APPLICATIONS OF
4
THE DERIVATIVE
4.1 Linear Approximation and Applications
Preliminary Questions
1. True or False? The Linear Approximation says that the vertical change in the graph is approximately equal to the vertical
change in the tangent line.
SOLUTION This statement is true. The linear approximation does say that the vertical change in the graph is approximately equal
to the vertical change in the tangent line.
2. Estimate g.1:2/ ! g.1/ if g 0 .1/ D 4.
SOLUTION
Using the Linear Approximation,
g.1:2/ ! g.1/ " g 0 .1/.1:2 ! 1/ D 4.0:2/ D 0:8:
3. Estimate f .2:1/ if f .2/ D 1 and f 0 .2/ D 3.
SOLUTION
Using the Linearization,
f .2:1/ " f .2/ C f 0 .2/.2:1 ! 2/ D 1 C 3.0:1/ D 1:3
4. Complete the sentence: The Linear Approximation shows that up to a small error, the change in output !f is directly proportional to . . . .
SOLUTION The Linear Approximation tells us that up to a small error, the change in output !f is directly proportional to the
change in input !x when !x is small.
Exercises
In Exercises 1–6, use Eq. (1) to estimate !f D f .3:02/ ! f .3/.
1. f .x/ D x 2
SOLUTION
Let f .x/ D x 2 . Then f 0 .x/ D 2x and !f " f 0 .3/!x D 6.0:02/ D 0:12.
2. f .x/ D x 4
SOLUTION
Let f .x/ D x 4 . Then f 0 .x/ D 4x 3 and !f " f 0 .3/!x D 4.27/.0:02/ D 2:16:
3. f .x/ D x !1
1
Let f .x/ D x !1 . Then f 0 .x/ D !x !2 and !f " f 0 .3/!x D ! .0:02/ D !0:00222:
9
1
4. f .x/ D
xC1
1
SOLUTION Let f .x/ D .x C 1/!1 . Then f 0 .x/ D !.x C 1/!2 and !f " f 0 .3/!x D !
.0:02/ D !0:00125:
16
p
5. f .x/ D x C 6
p
1
SOLUTION Let f .x/ D x C 6. Then f 0 .x/ D 2 .x C 6/!1=2 and
SOLUTION
"x
3
SOLUTION Let f .x/ D tan
!f " f 0 .3/!x D
1 !1=2
9
.0:02/ D 0:003333:
2
6. f .x/ D tan
!x
3 .
Then f 0 .x/ D
!
3
sec2
!x
3
and
"
.0:02/ D 0:020944:
3
7. The cube root of 27 is 3. How much larger is the cube root of 27.2? Estimate using the Linear Approximation.
!f " f 0 .3/!x D
Let f .x/ D x 1=3 , a D 27, and !x D 0:2. Then f 0 .x/ D
Approximation is
SOLUTION
!f " f 0 .a/!x D
1 !2=3
3x
1
.0:2/ D 0:0074074
27
and f 0 .a/ D f 0 .27/ D
1
27 .
The Linear
356
APPLICATIONS OF THE DERIVATIVE
CHAPTER 4
8. Estimate ln.e 3 C 0:1/ ! ln.e 3 / using differentials.
SOLUTION
Let f .x/ D ln x, a D e 3 , and !x D 0:1. Then f 0 .x/ D x !1 and f 0 .a/ D e !3 . Thus,
ln.e 3 C 0:1/ ! ln.e 3 / D !f " f 0 .a/!x D e !3 .0:1/ D 0:00498:
In Exercises 9–12, use Eq. (1) to estimate !f . Use a calculator to compute both the error and the percentage error.
