x ) d x. e2x38. 40.ex [1 + cos (ex )] d x.42.Exercises 4352. Evaluate. 11e x d x.43. 044. 0ekx d x.1e6x d x.46.0xex d x. 20 147. 0ex + 1 d x. exln 249. 012. y = (3 2ex )3 .13. y = (e x + 1)2 . 2y y y y y45.151.14 ex d x. ex1ex d x. 4 ex48. 0xe d x. ex + 1 2x(e x + 2) d x.050. 0ln /452.e x sec e x d x.053. Let a be a positive constant. (a) Find a formula for the nth derivative of f (x) = eax . (b) Find a formula for the nth derivative of f (x) = eax . 54. A particle moves along a coordinate line, its position at time t given by the function x(t) = Aekt + Bekt .(A > 0, B > 0, k > 0)(a) Find the times t at which the particle is closest to the origin. (b) Show that the acceleration of the particle is proportional to the position coordinate. What is the constant of proportionality? 55. A rectangle has one side on the x-axis and the upper two ver2 tices on the graph of y = ex . Where should the vertices be placed so as to maximize the area of the rectangle? 56. A rectangle has two sides on the positive x- and y-axes and one vertex at a point P that moves along the curve y = e x in such a way that y increases at the rate of 1 unit per minute. 2 How is the area of the rectangle changing when y = 3? f (x) = ex . What is the symmetry of the graph? On what intervals does the function increase? decrease? What are the extreme values of the function? Determine the concavity of the graph and nd the points of inection. (e) The graph has a horizontal asymptote. What is it? (f) Sketch the graph. 58. Let be the region below the graph of y = e x from x = 0 to x = 1. (a) Find the volume of the solid generated by revolving about the x-axis. 57. Set (a) (b) (c) (d)213:36 387. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.4 THE EXPONENTIAL FUNCTIONx2 x3 xn + + + . 2! 3! n! Recall that n! = n(n 1)(n 2) 3 2 1. ex > 1 + x +59. Let be the region below the graph of y = ex from x = 0 to x = 1. (a) Find the volume of the solid generated by revolving about the y-axis. (b) Form the denite integral that gives the volume of the solid generated by revolving about the x-axis using the disk method. (At this point we cannot carry out the integration.) 2HINT:y0ex > x n CC CCB Cy = eaxy = ea xIx tangenttangent(a) Find the points of tangency, marked A and B. (b) Find the area of region I. (c) Find the area of region II.xet dt > 1 +(1 + t)dt0x2 , and so on. 2 73. Prove that, if n is a positive integer, thenCIIdt = 1 + x= 1+x +CAxet dt > 1 + 0xex = 1 +2f (x) = (1 x)e x . 65. f (x) = e(1/x ) . 67. f (x) = x 2 ln x. f (x) = x 2 ex . 2 x f (x) = (x x )e For each positive integer n nd the number xn for which xn x 0 e d x = n. 70. Find the critical points and the extreme values. Take k as a positive integer. x > 0. (a) f (x) = x k ln x, x real. (b) f (x) = x k ex , 71. Take a > 0 and refer to the gure.xex = 1 + 0Exercises 6468. Determine the following: (i) the domain; (ii) the intervals on which f increases, decreases; (iii) the extreme values; (iv) the concavity of the graph and the points of inection. Then sketch the graph, indicating all asymptotes. 64. 66. 68. 69.36372. Prove that for all x > 0 and all positive integers n(b) Set up the denite integral that gives the volume of the solid generated by revolving about the y-axis using the shell method. (You will see how to evaluate this integral in Section 8.2.)Exercises 6063. Sketch the region bounded by the curves and nd its area. 60. x = e2y , x = ey , x = 4. 61. y = e x , y = e2x , y = e4 . 62. y = e x , y = e, y = x, x = 0. 63. x = e y , y = 1, y = 2, x = 2.Cfor all x sufciently large.HINT: Exercise 72. 2 74. Set f (x) = ex and g(x) = x 2 . (a) Use a graphing utility to draw a gure that displays the graphs of f and g. (b) Estimate the x-coordinates a and b (a < b) of the two points where the curves intersect. Use four decimal place accuracy. (c) Estimate the area between the two curves from x = a to x = b. 75. Exercise 74 with f (x) = e x and g(x) = 4 x 2 . Exercises 7678. Use a graphing utility to draw a gure that displays the graphs of f and g. The gure should suggest that f and g are inverses. Show that this is true by verifying that f (g(x)) = x for each x in the domain of g. 76. f (x) = e2x , g(x) = ln x; x > 0. 2 77. f (x) = e x , g(x) = ln x; x 1. 78. f (x) = e x2 , g(x) = 2 + ln x; x > 0. 79. Set f (x) = sin e x . (a) Find the zeros of f. (b) Use a graphing utility to graph f. 80. Exercise 79 with f (x) = esin x 1. 81. Set f (x) = ex and g(x) = ln x. (a) Use a graphing utility to draw a gure that displays the graphs of f and g. (b) Estimate the x-coordinate of the point where the two graphs intersect. (c) Estimate the slopes at the point of intersection. (d) Are the curves perpendicular to each other? 82. (a) Use a graphing utility to draw a gure that displays the graphs of f (x) = 10ex and g(x) = 7 e x . (b) Find the x-coordinates a and b (a < b) of the two points where the curves intersect. (c) Use a CAS to nd the area between the two curves from x = a to x = b. 83. Use a CAS to calculate the integral. 1 ex 4 1 (a) d x. (b) ex d x. x 1e ex (c)etan x d x. cos2 x13:36 388. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006364 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS PROJECT 7.4 Some Rational Bounds for the Number e The purpose of this project is to lead you through a proof that, for each positive integer n(7.4.11)1+1 nne 1+1 nHINT:For all numbers t in 1, 1 + 1 1+n+1.1 3 2 1 4 3,(1 + 1 )4 e 1 + 41 5 4,x 111 dt, tx >0Step 1. Show that for each positive integer n, 1 n1 nn+11+1 100100 2.7048138 =and1+1 100101 2.7318620. =The proof has two steps.1+e 1+The bounds that we have derived for e are simple, elegant, and easy to remember, but they do not provide a very efcient method for calculating e. For example, rounded off to seven decimal places,1 dt = 1. t1 ln n+1nand so on.and on the characterization of e as the unique number for which e1 1. tby applying the exponential function to each entry in the inequality derived in Step 1.The proof outlined below is based directly on the denition of the logarithm function ln x =1 n1+,(1 + 1 )3 e 1 + 3,Step 2. Show thatIt will follow that (1 + 1 )2 e 1 + 21 n1 n1 . nApparently a lot of accuracy here, but it doesnt help us much in nding a decimal expansion for e. It tells us only that, rounded off to one decimal place, e 2.7. For a more accurate decimal = expansion of e, we need to resort to very large values of n. A much more efcient way of calculating e is given in Section 12.6. 7.5 ARBITRARY POWERS; OTHER BASES Arbitrary Powers: The Function f (x) = xr The elementary notion of exponent applies only to rational numbers. Expressions such as 21/3 , 74/5 , 1/2 105 , make sense, but so far we have attached no meaning to expressions such as 10 2 , 2 , 7 3 , e. The extension of our sense of exponent to allow for irrational exponents is conveniently done by making use of the logarithm function and the exponential function. The heart of the matter is to observe that for x > 0 and p/q rational, x p/q = e( p/q) ln x . (To verify this, take the logarithm of both sides.) We dene x z for irrational z by setting x z = e z ln x . We can now state that(7.5.1)if x > 0, then x r = er ln x for all real numbers r.13:36 389. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.5 ARBITRARY POWERS; OTHER BASESIn particular, 210=e 2 ln 10, 32 = e ln 2 ,7= e3 ln 7, e = ee ln .With this extended sense of exponent, the usual laws of exponents still hold:(7.5.2)x r +s = x r x s ,x r s =xr , xs(x r )s = x r sPROOFx r +s = e(r +s) ln x = er ln x es ln x = x r x s , x r s = e(r s) ln x = er ln x es ln x =er ln x xr = s, es ln x xr(x r )s = es ln x = er s ln x = x r s . The differentiation of arbitrary powers follows the pattern established for rational powers; namely, for each real number r and each x > 0 d r (x ) = r x r 1 . dx(7.5.3)PROOFd r d d r ln x r (x ) = (e ) = er ln x (r ln x) = x r = r x r 1 . dx dx dx x Another way to see this is to write f (x) = x r and use logarithmic differentiation: ln f (x) = r ln x f (x) r = f (x) x f (x) = Thusr xr r f (x) = = r x r 1 . x x d x 2 = 2 x 21 dxandd (x ) = x 1 . dxAs usual, we differentiate compositions by the chain rule. Thus d d (x 2 + 5) 3 = 3(x 2 + 5) 31 (x 2 + 5) = 2 3 x(x 2 + 5) 31 . dx dx d (x 2 + 1)3x . dx 2 SOLUTION One way to nd this derivative is to observe that (x 2 + 1)3x = e3x ln(x +1) and then differentiate: 2x d d 2 2 (x 2 + 1)3x = e3x ln (x +1) = e3x ln (x +1) 3x 2 + 3 ln (x 2 + 1) dx dx x +1Example 1 Find= (x 2 + 1)3x6x 2 + 3 ln (x 2 + 1) . x2 + 136513:36 390. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006366 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Another way to nd this derivative is to set f (x) = (x 2 + 1)3x , take the logarithm of both sides, and proceed from there ln f (x) = 3x ln (x 2 + 1) 6x 2 2x f (x) = 3x 2 + [ln (x 2 + 1)](3) = 2 + 3 ln (x 2 + 1) f (x) x +1 x +1 6x 2 + 3 ln (x 2 + 1) x2 + 1f (x) = f (x)= (x 2 + 1)3x6x 2 + 3 ln (x 2 + 1) . x2 + 1Each derivative formula gives rise to a companion integral formula. The integral version of (7.5.3) takes the formxr d x =(7.5.4)x r +1 + C, r +1for r = 1.Note the exclusion of r = 1. What is the integral if r = 1?Example 2 Findx3 d x. (2x 4 + 1)SOLUTION Set u = 2x 4 + 1,x3 1 dx = 4 + 1) (2x 8du = 8x 3 d x.u du =1 8u 1 1+C =(2x 4 + 1)1 + C. 8(1 )Base p: The Function f (x) = p x To form the function f (x) = x r , we take a positive variable x and raise it to a constant power r. To form the function f (x) = p x , we take a positive constant p and raise it to a variable power x. Since 1x = 1 for all x, the function is of interest only if p = 1. Functions of the form f (x) = p x are called exponential functions with base p. The high status enjoyed by Eulers number e comes from the fact that d x (e ) = e x . dx For other bases the derivative has an extra factor:(7.5.5)d x ( p ) = p x ln p. dxPROOFd x ln p d x ) = e x ln p ln p = p x ln p. (p ) = (e dx dx For example, d x (2 ) = 2x ln 2 dxandd (10x ) = 10x ln 10. dx13:36 391. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.5 ARBITRARY POWERS; OTHER BASESThe next differentiation requires the chain rule: d d 2 2 2 23x = 23x (ln 2) (3x 2 ) = 6x23x ln 2. dx dx The integral version of (7.5.5) reads px d x =(7.5.6)1 x p + C. ln pThe formula holds for all positive numbers p different from 1. For example, 2x d x =1 x 2 + C. ln 2x5x d x. 2Example 3 FindSOLUTION Set u = x 2 ,du = 2x d x.x5x d x = 2=1 25u du = 1 x 2 + C. 5 2 ln 51 21 5u + C ln 5232x1 d x.Example 4 Evaluate 1SOLUTION Set u = 2x 1,Atx = 1, 2u = 1;at x = 2,32x1 d x =1du = 2 d x.1 23u = 3. Thus3u du =11 1 3u 2 ln 33= 1Base p: The Function f (x) = log p x If p > 0, then ln p t = t ln pfor all t.If p is also different from 1, then ln p = 0, and we have ln pt = t. ln p This indicates that the function ln x ln pf (x) = satises the relation f ( pt ) = tfor all real t.In view of this, we call ln x ln p12 = 10.923. ln 336713:36 392. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006368 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS the logarithm of x to the base p and write log p x =(7.5.7)ln x . ln pThe relation holds for all x > 0 and assumes that p is a positive number different from 1. For example, log2 32 =ln 32 ln 25 5 ln 2 = = =5 ln 2 ln 2 ln 2and log100=1 101 ln 10 ln 101 1 ln 10 = = . = 2 ln 100 ln 10 2 ln 10 2We can obtain these same results more directly from the relation log p p t = t.(7.5.8)Accordingly log2 32 = log2 25 = 5andlog1001 10= log100 (1001/2 ) = 1 . 2Since log p x and ln x differ only by a constant factor, there is no reason to introduce new differentiation and integration formulas. For the record, we simply point out that d d (log p x) = dx dxln x ln p=1 . x ln pIf p is e, the factor ln p is 1 and we have 1 d (loge x) = . dx x The logarithm to the base e, ln = loge , is called the natural logarithm (or simply the logarithm) because it is the logarithm with the simplest derivative.Example 5 Calculate (a)d (log5 x ), dx(b)d [log2 (3x 2 + 1)], dx(c)1 d x. x ln 10SOLUTION(a)(b)d d (log5 x ) = dx dxln x 1 = . ln 5 x ln 5d d [log2 (3x 2 + 1)] = dx dxln (3x 2 + 1) ln 21 d 6x = (3x 2 + 1) = . 2 + 1) ln 2 d x 2 + 1) ln 2 (3x (3x by the chain rule (c)1 1 dx = x ln 10 ln 101 ln x dx = + C = log10 x + C. x ln 1013:36 393. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.5 ARBITRARY POWERS; OTHER BASES369EXERCISES 7.5 d sin x [x ]. dx d 49. [(sin x)cos x ]. dx d (2x ) [x ]. 51. dx 53. Show thatExercises 18. Evaluate. 1. log2 64.2. log21 . 641 log64 2 .3. 4. log10 0.01. 6. log5 0.2. 5. log5 1. 8. log2 43 . 7. log5 125. Exercises 912. Show that the identity holds. 9. log p x y = log p x + log p y. 1 11. log p x y = y log p x. 10. log p = log p x. x x 12. log p = log p x log p y. y23. g(x) =2+xas h 0,25. f (x) = tan(log5 x). 27. F(x) = cos(2x + 2x )..24. h(x) = 7 . log10 x 26. g(x) = . x2 28. h(x) = a x cos bx.259. e.ln (1 + h) ln 1 ln (1 + h) = 1. h h2x d x.1 1 x29.3 d x.30.2 d x.31.(x 3 + 3x ) d x.32.x10x d x.4x d x.60. 61.0 421dx . x ln 22dx log5 x 33. . 34. d x. x ln 5 x log2 x 3 d x. 35. x 36. Show that, if a, b, c are positive, thenp x/2 d x.62. 0 1 0 164. 0 1provided that a and b are both different from 1.2x101+x d x.63.loga c = loga b logb c Exercises 3740. Find f (e). 37. f (x) = log3 x. 39. f (x) = ln (ln x).xExercises 5965. Evaluate.Exercises 2935. Calculate. x1 xExercises 5458. Draw a gure that displays the graphs of both functions. 54. f (x) = e x and g(x) = 3x . 55. f (x) = e x and g(x) = 2x . 56. f (x) = ln x and g(x) = log3 x. 57. f (x) = 2x and g(x) = log2 x. 58. f (x) = ln x and g(x) = log2 x.sin x 2log3 x.1+HINT: Since the logarithm function has derivative 1 at x = 1,Exercises 1928. Differentiate. 2 20. g(x) = 43x . 19. f (x) = 32x . 22. F(x) = 52x48.as x ,Exercises 1316. Find the numbers x which satisfy the equation. 14. log5 x = 0.04. 13. 10x = e x . 16. logx 2 = log3 x. 15. logx 10 = log4 100. 17. Estimate ln a given that et1 < a < et2 . 18. Estimate eb given that ln x1 < b < ln x2 .21. f (x) = 25x 3ln x .d 2 [(cos x)(x +1) ]. dx d (x 2 ) 50. [x ]. dx d 52. [(tan x)sec x ]. dx47.65.5 p x+1 d x. x +1 (2x + x 2 ) d x.038. f (x) = x log3 x. 40. f (x) = log3 (log2 x).Exercises 4142. Calculate f (x) by rst taking the logarithm of both sides. 42. f (x) = p g(x) . 41. f (x) = p x . Exercises 4352. Calculate. d d 43. 44. [(x + 1)x ]. [(ln x)x ]. dx dx d 1 x d . 46. [(ln x)ln x ]. 45. dx dx xCCExercises 6668. Give the exact value. 67. 5(ln 17)/(ln 5) . 66. 71/ ln 7 . 1/ ln 2 . 68. (16) 69. (a) Use a graphing utility to draw a gure that displays the graphs of both f (x) = 2x and g(x) = x 2 1. (b) Use a CAS to nd the x-coordinates of the three points where the curves intersect. (c) Use a CAS to nd the area of the bounded region that lies between the two curves. 70. Exercise 69 for f (x) = 2x and g(x) = 1/x 2 .13:36 394. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006370 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 7.6 EXPONENTIAL GROWTH AND DECAY We begin by comparing exponential change to linear change. Let y = y(t) be a function of time t. If y is a linear function, a function of the form y(t) = kt + C,(k, C constants)then y changes by the same additive amount during all periods of the same duration: y(t +t) = k(t +t) + C = (kt + C) + k t = y(t) + k t.During every period of length t, y changes by the same amount k t. If y is a function of the form y(t) = Cekt ,(k, C constants)then y changes by the same multiplicative factor during all periods of the same duration: y(t +t) = Cek(t+t)= Cekt ekt= ek t y(t).t, y changes by the factor ek t .During every period of length Functions of the formf (t) = Cekt have the property that the derivative f (t) is proportional to f (t): f (t) = Ckekt = kCekt = k f (t). Moreover, they are the only such functions:THEOREM 7.6.1If f (t) = k f (t)for all t in some interval,then there is a constant C such that f (t) = CektPROOFfor all t in that interval.We assume that f (t) = k f (t)and write f (t) k f (t) = 0. Multiplying this equation by e ()kte, we havektf (t) k ekt f (t) = 0.Observe now that the left side of this equation is the derivative d kt [e f (t)]. dt Equation () can therefore be written d kt [e f (t)] = 0. dt It follows that ekt f (t) = Cfor some constant C.(Verify this.)13:36 395. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.6 EXPONENTIAL GROWTH AND DECAYMultiplication by ekt gives f (t) = Cekt . Remark In the study of exponential growth or decay, time is usually measured from time t = 0. The constant C is the value of f at time t = 0: f (0) = Ce0 = C. This is called the initial value of f. Thus the exponential f (t) = Cekt can be written f (t) = f (0)ekt .Example 1 Find f (t) given that f (t) = 2 f (t) for all t and f (0) = 5. SOLUTION The fact that f (t) = 2 f (t) tells us that f (t) = Ce2t where C is some constant. Since f (0) = C = 5, we have f (t) = 5e2t . Population Growth Under ideal conditions (unlimited space, adequate food supply, immunity to disease, and so on), the rate of increase of a population P at time t is proportional to the size of the population at time t. That is, P (t) = k P(t) where k > 0 is a constant, called the growth constant. Thus, by our theorem, the size of the population at any time t is given by P(t) = P(0)ekt , and the population is said to grow exponentially. This is a model of uninhibited growth. In reality, the rate of increase of a population does not continue to be proportional to the size of the population. After some time has passed, factors such as limitations on space or food supply, diseases, and so forth set in and affect the growth rate of the population.Example 2 In 1980 the world population was approximately 4.5 billion and in the year 2000 it was approximately 6 billion. Assume that the world population at each time t increases at a rate proportional to the world population at time t. Measure t in years after 1980. (a) Determine the growth constant and derive a formula for the population at time t. (b) Estimate how long it will take for the world population to reach 9 billion (double the 1980 population). (c) The world population for 2002 was reported to be about 6.2 billion. What population did the formula in part (a) predict for the year 2002? SOLUTION Let P(t) be the world population in billions t years after 1980. SinceP(0) = 4.5 = 9 , the basic equation P (t) = k P(t) gives 2 P(t) = 9 ekt . 2(a) Since P(20) = 6, we have 9 20k e 2= 6,20k = ln 12 = ln 4 , 9 3k=1 20ln 4 0.0143. 3 =The growth constant k is approximately 0.0143. The population t years after 1980 is P(t) 9 e0.0143t . =237113:36 396. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006372 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS (b) To nd the value of t for which P(t) = 9, we set 9 e0.0143t = 9: 2 ln 2 = 48.47. 0.0143 Based on the data given, the world population should reach 9 billion approximately 48 1 years after 1980around midyear 2028. (As of January 1, 2002, demographers 2 were predicting that the world population would peak at 9 billion in the year 2070 and then start to decline.) (c) The population predicted for the year 2002 is P(22) 9 e0.0143(22) = 9 e0.3146 6.164 = = e0.0143t = 2,0.0143t = ln 2,t=and22billion, not far off the reported gure of 6.2 billion. Bacterial Colonies Example 3 The size of a bacterial colony increases at a rate proportional to the size of the colony. Suppose that when the rst measurement is taken, time t = 0, the colony occupies an area of 0.25 square centimeters and 8 hours later the colony occupies 0.35 square centimeters. (a) Estimate the size of the colony t hours after the initial measurement is taken. What is the expected size of the colony at the end of 12 hours? (b) Find the doubling time, the time it takes for the colony to double in size. SOLUTION Let S(t) be the size of the colony at time t, size measured in square centimeters, t measured in hours. The basic equation S (t) = k S(t) givesS(t) = S(0) ekt . Since S(0) = 0.25, we have S(t) = (0.25) ekt . We can evaluate the growth constant k from the fact that S(8) = 0.35: 0.35 = (0.25) e8k ,e8k = 1.4,8k = ln (1.4)and therefore k=1 8ln (1.4) 0.042. =(a) The size of the colony at time t is S(t) = (0.25) e0.042tsquare centimeters.The expected size of the colony at the end of 12 hours is S(12) (0.25) e0.042(12) (0.25) e0.504 0.41 square centimeters. = = = (b) To nd the doubling time, we seek the value of t for which S(t) = 2(0.25) = 0.50. Thus we set (0.25) e0.042t = 0.50 and solve for t: e0.042t = 2,0.042t = ln 2,t=The doubling time is approximately 16 1 hours. 2ln 2 = 16.50. 0.04213:36 397. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.6 EXPONENTIAL GROWTH AND DECAYRemark There is a way of expressing S(t) that uses the exact value of k. We have seen that k = 1 ln (1.4). Therefore 8 t/8S(t) = (0.25)e(t/8) ln (1.4) = (0.25) eln (1.4)= (0.25)(1.4)t/8 .We leave it to you as an exercise to verify that the population function derived in t/20 Example 2 can be written P(t) = 9 4 . 2 3Radioactive Decay Although different radioactive substances decay at different rates, each radioactive substance decays at a rate proportional to the amount of the substance present: if A(t) is the amount present at time t, then A (t) = k A(t)for some constant k.Since A decreases, the constant k, called the decay constant, is a negative number. From general considerations already explained, we know that A(t) = A(0) ekt where A(0) is the amount present at time t = 0. The half-life of a radioactive substance is the time T it takes for half of the substance to decay. The decay constant k and the half-life T are related by the equation kT = ln 2.(7.6.2)PROOFThe relation A(T ) = 1 A(0) gives 2 1 2A(0) = A(0)ekT ,ekT = 1 , 2kT = ln 2. Example 4 Today we have A0 grams of a radioactive substance with a half-life of 8 years. (a) (b) (c) (d)How much of this substance will remain in 16 years? How much of the substance will remain in 4 years? What is the decay constant? How much of the substance will remain in t years?SOLUTION We know that exponentials change by the same factor during all time periods of the same length.(a) During the rst 8 years A0 will decrease to 1 A0 , and during the following 8 years 2 it will decrease to 1 ( 1 A0 ) = 1 A0 . Answer: 1 A0 grams. 2 2 4 4 (b) In 4 years A0 will decrease to some fractional multiple A0 and in the following 4 years to 2 A0 . Since 2 = 1 , = 2/2. Answer: ( 2/2)A0 grams. 2 (c) In general, kT = ln 2. Here T = 8 years. Answer: k = 1 ln 2. 8 (d) In general, A(t) = A(0)ekt . Here A(0) = A0 and k = 1 ln 2. Answer: A(t) = 8 1 A0 e 8 (ln 2)t . Example 5 Cobalt-60 is a radioactive substance used extensively in radiology. It has a half-life of 5.3 years. Today we have a sample of 100 grams.37313:36 398. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006374 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS (a) Determine the decay constant of cobalt-60. (b) How much of the 100 grams will remain in t years? (c) How long will it take for 90% of the sample to decay? SOLUTION(a) Equation (7.6.2) gives ln 2 ln 2 = = 0.131. T 5.3 (b) Given that A(0) = 100, the amount that will remain in t years is k=A(t) = 100e0.131t . (c) If 90% of the sample decays, then 10%, which is 10 grams, remains. We seek the time t at which 100e0.131t = 10. We solve this equation for t: ln (0.1) = 17.6. 0.131 It will take approximately 17.6 years for 90% of the sample to decay. e0.131t = 0.1,0.131t = ln (0.1),t=Compound Interest Consider money invested at annual interest rate r. If the accumulated interest is credited once a year, then the interest is said to be compounded annually; if twice a year, then semiannually; if four times a year, then quarterly. The idea can be pursued further. Interest can be credited every day, every hour, every second, every half-second, and so on. In the limiting case, interest is credited instantaneously. Economists call this continuous compounding. The economists formula for continuous compounding is a simple exponential: A(t) = A0 er t .(7.6.3)Here t is measured in years, A0 = A(0) = the initial investment, r = the annual interest rate expressed as a decimal, A(t) = the principal at time t. A DERIVATION OF THE COMPOUND INTEREST FORMULAFix t and take h as a small time incre-ment. Then A(t + h) A(t) = interest earned from time t to time t + h. Had the principal remained A(t) from time t to time t + h, the interest earned during this time period would have been rh A(t). Had the principal been A(t + h) throughout the time interval, the interest earned would have been rh A(t + h).13:36 399. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.6 EXPONENTIAL GROWTH AND DECAYThe actual interest earned must be somewhere in between: rh A(t) A(t + h) A(t) rh A(t + h). Dividing by h, we get A(t + h) A(t) rA(t + h). h If A varies continuously, then, as h tends to zero, rA(t + h) tends to rA(t) and (by the pinching theorem) the difference quotient in the middle must also tend to rA(t): rA(t) limh0A(t + h) A(t) = rA(t). hThis says that A (t) = rA(t). Thus, with continuous compounding, the principal increases at a rate proportional to the amount present and the growth constant is the interest rate r. Now, it follows that A(t) = Cer t . If A0 is the initial investment, we have C = A0 and therefore A(t) = A0 er t .Remark Frequency of compounding affects the return on principal, but (on modest sums) not very much. Listed below are the year-end values of $1000 invested at 6% under various forms of compounding: (a) (b) (c) (d)Annual compounding: 1000(1 + 0.06) = $1060. Quarterly compounding: 1000[1 + (.06/4)]4 = $1061.36. Monthly compounding: 1000[1 + (.06/12)]12 = $1061.67. Continuous compounding: 1000 e0.06 $1061.84. =Example 6 $1000 is deposited in a bank account that yields 5% compounded continuously. Estimate the value of the account 6 years later. How much interest will have been earned during that 6-year period? SOLUTION Here A0 = 1000 and r = 0.05. The value of the account t years after thedeposit is made is given by the function A(t) = 1000 e0.05t . At the end of the sixth year, the value of the account will be A(6) = 1000 e0.05(6) = 1000 e0.3 1349.86. = Interest earned: $349.86. Example 7 How long does it take to double your money at interest rate r compounded continuously? SOLUTION During t years an initial investment A0 grows in value toA(t) = A0 er t . You double your money once you have reached the time period t for which A0 er t = 2A0 . Solving this equation for t, we have er t = 2,r t = ln 2,t=ln 2 0.69 . = r r37513:36 400. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006376 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS For example, at 8% an investment doubles in value in Remark of 72:0.69 0.08= 8.625 years.A popular estimate for the doubling time at an interest rate % is the rule72 doubling time . = According to this rule, the doubling time at 8% is approximately how the rule originated: 0.69 69 72 = . = /100 72 8= 9 years. Here isFor rough calculations 72 is preferred to 69 because 72 has more divisors.This way of calculating doubling time is too inaccurate for our purposes. We will not use it.EXERCISES 7.6 CNOTE: Some of these exercises require a calculator or graphing utility. 1. Find the amount of interest earned by $500 compounded continuously for 10 years: (a) at 6%, (b) at 8%, (c) at 10%. 2. How long does it take for a sum of money to double if compounded continuously: (a) at 6%? (b) at 8%? (c) at 10%? 3. At what rate r of continuous compounding does a sum of money triple in 20 years? 4. At what rate r of continuous compounding does a sum of money double in 10 years? 5. Show that the population function derived in Example 2 can t/20 . be written P(t) = 9 4 2 3 6. A biologist observes that a certain bacterial colony triples every 4 hours and after 12 hours occupies 1 square centimeter. (a) How much area was occupied by the colony when rst observed? (b) What is the doubling time for the colony? 7. A population P of insects increases at a rate proportional to the current population. Suppose there are 10,000 insects at time t = 0 and 20,000 insects a week later. (a) Find an expression for the number P(t) of insects at each time t > 0. (b) How many insects will there be in 1 year? In 1 year? 2 8. Determine the time period in which y = Cekt changes by a factor of q. 9. The population of a certain country increases at the rate of 3.5% per year. By what factor does it increase every 10 years? What percentage increase per year will double the population every 15 years? 10. According to the Bureau of the Census, the population of the United States in 1990 was approximately 249 million11.12. 13.14.and in 2000, 281 million. Use this information to estimate the population in 1980. (The actual gure was about 227 million.) Use the data of Exercise 10 to predict the population for 2010. Compare the prediction for 2001 with the actual reported gure of 284.8 million. Use the data of Exercise 10 to estimate how long it will take for the U.S population to double. It is estimated that the arable land on earth can support a maximum of 30 billion people. Extrapolate from the data given in Example 2 to estimate the year when the food supply will become insufcient to support the world population. (Rest assured that there are strong reasons to believe that such extrapolations are invalid. Conditions change.) Water is pumped into a tank to dilute a saline solution. The volume of the solution, call it V, is kept constant by continuous outow. The amount of salt in the tank, call it s, depends on the amount of water that has been pumped in; call this x. Given that s ds = , dx V nd the amount of water that must be pumped into the tank to eliminate 50% of the salt. Take V as 10,000 gallons.15. A 200-liter tank initially full of water develops a leak at the bottom. Given that 20% of the water leaks out in the rst 5 minutes, nd the amount of water left in the tank t minutes after the leak develops if the water drains off at a rate proportional to the amount of water present. 16. What is the half-life of a radioactive substance if it takes 5 years for one-third of the substance to decay? 17. Two years ago there were 5 grams of a radioactive substance. Now there are 4 grams. How much will remain 3 years from now?13:36 401. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.6 EXPONENTIAL GROWTH AND DECAY18. A year ago there were 4 grams of a radioactive substance. Now there are 3 grams. How much was there 10 years ago? 19. Suppose the half-life of a radioactive substance is n years. What percentage of the substance present at the start of a year will decay during the ensuing year? 20. A radioactive substance weighed n grams at time t = 0. Today, 5 years later, the substance weighs m grams. How much will it weigh 5 years from now?dp = kp dhwhere k is a constant.Given that p is 15 pounds per square inch at sea level and 10 pounds per square inch at 10,000 feet, nd p at: (a) 5000 feet; (b) 15,000 feet. 26. The compound interest formula Q = Per t can be written P = Qer t .22. Cobalt-60 has a half-life of 5.3 years. What percentage of a given amount of cobalt will remain after 8 years? If you have 100 grams of cobalt now, how much was there 3 years ago?24. (The weakness of logarithmic growth) Having been soundly beaten in the race of Exercise 23, LIN nds an opponent she can beat, LOG, the logarithmic racer [position function x3 (t) = k ln (t + 1) + C]. Once again the racetrack is the xaxis calibrated in meters. Both racers start out at the origin, LOG at 1 million meters per second, LIN at only 1 meter per second. (LIN is tired from the previous race.) In this race LOG will shoot ahead and remain ahead for a long time, but eventually LIN will catch up to LOG, pass her, and leave her permanently behind. Show that this is true as follows: (a) Express the position of each racer as a function of time t, measuring t in seconds. (b) Show that LOGs lead over LIN starts to decline 106 1 seconds into the race. (c) Show that LOG is still ahead of LIN 107 1 seconds into the race but behind LIN 108 1 seconds into the race. (d) Show that, once LIN passes LOG, LOG can never catch up.37725. Atmospheric pressure p varies with altitude h according to the equation21. The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 500 years? How long will it take for the original amount to be reduced by 75%?23. (The power of exponential growth) Imagine two racers competing on the x-axis (which has been calibrated in meters), a linear racer LIN [position function of the form x1 (t) = kt + C] and an exponential racer EXP [position function of the form x2 (t) = ekt + C]. Suppose that both racers start out simultaneously from the origin, LIN at 1 million meters per second, EXP at only 1 meter per second. In the early stages of the race, fast-starting LIN will move far ahead of EXP, but in time EXP will catch up to LIN, pass her, and leave her hopelessly behind. Show that this is true as follows: (a) Express the position of each racer as a function of time, measuring t in seconds. (b) Show that LINs lead over EXP starts to decline about 13.8 seconds into the race. (c) Show that LIN is still ahead of EXP some 15 seconds into the race but far behind 3 seconds later. (d) Show that, once EXP passes LIN, LIN can never catch up.27. 28.29.30.31.32.33.In this formulation we have P as the investment required today to obtain Q in t years. In this sense P dollars is present value of Q dollars to be received t years from now. Find the present value of $20,000 to be received 4 years from now. Assume continuous compounding at 4%. Find the interest rate r needed for $6000 to be the present value of $10,000 8 years from now. You are 45 years old and are looking forward to an annual pension of $50,000 per year at age 65. What is the presentday purchasing power (present value) of your pension if money can be invested over this period at a continuously compounded interest rate of: (a) 4%? (b) 6%? (c) 8%? The cost of the tuition, fees, room, and board at XYZ College is currently $25,000 per year. What would you expect to pay 3 years from now if the costs at XYZ are rising at the continuously compounded rate of: (a) 5%? (b) 8%? (c) 12%? A boat moving in still water is subject to a retardation proportional to its velocity. Show that the velocity t seconds after the power is shut off is given by the formula v = ekt where is the velocity at the instant the power is shut off. A boat is drifting in still water at 4 miles per hour; 1 minute later, at 2 miles per hour. How far has the boat drifted in that 1 minute? (See Exercise 30.) During the process of inversion, the amount A of raw sugar present decreases at a rate proportional to A. During the rst 10 hours, 1000 pounds of raw sugar have been reduced to 800 pounds. How many pounds will remain after 10 more hours of inversion? The method of carbon dating makes use of the fact that all living organisms contain two isotopes of carbon, carbon12, denoted 12 C (a stable isotope), and carbon-14, denoted 14 C (a radioactive isotope). The ratio of the amount of 14 C to the amount of 12 C is essentially constant (approximately 1/10,000). When an organism dies, the amount of 12 C present remains unchanged, but the 14 C decays at a rate proportional to the amount present with a half-life of approximately 5700 years. This change in the amount of 14 C relative to the amount of 12 C makes it possible to estimate the time at which the organism lived. A fossil found in an archaeological dig was found to contain 25% of the original amount of 14 C. What is the approximate age of the fossil?13:36 402. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006378 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 37. f (t) = cos t f (t). 36. f (t) = sin t f (t). 38. Let g be a function everywhere continuous and not identically zero. Show that if f (t) = g(t) f (t) for all real t, then either f is identically zero or f does not take on the value zero.34. The Dead Sea Scrolls are approximately 2000 years old. How much of the original 14 C remains in them? Exercises 3537. Find all the functions f that satisfy the equation for all real t. 35. f (t) = t f (t). HINT: Write f (t) t f (t) = 0 and multiply 2 the equation by et /2 . 7.7 THE INVERSE TRIGONOMETRIC FUNCTIONS Arc Sine yy=1 5 232 211 223 225 2x y = 1y = sin xFigure 7.7.1Figure 7.7.1 shows the sine wave. Clearly the sine function is not one-to-one: it takes on every value from 1 to 1 an innite number of times. However, on the interval [ 1 , 1 ] it takes on every value from 1 to 1, but only once. (See the solid part of 2 2 the wave.) Thus the function x 1 , 1 2 2y = sin x,maps the interval [ 1 , 1 ] onto [1, 1] in a one-to-one manner and has an inverse 2 2 that maps [1, 1] back to [ 1 , 1 ], also in a one-to-one manner. The inverse is called 2 2 the arc sine function: y = arcsin x,x [1, 1]is the inverse of the function x 1 , 1 . 2 2y = sin x,These functions are graphed in Figure 7.7.2. Each graph is the reection of the other in the line y = x. yy 1 211 21 2 x111 1 2 [1 2y = sin x, x ,1 2]y = arcsin x, x [1, 1]Figure 7.7.2x13:36 403. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.7 THE INVERSE TRIGONOMETRIC FUNCTIONSSince these functions are inverses, Table 7.7.1xfor all x [1, 1],(7.7.1)sin (arcsin x) = x1 21 1 3 2 1 2 21 3 1 4 1 6and for all x 1 , 1 , 2 2(7.7.2)arcsin (sin x) = x.Table 7.7.1 gives some representative values of the sine function from x = 1 2 to x = 1 . Reversing the order of the columns, we have a table for the arc sine. (Table 2 7.7.2.) On the basis of Table 7.7.2 one could guess that for all x [1, 1] arcsin (x) = arcsin (x).Example 1 Calculate if dened: 1 (a) arcsin (sin 16 )001 6 1 4 1 3 1 21 2 1 2 2 1 3 2 1 Table 7.7.2(c) sin (arcsin 1 ) 3 (e) sin (arcsin 2).xarcsin1 1 3 2 1 2 21 21 61 4 01 2(d) arcsin (sin 9 ) 51 6 1 4 1 3 1 21 2 2 1 3 2SOLUTION1is in the interval [ 1 , 1 ], we know from (7.7.2) that 2 2 1 arcsin sin 16 =1 30(b) arcsin (sin 7 ) 31 161 21 2This is indeed the case. Being the inverse of an odd function (sin (x) = sin x for all x [ 1 , 1 ]), the arc sine is itself an odd function. (We leave it to you to verify 2 2 this.)(a) Sincesin x1 . 16(b) Since 7 is not in the interval [ 1 , 1 ], we cannot apply (7.7.2) directly. However, 3 2 2 7 = 1 + 2 and sin ( 1 + 2 ) = sin ( 1 ). Therefore 3 3 3 3 arcsin sin 7 = arcsin sin 1 = 1 . 3 3 3 from (7.7.2) (c) From (7.7.1), sin arcsin 1 = 1 . 3 3 (d) Since 9 is not within the interval [ 1 , 1 ], we cannot apply (7.7.2) directly. 