p
9. f .x/ D 1 C x, a D 3, !x D 0:2
SOLUTION Let f .x/ D
!f " f 0 .a/!x D 14 .0:2/
.1 C x/1=2 , a D 3, and !x D 0:2. Then f 0 .x/ D 12 .1 C x/!1=2 , f 0 .a/ D f 0 .3/ D
D 0:05. The actual change is
p
!f D f .a C !x/ ! f .a/ D f .3:2/ ! f .3/ D 4:2 ! 2 " 0:049390:
1
4
and
The error in the Linear Approximation is therefore j0:049390 ! 0:05j D 0:000610; in percentage terms, the error is
0:000610
# 100% " 1:24%:
0:049390
10. f .x/ D 2x 2 ! x, a D 5,
!x D !0:4
Let f .x/ D 2x 2 ! x, a D 5 and !x D !0:4. Then f 0 .x/ D 4x ! 1, f 0 .a/ D 19 and !f " f 0 .a/!x D
19.!0:4/ D !7:6. The actual change is
SOLUTION
!f D f .a C !x/ ! f .a/ D f .4:6/ ! f .5/ D 37:72 ! 45 D !7:28:
The error in the Linear Approximation is therefore j ! 7:28 ! .!7:6/j D 0:32; in percentage terms, the error is
0:32
# 100% " 4:40%:
7:28
11. f .x/ D
SOLUTION
f 0 .a/!x D
1
, a D 3, !x D 0:5
1 C x2
Let f .x/ D
1
,
1Cx 2
2x
0
0
a D 3, and !x D :5. Then f 0 .x/ D ! .1Cx
2 /2 , f .a/ D f .3/ D !0:06 and !f "
!0:06.0:5/ D !0:03. The actual change is
!f D f .a C !x/ ! f .a/ D f .3:5/ ! f .3/ " !0:0245283:
The error in the Linear Approximation is therefore j ! 0:0245283 ! .!0:03/j D 0:0054717; in percentage terms, the error is
ˇ
ˇ
ˇ 0:0054717 ˇ
ˇ
ˇ
ˇ !0:0245283 ˇ # 100% " 22:31%
12. f .x/ D ln.x 2 C 1/,
a D 1,
!x D 0:1
Let f .x/ D
C 1/, a D 1, and !x D 0:1. Then f 0 .x/ D
1.0:1/ D 0:1. The actual change is
SOLUTION
ln.x 2
2x
,
x 2 C1
f 0 .a/ D f 0 .1/ D 1, and !f " f 0 .a/!x D
!f D f .a C !x/ ! f .a/ D f .1:1/ ! f .1/ D 0:099845:
The error in the Linear Approximation is therefore j0:099845 ! 0:1j D 0:000155; in percentage terms, the error is
0:000155
# 100% " 0:16%:
0:099845
In Exercises 13–16, estimate !y using differentials [Eq. (3)].
13. y D cos x,
SOLUTION
aD
!
6,
dx D 0:014
Let f .x/ D cos x. Then f 0 .x/ D ! sin x and
!y " dy D f 0 .a/dx D ! sin
14. y D tan2 x,
SOLUTION
aD
!
4,
dx D !0:02
!" "
6
.0:014/ D !0:007:
Let f .x/ D tan2 x. Then f 0 .x/ D 2 tan x sec2 x and
!y " dy D f 0 .a/dx D 2 tan
"
"
sec2 .!0:02/ D !0:08:
4
4
Linear Approximation and Applications
S E C T I O N 4.1
15. y D
10 ! x 2
, a D 1,
2 C x2
SOLUTION
Let f .x/ D
357
dx D 0:01
10 ! x 2
. Then
2 C x2
f 0 .x/ D
.2 C x 2 /.!2x/ ! .10 ! x 2 /.2x/
24x
D!
2
2
.2 C x /
.2 C x 2 /2
and
!y " dy D f 0 .a/dx D !
16. y D x 1=3 e x!1 , a D 1,
24
.0:01/ D !0:026667:
9
dx D 0:1
x 1=3 e x!1 ,
Let y D
a D 1, and dx D 0:1. Then y 0 .x/ D
4
0
!y " dy D y .a/ dx D 3 .0:1/ D 0:133333.
SOLUTION
1 !2=3 x!1
e
.3x
3x
C 1/, y 0 .a/ D y 0 .1/ D
In Exercises 17–24, estimate using the Linear Approximation and find the error using a calculator.
p
p
17. 26 ! 25
p
1
SOLUTION Let f .x/ D x, a D 25, and !x D 1. Then f 0 .x/ D 2 x !1=2 and f 0 .a/ D f 0 .25/ D
"
"
"
SOLUTION
"
and
1
10 .
1
The Linear Approximation is !f " f 0 .a/!x D 10
.1/ D 0:1.