5 2 2 However, 9 = 2 1 . Therefore 5 5 arcsin sin 9 = arcsin sin 1 5 5= 1 . 5 from (7.7.2) (e) The expression sin (arcsin 2) makes no sense since 2 is not in the domain of the arc sine. (There is no angle with sine 2.) The arc sine is dened only on [1, 1]. 37913:36 404. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006380 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Since the derivative of the sine function, d (sin x) = cos x, dx is nonzero on ( 1 , 1 ), the arc sine function is differentiable on the open interval 2 2 (1, 1) . We can nd the derivative as follows: reading from the accompanying gure y = arcsin x 1sin y = xx (0 x < 1)y ycos y 1x (1 < x 0)dy =1 dx dy 1 1 = = . dx cos y 1 x2Thus 1 d . (arcsin x) = dx 1 x2(7.7.3)Example 2 1 d 6x d (arcsin 3x 2 ) = (3x 2 ) = . 2 )2 d x dx 1 9x 4 1 (3x the chain rule NOTE: We continue with the convention that if the domain of a function f is not specied explicitly, then it is understood to be the maximal set of real numbers x for which f (x) is a real number. In this case, the domain is the set of real numbers x for which 1 3x 2 1. This is the interval [1/ 3, 1/ 3]. The integral counterpart of (7.7.3) reads(7.7.4)dx = arcsin x + C. 1 x2Example 3 Show that for a > 0(7.7.5)dx x = arcsin + C. a a2 x 2SOLUTION We change variables so that a 2 is replaced by 1 and we can use (7.7.4).To this end we set au = x, Section 7.1.a du = d x.13:36 405. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.7 THE INVERSE TRIGONOMETRIC FUNCTIONSThen dx = 2 x2 aa du a du = 2 a2u2 a a 1 u2 since a > 0 du x = arcsin u + C = arcsin + C. 2 a 1u= 3Example 4 Evaluate 0dx d x. 4 x2SOLUTION By (7.7.5),dx x = arcsin + C. 2 2 4x It follows that 03dxx d x = arcsin 2 2 4x 3 0 3 arcsin 0 = 0 = . = arcsin 2 3 3Arc Tangent Although not one-to-one on its full domain, the tangent function is one-to-one on the open interval ( 1 , 1 ) and on that interval the function takes on as a value every real 2 2 number. (See Figure 7.7.3.) Thus the function x 1 , 1 2 2y = tan x,maps the interval ( 1 , 1 ) onto (, ) in a one-to-one manner and has an inverse 2 2 that maps (, ) back to ( 1 , 1 ), also in a one-to-one manner. This inverse is 2 2 called the arc tangent: the arc tangent function y = arctan x,x (, )y x = x=2y2y= 1 21 22xO 1 2x 1 2 y= y = tan1x, x realy = tan x, x(1 , 1 2 2) Figure 7.7.3 238113:36 406. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006382 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS is the inverse of the function y = tan x,x 1 , 1 . 2 2These functions are graphed in Figure 7.7.3. Each graph is the reection of the other in the line y = x. While the tangent has vertical asymptotes, the inverse tangent has horizontal asymptotes. Both functions are odd functions. Since these functions are inverses, (7.7.6)for all real numbers xtan (arctan x) = xand for all x 1 , 1 , 2 2(7.7.7)arctan (tan x) = x.It is hard to make a mistake with (7.7.6) since that relation holds for all real numbers, but the application of (7.7.7) requires some care since it applies only to x in ( 1 , 1 ). 2 2 Thus, while arctan (tan 1 ) = 1 , arctan (tan 7 ) = 7 . To calculate arctan (tan 7 ), 4 4 5 5 5 we use the fact that the tangent function has period . Therefore arctan tan 7 = arctan tan 2 = 2 . 5 5 5 The nal equality holds since 2 ( 1 , 1 ). 5 2 2 Since the derivative of the tangent function, 1 d (tan x) = sec2 x = , dx cos2 x is never 0 on ( 1 , 1 ), the arc tangent function is everywhere differentiable. (Section 2 2 7.1) We can nd the derivative as we did for the arc sine: reading from the gure y = arctan x tan y = x21+xy y1x (x 0)sec2 y1+ x2x (x 0)dy =1 dx dy 1 1 = = cos2 y = . 2y dx sec 1 + x2We have found that 1 d . (arctan x) = dx 1 + x2(7.7.8)Example 5 d 1 d [arctan (ax 2 + bx + c)] = (ax 2 + bx + c) 2 + bx + c)2 d x dx 1 + (ax by the chain rule 2ax + b = . 1 + (ax 2 + bx + c)213:36 407. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.7 THE INVERSE TRIGONOMETRIC FUNCTIONSThe integral counterpart of (7.7.8) reads dx = arctan x + C. 1 + x2(7.7.9)Example 6 Show that, for a = 0, (7.7.10)a2dx 1 x = arctan + C. 2 +x a aSOLUTION We change variables so that a 2 is replaced by 1 and we can use (7.7.9).We set au = x,a du = d x.Then dx = + x2a du 1 = 2u2 +a adu 1 + u2 1 1 x = arctan u + C = arctan + C. a a a (7.7.9) a22Example 7 Evaluate 0a2dx . 4 + x2SOLUTION By (7.7.10),dx = 4 + x2dx 1 x = arctan + C, 2 +x 2 222and therefore 2 0dx x 1 arctan = 2 4+x 2 22= 01 1 arctan 1 arctan 0 = . 2 2 8Arc Cosine, Arc Cotangent, Arc Secant, Arc Cosecant These functions are not as important to us as the arc sine and arc tangent, but they do deserve some attention. While the cosine function is not one-to-one, it is one-to-one on [0, ] and maps that interval onto [1, 1]. (Figure 1.6.13) The arc cosine functionArc Cosiney = arccos x,x [1, 1]is the inverse of the function y = cos x,x [0, ].Arc Cotangent The cotangent function is one-to-one on (0, ) and maps that interval onto (, ). The arc cotangent functiony = arccot x,x (, )is the inverse of the function y = cot x,x (0, ).38313:36 408. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006384 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Arc Secant, Arc Cosecant These functions can be dened explicitly in terms of the arc cosine and the arc sine. For x 1, we setarcsec x = arccos (1/x),arccsc x = arcsin (1/x).In the Exercises you are asked to show that for all x 1 sec (arcsec x) = xcsc (arccsc x) = x.andRelations to 1 2 Where dened arcsin x + arccos x = 1 , 2 arctan x + arccot x = 1 , 2(7.7.11)arcsec x + arccsc x = 1 . 2 We derive the rst relation; the other two we leave to you. (Exercises 73, 74.) Our derivation is based on the identity cos = sin1 2 .(Section 1.6)Suppose that y = arccos x. Then cos y = xy [0, ]withand therefore sin1 2y =xwith1 2 y 1 , 1 . 2 2It follows that arcsin x = 1 y, 2arcsin x + y = 1 , 2arcsin x + arccos x = 1 2as asserted.Derivatives(7.7.12)1 d (arcsin x) = , dx 1 x2 1 d (arctan x) = , dx 1 + x2 d 1 , (arcsec x) = dx x x 2 11 d (arccos x) = dx 1 x2 1 d (arccot x) = dx 1 + x2 d 1 . (arccsc x) = dx x x 2 1VERIFICATION The derivatives of the arc sine and the arc tangent were calculated earlier. That the derivatives of the arc cosine and the arc cotangent are as stated follows immediately from (7.7.11). Once we show that1 d , (arcsec x) = dx x x 2 113:36 409. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.7 THE INVERSE TRIGONOMETRIC FUNCTIONS385the last formula will follow from (7.7.11). Hence we focus on the arc secant. Since arcsec x = arccos (1/x), the chain rule gives d 1 d 1 (arcsec x) = 2 dx dx x 1 (1/x) x2 x2 1 2 = . = x x2 1 x2 x2 1 This tells us that 1 , d x x2 1 (arcsec x) = 1 dx , x x 2 1for x > 1 for x < 1.The statement 1 d (arcsec x) = dx x x 2 1 is just a summary of this result.Remark on NotationThe expressions arcsin x, arctan x, arccos x, and so on are sometimes written sin1 x, tan1 x, cos1 x, and so on. EXERCISES 7.7 Exercises 19. Determine the exact value. 1. (a) arctan 0; 2. (a) arcsec 2; 3. (a) arccos ( 1 ); 2 (b) arcsin ( 3/2). (b) arctan ( 3). (b) arcsec ( 2).4. 5. 6. 7. 8.(b) sec (arccos [ 1 ]). 2 (b) arctan (sec 0). (b) arctan (tan [11/4]). (b) arcsec (sin [13/6]). (b) sec arctan [ 4 ] . 3 (a) sec (arcsec [2/ 3]); (a) cos (arcsec 2); (a) arcsin (sin [11/6]); (a) arccos (sec [7]/6); (a) cos (arcsin [ 3 ]); 59. (a) sin (2 arccos [ 1 ]); (b) cos (2 arcsin [ 4 ]). 2 5 10. (a) What are the domain and range of the arc cosine? (b) What are the domain and range of the arc cotangent? Exercises 1132. Differentiate. 11. y = arctan (x + 1). 12. y = arctan x. 14. f (x) = e x arcsin x. 13. f (x) = arcsec (2x 2 ). 15. f (x) = x arcsin 2x. 16. f (x) = earctan x . 18. v = arctan e x . 17. u = (arcsin x)2 . arctan x . 20. y = arcsec x 2 + 2. 19. y = x 22. f (x) = ln (arctan x). 21. f (x) = arctan 2x. 23. y = arctan (ln x). 24. g(x) = arcsec (cos x + 2). r 26. = arcsin . 25. = arcsin( 1 r 2 ). r +127. g(x) = x 2 arcsec11 . x1 . 1 + r228. = arctan 129. y = sin [arcsec (ln x)]. 30. f (x) = esec x . x 31. f (x) = c2 x 2 + c arcsin . Take c > 0. c x x 32. y = . Take c > 0. arcsin c c2 x 2 33. If 0 < x < 1, then arcsin x is the radian measure of the acute angle that has sine x. We can construct an angle of radian measure arcsin x by drawing a right triangle with a side of length x and hypotenuse of length 1. Use the accompanying gure to determine the following:1 x arcsin x(a) sin (arcsin x). (b) cos (arcsin x). (c) tan (arcsin x) (d) cot (arcsin x) (e) sec (arcsin x) (f) csc (arcsin x). 34. If 0 < x < 1, then arctan x is the radian measure of the acute angle with tangent x. We can construct an angle of radian13:36 410. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006386 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS measure arctan x by drawing a right triangle with legs of length x and 1. Use the accompanying gure to determine the following:55. 57. 59.x61.arctan x 1(a) tan (arctan x) (c) sin (arctan x) (e) sec (arctan x)(b) cot (arctan x) (d) cos (arctan x) (f) csc (arctan x). 135. Calculate63. 64. 65.taking a > 0.dx (x + HINT: Set au = x + b. 1 dx taking a > 0. 36. Calculate 2 + (x + b)2 a 1 1 x 37. Show that dx = arcsec + C, a a x x 2 a 2 taking a > 0. 38. (a) Verify, without reference to right triangles, that for all x 1 a2b)2sec (arcsec x) = xcsc (arcsec x) = x.and(b) What is the range of the arc secant? (The arc secant is the inverse of the secant restricted to this set.) (c) What is the range of the arc cosecant? (The arc cosecant is the inverse of the cosecant restricted to this set.)66.67. 68.69.x d x. 1 + x4 sec2 x d x. 9 + tan2 x arcsin x d x. 1 x2 dxdx . 4x x 2 cos x 58. d x. 3 + sin2 x arctan x 60. d x. 1 + x2 dx . 62. . 2 x[1 + (ln x)2 ] x 1 (ln x) Find the area below the curve y = 1/ 4 x 2 from x = 1 to x = 1. Find the area below the curve y = 3/(9 + x 2 ) from x = 3 to x = 3. Sketch the region bounded above by y = 8/(x 2 + 4) and bounded below by 4y = x 2 . What is the area of this region? The region below the curve y = 1/ 4 + x 2 from x = 0 to x = 2 is revolved about the x-axis. Find the volume of the resulting solid. The region in Exercise 66 is revolved about the y-axis. Find the volume of the resulting solid. The region below the curve y = 1/x 2 x 2 9 from x = 2 3 to x = 6 is revolved about the y-axis. Find the volume of the resulting solid. A billboard k feet wide is perpendicular to a straight road and is s feet from the road. From what point on the road would a motorist have the best view of the billboard; that is, at what point on the road (see the gure) is the angle subtended by the billboard a maximum?sExercises 3952. Evaluate. 1 dx . 39. 1 + x2 0 1/ 241. 0 543. 045. 0 347. 3/2x2.42.2 3 ln 2dx 4 (x + 3)2dx 4 x2.8dx . x x 2 165dx . 9 + (x 2)244. 5dx . 9 + 4x 246. 2 648.ln 3ex d x. 1 + e2xExercise 5362. Calculate. x 53. d x. 1 x4dx(x 3) x 2 6x + 84.kdx . 1 + x20dx . x 16x 2 949.011 1dxdx . 25 + x 23/251.140.56.50. ln 2 1/252. 0ex 1 e2x1 3 4x 2d x. d x..70. A person walking along a straight path at the rate of 6 feet per second is followed by a spotlight that is located 30 feet from the path. How fast is the spotlight turning at the instant the person is 50 feet past the point on the path that is closest to the spotlight? 71. (a) Show that F(x) =a2 x x a2 x 2 + arcsin ,a > 0 2 2 ais an antiderivative for f (x) = 54.sec2 x 9 tan2 xd x. a2 x 2. a(b) Use the result in part (a) to calculate and interpret your result as an area.aa2 x 2 d x13:36 411. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.7 THE INVERSE TRIGONOMETRIC FUNCTIONS C72. Set a+x f (x) = arctan 1 ax1 , x = 1/a. 1 + x2 (b) Show that there is no constant C such that f (x) = arctan x + C for all x = 1/a. (c) Find constants C1 and C2 such that f (x) = arctan x + C1 f (x) = arctan x + C2limx0C x1 = 0.05,3 01 1 x2d x = arcsin 3 arcsin 0 = arcsin 3is nonsensical since the sine function does not take on the value 3. Where did we go wrong here?1 x2 x2 = 0.15, x4 = 0.35,HINT: Use the identity cot = tan ( 1 ). 2 74. Show, without reference to right triangles, that75. The statement1dxby using the partition {0, 0.1, 0.2, 0.3, 0.4, 0.5} and the intermediate pointsfor all real x.for x 1.0for x < 1/a for x > 1/a.arcsec x + arccsc x = 1 2arcsin x xnumerically. Justify your answer by other means. 77. Estimate the integral 0.573. Show, without reference to right triangles, that arctan x + arccot x = 1 238776. Evaluate, x = 1/a.(a) Show that f (x) =C x3 = 0.25, x5 = 0.45.Note that the sine of your estimate is close to 0.5. Explain the reason for this. 1 78. Use a graphing utility to draw the graph of f (x) = 1 + x2 on [0, 10]. n (a) Calculate 0 f (x) d x for n = 1000, 2500, 5000, 10,000. (b) What number are these integrals approaching? (c) Determine the value of tlimt 01 d x. 1 + x2 PROJECT 7.7 Refraction Dip a straight stick in a pool of water and it appears to bend. Only in a vacuum does light travel at speed c (the famous c of E = mc2 ). Light does not travel as fast through a material medium. The index of refraction n of a medium relates the speed of light in that medium to c: n=c speed of light in the mediumLike the law of reection (see Example 5, Section 4.5), Snells law of refraction can be derived from Fermats principle of least time. Problem 1. A light beam passes from a medium with index of refraction n 1 through a plane sheet of material with top and bottom faces parallel and then out into some other medium with index of refraction n 2 . Show that Snells law implies the n 1 sin 1 = n 2 sin 2 regardless of the thickness of the sheet or its index of refraction.index of refraction ni 1n1 ir2n2index of refraction nrWhen light travels from one medium to another, it changes direction; we say that light is refracted. Experiment shows that the angle of refraction r is related to the angle of incidence i by Snells law: n i sin i = nr sin r .A star is not where it is supposed to be. The index of refraction of the atmosphere varies with height above the earths surface, n = n(y), and light that passes through the atmosphere follows some curved path, y = y(x). Think of the atmosphere as a succession of thin parallel slabs. When a light ray strikes a slab at height y, it is traveling at some angle to the vertical; when it emerges at height y + y, it is traveling at a slightly different angle, + .13:36 412. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006388 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS (b) Verify that the slope of the light path must vary in such a way thatProblem 2. (a) Use the result in Problem 1 to show that1 + (dy/d x)2 = (constant) [n(y)]2 . 22d d y/d x 1 dn . = cot = n dy dy 1 + (dy/d x)2(c) How must n vary with height y for light to travel along a circular arc? 7.8 THE HYPERBOLIC SINE AND COSINE Certain combinations of the exponentials e x and ex occur so frequently in mathematical applications that they are given special names. The hyperbolic sine (sinh) and hyperbolic cosine (cosh) are the functions dened as follows: sinh x = 1 (e x ex ), 2(7.8.1)cosh x = 1 (e x + ex ). 2The reason for these names will become apparent as we go on. Since d d 1 x (sinh x) = (e ex ) = 1 (e x + ex ) 2 dx dx 2 and d d 1 x (e + ex ) = 1 (e x ex ), (cosh x) = 2 dx dx 2 we have d (sinh x) = cosh x, dx(7.8.2)d (cosh x) = sinh x. dxIn short, each of these functions is the derivative of the other.The Graphs We begin with the hyperbolic sine. Since sinh (x) = 1 (ex e x ) = 1 (e x ex ) = sinh x, 2 2 the hyperbolic sine is an odd function. The graph is therefore symmetric about the origin. Since d (sinh x) = cosh x = 1 (e x + ex ) > 0 2 dx the hyperbolic sine increases everywhere. Sincefor all real x,d d2 (cosh x) = sinh x = 1 (e x ex ), (sinh x) = 2 dx2 dx you can see that d2 (sinh x) dx2is negative, 0, positive,for at forx 0.13:36 413. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.8 THE HYPERBOLIC SINE AND COSINEThe graph is therefore concave down on (, 0) and concave up on (0, ). The point (0, sinh 0) = (0, 0) is a point of inection, the only point of inection. The slope at the origin is cosh 0 = 1. A sketch of the graph appears in Figure 7.8.1. We turn now to the hyperbolic cosine. Since389ycosh (x) = 1 (ex + e x ) = 1 (e x + ex ) = cosh x, 2 2 xOthe hyperbolic cosine is an even function. The graph is therefore symmetric about the y-axis. Since d (cosh x) = sinh x, dx you can see that d (cosh x) dxis negative, 0, positive,for at fory = sinh xx 0.Figure 7.8.1 yThe function therefore decreases on (, 0] and increases on [0, ). The number cosh 0 = 1 (e0 + e0 ) = 1 (1 + 1) = 1 2 2 is a local and absolute minimum. There are no other extreme values. Since d2 d (cosh x) = (sinh x) = cosh x > 0 for all real x. dx2 dx the graph is everywhere concave up. (See Figure 7.8.2.) Figure 7.8.3 shows the graphs of three functions: y = sinh x = 1 (e x ex ), 2(0,1) x xy = 1 ex , 2y = cosh x = 1 (e x +e ). 2y = cosh xSince ex > 0, it follows that, for all real x,Figure 7.8.2sinh x < 1 e x < cosh x. 2for all real x.yAlthough markedly different for negative x, these functions are almost indistinguishable for large positive x. This follows from the fact that, as x , ex 0.y = cosh xThe Catenary(0, 1)A preliminary point: in what follows we use the fact that for all real numbers t cosh2 t = 1 + sinh2 t.y=1 2ex xThe verication of this identity is left to you as an exercise. Figure 7.8.4 depicts a exible cable of uniform density supported from two points of equal height. The cable sags under its own weight and so forms a curve. Such a curve is called a catenary. (After the Latin word for chain.) To obtain a mathematical characterization of the catenary, we introduce an x,ycoordinate system so that the lowest point of the chain falls on the positive y-axis (Figure 7.8.5). An engineering analysis of the forces that act on the cable shows that the shape of the catenary, call it y = f (x), is such that ()1 d2 y = 1+ dx2 ady dxy = sinh xFigure 7.8.3 y2where a is a positive constant that depends on the length of the cable and on its mass density. As we show below, curves of the form x () y = a cosh + C (C constant) axFigure 7.8.413:36 414. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006390 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS ymeet this condition exactly: dy x 1 x = a sinh = sinh dx a a a d2 y 1 x x 1 = cosh = cosh dx2 a a a a(0, h)d2 y x 1 1 1 dy x = cosh = 1 + sinh2 = 1+ 2 dx a a a a a dx follows from cosh2 t = 1 + sinh2 t xFigure 7.8.52.The cable of Figure 7.8.5 is of the formyx + (h a). a [This assertion is based on the fact that only curves of the form () satisfy () and the conditions imposed by Figure 7.8.5. This can be proven.] The Gateway Arch in St. Louis, Missouri, is in the shape of an inverted catenary (see Figure 7.8.6). This arch is 630 feet high at its center, and it measures 630 feet across the base. The value of the constant a for this arch is approximately 127.7, and its equation takes the form y = a cosh500y = 127.7 cosh (x/127.7) + 757.7. x 300 300Identities The hyperbolic sine and cosine functions satisfy identities similar to those satised by the circular sine and cosine.Figure 7.8.6cosh2 t sinh2 t = 1, sinh (t + s) = sinh t cosh s + cosh t sinh s, (7.8.3)cosh (t + s) = cosh t cosh s + sinh t sinh s, sinh 2t = 2 sinh t cosh t, cosh 2t = cosh2 t + sinh2 t.The verication of these identities is left to you as a collection of exercises.Relation to the Hyperbola x 2 y 2 = 1 The hyperbolic sine and cosine are related to the hyperbola x 2 y 2 = 1 much as the circular sine and cosine are related to the circle x 2 + y 2 = 1:y x2+y2=1 (cos t, sin t)1. For each real t, cos2 t + sin2 t = 1,O(1, 0)xand thus the point (cos t, sin t) lies on the circle x 2 + y 2 = 1. For each real t, cosh2 t sinh2 t = 1,area of circular sector = 1 t 2Figure 7.8.7and thus the point (cosh t, sinh t) lies on the hyperbola x 2 y 2 = 1. 2. For each t in [0, 2 ] (see Figure 7.8.7), the number 1 t gives the area of the circular 2 sector generated by the circular arc that begins at (1, 0) and ends at (cos t, sin t). As13:36 415. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 20067.8 THE HYPERBOLIC SINE AND COSINEwe prove below, for each t > 0 (see Figure 7.8.8), the number 1 t gives the area of 2 the hyperbolic sector generated by the hyperbolic arc that begins at (1, 0) and ends at (cosh t, sinh t).391y x2 y2 = 1 P(cosh t, sinh t)PROOFLet A(t) be the area of the hyperbolic sector. Then, A(t) =cosh tcosh t sinh t 1 2x 2 1 d x.O(1, 0) Qx1The rst term, 1 cosh t sinh t, gives the area of the triangle OPQ, and the integral 2 cosh tx2 1 dxarea of hyperbolic sector =1gives the area of the unshaded portion of the triangle. We wish to show that A(t) =1 t 21 t 2Figure 7.8.8for all t 0.We will do so by showing that A (t) =for all t > 01 2A(0) = 0.andDifferentiating A(t), we have A (t) =1 2cosh td d d (sinh t) + sinh t (cosh t) dt dt dtcosh tx2 1 dx ,1and therefore A (t) = 1 (cosh2 t + sinh2 t) 2(1)d dtcosh tx2 1 dx ,1Now we differentiate the integral: d dtcosh tx2 1 dx=1cosh2 t 1 (5.8.7) d (cosh t) = sinh t sinh t = sinh2 t. dtSubstituting this last expression into (1), we have A (t) = 1 (cosh2 t + sinh2 t) sinh2 t = 1 (cosh2 t sinh2 t) = 1 . 2 2 2 It is not hard to see that A(0) = 0: A(0) =1 2cosh 0cosh 0 sinh 0 1x 2 1d x = 1 (1)(0) 21x 2 1 d x = 0. 1EXERCISES 7.8 Exercises 118. Differentiate. 2. y 1. y = sinh x 2 . 4. y 3. y = cosh ax. sinh x 5. y = . 6. y cosh x 1 7. y = a sinh bx b cosh ax. 8. y = e x (cosh x + sinh x). 9. y = ln sinh ax . 10. y 2x 12. y 11. y = sinh (e ).= cosh (x + a). = (sinh ax)(cosh ax). sinh x = . x= ln 1 cosh ax . = cosh (ln x 3 ).14. y = arctan (sinh x). 13. y = ex cosh 2x. 15. y = ln (cosh x). 16. y = ln (sinh x). 18. y = x cosh x . 17. y = (sinh x)x . Exercises 1925. Verify the identity. 19. cosh2 t sinh2 t = 1. 20. sinh (t + s) = sinh t cosh s + cosh t sinh s. 21. cosh (t + s) = cosh t cosh s + sinh t sinh s. 22. sinh 2t = 2 sinh t cosh t. 23. cosh 2t = cosh2 t + sinh2 t = 2 cosh2 t 1 = 2 sinh2 t + 1.13:36 416. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006392 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 24. cosh (t) = cosh t; the hyperbolic cosine function is even. 25. sinh (t) = sinh t; the hyperbolic sine function is odd. Exercises 2628. Find the absolute extreme values. 26. y = 5 cosh x + 4 sinh x. 27. y = 5 cosh x + 4 sinh x. 28. y = 4 cosh x + 5 sinh x. 29. Show that for each positive integer n45. f (x) = cosh x, x [1, 1]. 46. f (x) = sinh 2x, x [0, 4]. 47. Find the area below the curve y = sinh x from x = 0 to x = ln 10. 48. Find the area below the curve y = cosh 2x from x = ln 5 to x = ln 5. 49. Find the volume of the solid generated by revolving about the x-axis the region between y = cosh x and y = sinh x from x = 0 to x = 1. 50. The region below the curve y = sinh x from x = 0 to x = ln 5 is revolved about the x-axis. Find the volume of the resulting solid. 51. The region below the curve y = cosh 2x from x = ln 5 to x = ln 5 is revolved about the x-axis. Find the volume of the resulting solid. 52. (a) Evaluate(cosh x + sinh x)n = cosh nx + sinh nx. 30. Verify that y = A cosh cx + B sinh cx satises the equation y c2 y = 0. 31. Determine A, B, and c so that y = A cosh cx + B sinh cx satises the conditions y 9y = 0, y(0) = 2, y (0) = 1. Take c > 0. 32. Determine A, B, and c so that y = A cosh cx + B sinh cx satises the conditions 4y y = 0, y(0) = 1, y (0) = 2. Take c > 0. Exercises 3344. Calculate. 33.cosh ax d x.35.sinh ax cosh2 ax d x.37. 39.sinh ax d x. cosh ax sinh ax d x. cosh2 axcosh x eax(b) Evaluatesinh2 ax cosh ax d x.36.sinh x . exlim34.limxsinh ax d x.xC38.cosh ax d x. sinh ax40.sinh2 x d x.cosh2 x d x. 42. sinh 2x ecosh 2x d x. sinh x sinh x 43. d x. d x. 44. 1 + cosh x x Exercises 45 and 46. Find the average value of the function on the interval indicated. 41.Cif 0 < a < 1 and if a > 1. 53. Use a graphing utility to sketch in one gure the graphs of f (x) = 2 sinh x and g(x) = cosh x. (a) Use a CAS to nd the point in the rst quadrant where the two graphs intersect. (b) Use a CAS to nd the area of the region in the rst quadrant bounded by the graphs of f and g and the y-axis. 54. Use a graphing utility to sketch in one gure the graphs of f (x) = cosh x 1 and g(x) = 1/ cosh x. (a) Use a CAS to nd the points where the two graphs intersect. (b) Use a CAS to nd the area of the region bounded by the graphs of f and g. *7.9 THE OTHER HYPERBOLIC FUNCTIONS The hyperbolic tangent is dened by setting e x ex sinh x . = x tanh x = cosh x e + ex There is also a hyperbolic cotangent, a hyperbolic secant, and a hyperbolic cosecant: cosh x 1 1 coth x = , sech x = , csch x = . sinh x cosh x sinh x The derivatives are as follows:(7.9.1)d (tanh x) = sech2 x, dx d (sech x) = sech x tanh x, dxd (coth x) = csch2 x, dx d (csch x) = csch x coth x. dx13:36 417. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006*7.9 THE OTHER HYPERBOLIC FUNCTIONSThese formulas are easy to verify. For instance, d d (tanh x) = dx dxsinh x cosh x= =cosh xd d (sinh x) sinh x (cosh x) dx dx cosh2 x1 cosh2 x sinh2 x = = sech2 x. 2 cosh x cosh2 xWe leave it to you to verify the other formulas. Lets examine the hyperbolic tangent a little further. Since tanh (x) = sinh x sinh (x) = = tanh x, cosh (x) cosh xthe hyperbolic tangent is an odd function and thus the graph is symmetric about the origin. Since d (tanh x) = sech2 x > 0 dxfor all real x,the function is everywhere increasing. From the relation tanh x =e x ex e x e2x 1 e2x + 1 2 2 e x ex = x x = 2x = = 1 2x , x + ex x 2x + 1 e e +e e e +1 e e +1you can see that tanh x always remains between 1 and 1. Moreover, as x ,tanh x 1andas x ,tanh x 1.The lines y = 1 and y = 1 are horizontal asymptotes. To check on the concavity of the graph, we take the second derivative: d2 d d (tanh x) = (sech2 x) = 2 sech x (sech x) 2 dx dx dx = 2 sech x (sech x tanh x) = 2 sech2 x tanh x. Since e x ex tanh x = x e + exis negative, for 0, at positive, forx 0,you can see that d2 (tanh x) dx2is positive, 0, negative,for at forx 0.The graph is therefore concave up on (, 0) and concave down on (0, ). The point (0, tanh 0) = (0, 0) is a point of inection. At the origin the slope is sech2 0 = The graph is shown in Figure 7.9.1.1 = 1. cosh2 039313:36 418. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006394 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS y horizontal asymptotey=1 x y = 1horizontal asymptote y = tanh xFigure 7.9.1The Hyperbolic Inverses Of the six hyperbolic functions, only the hyperbolic cosine and its reciprocal, hyperbolic secant, fail to be one-to-one (refer to the graphs of y = sinh x, y = cosh x, and y = tanh x). Thus, the hyperbolic sine, hyperbolic tangent, hyperbolic cosecant, and hyperbolic cotangent functions all have inverses. If we restrict the domains of the hyperbolic cosine and hyperbolic secant functions to x 0, then these functions will also have inverses. The hyperbolic inverses that are important to us are the inverse hyperbolic sine, the inverse hyperbolic cosine, and the inverse hyperbolic tangent. These functions, y = sinh1 x,y = cosh1 x,y = tanh1 x,are the inverses of y = sinh x,y = cosh x(x 0),y = tanh xrespectively.THEOREM 7.9.2(i) sinh1 x = ln (x +x 2 + 1), x real 1 2 1), x 1 (ii) cosh x = ln (x + x 1+x (iii) tanh1 x = 1 ln , 1 < x < 1. 2 1xPROOFTo prove (i), we set y = sinh1 x and note that sinh y = x.This gives in sequence: 1 y (e 2 ey ) = x,e y ey = 2xe y 2x ey = 0,e2y 2x e y 1 = 0.This last equation is a quadratic equation in e y . From the general quadratic formula, we nd that e y = 1 (2x 4x 2 + 4) = x x 2 + 1. 2 Since e y > 0, the minus sign on the right is impossible. Consequently, we have ey = x +x 2 + 1, The expressions sinh1 x, cosh1 x, tan1 x can be written arcsinh x, arccosh x, arctanh x. However the 1 notation is more common.13:36 419. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006*7.9 THE OTHER HYPERBOLIC FUNCTIONSand, taking the natural log of both sides, y = ln (x +x 2 + 1).To prove (ii), we set y = cosh1 x,x 1and note that cosh y = xandy 0.This gives in sequence: 1 y (e 2+ ey ) = x,e y + ey = 2x,e2y 2xe y + 1 = 0.Again we have a quadratic in e y . Here the general quadratic formula gives e y = 1 (2x 4x 2 4) = x x 2 1. 2 Since y is nonnegative, ey = x x2 1cannot be less than 1. This renders the negative sign impossible (check this out) and leaves ey = x +x2 1as the only possibility. Taking the natural log of both sides, we get y = ln (x +x 2 1).The proof of (iii) is left as an exercise. EXERCISES 7.9 Exercises 110. Differentiate. = tanh2 3x. = tanh (ln x). = sech (3x 2 + 1). = ln (sech x). cosh x 10. y = . 1 + sech x= tanh2 x. = ln (tanh x). = sinh (arctan e2x ). = coth ( x 2 + 1). sech x 9. y = . 1 + cosh x 1. 3. 5. 7.2. 4. 6. 8.y y y yy y y yExercises 1113. Verify the formula. d (coth x) = csch2 x. 11. dx d (sech x) = sech x tanh x. 12. dx d (csch x) = csch x coth x. 13. dx 14. Show that tanh t + tanh s . tanh (t + s) = 1 + tanh t tanh s 15. Given that tanh x0 = nd (a) sech x0 . HINT: 1 tanh2 x = sech2 x. Then nd (b) cosh x0 , (c) sinh x0 , (d) coth x0 , (e) csch x0 . 5 16. Given that tanh t0 = 12 , evaluate the remaining hyperbolic functions at t0 .17. Show that, if x 2 1, then x 18. Show that tanh1 x =1+x 1 ln 2 1xx 2 1 1.,1 < x < 1.19. Show that(7.9.3)1 d , (sinh1 x) = 2+1 dx xx real.1 d , (cosh1 x) = dx x2 1x > 1.20. Show that(7.9.4)21. Show that4 , 5(7.9.5)1 d , (tanh1 x) = dx 1 x21 < x < 1.39513:36 420. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006396 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS 22. Show that 1 d , 0 < x < 1. (sech1 x) = dx x 1 x2 23. Show that 1 d , x = 0. (csch1 x) = dx x 1 + x 2 24. Show that25. 26. 27. 28. 29.30.1 d (coth1 x) = , x > 1. dx 1 x2 Sketch the graph of y = sech x, giving: (a) the extreme values; (b) the points of inection; and (c) the concavity. Sketch the graphs of (a) y = coth x, (b) y = csch x. Graph y = sinh x and y = sinh1 x in the same coordinate system. Find all points of inection. Sketch the graphs of (a) y = cosh1 x, (b) y = tanh1 x. Given that tan = sinh x, show that d = sech x. (a) dx (b) x = ln (sec + tan ). dx (c) = sec . d The region bounded by the graph of y = sech x between x = 1 and x = 1 is revolved about the x-axis. Find the volume of the solid generated.37.tanh x ln (cosh x) d x.38.1 + tanh x d x. cosh2 x39.sech2 x d x. 1 + tanh x40.tanh5 x sech2 x d x.Exercises 4143. Verify the formula. In each case, take a > 0. x 1 d x = sinh1 + C. 41. a a2 + x 2 1 x 42. d x = cosh1 + C. 2 a2 a x 1 tanh1 x + C if x < a. 1 a a dx = 43. 1 a2 x 2 x coth1 +C if x > a. a a 44. If an object of mass m falling from rest under the action of gravity encounters air resistance that is proportional to the square of its velocity, then the velocity v(t) of the object at time t satises the equation md = mg kv 2 dtwhere k > 0 is the constant of proportionality and g is the gravitational constant. (a) Show that v(t) =Exercises 3140. Calculate. 31.tanh x d x.32.coth x d x.33.sech x d x.34.mg tanh kgk t mis a solution of the equation which satises v(0) = 0. (b) Findcsch x d x.lim v(t).t35.sech3 x tanh x d x.36.x sech2 x 2 d x.This limit is called the terminal velocity of the body. CHAPTER 7. REVIEW EXERCISES Exercises 18. Determine whether the function f is one-to-one and, if so, nd the inverse. 1. f (x) = x 1/3 + 2. x +1 . 3. f (x) = x 1 5. f (x) = e1/x . 7. f (x) = x ln x.2. f (x) = x 2 x 6.10. f (x) = 3x x11. f (x) =1 , x > 0; x3 4 + t 2 dt;c = 2. c = 0.04. f (x) = (2x + 1)3 .12. f (x) = x + cos x;6. f (x) = sin 2x + cos xExercises 1322. Calculate the derivative. 13. f (x) = (ln x 2 )3 . 14. y = 2 sin (e3x ).8. f (x) =2x + 1 . 3 2xExercises 912. Show that f has an inverse and nd ( f 1 ) (c). 1 9. f (x) = ; c = 1. 2 1 + exex . 1 + e2x 17. y = ln (x 3 + 3x ).15. g(x) =19. f (x) = (cosh x)1/xc = 1.16. f (x) = (x 2 + 1)sinh x . 18. g(x) = arctan (cosh x). .20. f (x) = 2x 3 arcsin (x 2 ).13:36 421. P1: PBU/OVYP2: PBU/OVYJWDD023-07JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 25, 2006*7.9 THE OTHER HYPERBOLIC FUNCTIONS21. f (x) = log31+x . 1x22. f (x) = arcsecx 2 + 4.the solid generated by revolving (b) about the y-axis.397(a) about the x-axis;49. Exercise 48 for the function f (x) = x 2 ex . ln x 50. Let f (x) = on (0, ). (a) Find the intervals where f x increases and the intervals where it decreases; (b) nd the extreme values; (c) determine the concavity of the graph and nd the points of inection; (d) sketch the graph, including all asymptotes. 51. Given that a < 1, nd the value of b for which 2Exercises 2338. Calculate. ex d x. 23. 1 e2x cos x d x. 25. 4 + sin2 x sec x 27. d x. x 5ln x d x. x 8 x 1/3 31. d x. 4/3 + 1 1 x 29.xx2 sinh 2 d x.33.535. 21 d x. 2 4x + 13 xe24. 126.ln x d x. xtan x ln (cos x) d x. 28.1x x4 9 2d x.32. 34. 36.sec x tan x d x. 1 + sec2 x ex d x. x + ex e 1 d x. 15 + 2x x 2x d x. 38. tanh2 2x d x. 2 0 Exercises 3942. Find the area below the graph. x 1 39. y = 2 , x [0, 1]. 40. y = 2 , x [0, 1]. x +1 x +1 1 41. y = , x 0, 1 . 2 1 x2 x 42. y = , x 0, 1 . 2 1 x2 43. (a) Apply the mean-value theorem to the function f (x) = ln (1 + x) to show that for all x > 1 x < ln (1 + x) < x. 1+x ln (1 + x) (b) Use the result in part (a) to show that lim = 1. x0 x 44. Show that for all positive integers m and n with 1 < m < n sech2ln00237.13x 2 e x d x.30.1 1 1 n n+1 < + + + < ln . m m m+1 n m1k+1 1 dx 1 < < . k+1 x k k 45. Find the area of the region between the curve x y = a 2 , the x-axis, and the vertical lines x = a, x = 2a. 46. Find the area of the region between the curve y = sec 1 x 2 and the x-axis from x = 1 and x = 1 . 2 2HINT:47. Let be the region between the graph of y = (1 + x 2 )1/2 and the x-axis, from x = 0 to x = 3. Find the volume of the solid generated by revolving (a) about the x-axis; (b) about the y-axis. 48. Let be the region between the graph of y = (1 + x 2 )1/4 and the x-axis, from x = 0 to x = 1 . Find the volume of 2b 1 b2 x 2adx =01 1 x2d x.52. Show that 1 0a dx = 1 + a2 x 2a 01 dx 1 + x2for all real numbers a.53. A certain bacterial culture, growing exponentially, increases from 20 grams to 40 grams in the period from 6 a.m. to 8 a.m. (a) How many grams will be present at noon? (b) How long will it take for the culture to reach 200 grams? 54. A certain radioactive substance loses 20% of its mass per year. What is the half-life of the substance? 55. Polonium-210 decays exponentially with a half-life of 140 days. (a) At time t = 0 a sample of polonium-210 has a mass of 100 grams. Find an expression that gives the mass at an arbitrary time t. (b) How long will it take for the 100-gram mass to decay to 75 grams? 56. From 1980 to 1990 the population of the United States grew from 227 million to 249 million. During that same period the population of Mexico grew from 62 million to 79 million. If the populations of the United States and Mexico continue to grow at these rates, when will the two populations be equal? 57. The population of a suburb of a large city is increasing at a rate proportional to the number of people currently living in the suburb. If, after two years, the population has doubled and after four years the population is 25,000, nd: (a) the number of people living in the suburb initially; (b) the length of time for the population to quadruple. 58. Let p and q be numbers greater than 1 which satisfy the condition 1/ p + 1/q = 1. Show that for all positive a and b ab bq ap + . p qHINT: Let C be the curve y = x p1 , x 0. Let 1 be the region between C and the x-axis from x = 0 to x = a. Let 2 be the region between C and the y-axis from y = 0 to y = b. Argue that ab area of1+ area of2.13:36 422. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER8 TECHNIQUES OF INTEGRATION 8.1 INTEGRAL TABLES AND REVIEW We begin by listing the more important integrals with which you are familiar. 1.k du = ku + C,2.u r du =3. 5.k constant.u r +1 + C, r +11 du = ln u + C. u pu p u du = + C, ln pr constant, r = 1. 4.eu du = eu + C.p > 0 constant, p = 1.6.sin u du = cos u + C.7.cos u du = sin u + C.8.tan u du = ln sec u + C.9.cot u du = ln sin u + C.10.sec u du = ln sec u + tan u + C.11.csc u du = ln csc u cot u + C.12.sec u tan u du = sec u + C.13.csc u cot u du = csc u + C.14.sec2 u du = tan u + C.15.csc2 u du = cot u + C.16. 17.398u du du = arcsin + C, a a2 u2 u 1 du = arctan + C, 2 + u2 a a aa > 0 constant. a > 0 constant.13:29 423. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.1 INTEGRAL TABLES AND REVIEW18.du 1 u du = arcsec + C, 2 a2 a a u ua > 0 constant.19.sinh u du = cosh u + C.cosh u du = sinh u + C.20.For review we work out a few integrals by u-substitution. x tan x 2 d x.Example 1 CalculateSOLUTION Set u = x 2 , du = 2x d x. Thenx tan x 2 d x =tan u du =1 21 2ln sec u + C =1 2ln sec x 2 + C. Formula 8 1Example 2 Calculate 0ex d x. ex + 2SOLUTION Setu = e x + 2,du = e x dx.At x = 0, u = 3; at x = 1, u = e + 2.Thus 1 0ex dx = ex + 2e+2 3du = ln u ue+2 3Formula 3 = ln (e + 2) ln 3 = ln+ 2) 0.45. =cos 2x dx. (2 + sin 2x)1/3Example 3 CalculateSOLUTION Set u = 2 + sin 2x,cos 2x dx = (2 + sin 2x)1/31 (e 3du = 2 cos 2x d x. Then1 du = u 1/31 21 2u 1/3 du =Formula 2 1 23 2u 2/3 + C= 3 (2 + sin 2x)2/3 + C. 4The nal example requires a little algebra.Example 4 Calculatedx . x 2 + 2x + 5SOLUTION First we complete the square in the denominator:x2dx = + 2x + 5(x 2dx = + 2x + 1) + 4We know that 1 u du = arctan + C. +4 2 2u2 Settingu = x + 1,du = dx,dx . (x + 1)2 + 439913:29 424. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006400 CHAPTER 8 TECHNIQUES OF INTEGRATION we have du u 1 x +1 1 = arctan + C = arctan u 2 + 22 2 2 2 2dx = x 2 + 2x + 5+ C.Using a Table of Integrals A table of over 100 intergrals, including those listed at the beginning of this section, appears on the inside covers of this text. This is a relatively short list. Mathematical handbooks such as Buringtons Handbook of Mathematical Tables and Formulas and CRC Standard Mathematical Tables contain extensive tables; the table in the CRC reference lists 600 integrals. The entries in a table of integrals are grouped by the form of the integrand: forms containing a + bu, forms containing a 2 u 2 , trigonometric forms, and so forth. The table on the inside covers is grouped in this manner. This is the only table of integrals well refer to in this text.Example 5 We use the table to calculate dx . 4 + x2 SOLUTION Of the integrals containing a 2 + u 2 , the one that ts our needs isFormula 77: du = ln u + 2 + u2 aa 2 + u 2 + C.In our case, a = 2 and u = x. Therefore dx = ln x + 4 + x24 + x 2 + C.Example 6 We use the table to calculate dx . 