The actual change is !f D f .a C !x/ ! f .a/ D f .26/ ! f .25/ " 0:0990195.
The error in this estimate is j0:0990195 ! 0:1j D 0:000980486.
18. 16:51=4 ! 161=4
"
4
3,
Let f .x/ D x 1=4 , a D 16, and !x D :5. Then f 0 .x/ D 14 x !3=4 and f 0 .a/ D f 0 .16/ D
The Linear Approximation is !f " f 0 .a/!x D
The actual change is
1
32 .0:5/
1
32 .
D 0:015625.
!f D f .a C !x/ ! f .a/ D f .16:5/ ! f .16/ " 2:015445 ! 2 D 0:015445
"
The error in this estimate is j0:015625 ! 0:015445j " 0:00018.
19. p
1
101
!
SOLUTION
!0:0005.
"
"
1
10
Let f .x/ D
p1 ,
x
1
D ! 12 x !3=2 and f 0 .a/ D ! 12 . 1000
/ D
The Linear Approximation is !f " f 0 .a/!x D !0:0005.1/ D !0:0005.
The actual change is
!f D f .a C !x/ ! f .a/ D p
"
d
.x !1=2 /
dx
a D 100, and !x D 1. Then f 0 .x/ D
1
101
!
1
D !0:000496281:
10
The error in this estimate is j!0:0005 ! .!0:000496281/j D 3:71902 # 10!6 .
1
1
20. p !
98 10
SOLUTION
!0:0005.
"
"
"
SOLUTION
"
"
p1 ,
x
a D 100, and !x D !2. Then f 0 .x/ D
d
.x !1=2 /
dx
1
D ! 12 x !3=2 and f 0 .a/ D ! 12 . 1000
/D
The Linear Approximation is !f " f 0 .a/!x D !0:0005.!2/ D 0:001:
The actual change is !f D f .a C !x/ ! f .a/ D f .98/ ! f .100/ D 0:00101525.
The error in this estimate is j0:001 ! 0:00101525j " 0:00001525.
21. 91=3 ! 2
"
Let f .x/ D
Let f .x/ D x 1=3 , a D 8, and !x D 1. Then f 0 .x/ D 13 x !2=3 and f 0 .a/ D f 0 .8/ D
1
The Linear Approximation is !f " f 0 .a/!x D 12
.1/ D 0:083333.
The actual change is !f D f .a C !x/ ! f .a/ D f .9/ ! f .8/ D 0:080084.
The error in this estimate is j0:080084 ! 0:083333j " 3:25 # 10!3 .
1
12 .
358
CHAPTER 4
22. tan!1 .1:05/ !
SOLUTION
"
"
"
APPLICATIONS OF THE DERIVATIVE
!
4
Let f .x/ D tan!1 x, a D 1, and !x D 0:05. Then f 0 .x/ D .1 C x 2 /!1 and f 0 .a/ D f 0 .1/ D 12 .
The Linear Approximation is !f " f 0 .a/!x D 12 .0:05/ D 0:025.
The actual change is !f D f .a C !x/ ! f .a/ D f .1:05/ ! f .1/ D 0:024385.
The error in this estimate is j0:024385 ! 0:025j " 6:15 # 10!4 .
23. e !0:1 ! 1
SOLUTION
"
"
"
Let f .x/ D e x , a D 0, and !x D !0:1. Then f 0 .x/ D e x and f 0 .a/ D f 0 .0/ D 1.
The Linear Approximation is !f " f 0 .a/!x D 1.!0:1/ D !0:1.
The actual change is !f D f .a C !x/ ! f .a/ D f .!0:1/ ! f .0/ D !0:095163.
The error in this estimate is j ! 0:095163 ! .!0:1/j " 4:84 # 10!3 .
24. ln.0:97/
SOLUTION
"
"
Let f .x/ D ln x, a D 1, and !x D !0:03. Then f 0 .x/ D
1
x
and f 0 .a/ D f 0 .1/ D 1.
The Linear Approximation is !f " f 0 .a/!x D .1/.!0:03/ D !0:03, so ln.0:97/ " ln 1 ! 0:03 D !0:03.
The actual change is
!f D f .a C !x/ ! f .a/ D f .0:97/ ! f .1/ " !0:030459 ! 0 D !0:030459:
"
The error is j!f ! f 0 .a/!xj " 0:000459.