3x 2 (2x 1) SOLUTION The presence of the linear expression 2x 1 prompts us to look in the a + bu grouping. The formula that applies is Formula 109:a + bu 1 b du = + 2 ln + C. + bu) au a uu 2 (aIn our case a = 1, b = 2, u = x. Therefore dx 3x 2 (2x 1)=1 3dx 1 1 2x 1 = + 2 ln 1) 3 x xx 2 (2xExample 7 We use the table to calculate 9 4x 2 dx. x2 SOLUTION Closest to what we need is Formula 90: 1 u a2 u2 du = a 2 u 2 arcsin + C. 2 u u a+ C.13:29 425. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.1 INTEGRAL TABLES AND REVIEWWe can write our integral to t the formula by setting u = 2x, Doing this, we have 9 4x 2 dx = 2 x2 Check this out. du = 2 dx. 1 u 9 u2 du = 2 9 u 2 arcsin +C u2 u 3 =2 1 2x2x + C. 39 4x 2 arcsinEXERCISES 8.1 Exercises 138. Calculate. e2x dx.1.13.sin x dx.14.0/37.cot x dx. /69./411.dx . 5x 1 x3 8. dx. 4 0 1+x6.1 x2dx.sin x dx. cos2 x10. /4 15. 17. 19. 21. 23. 25.e dx. x2 1 c dx . 2 2 0 x +c sec2 d. 3 tan + 1 ex dx. a ex b x dx. (x + 1)2 + 4 x dx. 1 x4 dx . x 2 + 6x + 10 227.x sin x dx. tan2 x dx.29.e x dx. x14.133.ln x 3 dx. x arcsin x dx. 1 x2x3 1x416.dx.a e dx. sin d. 3 2 cos dx . x 2 4x + 13 ln x dx. x ex dx. 1 + e2x18. 20. 22. 24. 26.e x tan e x dx.28.x dx. 9 + x4 cosh 2x sinh3 2x dx.30./432. 034.036.1 + sin x dx. cos2 xx2 dx. x2 + 11 1/238. 01+x dx. 1 x2Exercises 3948. Calculate using our table of integrals. x 2 4 dx.39.42.49. Evaluatex dx . 2 + 3x46.x 3 ln x dx.47.sec4 t dt.44.dx . x(2x + 3) x2 + 9 dx. x243.4 x 2 dx.40.cos3 2t dt.41.45.x xc31.dx . cos2 x12./4 2 1/x13.sec x tan x dx./4x/437.0sec2 (1 x) dx.5.cos 2 x dx. 32.11 dx. x ln x35.48.x2dx . x2 2x 3 sin x dx. 1 + cos x dx.01 HINT: cos x = 2 cos2 2 x 1.50. Calculatesec2 x tan x d x in two ways.(a) Set u = tan x and verify that sec2 x tan x d x =1 2tan2 x + C1 .1 2sec2 x + C2 .(b) Set u = sec x and verify that sec2 x tan x d x =(c) Reconcile the results in parts (a) and (b). 51. Verify that, for each positive integer n: (a) 0sin2 nx d x = 1 . 2HINT: sin2 = 1 (1 cos 2 ) 2 arctan x dx. 1 + x2(b)e x cosh (2 e x ) dx.(c)00sin nx cos nx d x = 0./nsin nx cos nx d x = 0.40113:29 426. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006402 CHAPTER 8 TECHNIQUES OF INTEGRATIONC52. (a) Calculate sin3 x d x. HINT: sin2 x = 1 cos2 x. (b) Calculate sin5 x d x. (c) Explain how to calculate sin2k1 x d x for an arbitrary positive integer k. 53. (a) Calculate tan3 x d x. HINT: tan2 x = sec2 x 1. (b) Calculate tan5 x d x. (c) Calculate tan7 x d x. (d) Explain how to calculate tan2k+1 x d x for an arbitrary positive integer k. 54. (a) Sketch the region between the curves y = csc x and y = sin x over the interval [ 1 , 1 ]. 6 2 (b) Calculate the area of that region. (c) The region is revolved about the x-axis. Find the volume of the resulting solid. 55. (a) Use a graphing utility to sketch the graph of f (x) =1 sin x + cos xfor0x . 2CC56. (a) Use a graphing utility to sketch the graph of f (x) = 2 ex . (b) Let a > 0. The region between the graph of f and the y-axis from x = 0 to x = a is revolved about the y-axis. Find the volume of the resulting solid. (c) Find the value of a for which the solid in part (b) has a volume of 2 cubic units. 57. (a) Use a graphing utility to draw the curves y=Cx2 + 1 x +1forx > 1andx + 2y = 16in the same coordinate system. (b) These curves intersect at two points and determine a bounded region . Estimate the x-coordinates of the two points of intersection accurate to two decimal places. (c) Determine the approximate area of the region . 58. (a) Use a graphing utility to draw the curve y 2 = x 2 (1 x).(b) Find A and B such that sin x + cos x = A sin (x + B). (c) Find the area of the region between the graph of f and the x-axis.(b) Your drawing in part (a) should show that the curve forms a loop for 0 x 1. Calculate the area of the loop. HINT: Use the symmetry of the curve. 8.2 INTEGRATION BY PARTS We begin with the formula for the derivative of a product: u(x)v (x) + v(x)u (x) = (u v) (x). Integrating both sides, we get u(x)v (x) dx +v(x)u (x) dx =(u v) (x) dx.Since (u v) (x) dx = u(x)v(x) + C, we have u(x)v (x) dx +v(x)u (x) dx = u(x)v(x) + Cand therefore u(x)v (x) dx = u(x)v(x) v(x)u (x) dx + C.Since the calculation of v(x)u (x) dx will yield its own arbitrary constant, there is no reason to keep the constant C. We therefore drop it and write(8.2.1)u(x)v (x) dx = u(x)v(x) v(x)u (x) dx.13:29 427. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.2 INTEGRATION BY PARTSThe process of nding u(x)v (x) dx by calculating v(x)u (x) dx and then using (8.2.1) is called integration by parts. Usually we write u = u(x),dv = v (x) dxdu = u (x) dx,v = v(x).Then the formula for integration by parts readsu dv = uv (8.2.2)v du.Integration by parts is a very versatile tool. According to (8.2.2) we can calculate v du instead. The payoff is immediate in those cases whereu dv by calculatingwe can choose u and v so that v duExample 1 Calculateis easier to calculate thanu dv.x e x dx.SOLUTION We want to separate x from e x . Settingu = x,dv = e x dxdu = dx,v = ex .we have Accordingly, x ex d x =u dv = uv v du = xe x e x dx = xe x e x + C.Our choice of u and dv worked out well. Does the choice of u and dv make a difference? Suppose we had set u = ex ,dv = x dx.Then we would have had du = e x d x,v = 1 x 2. 2In this case integration by parts would have led to x e x dx =u dv = uv v du = 1 x 2 e x 21 2x 2 e x dx,giving us an integral which at this stage is difcult for us to deal with. This choice of u and dv would not have been helpful. 40313:29 428. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006404 CHAPTER 8 TECHNIQUES OF INTEGRATION Example 2 Calculatex sin 2x dx.SOLUTION Settingu = x,dv = sin 2x dx,we have du = dx,v = 1 cos 2x. 2Therefore, x sin 2x dx = 1 x cos 2x 2 1 cos 2x dx = 1 x cos 2x + 1 sin 2x + C. 2 2 4As you can verify, had we set u = sin 2x,dv = x dx,then we would have run into an integral more difcult to evaluate than the integral with which we started. In Examples 1 and 2 there was only one effective way of choosing u and dv. With some integrals we have more latitude.Example 3 Calculatex ln x dx.SOLUTION Settingu = ln x,dv = x dx,we have du =1 dx, xv=x2 . 2The substitution gives x ln x dx = =u dv = uv x2 ln x 2v du1 x2 dx = 1 x 2 ln x 2 x 21 2x dx = 1 x 2 ln x 1 x 2 + C. 2 4ANOTHER APPROACH This time we setu = x ln x,dv = d xso that du = (1 + ln x) dx,v = x.In this case the relation u dv = uv v dugives x ln x dx = x 2 ln x x(1 + ln x) dx.The new integral is more complicated than the one with which we started. It may therefore look like we are worse off than when we began, but that is not the case. Going13:29 429. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.2 INTEGRATION BY PARTSon, we have x ln x dx = x 2 ln x x ln x d x = x 2 ln x 2x dx x dxx ln x dx= x 2 ln x 1 x 2 + C 2 x ln x dx = 1 x 2 ln x 1 x 2 + C. 2 4 This is the result we obtained before. In the last step we wrote C/2 as C. We can do this because C represents an arbitrary constant. RemarkAs you just saw, integration by parts can be useful even ifeasier to calculate thanv du is notu dv. What matters to us is the interplay between the twointegrals. To calculate some integrals, we have to integrate by parts more than once.Example 4 Calculatex 2 ex d x.SOLUTION Settingu = x 2,dv = ex dx,du = 2x dxv = ex .we haveThis gives x 2 ex d x =u dv = uv = x 2 ex +v du = x 2 ex 2xex d x2xex dx.We now calculate the integral on the right, again by parts. This time we set u = 2x,dv = ex dx.du = 2 dx,v = exThis givesand thus 2xex dx =u dv = uv = 2xex +v du = 2xex 2ex dx2ex dx = 2xex 2ex + C.Combining this with our earlier calculations, we have x 2 ex dx = x 2 ex 2xex 2ex + C = (x 2 + 2x + 2)ex + C.40513:29 430. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006406 CHAPTER 8 TECHNIQUES OF INTEGRATION e x cos x dx.Example 5 CalculateSOLUTION Once again well need to integrate by parts twice. First we writeu = ex , du = e x d x,dv = cos x d x v = sin x.This gives (1)e x cos x dx =u dv = uv v du = e x sin x e x sin x dx.Now we work with the integral on the right. Setting u = ex , du = e x d x,dv = sin x d x v = cos x,we have (2)e x sin x dx =u dv = uv v du = e x cos x +e x cos x dx.Substituting (2) into (1), we get e x cos x d x = e x sin x + e x cos x e x cos x d xe x cos x d x = e x (sin x + cos x)2e x cos x d x = 1 e x (sin x + cos x). 2 Since this is an indenite integral, we add an arbitrary constant C: e x cos x d x = 1 e x (sin x + cos x) + C. 2 (We began this example by setting u = e x , dv = cos x dx. As you can check, the substitution u = cos x, dv = e x dx would have worked out just as well.) Integration by parts is often used to calculate integrals where the integrand is a mixture of function types; for example, polynomials mixed with exponentials, polynomials mixed with trigonometric functions, and so forth. Some integrands, however, are better left as mixtures; for example, 222xe x dx = e x + Cand3x 2 cos x 3 dx = sin x 3 + C.Any attempt to separate these integrands for integration by parts is counterproductive. The mixtures in these integrands arise from the chain rule, and we need these mixtures to calculate the integrals.Example 6 Calculatex 5 cos x 3 dx.SOLUTION To integrate cos x 3 , we need an x 2 factor. So well keep x 2 together withcos x 3 and set u = x 3,dv = x 2 cos x 3 dx.13:29 431. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.2 INTEGRATION BY PARTSThen v=du = 3x 2 dx,1 3sin x 3and x 5 cos x 3 dx = 1 x 3 sin x 3 3x 2 sin x 3 dx= 1 x 3 sin x 3 + 1 cos x 3 + C. 3 3The counterpart to (8.2.1) for denite integrals readsb(8.2.3)u(x)v (x) dx = u(x)v(x)ab abv(x)u (x) dx.aThis follows directly from writing the product rule as u(x)v (x) = (u v) (x) v(x)u (x). Just integrate from x = a to x = b. We can circumvent this formula by working with indenite integrals and bringing in the limits of integration only at the end. This is the course we will follow in the next example. 2x 3 ln x dx.Example 7 Evaluate 1SOLUTION First we calculate the indenite integral, proceeding by parts. We setu = ln x, 1 du = dx, xdv = x 3 dx 1 v = x 4. 4This gives x 3 ln x dx = 1 x 4 ln x 41 4x 3 d x = 1 x 4 ln x 41 4 x 16+ C.To evaluate the denite integral, we need only one antiderivative. We choose the one with C = 0. This gives 2 1x 3 ln x d x =1 4 x 4ln x 2 1 4 x 16 1= 4 ln 2 15 . 16Through integration by parts, we construct an antiderivative for the logarithm, for the arc sine, and for the arc tangent.(8.2.4)(8.2.5)ln x d x = x ln x x + C.arcsin x d x = x arcsin x +1 x 2 + C.40713:29 432. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006408 CHAPTER 8 TECHNIQUES OF INTEGRATIONarctan x d x = x arctan x 1 ln (1 + x 2 ) + C. 2(8.2.6)We will work with the arc sine. The logarithm and the arc tangent formulas are left to the Exercises. To nd the integral of the arc sine, we set u = arcsin x, 1 dx, du = 1 x2dv = d x v = x.This gives x dx = x arcsin x + 1 x2arcsin x d x = x arcsin x 1 x 2 + C.EXERCISES 8.2 1/4Exercises 140. Calculate. 2xex dx.1.10x 2 ex dx. 33.15.0x 2 ex dx.x235.HINT: Set u =x 2 cosh 2x dx.212.213.(ln x) dx. x 3 3x dx. x(x + 5)14 dx.x(x + 5)16.17.1 arcsin (ln x) dx. x cos (ln x) dx. sin (ln x) dx.HINT: Integrate by parts twice.x1414.15.37.x (e 1) dx. 218. cos x dx. x138.0ln (x + 1) dx. x +1x, dv =x sinh 2x 2 dx.36.39.10.1dx . x(ln x)3 3 x x + 1 dx.8.111.x d x.cosx 3 ex dx.6.dx. 1x c2 x ln x dx.9.34.07.x arctan x 2 dx.33. x ln x 2 dx.4.arcsin 2x dx. 1 4x 232.0x2x dx.2.arcsin 2x dx.31.2ex 2 (ln x)2 dx.40. 1dx. 41. Derive (8.2.4):ln x dx = x ln x x + C.x ln x dx. 42. Derive (8.2.6):1/219.x cos x dx.(2x + x 2 )2 dx. /220.0x 2 sin x dx.021.x 2 (x + 1)9 dx.22.x 2 (2x 1)7 dx.23.e x sin x dx.24.(e x + 2x)2 dx.26.x ln (x + 1) dx.28.e3x cos 2x dx.30.x 3 sin x dx.125.ln (1 + x 2 ) dx.027.x n ln x d x,29.x 3 sin x 2 dx.n = 1.arctan x d x = x arctan x 1 ln (1 + x 2 ) + C. 2 Derive the following three formulas. x k+1 x k+1 ln x + C, k = 1. 43. x k ln x d x = k+1 (k + 1)2 eax (a cos bx + b sin bx) 44. eax cos bx d x = + C. a 2 + b2 ax e (a sin bx b cos bx) 45. eax sin bx d x = + C. a 2 + b2 46. What happens if you try integration by parts to calculate eax cosh ax d x? Calculate this integral by some other method. 47. Set f (x) = x sin x. Find the area between the graph of f and the x-axis from x = 0 to x = .13:29 433. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.2 INTEGRATION BY PARTS48. Set g(x) = x cos 1 x. Find the area between the graph of g 2 and the x-axis from x = 0 to x = . Exercises 4950. Find the area between the graph of f and the x-axis. 49. f (x) = arcsin x, x 0, 1 . 2 50. f (x) = xe2x , x 0, 2 . 51. Let be the region between the graph of the logarithm function and the x-axis from x = 1 to x = e. (a) Find the area of . (b) Find the centroid of . (c) Find the volume of the solids generated by revolving about each of the coordinate axes. ln x , x [1, 2e]. 52. Let f (x) = x (a) Find the area of the region bounded by the graph of f and the x-axis. (b) Find the volume of the solid generated by revolving about the x-axis. Exercises 5356. Find the centroid of the region under the graph. 53. f (x) = e x , x [0, 1]. 54. f (x) = ex , x [0, 1]. 55. f (x) = sin x, x [0, ]. 56. f (x) = cos x, x 0, 1 . 2 57. The mass density of a rod that extends from x = 0 to x = 1 is given by the function (x) = ekx where k is a constant. (a) Calculate the mass of the rod. (b) Find the center of mass of the rod. 58. The mass density of a rod that extends from x = 2 to x = 3 is given by the logarithm function f (x) = ln x. (a) Calculate the mass of the rod. (b) Find the center of mass of the rod. Exercises 5962. Find the volume generated by revolving the region under the graph about the y-axis. 59. f (x) = cos 1 x, x [0, 1]. 2 60. f (x) = x sin x, x [0, ]. 61. f (x) = x e x , x [0, 1]. 62. f (x) = x cos x, x 0, 1 . 2 63. Let be the region under the curve y = e x , x [0, 1]. Find the centroid of the solid generated by revolving about the x-axis. (For the appropriate formula, see Project 6.4.) 64. Let be the region under the graph of y = sin x, x 0, 1 . Find the centroid of the solid generated by revolving 2 about the x-axis. (For the appropriate formula, see Project 6.4.) 65. Let be the region between the curve y = cosh x and the xaxis from x = 0 to x = 1. Find the area of and determine the centroid. 66. Let be the region given in Exercise 65. Find the centroid of the solid generated by revolving : (a) about the x-axis; (b) about the y-axis 67. Let n be a positive integer. Use integration by parts to show that x n eax n x n1 eax d x, a = 0. x n eax d x = a a40968. Let n be a positive integer. Show that (ln x)n d x = x(ln x)n n(ln x)n1 d x.The formula given in Exercise 67 reduces the calculation of x n eax d x to the calculation of x n1 eax d x. The formula given in Exercise 68 reduces the calculation of (ln x)n d x to the calculation of (ln x)n1 d x. Formulas (such as these) which reduce the calculation of an expression in n to the calculation of the corresponding expression in n 1 are called reduction formulas. Exercises 6972. Calculate the following integrals by using the appropriate reduction formulas. 69.x 3 e2x d x.70.x 2 ex d x.71.(ln x)3 d x.72.(ln x)4 d x.73. (a) As you can probably see, were you to integrate by parts, the result would be of the formx 3 ex d xx 3 e x d x = Ax 3 e x + Bx 2 e x + C xe x + De x + E. Differentiate both sides of this equation and solve for the coefcients A, B, C, D. In this manner you can calculate the integral without actually carrying out the integration. (b) Calculate x 3 e x d x by using the appropriate reduction formula. 74. If P is a polynomial of degree k, then P(x) e x d x = [P(x) P (x) + P (k) (x)]e x + C. Verify this statement. For simplicity, take k = 4. 75. Use the statement in Exercise 74 to calculate: (a)(x 2 3x + 1)e x d x.(x 3 2x)e x d x.(b)76. Use integration by parts to show that if f has an inverse with continuous rst derivative, then f 1 (x) dx = x f 1 (x) x( f 1 ) (x) dx.77. Show that if f and g have continuous second derivatives and f (a) = g(a) = f (b) = g(b) = 0, then bbf (x)g (x) dx =ag(x) f (x) dx. a78. You are familiar with the identity bf (b) f (a) =f (x) dx. a(a) Assume that f has a continuous second derivative. Use integration by parts to derive the identity bf (b) f (a) = f (a)(b a) af (x) (x b) dx.13:29 434. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006410 CHAPTER 8 TECHNIQUES OF INTEGRATION (a) from x = 1 to x = 3 . 2 2(b) Assume that f has a continuous third derivative. Use the result in part (a) and integration by parts to derive the identity(b) from x = 3 to x = 5 . 2 2f (a) (b a)2 f (b) f (a) = f (a)(b a) + 2 bf (x) (x b)2 d x. 2aCCGoing on in this manner, we are led to what are called Taylor series (Chapter 12). 79. Use a graphing utility to draw the curve y = x sin x for x 0. Then use a CAS to calculate the area between the curve and the x-axis (a) from x = 0 to x = . (b) from x = to x = 2. (c) from x = 2 to x = 3. (d) What is the area between the curve and the x-axis from x = n to x = (n + 1) ? Take n an arbitrary nonnegative integer. 80. Use a graphing utility to draw the curve y = x cos x for x 0. Then use a CAS to calculate the area between the curve and the x-axisCC(c) from x = 5 to x = 7 . 2 2 (d) What is the area between the curve and the x-axis from 1 x = 2 (2n 1) to x = 1 (2n + 1)? Take n an arbi2 trary positive integer. 81. Use a graphing utility to draw the curve y = 1 sin x from x = 0 to x = . Then use a CAS (a) to nd the area of the region between the curve and the x-axis. (b) to nd the volume of the solid generated by revolving about the y-axis. (c) to nd the centroid of . 82. Use a graphing utility to draw the curve y = xe x from x = 0 to x = 10. Then use a CAS (a) to nd the centroid of the region between the curve and the x-axis. (b) to nd the volume of the solid generated by revolving about the x-axis. (c) to nd the volume of the solid generated by revolving about the y-axis. PROJECT 8.2 Sine Waves y = sin nx and Cosine Waves y = cos nx Problem 1. Show that for each positive integer n, 2sin2 nx d x = 2and0to show that 2cos2 nx d x = .001 2 1 cos 2, 2cos2 =1 2+ 1 cos 2. 2Problem 3. Show that for m = n, 2Problem 2. Show that for m = n, 2sin mx cos nx dx = 0.0HINT: Verify that sin mx sin nx dx = 002and 2sin mx sin nx d x. 0Evaluate the cosine integral by parts.HINT: Use the identities sin2 =2cos mx cos nx d x =sin [(m + n)x] dx = 0.0cos mx cos nx dx = 0.0Then use the addition formula sin ( + ) = sin cos + cos sin HINT: Verify that 2cos [(m + n)x] dx = 0.to show that0 2Then use the addition formula cos ( + ) = cos cos sin sin 02sin mx cos nx d x +cos mx sin nx d x = 0.0Evaluate the second integral by parts.13:29 435. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.3 POWERS AND PRODUCTS OF TRIGONOMETRIC FUNCTIONSFor each positive integer n sin2 nx d x = f (x) = a1 sin x + a2 sin 2x + + an sin nx + b1 cos x and+ b2 cos 2x + + bn cos nx0 2is called a trigonometric polynomial, and the coefcients ak , bk are called the Fourier coefcients. Determine the ak and bk from k = 1 to k = n. HINT: Evaluatecos2 nx d x = ,0(8.2.6)and for positive integers m = n 22sin mx sin nx d x = 0,2f (x) sin kx d x0and0 2411These relations lie at the heart of wave theory. Problem 4. (The superposition of waves) A function of the formSummary2f (x) cos kx d x 0using the relations just summarized.cos mx cos nx d x = 0,0 2sin mx cos nx d x = 0. After the French mathematician J. B. J. Fourier (17681830), who was the rst to use such polynomials to closely approximate functions of great generality.0 8.3 POWERS AND PRODUCTS OF TRIGONOMETRIC FUNCTIONS Integrals of trigonometric powers and products can usually be reduced to elementary integrals by the imaginative use of the basic trigonometric identities and, here and there, some integration by parts. These are the identities that well rely on: Unit circle relations sin2 + cos2 = 1,tan2 + 1 = sec2 ,cot2 + 1 = csc2 Addition formulas sin ( + ) = sin cos + cos sin sin ( ) = sin cos cos sin cos ( + ) = cos cos sin sin cos ( ) = cos cos + sin sin Double-angle formulas sin 2 = 2 sin cos ,cos 2 = 1 2 sin2 Half-angle formulas sin2 =1 2 1 cos 2, 2cos2 =Sines and Cosines Example 1 Calculate sin2 x cos5 x dx.1 2+ 1 cos 2. 213:29 436. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006412 CHAPTER 8 TECHNIQUES OF INTEGRATION SOLUTION The relation cos2 x = 1 sin2 x enables us to express cos4 x in terms ofsin x. The integrand then becomes (a polynomial in sin x) cos x, an expression that we can integrate by the chain rule. sin2 x cos5 x d x =sin2 x cos4 x cos x d x=sin2 x(1 sin2 x)2 cos x d x=(sin2 x 2 sin4 x + sin6 x) cos x d x=1 3sin3 x 2 sin5 x + 1 sin7 x + C. 5 7Example 2 Calculate sin5 x dx SOLUTION The relation sin2 x = 1 cos2 x enables us to express sin4 x in terms ofcos x. The integrand then becomes (a polynomial in cos x) sin x, an expression that we can integrate by the chain rule: sin5 x dx =sin4 x sin x dx=(1 cos2 x)2 sin x dx=(1 2 cos2 x + cos4 x) sin x dx=sin x dx 2cos2 x sin x d x += cos x + 2 cos3 x 1 cos5 x + C. 3 5cos4 x sin x dx Example 3 Calculate sin2 x dx SOLUTION Since sin2 x =1 2andcos2 x dx. 1 cos 2x and cos2 x = 21 2+ 1 cos 2x, 2sin2 x dx =1 2 1 cos 2x dx = 1 x 1 sin 2x + C 2 2 4cos2 x dx =1 2+ 1 cos 2x dx = 1 x + 1 sin 2x + C. 2 2 4and 13:29 437. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.3 POWERS AND PRODUCTS OF TRIGONOMETRIC FUNCTIONSExample 4 Calculate sin2 x cos2 x dx. SOLUTION The relation 2 sin x cos x = sin 2x gives sin2 x cos2 x =that we can integrate: sin2 x cos2 x d x =1 4=1 4=1 81 4sin2 2x andsin2 2x dx 1 2 1 cos 4x dx 2dx cos 4x d x = 1 x 81 81 32sin 4x + C.Example 5 Calculate sin4 x d x. SOLUTIONsin4 x = (sin2 x)2 =1 2 1 cos 2x 2=1 4 1 cos 2x + 2=3 82=1 4 1 cos 2x + 1 cos2 2x 2 4 1 cos 2x + 1 cos 4x. 2 81 8+ 1 cos 4x 8Therefore sin4 x d x =3 8 1 cos 2x + 1 cos 4x d x 2 8= 3 x 1 sin 2x + 8 41 32sin 4x + C.Example 6 Calculate sin 5x sin 3x dx. SOLUTION The only identities that feature the product of sines with different arguments are the addition formulas for the cosine:cos ( ) = cos cos + sin sin ,cos ( + ) = cos cos sin sin .We can express sin sin in terms of something we can integrate by subtracting the second equation from the rst one. In our case we have cos 2x = cos (5x 3x) = cos 5x cos 3x + sin 5x sin 3x cos 8x = cos (5x + 3x) = cos 5x cos 3x sin 5x sin 3x and therefore sin 5x sin 3x = 1 (cos 2x cos 8x). 241313:29 438. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006414 CHAPTER 8 TECHNIQUES OF INTEGRATION Using this relation, we have sin 5x sin 3x d x =1 2=1 4cos 2x d x sin 2x 1 161 2cos 8x d xsin 8x + C.Tangents and Secants Example 7 Calculate tan4 x dx. SOLUTION The relation tan2 x = sec2 x 1 givestan4 x = tan2 x sec2 x tan2 x = tan2 x sec2 x sec2 x + 1. Therefore tan4 x d x = =(tan2 x sec2 x sec2 x + 1) dx 1 3tan3 x tan x + x + C.Example 8 Calculate sec3 x dx. SOLUTION The relation sec2 x = tan2 x + 1 givessec3 x dx =sec x (tan2 x + 1) dx =sec x tan2 x dx +sec x dx.We know the second integral, but the rst integral gives us problems. (Here the relation tan2 x = sec2 x 1 doesnt help, for, as you can check, that takes us right back to where we started.) Not seeing any other way to proceed, we try integration by parts on the original integral. Setting u = sec x, du = sec x tan x d x,dv = sec2 x d x v = tan x,we have sec3 x d x = sec x tan x = sec x tan x 2 2 tan x = sec x 1 2sec3 x d x = sec x tan x + sec3 x d x =1 2sec x tan2 x d x sec3 x d x +sec x d xsec x d xsec x tan x + 1 ln sec x + tan x + C. 213:29 439. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.3 POWERS AND PRODUCTS OF TRIGONOMETRIC FUNCTIONSExample 9 Calculate sec6 x dx. SOLUTION Applying the relation sec2 x = tan2 x + 1 to sec4 x, we can express theintegrand as (a polynomial in tan x) sec2 x. We can integrate this by the chain rule: sec6 x = sec4 x sec2 x = (tan2 x + 1)2 sec2 x = (tan4 x + 2 tan2 x + 1) sec2 x. Therefore sec6 x dx = =(tan4 x sec2 x + 2 tan2 x sec2 x + sec2 x) dx 1 5tan5 x + 2 tan3 x + tan x + C. 3Example 10 Calculate tan5 x sec3 x dx. SOLUTION Applying the relation tan2 x = sec2 x 1 to tan4 x, we can express theintegrand as (a polynomial in sec x) sec x tan x. We can integrate this by the chain rule: tan5 x sec3 x d x =tan4 x sec2 x sec x tan x d x=(sec2 x 1)2 sec2 x sec x tan x d x=(sec6 x 2 sec4 x + sec2 x) sec x tan x d x=1 7sec7 x 2 sec5 x + 1 sec3 x + C. 5 3Cotangents and Cosecants The integrals in Examples 710 featured tangents and secants. In carrying out the integrations, we relied on the identity tan2 x + 1 = sec2 x and in one instance resorted to integration by parts. To calculate integrals that feature cotangents and cosecants, use the identity cot2 x + 1 = csc2 x and, if necessary, integration by parts. EXERCISES 8.3 Exercises 144. Calculate. (If you run out of ideas, use the examples as models.) sin3 x dx.1./82.cos2 4x dx.5.cos4 x sin3 x dx.6.sin3 x cos2 x dx.7.sin3 x cos3 x dx.8.sin2 x cos4 x dx.9.sec2 x dx.0 /63.2sin 3x dx. 04.cos3 x dx.10.csc2 2x dx.41513:29 440. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006416 CHAPTER 8 TECHNIQUES OF INTEGRATION tan3 x dx.11.13.sin4 x dx.12. 14.49. The region bounded by the y-axis, the line y = 1, and the curve y = tan x, x [0, /4], is revolved about the x-axis. Find the volume of the resulting solid. 50. The region between the curve y = tan2 x and the x-axis from x = 0 to x = /4 is revolved about the x-axis. Find the volume of the resulting solid. 51. The region between the curve y = tan x and the x-axis from x = 0 to x = /4 is revolved about the line y = 1. Find the volume of the resulting solid. 52. The region between the curve y = sec2 x and the x-axis from x = 0 to x = /4 is revolved about the x axis. Find the volume of the resulting solid.cot3 x dx. cos3 x cos 2x dx.0 /215.sin 2x cos 3x dx.16.cos 2x sin 3x dx. 017.tan2 x sec2 x dx.18.19.sin2 x sin 2x dx.cot2 x csc2 x dx.20./2cos4 x dx.0 621.sin x dx. /223.2cos5 x sin5 x dx.22.53. (a) Use integration by parts to show that for n > 2 1 n1 sinn x dx = sinn1 x cos x + n n6cot x dx.24.tan x dx.cot3 x csc3 x dx.26.tan3 x sec3 x dx./625.(b) Then show that /2427.sin 5x sin 2x dx.28.sec 3x dx.29.sin5/2 x cos3 x dx.30.sin3 x dx. cos x5tan 3x dx.31.0/2cot 2x dx.0/2/2/2(a) 0cot4 x csc4 x dx.38.C 2sin 1 x cos 2x dx. 239./441.tan3 x sec2 x dx.40./242.csc3 x cot x dx./4 /60sin2 ax dx, a = 0.0043.tan2 2x dx./2sinn x dx.055. Evaluate by the Wallis formulas:(sin 3x sin x)2 dx. tan4 x sec4 x dx.(n 1) 4 2 . n5 3cosn x dx =0037.sinn x dx =54. Use Exercise 53 to show thatcos 4x sin 2x dx.36.(n 1) 5 3 1 ; n6 4 2 20cos x cos 1 x dx. 2/435.sinn2 x dx.0for odd n 3,1/634.sinn x dx =0sin4 3 x cos3 3 x dx. 1/2/2n1 n(c) Verify the Wallis sine formulas: for even n 2,1/333.sinn x dx =532.sinn2 x dx./344.tan x sec3/2 x dx.C045. Find the area between the curve y = sin2 x and the x-axis from x = 0 to x = . 46. The region between the curve y = cos x and the x-axis from x = /2 to x = /2 is revolved about the x-axis. Find the volume of the resulting solid. 47. The region of Exercise 45 is revolved about the x-axis. Find the volume of the resulting solid. 48. The region bounded by the y-axis and the curves y = sin x and y = cos x, 0 x /4, is revolved about the x-axis. Find the volume of the resulting solid.Csin7 x dx./2(b)cos6 x dx.056. Use a graphing utility to draw the graph of the function f (x) = x + sin 2x, x [0, ]. The region between the graph of f and the x-axis is revolved about the x-axis. (a) Use a CAS to nd the volume of the resulting solid. (b) Calculate the volume exactly by carrying out the integration. 57. Use a graphing utility draw the graph of the function to g(x) = sin2 x 2 , x [0, ]. The region between the graph of g and the x-axis is revolved about the y-axis. (a) Use a CAS to nd the volume of the resulting solid. (b) Calculate the volume exactly by carrying out the integration. 58. Use a graphing utility to draw in one gure the graphs of both f (x) = 1 + cos x and g(x) = sin 1 x from x = 0 to x = 2. 2 (a) Use a CAS to nd the points where the two curves intersect; then nd the area between the two curves. (b) The region between the two curves is revolved about the x-axis. Use a CAS to nd the volume of the resulting solid.13:29 441. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.4 INTEGRALS FEATURING 8.4 INTEGRALS FEATURING a 2 x2 , a 2 + x2 , x2 a 2417 a 2 x2, a 2 + x2, x2 a 2Preliminary Remark By reversing the roles played by x and u in the statement of Theorem 5.7.1, we can conclude thatif F = f, (8.4.1)f (x(u))x (u) du = F(x(u)) + Cthenand x(b)bf (x) dx =x(a)f (x(u))x (u) du. aThese are the substitution rules that we will follow in this section. Integrals that feature a 2 x 2 , a 2 + x 2 , or x 2 a 2 can often be calculated by a trigonometric substitution. Taking a > 0, we proceed as follows: fora2 x 2we setx = a sin u;fora2 + x 2we setx = a tan u;forx 2 a2we setx = a sec u.In making such substitutions, we must make clear exactly what values of u we are using. Failure to do this can lead to nonsensical results. We begin with a familiar integral. yExample 1 You know that aa 2 x 2 dxarepresents the area of the half-disk of radius a and is therefore 1 a 2 . (Figure 8.4.1) We 2 conrm this by a trigonometric substitution. For x from a to a, we set x = a sin u,dx = a cos u du,taking u from 1 to 1 . For such u, cos u 0 and 2 2 a2 x 2 = Atu = 1 , x = a; 2 a aa 2 x 2 dx =a 2 a 2 sin2 u = a cos u.at u = 1 , x = a. Therefore 2/2 /2a 2 cos2 u du = a 2/2 /21 2+ 1 cos 2u du = 1 a 2 . 2 2To a limited extent we can give a geometric view of a trigonometric substitution by drawing a suitable right triangle. Since the right triangle interpretation applies only to u between 0 and 1 , we will not use it as the basis of our calculations. 2aaFigure 8.4.1x13:29 442. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006418 CHAPTER 8 TECHNIQUES OF INTEGRATION Example 2 To calculate (a 2dx dx x 2 )3/2we note that the integral can be written 1 2 x2 aa x uThis integral featuresa2 x2dx. a 2 x 2 . For each x between a and a, we set x = a sin u,taking u between31 2and1 . 2dx = a cos u du,For such u, cos u > 0 and a 2 x 2 = a cos u.Therefore (a 2dx = x 2 )3/2a cos u du (a cos u)31 1 du a2 cos2 u 1 = 2 sec2 u du a x 1 + C. = 2 tan u + C = 2 a2 x 2 a a sin u tan u = =cos uCheck the result by differentiation.Before giving the next example, we point out that, at those numbers u where cos u = 0, sec u and cos u have the same sign (sec u = 1/ cos u).Example 3 We calculate a 2 + x 2 dx. For each real number x, we set 2a2 +xx = a tan u,xdx = a sec2 u du,taking u between 1 and 1 . For such u, sec u > 0 and 2 2u aa 2 + x 2 = a sec u. Check this out. Therefore a 2 + x 2 dx = = a2(a sec u) a sec2 u du sec3 u du13:29 443. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.4 INTEGRALS FEATURING= a 2 x2 , a 2 + x2 , x2 a 2419a2 (sec u tan u + ln sec u + tan u ) + C 2Example 7, Section 8.3 a2 + x 2 x a2 + x 2 x + ln + +C a a a a = 1 x a 2 + x 2 + 1 a 2 ln x + a 2 + x 2 1 a 2 ln a + C. 2 2 2 a2 = 2Noting that x + write(8.4.2) a 2 + x 2 > 0 and, absorbing the constant 1 a 2 ln a in C, we can 2 a 2 + x 2 dx = 1 x a 2 + x 2 + 1 a 2 ln x + a 2 + x 2 + C. 2 2 This is a standard integration formula. (Formula 78)Example 4 We calculate dx. x2 1 The domain of the integrand consists of two separated sets: all x > 1 and all x < 1. Both for x > 1 and x < 1, we set x = sec u,dx = sec u tan u du.For x > 1 we take u between 0 and 1 ; for x < 1 we take u between and 3 . For 2 2 such u, tan u > 0 and x 2 1 = tan u. dx x21=sec u tan u du = tan usec u du= ln sec u + tan u + C = ln x +x 2 1 + C.Check the result by differentiation.Example 5 We calculate x2dx . x2 4For x > 2 and x < 2, we set x = 2 sec u,x2 1u 1Therefore, xdx = 2 sec u tan u du.For x > 2 we take u between 0 and 1 ; for x < 2 we take u between and 3 . For 2 2 such u, tan u > 0 and x 2 4 = 2 tan u.13:29 444. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006420 CHAPTER 8 TECHNIQUES OF INTEGRATION With this substitution xx2x2 42 sec u tan u du 4 sec2 u 2 tan udx = x2 41 du sec u1 4 1 = 4=u 2cos u du 1 = sin u + C = 4 Check the result by differentiation. x2 4 + C. 4xRemark Before rushing into a trigonometric substitution, look at the integral carefully to see whether there is a simpler way to proceed. For instance, you can calculate x dx 2 x2 a by setting x = a sin u, and so on, but you can also carry out the integration by setting u = a 2 x 2 . Try both substitutions and decide which is the more effective. Example 6 Calculate2 xx++2dx. + 2x + 5 By completing the square under the radical, we can write the integral as 5x2x+1dx (x + 1)2 + 4u2.For each real number x, we set x + 1 = 2 tan u,dx = 2 sec2 u du,taking u between 1 and 1 . For such u, sec u > 0 and 2 2 (x + 1)2 + 4 =4 tan2 u + 4 = 2 tan2 u + 1 = 2 sec u.Therefore dx x 2 + 2x + 5=2 sec2 u du 2 sec u=sec u du = ln sec u + tan u + C = ln 1 x 2 + 2x + 5 + 1 (x + 1) + C 2 2 1 2 + 2x + 5 + x + 1 + C. = ln 2 + ln x Absorbing the constant ln 1 in C and noting that the expression within the absolute 2 value signs is positive, we have dx= ln x 2 + 2x + 5 Check the result by differentiation. x 2 + 2x + 5 + x + 1 + C.13:29 445. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.4 INTEGRALS FEATURING a 2 x2 , a 2 + x2 , x2 a 2421Trigonometric substitutions can be effective in cases where the quadratic is not under a radical sign. In particular, the reduction formula1 dx dx = 2n1 (x 2 + a 2 )n a(8.4.3)cos2(n1) u du(a very useful little formula) can be obtained by setting x = a tan u, taking u between 1 and 1 . The derivation of this formula is left to you as an exercise. 2 2EXERCISES 8.4 Exercises 134. Calculate. dx. a2 x 21.2.x 2 1 dx.3.5.x2 4 x24.1/29. 0x2 dx. (1 x 2 )3/28.x13.25 2x212. dx.14.015.(x 2dx . x a2 x 217.3022.123. 09+x2dx . (5 x 2 )3/2 dx.x 2 x 2 a2 dx . x e2x 9 e dx . (x 2 4x + 4)3/2x24. 26. 28. 30.dx.dx.x2 16 x 2dx.a 2 x 2 dx. x2 1 dx. xx2x +2x2x 2 1 + C.1 a2 x 2 d x x (a) by setting u = a 2 x 2 . (b) by a trigonometric substitution. (c) Then reconcile the results. 37. Verify (8.4.3). Exercises 3839. Use (8.4.3) to calculate the integral. 1 1 dx. 39. dx. 38. 2 + 1)2 2 + 1)3 (x (x Exercises 4041. Calculate the integral: (a) by integrating by parts, (b) by applying a trigonometric substitution.36. Calculate 1 x2 dx. x432.arcsec x dx = x arcsec x ln x +dx.24 + x218.dx . x 2 a2 + x 221.29.x3x2 4a20.027.x216.dx.19.25.x2 dx. + 8)3/24 x2dx.x dx. a2 + x 210.0 5x2 4 x2x 4 x 2 dx.11.6.dx.x dx. (1 x 2 )3/27.5/2xx 6x x 2 8 dx.dx. + 4x + 13 x x dx. dx. 34. 33. 2 2x + 3 (x 2 + 2x + 5)2 x 35. Use integration by parts to derive the formula31. 4dx . a2 x 2dx . (x 2 + 2)3/2 dx . 4 + e2x ex dx. 9 e2x dx . 2 2x 3 x x dx. 6x x 240.x arctan x dx.41.x arcsin x dx. 42. Find the area under the curve y = ( x 2 9)/x from x = 3 to x = 5. 43. The region under the curve y = 1/(1 + x 2 ) from x = 0 to x = 1 is revolved about the x-axis. Find the volume of the resulting solid. 44. The shaded part in the gure is called a circular segment. Calculate the area of the segment:exh r45. Show that in a disk of radius r a sector with central angle of radian measure has area A = 1 r 2 . HINT: Assume rst 213:29 446. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006422 CHAPTER 8 TECHNIQUES OF INTEGRATION that 0 < < 1 and subdivide the region as indicated in the 2 gure. Then verify that the formula holds for any sector.51. Sketch , nd the area of , and locate the centroid. 52. Find the volume of the solid generated by revolving about the x-axis and determine the centroid of that solid. 53. Find the volume of the solid generated by revolving about the y-axis and determine the centroid of that solid. 54. Use a trigonometric substitution to derive the formulay1 dx = ln x + 2 + x2 a (r, 0)a 2 + x 2 + C.55. Use a trigonometric substitution to derive the formulax46. Find the area of the region bounded on the left and right by the two branches of the hyperbola (x 2 /a 2 ) (y 2 /b2 ) = 1, and above and below by the lines y = b. 47. Find the area between the right branch of the hyperbola (x 2 /9) (y 2 /16) = 1 and the line x = 5. 48. If the circle (x b)2 + y 2 = a 2 , b > a > 0, is revolved about the y-axis, the resulting doughnut-shaped solid is called a torus. Use the shell method to nd the formula for the volume of the torus. 49. Calculate the mass and the center of mass of a rod that extends from x = 0 to x = a > 0 and has mass density (x) = (x 2 + a 2 )1/2 . 50. Calculate the mass and center of mass of the rod of Exercise 49 given that the rod has mass density (x) = (x 2 + a 2 )3/2 . For Exercises 5153, let be the region under the curve y = x 2 a 2 from x = a to x = 2 a.C56. Set f (x) = (a) (b) (c)C1 x 2 a257. Set (a) (b) (c)dx = ln x +x 2 a 2 + C.x2. Use a CAS 1 x2 to draw the graph of f ; to nd the area between the graph of f and the x-axis from x = 0 to x = 1 ; 2 to nd the volume of the solid generated by revolving about the y-axis the region described in part (b). x2 9 , x 3. Use a CAS f (x) = x2 to draw the graph of f ; to nd the area between the graph of f and the x-axis from x = 3 to x = 6; to locate the centroid of the region described in part (b). 8.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONS In this section we present a method for integrating rational functions. Recall that a rational function is, by denition, the quotient of two polynomials. For example, f (x) =1 , x2 4g(x) =2x 2 + 3 , x(x 1)2h(x) =3x 4 20x 2 + 17 x 3 + 2x 2 7are rational functions, but 1 f (x) = , xg(x) =x2 + 1 , ln xh(x) =sin x xare not rational functions. A rational function R(x) = P(x)/Q(x) is said to be proper if the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, then the rational function is called improper. We will focus our attention on proper rational functions because any improper rational function can be written as a sum of a polynomial and a proper rational function: r (x) P(x) = p(x) + . Q(x) Q(x) These terms are taken from the familiar terms used to describe rational numbers p/q. This is analogous to writing an improper fraction as a so-called mixed number.13:29 447. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONSThis is accomplished simply by dividing the denominator into the numerator. As is shown in algebra, every proper rational function can be written as the sum of partial fractions, fractions of the form(8.5.1)A (x )kand(x 2Bx + C + x + )kwith x 2 + x + irreducible. Such a sum is called a partial fraction decomposition. We begin by calculating some partial fraction decompositions. Later we will use these decompostions to calculate integrals.Example 1 (The denominator splits into distinct linear factors.) For 2x 2x = 2x 2 x (x 2)(x + 1) we write A B 2x = + , x 2 x 2 x +1 with the constants A and B to be determined. Multiplication by (x 2)(x + 1) yields the equation x22x = A(x + 1) + B(x 2)(1)We illustrate two methods for nding A and B. METHOD 1 We substitute numbers for x in (1):Setting x = 2, we get 4 = 3A, which gives A = 4 , 3 Setting x = 1, we get 2 = 3B, which gives B = 2 . 3 The desired decomposition reads 4 2 2x = + . x2 x 2 3(x 2) 3(x + 1) You can verify this by carrying out the addition on the right. METHOD 2 (This method is based on the observation that two polynomials are equaliff their coefcients are equal.) We rewrite (1) as 2x = (A + B)x + (A 2B). Equating coefcients, we have A+B =2 A 2B = 0. We can nd A and B by solving these equations simultaneously. The solutions are again: A = 4, B = 2. 3 3Not factorable into linear expressions with real coefcients. This is the case if 2 4 < 0.42313:29 448. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006424 CHAPTER 8 TECHNIQUES OF INTEGRATION In general, each distinct linear factor x in the denominator gives rise to a term of the form A . x Example 2 (The denominator has a repeated linear factor.) For 2x 2 + 3 , x(x 1)2 we write A B C 2x 2 + 3 = + . + 2 x(x 1) x x 1 (x 1)2 This leads to 2x 2 + 3 = A(x 1)2 + Bx(x 1) + C x. To determine A, B, C, we substitute three values for x. (Method 1.) We select 0 and 1 because for these values of x several terms on the right will drop out. As a third value of x, any other number will do; we select 2 just to keep the arithmetic simple. Setting x = 0, we get 3 = A. Setting x = 1, we get 5 = C. Setting x = 2, we get 11 = A + 2B + 2C, which, with A = 3 and C = 5, gives B = 1. This gives us 2x 2 + 3 1 5 3 + = . 2 x(x 1) x x 1 (x 1)2In general, each factor of the form (x a)k in the denominator gives rise to an expression of the form A1 Ak A2 + + . + 2 x (x ) (x )kExample 3 (The denominator has an irreducible quadratic factor.) For x 2 + 5x + 2 , (x + 1)(x 2 + 1) we write x 2 + 5x + 2 A Bx + C = + 2 (x + 1)(x 2 + 1) x +1 x +1 and obtain x 2 + 5x + 2 = A(x 2 + 1) + (Bx + C)(x + 1) = (A + B)x 2 + (B + C)x + A + C.13:29 449. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONSEquating coefcients (Method 2), we have A+B =1 B +C = 5 A + C = 2. This system of equations is satised by A = 1, B = 2, C = 3. (Check this out.) The decomposition reads x 2 + 5x + 2 1 2x + 3 = + 2 . 2 + 1) (x + 1)(x x +1 x +1 (We could have obtained this result by using Method 1; for example, by setting x = 1, x = 0, x = 1.) In the examples that follow, well use Method 1.Example 4 (The denominator has an irreducible quadratic factor.) For 1 , x(x 2 + x + 1) we write x(x 2A Bx + C 1 = + 2 + x + 1) x x +x +1and obtain 1 = A(x 2 + x + 1) + (Bx + C)x. Again, we select values of x that produce simple arithmetic. 1= A 1 = 3A + B + C 1= A+ B C(x = 0) (x = 1) (x = 1).From this we nd that A = 1,B = 1,C = 1,and therefore x(x 21 1 x +1 = 2 . + x + 1) x x +x +1In general, each irreducible quadratic factor x 2 + x + in the denominator gives rise to a term of the form x2Ax + B . + x + Example 5 (The denominator has a repeated irreducible quadratic factor.) For 3x 4 + x 3 + 20x 2 + 3x + 31 , (x + 1)(x 2 + 4)242513:29 450. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006426 CHAPTER 8 TECHNIQUES OF INTEGRATION we write 3x 4 + x 3 + 20x 2 + 3x + 31 A Bx + C Dx + E = . + 2 + 2 2 + 4)2 (x + 1)(x x +1 x +4 (x + 4)2 This gives 3x 4 + x 3 + 20x 2 + 3x + 31 = A(x 2 + 4)2 + (Bx + C)(x + 1)(x 2 + 4) + (Dx + E)(x + 1). This time we use x = 1, 0, 1, 2, 2: 50 = 25A(x = 1)31 = 16A+ 4C+ E(x = 0)58 = 25A + 10B + 10C + 2D + 2E(x = 1)173 = 64A + 48B + 24C + 6D + 3E(x = 2)145 = 64A + 16B 8C + 2D E(x = 2).With a little patience, you can determine that A = 2,B = 1,C = 0,D = 0,E = 1.This gives the decomposition 3x 4 + x 3 + 20x 2 + 3x + 31 x 1 2 + = . (x + 1)(x 2 + 4)2 x + 1 x 2 + 4 (x 2 + 4)2In general, each multiple irreducible quadratic factor (x 2 + x + )k in the denominator gives rise to an expression of the form A2 x + B2 Ak x + Bk A1 x + B1 + 2 + + 2 . x 2 + x + (x + x + )2 (x + x + )kAs indicated at the beginning of this section, if the rational function is improper, then a polynomial will appear in the decomposition.Example 6 (An improper rational function.) The quotient x5 + 2 x2 1 is improper. Dividing the denominator into the numerator, we nd that x5 + 2 x +2 = x3 + x + 2 . 21 x x 1(Verify this.)The decomposition of the remaining fraction reads A B x +2 = + . x2 1 x +1 x 1 As you can verify, A = 1 , B = 3 . Therefore 2 2 3 1 x5 + 2 = x3 + x + . 21 x 2(x + 1) 2(x 1)13:29 451. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONSWe have been decomposing quotients into partial fractions in order to integrate them. Here we carry out the integrations, leaving some of the details to you.Example 1 x22x dx = x 24 2 + dx 3(x 2) 3(x + 1)=4 3ln x 2 + 2 ln x + 1 + C 3=1 3ln [(x 2)4 (x + 1)2 ] + C.Example 2 2x 2 + 3 dx = x(x 1)23 1 5 + x x 1 (x 1)2= 3 ln x ln x 1 = lndx5 +C x 1x3 5 +C x 1 x 1Example 3 x 2 + 5x + 2 dx = (x + 1)(x 2 + 1)1 2x + 3 + 2 x +1 x +1 1 dx + x +1=dx2x + 3 dx. x2 + 1Since 1 dx = ln x + 1 + C1 x +1and 2x + 3 dx = x2 + 12x dx + 3 +1x2x21 dx = ln (x 2 + 1) + 3 arctan x + C2 , +1we have x 2 + 5x + 2 dx = ln x + 1 + ln (x 2 + 1) + 3 arctan x + C (x + 1)(x 2 + 1) = lnx2 + 1 + 3 arctan x + C. x 1Example 4 I =x(x 2dx = + x + 1)1 x +1 2 dx = ln x x x +x +1x2x +1 dx. +x +1To calculate the remaining integral, note that (d/d x)(x 2 + x + 1) = 2x + 1. We can manipulate the integrand to get a term of the form du/u with u = x 2 + x + 1: 1 [2x + 1] + 1 1 x +1 2 = 2 2 = 2+x +1 x x +x +1 22x + 1 1 + 2 . +x +1 x +x +1x242713:29 452. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006428 CHAPTER 8 TECHNIQUES OF INTEGRATION This gives us x22x + 1 1 dx + +x +1 21 x +1 dx = +x +1 2x2x21 d x. +x +1The rst integral is a logarithm: 1 22x + 1 1 d x = ln (x 2 + x + 1) + C1 . +x +1 2x2(x 2 + x + 1 > 0 for all x)The second integral is an arc tangent: 1 21 1 dx = x2 + x + 1 2dx x+1 2 2+(3/2)21 2 1 = arctan x + 2 3 3+ C2 .Combining the results, we have the integral we want: I = ln x 1 1 2 1 ln (x 2 + x + 1) arctan x + 2 2 3 3+ C.Example 5 3x 4 + x 3 + 20x 2 + 3x + 31 dx = (x + 1)(x 2 + 4)2x 1 2 + 2 2 d x. x + 1 x + 4 (x + 4)2The rst two fractions are easy to integrate: 2 d x = 2 ln x + 1 + C1 , x +1 2x x dx = 1 d x = 1 ln (x 2 + 4) + C2 . 2 2 2+4 2+4 x x The integral of the last fraction is of the form (x 22xAs you saw in the preceding section, such integrals can be calculated by setting x = a tan u, u ( 1 , 1 ). [(8.4.3).] In this case 2 2+4xudx . + a 2 )n1 dx cos2 u du = 2 + 4) 8 x = 2 tan u 1 = (1 + cos 2u) du 16 half-angle formula 1 1 = u+ sin 2u + C3 16 32 1 1 u+ sin u cos u + C3 = 16 16 sin 2u = 2 sin u cos u 1 x 1 x = arctan + 2+4 16 2 16 x (x 22=x 1 1 arctan + 16 2 8x2x +4+ C3 .2 x2+4+ C313:29 453. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.5 RATIONAL FUNCTIONS; PARTIAL FRACTIONS429The integral we want is equal to 2 ln x + 1 +1 1 ln (x 2 + 4) 2 8x2x +41 x arctan + C. 16 2Example 6 x5 + 2 dx = x2 1x3 + x 3 1 + dx 2(x + 1) 2(x 1)= 1 x 4 + 1 x 2 1 ln x + 1 + 3 ln x 1 + C. 4 2 2 2 EXERCISES 8.5 Exercises 18. Decompose into partial fractions. 1 . + 7x + 6 x . 3. 4 x 1 x 2 3x 1 . 5. 3 x + x 2 2x 2x 2 + 1 7. 3 . x 6x 2 + 11x 6 1.x2.27.2(x 1)(x 2 + 4x + 5) x4 . 4. (x 1)3 x3 + x2 + x + 2 6. . x 4 + 3x 2 + 2 1 . 8. 2 + 1)2 x(xx2.12. 13. 15. 17.2x 4 4x 3 + 4x 2 + 3 dx. x3 x2 x2 + 1 dx. x(x 2 1) x5 dx. (x 2)2 x +3 dx. 3x + 2 dx . (x 1)3 x2x5 dx. x 214.x2 + 3 dx. 3x + 2 dx . x 2 + 2x + 216.x218.19.x2 dx. (x 1)2 (x + 1)20.2x 1 dx. (x + 1)2 (x 2)221.dx . 4 16 x22.3x 5 3x 2 + x dx. x3 123. 25.x 3 + 4x 2 4x 1 dx. 24. (x 2 + 1)2 dx . 26. x4 + 4HINT: With a > 0, x 4 + a 2 = (x 2 +30.Exercises 3134. Evaluate. 2 x dx. 31. 2 + 5x + 6 0 x 3 x 2 4x + 3 33. dx. 3 2 1 x + 2x + x332. 1 234.2ax + a)(x 2 01 dx. +x x3 dx. (x 2 + 2)2 x3et dt. e2t + 5et + 6 sec2 d . 3 tan2 tanExercises 3945. Derive the formula. 1 u du = 2 (a + bu a ln a + bu ) + C. 39. a + bu b du 1 u 40. = ln + C. u(a + bu) a a + bu du a + bu 1 b 41. = + 2 ln + C. 2 (a + bu) u au a u du 1 a + bu 1 42. = ln + + C. u(a + bu)2 a(a + bu) a 2 u du 1 a+u 43. = ln + C. a2 u2 2a au u du = 1 ln a 2 u 2 + C. 44. 2 a2 u2 2 a a+u u du = u + ln + C. 45. a2 u2 2 au Exercises 4647. Calculatedx . (x 2 + 16)2 dx . x 4 + 16 1 dx. (x 1)(x 2 + 1)2 x3 + x2 + x + 3 dx. (x 2 + 1)(x 2 + 3)28.Exercises 3538. Calculate. cos d. 36. 35. sin2 2 sin 8 1 dt. 38. 37. t([ln t]2 4)Exercises 930. Calculate. 7 dx. 9. (x 2)(x + 5) x 10. dx. (x + 1)(x + 2)(x + 3) 11.29.x 3 dx. x3 + x2 x +1 dx. 3 + x 2 6x xdu (a + bu)(c + du)2ax + a).with the coefcients as stipulated. 46. a, b, c, d all different from 0, 47. a, b, c, d all different from 0,ad = bc. ad = bc.13:29 454. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006430 CHAPTER 8 TECHNIQUES OF INTEGRATION 48. Show that for y =x21 , 155.dn y (1)n n! 1 1 . n = n+1 dx 2 (x 1) (x + 1)n+1 49. Find the volume of the solid generated by revolving the re gion between the curve y = 1/ 4 x 2 and the x-axis from x = 0 to x = 3/2: (a) about the x-axis; (b) about the y-axis. 50. CalculateCC(b) CCx 3 + 20x 2 + 4x + 93 . (x 2 + 4)(x 2 9)x 2 + 7x + 12 . x(x 2 + 2x + 4) Exercises 5455. Use a CAS to decompose the integrand into partial fractions. Use the decomposition to evaluate the integral. 2x 6 13x 5 + 23x 4 15x 3 + 40x 2 24x + 9 54. dx. x 5 6x 4 + 9x 2Verify your results by differentiation. 57. Set x f (x) = 2 . x + 5x + 6 (a) Use a graphing utility to draw the graph of f. (b) Calculate the area of the region that lies between the graph of f and the x-axis from x = 0 to x = 4. 58. (a) The region of Exercise 57 is revolved about the y-axis. Find the volume of the solid generated. (b) Find the centroid of the solid described in part (a). 59. Set 9x f (x) = . (x + 3)2 (a) Use a graphing utility to draw the graph of f. (b) Find the area of the region that lies between the graph of f and the x-axis from x = 2 to x = 9. 60. (a) The region of Exercise 59 is revolved about the x-axis. Find the volume of the solid generated. (b) Find the centroid of the solid described in part (a).(c)C56. Use a CAS to calculate the integrals 1 dx, n = 0, 1, 2. x 2 + 2x + nx 3 arctan x dx. 51. Find the centroid of the region under the curve y = (x 2 + 1)1 from x = 0 to x = 1. 52. Find the centroid of the solid generated by revolving the region of Exercise 51 about: (a) the x-axis; (b) the y-axis. 53. Use a CAS to decompose into partial fractions. 6x 4 + 11x 3 2x 2 5x 2 (a) . x 2 (x + 1)3x 8 + 2x 7 + 7x 6 + 23x 5 + 10x 4 + 95x 3 19x 2 + 133x 52 dx. x 6 + 2x 5 + 5x 4 16x 3 + 8x 2 + 32x 48 *8.6 SOME RATIONALIZING SUBSTITUTIONS There are integrands which are not rational functions but can be transformed into rational functions by a suitable substitution. Such substitutions are known as rationalizing substitutions. First we consider integrals in which the integrand contains an expression of the form n f (x). In such cases, setting u = n f (x), which is equivalent to setting u n = f (x), is sometimes effective. The idea is to replace fractional exponents by integer exponents; integer exponents are usually easier to work with.Example 1 Finddx . 1+ xSOLUTION To rationalize the integrand, we settaking u 0. Then u =dx = 1+ xu 2 = x,2u du = dx,x and 2 2u du = 2 du 1+u 1+u divide = 2u 2 ln (1 + u) + C 1+u >0 = 2 x 2 ln (1 + x) + C.13:29 455. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006*8.6 SOME RATIONALIZING SUBSTITUTIONSExample 2 Find 3dx . x+ x 3SOLUTION Here the integrand contains two distinct roots,x andx. We caneliminate both radicals by setting u 6 = x, 6u 5 du = dx, taking u > 0. This substitution gives 3 x = u 2 and x = u 3 . Therefore 3dx = x+ x6u 5 du = 6 u2 + u3 =6 divide u3 du 1+u u2 u + 1 1 1+udu= 6[ 1 u 3 1 u 2 + u ln (1 + u) + C] 3 2 1 + u > 0 = 2 x 3 3 x + 6 6 x 6 ln (1 + 6 x) + C.Example 3 Find 1 e x dx.SOLUTION To rationalize the integrand, we setu= 1 ex .Then 0 u < 1. To express dx in terms of u and du, we solve the equation for x: u 2 = 1 ex ,1 u 2 = ex ,The rest is straightforward: 1 e x dx =u ln(1 u 2 ) = x,2u 1 u22u 2 du = u2 1 divide; then use =2u du = dx. 1 u2du 2+1 1 du u1 u+1partial fractions= 2u + ln u 1 ln u + 1 + C u1 +C u+1 1 ex 1 + C. = 2 1 e x + ln 1 ex + 1 = 2u + lnNow we consider rational expressions in sin x and cos x. Suppose, for example, that we want to calculate 1 dx. 3 sin x 4 cos x We can convert the integrand into a rational function of u by setting u tan 1 x 2taking x between and .43113:29 456. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006432 CHAPTER 8 TECHNIQUES OF INTEGRATION This gives cos 1 x = 21 = sec 1 x 21 = 1 + u2 1 + tan2 1 x 2 1and u . sin 1 x = cos 1 x tan 1 x = 2 2 2 1 + u221+u1 x 2uThe right triangle illustrates these relations for x (0, ). Note that1sin x = 2 sin 1 x cos 1 x = 2 22u 1 + u2and cos x = cos2 1 x sin2 1 x = 2 21 u2 . 1 + u2Since u = tan 1 x with 1 x ( 1 , 1 ), 1 x = arctan u and x = 2 arctan u. 2 2 2 2 2 Therefore 2 dx = du. 1 + u2 In summary, if the integrand is a rational expression in sin x and cos x, then the substitution u = tan 1 x, x (, ), gives 2 sin x =2u , 1 + u2cos x =1 u2 , 1 + u2dx =2 du 1 + u2and converts the integrand into a rational function of u. The resulting integral can then be calculated by the methods of Section 8.5.Example 4 Find1 dx. 3 sin x 4 cos xSOLUTION Set u = tan 1 x with x (, ). Then 21 1 1 + u2 = = 2 2 )] [4(1 u 2 )/(1 + u 2 )] 3 sin x 4 cos x [6u/(1 + u 4u + 6u 4 and 1 dx = 3 sin x 4 cos x1 + u2 2 du = 4u 2 + 6u 4 1 + u 21 du. 2u 2 + 3u 2Since 2u 21 1/5 2/5 1 = = + , + 3u 2 (u + 2)(2u 1) u + 2 2u 12u 21 1 du = + 3u 2 5(partial fractions)we have 1 2 du + u+2 51 du 2u 1= 1 ln u + 2 + 1 ln 2u 1 + C 5 5 = 1 ln tan 1 x + 2 + 1 ln 2 tan 1 x 1 + C. 5 2 5 213:29 457. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.7 NUMERICAL INTEGRATION433EXERCISES *8.63.dx . 1 x 1 + e x dx.5. x 1 + x dx.1.(x + 2) x 1 dx.7.4. 6.13. 15. 17. 19.20.x dx. (ax + b)3/2 1 dx. 23. 1 + cos x sin x 1 25. dx. 2 + sin x 1 27. dx. sin x + tan x 1 cos x 29. dx. 1 + sin x Exercises 3136. Evaluate. 4 x 3/2 dx. 31. 0 x +1 /2 0sin 2x dx. 2 + cos x35.sec x dx =sec x dx =x(1 + x)1/3 dx.10.x dx. x +1 1 ex dx. 1 + ex dx . 1 + ex x +1 dx. x x 2 x 2 x 1 dx.16. 18.x dx. ax + b 1 dx. 2 + cos x sin x dx. 1 + sin2 x 1 dx. 1 + sin x + cos x 1 dx. 5 + 3 sin x22. 24. 26. 28. 30.832. 01 dx. 1+ 3 x/234. 01 dx. 1 + sin xcos x dx = cos2 xsec x dx = lncos x dx. 1 sin2 x1 + sin x + C. 1 sin x(b) Show that the result in part (a) is equivalent to the familiar formula sec x dx = ln sec x + tan x + C. 39. (a) Use the approach given in Exercise 38 (a) to show that csc x dx = ln1 cos x + C. 1 + cos x(b) Show that the result in part (a) is equivalent to the formula csc x dx = ln csc x cot x + C. 40. The integral of a rational function of sinh x and cosh x can be transformed into a rational function of u by means of the substitution u = tanh 1 x. Show that this substitution gives 2 2u 1 + u2 2 , cosh x = , dx = du. 1 u2 1 u2 1 u2 Exercises 4144. Integrate by setting u = tanh 1 x. 2 1 41. sech x dx. 42. dx. 1 + cosh x 1 ex 1 dx. dx. 44. 43. sinh x + cosh x 1 + ex sinh x = 8.7 NUMERICAL INTEGRATION To evaluate a denite integral of a continuous function by the formula bsec x dx is to writeUse the method of this section to show that12.1 + tan 1 x 1 2 dx = ln + C. cos x 1 tan 1 x 238. (a) Another way to calculate(x 1) x + 2 dx.14.21.33.8.2x 2 (4x + 1)5/2 dx.11.x dx. 1+x dx . x(x 1/3 1) x 2 1 + x dx.2.x3 dx. (1 + x 2 )3 x dx. x 1 x 1+1 dx. x 11 dx . 1 + ex x dx. x +49. 1 1 x dx. 36. dx. sin x cos x 1 x 0 0 1+ 37. Use the method of this section to show that /3Exercises 130. Calculate.f (x) dx = F(b) F(a),awe must be able to nd an antiderivative F and we must to able to evaluate this antiderivative both at a and at b. When this is not feasible, the method fails.13:29 458. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006434 CHAPTER 8 TECHNIQUES OF INTEGRATION The method fails even for such simple-looking integrals as 11x sin x dxor0ex dx. 20There are no elementary functions with derivative x sin x or ex . Here we take up some simple numerical methods for estimating denite integrals methods that you can use whether or not you can nd an antiderivative. All the methods we describe involve only simple arithmetic and are ideally suited to the computer. We focus now on 2bf (x) dx. aWe suppose that f is continuous on [a, b] and, for pictorial convenience, assume that f is positive. Take a regular partition P = {x0 , x1 , x2 , . . . , xn1 , xn } of [a, b], subdividing the interval into n subintervals each of length (b a)/n: [a, b] = [x0 , x1 ] [xi1 , xi ] [xn1 , xn ] with ba . n pictured in Figure 8.7.1, can be approximated in many ways. xi =The regionif f ixi 1xixba nFigure 8.7.1fxixi 1xFigure 8.7.2(1) By the left-endpoint rectangle (Figure 8.7.2): xixi 1area f (xi1 ) xi = f (xi1 ) =xFigure 8.7.3ba n.(2) By the right-endpoint rectangle (Figure 8.7.3): area f (xi ) xi = f (xi ) =fba n.(3) By the midpoint rectangle (Figure 8.7.4): area f = xi 1xi 1 + xi 2Figure 8.7.4xixxi1 + xi 2xi = fxi1 + xi 2ba n.(4) By a trapezoid (Figure 8.7.5): area =1 1 ba [ f (xi1 ) + f (xi )] xi = [ f (xi1 ) + f (xi )] 2 2 n.13:29 459. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006(5) By a parabolic region (Figure 8.7.6): take the parabola y = Ax 2 + Bx + C that passes through the three points indicated. area = =1 6xi1 + xi 2f (xi1 ) + 4 fxi1 + xi 2f (xi1 ) + 4 f+ f (xi )4358.7 NUMERICAL INTEGRATIONfxi ba 6n+ f (xi ).You can verify this formula for the area under the parabola by doing Exercises 11 and 12. (If the three points are collinear, the parabola degenerates to a straight line and the parabolic region becomes a trapezoid. The formula then gives the area of that trapezoid.) The approximations to i just considered yield the following estimates forxixi 1xFigure 8.7.5fbf (x) dx. a(1) The left-endpoint estimate: ba [ f (x0 ) + f (x1 ) + + f (xn1 )]. n (2) The right-endpoint estimate:xi 1Ln =Rn =2Figure 8.7.6ba [ f (x1 ) + f (x2 ) + + f (xn )]. n(3) The midpoint estimate: Mn =ba nfx0 + x1 2+ + fxn1 + xn 2.(4) The trapezoidal estimate (trapezoidal rule): Tn =ba nf (x0 ) + f (x1 ) f (x1 ) + f (x2 ) f (xn1 ) + f (xn ) + + + 2 2 2ba [ f (x0 ) + 2 f (x1 ) + + 2 f (xn1 ) + f (xn )]. 2n (5) The parabolic estimate (Simpsons rule): =Sn =ba 6nf (x0 ) + f (xn ) + 2[ f (x1 ) + + f (xn1 )] + 4 fx0 + x1 2+ + fxn1 + xn 2.The rst three estimates, L n , Rn , Mn , are Riemann sums (Section 5.2); Tn and Sn , although not explicitly written as Riemann sums, can be written as Riemann sums. (Exercise 26.) It follows from (5.2.6) that any one of these estimates can be used to approximate the integral as closely as we may wish. All we have to do is take n sufciently large. As an example, we will nd the approximate value of 2ln 2 = 1dx xby applying each of the ve estimates. Here f (x) =1 , x[a, b] = [1, 2].xi 1 + xixix13:29 460. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006436 CHAPTER 8 TECHNIQUES OF INTEGRATION Taking n = 5, we have 21 1 ba = = . n 5 5 The partition points are x0 = 5 , 5x1 = 6 , 5x2 = 7 , 5x3 = 8 , 5x4 = 9 , 5x5 =(Figure 8.7.7)10 . 5y f (x) =1 x6 517 58 59 52xFigure 8.7.7Using a calculator and rounding off to four decimal places, we have the following estimates: L 5 = 1 5 + 5 + 5 + 5 + 5 = 1 + 1 + 1 + 1 + 1 = 0.7456. 5 56R5 =1 5 5 6+M5 =1 10 5 11+T5 =1 5 10 5S5 =1 5 30 5795 8+5 910 13+10 15+10 17+10 6+10 7+10 8+5 10+25 7+85 6++5+=1 6+10 19=2+10 9+5 10+5 8+5 95 105 7671 7++1 8+1 1181 139 0.6456. =1 9+1 10+1 15+1 17+1 7+1 8+1 9++10 17+10 19=1 10+1 6+410 11+10 13+ +10 15 0.6919. =1 191 20 0.6956. = 0.6932. =Since the integrand 1/x decreases throughout the interval [1, 2], you can expect the left-endpoint estimate, 0.7456, to be too large, and you can expect the right-endpoint estimate, 0.6456, to be too small. The other estimates should be better. The value of ln 2 given on a calculator is ln 2 0.69314718, which is 0.6931 = rounded off to four decimal places. Thus S5 is correct to the nearest thousandth.y53Example 1 Find the approximate value of 4Take n = 6.3SOLUTION Each subinterval has length 2are x0 = 0,11 213 225 2Figure 8.7.83x4 + x 3 dx by the trapezoidal rule.0x1 = 1 , 2x2 = 1,x3 = 3 , 2ba 30 1 = = . The partition points n 6 2 x4 = 2,x5 = 5 , 2x6 = 3. (Figure 8.7.8)Then T6 1 [ f (0) + 2 f =41 2+ 2 f (1) + 2 f3 2+ 2 f (2) + 2 f5 2+ f (3)],13:29 461. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.7 NUMERICAL INTEGRATIONwith f (x) = have 4 + x 3 . Using a calculator and rounding off to three decimal places, wef (0) = 2.000,f1 2f (1) 2.236, = 2.031, =f (2) 3.464, =f5 2 4.430, =f3 2 2.716, =f (3) 5.568. =Thus T6 1 (2.000 + 4.062 + 4.472 + 5.432 + 6.928 + 8.860 + 5.568) 9.331. = =4Example 2 Find the approximate value of 34 + x 3 dx0by Simpsons rule. Take n = 3. SOLUTION There are three subintervals, each of lengthba 30 = = 1. n 3 Here x1 = 1, x2 = 2, x3 = 3, x0 = 0, 1 x1 + x2 3 x2 + x3 5 x0 + x1 = , = , = . 2 2 2 2 2 2 Simpsons rule yields S3 = 1 f (0) + f (3) + 2 f (1) + 2 f (2) + 4 f 1 + 4 f 3 + 4 f 5 , 6 2 2 2 3 . Taking the values of f as calculated in Example 1, we have with f (x) = 4 + x S3 = 1 (2.000 + 5.568 + 4.472 + 6.928 + 8.124 + 10.864 + 17.72) 9.279. = 6For comparison, the value of this integral accurate to ve decimal places is 9.27972. Error Estimates A numerical estimate is useful only to the extent that we can gauge its accuracy. When we use any kind of approximation method, we face two forms of error: the error inherent in the method we use (we call this the theoretical error) and the error that accumulates from rounding off the decimals that arise during the course of computation (we call this the round-off error). The effect of round-off error is obvious: if at each step we round off too crudely, then we can hardly expect an accurate nal result. We will examine theoretical error. We begin with a function f continuous and increasing on [a, b]. We subdivide [a, b] into n nonoverlapping intervals, each of length (b a)/n. We want to estimate bf (x) dx aby the left-endpoint method. What is the theoretical error? It should be clear from Figure 8.7.9 that the theoretical error does not exceed [ f (b) f (a)]ba n.43713:29 462. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006438 CHAPTER 8 TECHNIQUES OF INTEGRATION y f (b)f (b) f (a)f (a) ba nab each width isba nx; here n = 6Figure 8.7.9The error is represented by the sum of the areas of the shaded regions. These regions, when shifted to the right, all t together within a rectangle of height f (b) f (a) and base (b a)/n. Similar reasoning shows that, under the same circumstances, the theoretical error associated with the trapezoidal method does not exceed 1 [ 2()f (b) f (a)]ba n.In this setting, at least, the trapezoidal estimate does a better job than the left-endpoint estimate. The trapezoidal rule is more accurate than () suggests. As is shown in texts on numerical analysis, if f is continuous on [a, b] and twice differentiable on (a, b), then the theoretical error of the trapezoidal rule, T En =bf (x) dx Tn ,acan be written(8.7.1)T En = (b a)3 f (c), 12 n 2where c is some number between a and b. Usually we cannot pinpoint c any further. However, if f is bounded on [a, b], say f (x) M for a x b, then(8.7.2)T E n (b a)3 M. 12 n 2Recall the trapezoidal-rule estimate of ln 2 derived at the beginning of this section: 2ln 2 = 11 dx 0.696. = x13:29 463. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.7 NUMERICAL INTEGRATIONTo nd the theoretical error, we apply (8.72). Here 1 1 2 f (x) = 3 . , f (x) = 2 , x x x Since f is a decreasing function, it takes on its maximum value at the left endpoint of the interval. Thus f (x) f (1) = 2 on [1, 2]. Therefore, with a = 1, b = 2, and n = 5, we have f (x) =(2 1)3 1 2= < 0.007. 12 52 150 The estimate 0.696 is in theoretical error by less than 0.007. To get an estimate for T E 5 2ln 2 = 11 dx xthat is accurate to four decimal places, we need (1)(b a)3 M < 0.00005. 12n 2Since (2 1)3 1 (b a)3 M 2= , 2 12 n 12 n 2 6 n2 we can satisfy (1) by having 1 < 0.00005, 6n 2 which is to say, by having n 2 > 3333. As you can check, n = 58 is the smallest integer that satises this inequality. In this case, the trapezoidal rule requires a regular partition with at least 58 subintervals to guarantee four decimal place accuracy. Simpsons rule is more effective than the trapezoidal rule. If f is continuous on [a, b] and f (4) exists on (a, b), then the theoretical error for Simpsons rule, S En =bf (x) dx Sn ,acan be written(8.7.3)S En = (b a)5 (4) f (c), 2880n 4where, as before, c is some number between a and b. Whereas (8.7.1) varies as 1/n 2 , this quantity varies as 1/n 4 . Thus, for comparable n, we can expect greater accuracy from Simpsons rule. In addition, if we assume that f (4) (x) is bounded on [a, b], say f (4) (x) M for a x b, then(8.7.4)S E n (b a)5 M. 2880n 443913:29 464. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006440 CHAPTER 8 TECHNIQUES OF INTEGRATION To estimate 2ln 2 =1 dx x1by the trapezoidal rule with theoretical error less than 0.00005, we needed to subdivide the interval [1, 2] into at least fty-eight subintervals of equal length. To achieve the same degree of accuracy with Simpsons rule, we need only four subintervals: for f (x) = 1/x,f (4) (x) = 24/x 4 .Therefore f (4) (x) 24 for all x [1, 2] and S E n (2 1)5 1 24 = . 4 2880 n 120 n 4This quantity is less than 0.00005 provided only that n 4 > 167. This condition is met by n = 4. Remark If your calculus course is computer related, you will no doubt see these and other numerical methods more thoroughly applied. EXERCISES 8.7 CIn Exercises 110 round off your calculations to four decimal places.5. Estimate the value of by estimating the integral 11. Estimate 0 dx = arctan1 = 2 1+x 412 2x dx 0by: (a) the left-cndpoint estimate, n = 12; (b) the rightendpoint estimate, n = 12; (c) the midpoint estimate, n = 6; (d) the trapezoidal rule, n = 12; (e) Simpsons rule, n = 6. Check your results by performing the integration. 2. Estimate 1sin2 x dx0by: (a) the midpoint estimate, n = 3; (b) the trapezoidal rule, n = 6; (c) Simpsons rule, n = 3. Check your results by performing the integration. 3. Estimate 3 0dx 1 + x3by: (a) the left-endpoint estimate, n = 6; (b) the rightendpoint estimate, n = 6; (c) the midpoint estimate, n = 3; (d) the trapezoidal rule, n = 6; (e) Simpsons rule, n = 3. 4. Estimate 0sin x dx +xby: (a) the trapezoidal rule, n = 6; (b) Simpsons rule, n = 3. (Note the superiority of Simpsons rule.)by (a) the trapezoidal rule, n = 6; (b) Simpsons rule, n = 3. 6. Estimate 20dx 4 + x3by: (a) the trapezoidal rule, n = 4; (b) Simpsons rule, n = 2. 7. Estimate 1cos x 2 dx 1by: (a) the midpoint estimate, n = 4; (b) the trapezoidal rule, n = 8; (c) Simpsons rule, n = 4. 8. Estimate 2 1ex dx xby: (a) the midpoint estimate, n = 4; (b) the trapezoidal rule, n = 8; (c) Simpsons rule, n = 4. 9. Estimate 2ex dx 20by: (a) the trapezoidal rule, n = 10; (b) Simpsons rule, n = 5.13:29 465. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20068.7 NUMERICAL INTEGRATION10. Estimate 4 2g(x) dx =a441is replaced by equality if M is taken as the maximum value of f on [a, b]. (b) Set f (x) = x 4 . Let [a, b] = [0, 1] and take n = 1. Show that in this case the theoretical error inequality1 dx ln xby: (a) the midpoint estimate, n = 4; (b) the trapezoidal rule, n = 8; (c) Simpsons rule, n = 4. 11. Show that there is a unique parabola of the form y = Ax 2 + Bx + C through three distinct noncollinear points with different x-coordinates. 12. Show that the function g(x) = Ax 2 + Bx + C satises the condition ba+b 2ba g(a) + 4g 6(b a)5 M 2880n 4 is replaced by equality if M is taken as the maximum value of f (4) on [a, b]. 26. Show that, if f is continuous, then Tn and Sn can both be written as Riemann sums. 27. Let f be a function positive on [a, b]. Compare S E n + g(b)Mn ,bTn ,f (x) dx aCfor every interval [a, b]. Exercises 1322. Determine the values of n which guarantee a theoretical error less than if the integral is estimated by: (a) the trapezoidal rule; (b) Simpsons rule. 4 3 13. x dx; = 0.01. 14. x 5 dx; = 0.01. 115.x 5 dx; 1= 0.001. 18.sin x dx; 003= 0.00001. 16.x dx;17.= 0.00001.3e x dx;= 0.01.= 0.001.cos x dx; e20.1(b)= 0.01.ln x dx;Cex dx; 2= 0.0001. 22.0= 0.00001.e x dx; 023. Show that Simpsons rule is exact (theoretical error zero) for every polynomial of degree 3 or less. 24. Show that the trapezoidal rule is exact (theoretical error zero) if f is linear. 25. (a) Set f (x) = x 2 . Let [a, b] = [0, 1] and take n = 2. Show that in this case the theoretical error inequality T E n n = 50.(x 5 5x 4 + x 3 3x 2 x + 4) dx,/6(b)2(x + cos x) dx,n = 30.30. Use a CAS and Simpsons rule to estimate: 3 x2 (a) dx, n = 50. 2 4 x + 41 274019. 21.10(a)1 41Cgiven that the graph of f is: (a) concave up; (b) concave down. 28. Show that 1 Tn + 2 Mn = Sn . 3 3 29. Use a CAS and the trapezoidal rule to estimate:(x + tan x) dx,n = 25.031. Estimate the theoretical error if Simpsons rule with n = 20 is used to approximate 5 1x2 4 dx. x2 + 932. Estimate the theoretical error if the trapezoidal rule with n = 30 is used to approximate 7(b a)3 M 12n 22x2 dx. +1x2 CHAPTER 8. REVIEW EXERCISES 3Exercises 140. Calculate. 1.cos x dx. 4 + sin2 x/42. 04.9.2x dx. 1 + x2(tan x + cot x) dx.2x sinh x dx.5.x 3 dx. x 2 (x + 1)6.xarctan x dx.7.sin 2x cos x dx.8.3x e3x dx.x + 1 dx.10.2 dx. x(1 + x 2 )12.cos x dx. sin3 x023.lnsin3 x dx. cos x11.113.ex cosh x dx14.dx . e x 4ex16.015.x2 + 3 dx x2 + 9 1 dx, x3 113:29 466. P1: PBU/OVYP2: PBU/OVYJWDD023-08JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006442 CHAPTER 8 TECHNIQUES OF INTEGRATION x2x dx. a2 x 2 dx. x217. 19.222.sin5 x dx. cos7 x x 4 + x2 x 4 + x224./3x m (ln x)n dx =30x e dx.23.43. (a) Show thatx 2 e x dx.20.3 x221. ln(x x) dx.18.sin 2x sin 3x dx.x +3sin x dx. sin 2x26.27.x2 + x dx. 1 x228.x tan2 2x dx.29.cos4 x dx. sin2 x30.31.(sin 2x + cos 2x)2 dx.32.33.5x + 1 dx. (x + 2)(x 2 2x + 1)34.tan3/2 x sec4 x dx.35.25. /6x23x ln+ 2x 8dx.x 2 + 1 dx.0 cos x sin3 x dx.1/21 dx. x +1 x(b) Calculate dx.36. 0cos x cos 1 x dx. 2x m+1 (ln x)n n m+1 m+1 x m (ln x)n1 dx.x 4 (ln x)3 dx.44. Calculate the area of the region between the curve y = x 2 arctan x and the x-axis from x = 0 to x = 1. 45. Find the centroid of the region bounded by the graph of y = (1 x 2 )1/2 and the x-axis, x 0, 1 . 2 46. Let f (x) = x + sin x and g(x) = x both for x [0, ]. (a) Sketch the graphs of f and g in the same coordinate system. (b) Calculate the area of the region between the graphs of f and g. (c) Calculate the centroid of . 47. (a) Find the volume of the solid generated by revolving about the x-axis the region of Exercise 45. (b) Find the volume of the solid generated by revolving about the y-axis the region of Exercise 45. 48. The region between the curve y = ln 2x and the x-axis from x = 1 to x = e is revolved about the y-axis. Find the volume of the solid generated. 249. Estimatex 3 + x dx by: (a) the midpoint estimate,037.x 2 cos 2x dx.39.1 sin 2x dx. 1 + sin 2x40.5x + 3 dx. (x 1)(x 2 + 2x + 5)38.e2x sin 4x dx.041. Show that, for a = 0, (a)x n cos ax dx =n = 4; (b) the trapezoidal rule, n = 8; (c) Simpsons rule, n = 4. Round off your calculations to four decimal places. 2 50. Estimate 1 + 3x dx by: (a) the trapezoidal rule, n = 8;n x n sin ax a ax n1 sin ax dx,x n cos ax n + x n1 cos ax dx. a a 42. Use the formulas in Exercise 41 to calculate (b)x n sin ax dx = (a)x 2 cos 3x dx.(b)x 3 sin 4x dx.(b) Simpsons rule, n = 4. Round off your calculations to four decimal places. (c) Find the exact value of the integral and compare it to your results in (a) and (b). 51. For the integral of Exercise 50, determine the values of n which guarantee a theoretical error of less than 0.00001 if the integral is estimated by: (a) the trapezoidal rule; (b) Simpsons rule. 3 x e dx by: (a) the trapezoidal rule, n = 8; 52. Estimate x 1 (c) Simpsons rule, n = 4. Round off your calculations to four decimal places.13:29 467. P1: PBU/OVYP2: PBU/OVYJWDD023-09JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006CHAPTER9SOME DIFFERENTIAL EQUATIONSIntroduction An equation that relates an unknown function to one or more of its derivatives is called a differential equation. We have already introduced some differential equations. In Chapter 7 we used the differential equation (1)dy = ky dt[there written f (t) = k f (t)]to model exponential growth and decay. In various exercises (Section 3.6 and 4.9) we used the differential equation (2)d2 y + 2 y = 0, dt 2the equation of simple harmonic motion, to model the motion of a simple pendulum and the oscillation of a weight suspended at the end of a spring. The order of a differential equation is the order of the highest derivative that appears in the equation. Thus (1) is a rst-order equation and (2) is a second-order equation. A function that satises a differential equation is called a solution of the equation. Finding the solutions of a differential equation is called solving the equation. All functions y = Cekt where C is a constant are solutions of equation (1): dy = kCekt = ky. dt All functions of the form y = C1 cos t + C2 sin t, where C1 and C2 constants are solutions of equation (2): y = C1 cos t + C2 sin t dy = C1 sin t + C2 cos t dt d2 y = 2 C1 cos t 2 C2 sin t = 2 y dt 244313:30 468. P1: PBU/OVYP2: PBU/OVYJWDD023-09JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006444 CHAPTER 9 SOME DIFFERENTIAL EQUATIONS and therefore d2 y + 2 y = 0. dt 2 Remark Differential equations reach far beyond the boundaries of pure mathematics. Countless processes in the physical sciences, in the life sciences, in engineering, and in the social sciences are modeled by differential equations. The study of differential equations is a huge subject, certainly beyond the scope of this text or any text on calculus. In this little chapter we examine some simple, but useful, differential equations. We continue the study of differential equations in Chapter 19. One more point. In (1) and (2) we used the letter t to indicate the independent variable because were looking at changes with respect to time. In much of what follows, well use the letter x. Whether we use x or t doesnt matter. What matters is the structure of the equation 9.1 FIRST-ORDER LINEAR EQUATIONS A differential equation of the form(9.1.1)y + p(x)y = q(x)is called a rst-order linear differential equation. Here p and q are given functions dened and continuous on some interval I. [In the simplest case, p(x) = 0 for all x, the equation reduces to y = q(x). The solutions of this equation are the antiderivatives of q.] Solving Equationsy + p(x)y = q(x) First we calculate H (x) =p(x) dx,omitting the constant of integration. (We want one antiderivative for p, not a whole collection of them.) We form e H (x) , multiply the equation by e H (x) , and obtain e H (x) y + e H (x) p(x)y = e H (x) q(x). The left side of this equation is the derivative of e H (x) y. (Verify this.) Thus, we have d H (x) y = e H (x) q(x). e dx Integration gives e H (x) y =e H (x) q(x) dx + Cand yields(9.1.2)y(x) = eH (x)e H (x) q(x) dx + C .13:30 469. P1: PBU/OVYP2: PBU/OVYJWDD023-09JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20069.1 FIRST-ORDER LINEAR EQUATIONSRemark There is no reason to commit this formula to memory. What is important here is the method of solution. The key step is multiplication by e H (x) , where H (x) = p(x) dx. It is multiplication by this factor, called an integrating factor, that enables us to write the left side in a form that we can integrate directly. Note that (9.1.2) contains an arbitrary constant C. A close look at the steps taken to obtain this solution makes it clear that that this solution includes all the functions that satisfy the differential equation. For this reason we call it the general solution. By assigning a particular value to the constant C, we obtain what is called a particular solution.Example 1 Find the general solution of the equation y + ay = b,a = 0.a, b constants,SOLUTION First we calculate an integrating factor:H (x) =a dx = axe H (x) = eax .and thereforeMultiplying the differential equation by eax , we get eax y + a eax y = b eax . The left-hand side is the derivative of eax y. (Verify this.) Thus, we have d ax [e y] = b eax . dx We integrate this equation and nd that eax y =b ax e + C. aIt follows that y= This is the general solution.b + Ceax . aExample 2 Find the general solution of the equation y + 2x y = x and then nd the particular solution y for which y(0) = 2. SOLUTION This equation has the form (9.1.1). To solve the equation, we calculate the integrating factor e H (x) :H (x) =2x dx = x 22e H (x) = e x .and so2Multiplication by e x gives 22e x y + 2xe x y = xe x2d x2 2 e y = xe x . dx Integrating this equation, we get 22e x y = 1 e x + C, 244513:30 470. P1: PBU/OVYP2: PBU/OVYJWDD023-09JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 2006446 CHAPTER 9 SOME DIFFERENTIAL EQUATIONS which we write as y=1 2+ Cex . 2This is the general solution. To nd the solution y for which y(0) = 2, we set x = 0, y = 2 and solve for C: 2=1 2+ Ce0 =1 2+CC = 3. 2and soThe function y=1 2+ 3 ex 22is the particular solution that satises the given condition. Remark When a differential equation is used as a mathematical model in some application, there is usually an initial condition y(x0 ) = y0 that makes it possible to evaluate the arbitrary constant that appears in the general solution. The problem of nding a particular solution that satises a given condition is called an initial-value problem. Example 3 Solve the initial-value problem: x y 2y = 3x 4 ,y(1) = 2.SOLUTION This differential equation does not have the form of (9.1.1), but we can put it in that form by dividing the equation by x:y 2 y = 3x 3 . xNow we set 2 dx = 2 ln x = ln x 2 xH (x) = and get the integrating factore H (x) = eln x2= x 2 .Multiplication by x 2 gives x 2 y 2x 3 y = 3x d 2 [x y] = 3x. dx Integrating this equation, we get x 2 y = 3 x 2 + C, 2 which we write as y = 3 x 4 + C x 2. 2 This is the general solution. Applying the initial condition, we have y(1) = 2 = 3 (1)4 + C(1)2 = 23 2+ C.This gives C = 1 . The function y = 3 x 4 + 1 x 2 is the solution of the initial-value 2 2 2 problem. (Check this out.) 13:30 471. P1: PBU/OVYP2: PBU/OVYJWDD023-09JWDD023-Salas-v13QC: PBU/OVYT1: PBU September 22, 20069.1 FIRST-ORDER