25. Estimate f .4:03/ for f .x/ as in Figure 1.
y
y = f (x)
(10, 4)
Tangent line
(4, 2)
x
FIGURE 1
SOLUTION
Using the Linear Approximation, f .4:03/ " f .4/ C f 0 .4/.0:03/. From the figure, we find that f .4/ D 2 and
f 0 .4/ D
4!2
1
D :
10 ! 4
3
Thus,
1
f .4:03/ " 2 C .0:03/ D 2:01:
3
26.
At a certain moment, an object in linear motion has velocity 100 m/s. Estimate the distance traveled over the next
quarter-second, and explain how this is an application of the Linear Approximation.
SOLUTION
Because the velocity is 100 m/s, we estimate the object will travel
# $
!
m" 1
100
s D 25 m
s
4
in the next quarter-second. Recall that velocity is the derivative of position, so we have just estimated the change in position, !s,
using the product s 0 !t, which is just the Linear Approximation.
p
p
p
p
27. Which is larger: 2:1 ! 2 or 9:1 ! 9? Explain using the Linear Approximation.
p
1
SOLUTION Let f .x/ D x, and !x D 0:1. Then f 0 .x/ D 2 x !1=2 and the Linear Approximation at x D a gives
!f D
p
a C 0:1 !
p
1
0:05
a " f 0 .a/.0:1/ D a!1=2 .0:1/ D p
2
a
We see that !f decreases as a increases. In particular
p
p
0:05
2:1 ! 2 " p
2
is larger than
p
p
0:05
9:1 ! 9 "
3
S E C T I O N 4.1
Linear Approximation and Applications
28. Estimate sin 61ı ! sin 60ı using the Linear Approximation. Hint: Express !# in radians.
Let f .x/ D sin x, a D
Approximation is
SOLUTION
!
3,
and !x D
!
180 .
Then f 0 .x/ D cos x and f 0 .a/ D f 0 . !3 / D
1
2.
359
Finally, the Linear
1! " "
"
D
" 0:008727
2 180
360
29. Box office revenue at a multiplex cinema in Paris is R.p/ D 3600p ! 10p 3 euros per showing when the ticket price is p euros.
Calculate R.p/ for p D 9 and use the Linear Approximation to estimate !R if p is raised or lowered by 0:5 euros.
!f " f 0 .a/!x D
Let R.p/ D 3600p ! 10p 3 . Then R.9/ D 3600.9/ ! 10.9/3 D 25110 euros. Moreover, R0 .p/ D 3600 ! 30p 2 , so
by the Linear Approximation,
SOLUTION
!R " R0 .9/!p D 1170!p:
If p is raised by 0.5 euros, then !R " 585 euros; on the other hand, if p is lowered by 0.5 euros, then !R " !585 euros.
30. The stopping distance for an automobile is F .s/ D 1:1s C 0:054s 2 ft, where s is the speed in mph. Use the Linear Approximation to estimate the change in stopping distance per additional mph when s D 35 and when s D 55.
SOLUTION
"
Let F .s/ D 1:1s C 0:054s 2 .
The Linear Approximation at s D 35 mph is
!F " F 0 .35/!s D .1:1 C 0:108 # 35/!s D 4:88!s ft
"
The change in stopping distance per additional mph for s D 35 mph is approximately 4:88 ft.
The Linear Approximation at s D 55 mph is
!F " F 0 .55/!s D .1:1 C 0:108 # 55/!s D 7:04!s ft
The change in stopping distance per additional mph for s D 55 mph is approximately 7:04 ft.
31. A thin silver wire has length L D 18 cm when the temperature is T D 30ı C. Estimate !L when T decreases to 25ı C if the
coefficient of thermal expansion is k D 1:9 # 10!5ı C!1 (see Example 3).
SOLUTION
We have
dL
D kL D .1:9 # 10!5 /.18/ D 3:42 # 10!4 cm=ı C
dT
The change in temperature is !T D !5ı C, so by the Linear Approximation, the change in length is approximately
!L " 3:42 # 10!4 !T D .3:42 # 10!4 /.!5/ D !0:00171 cm
At T D 25ı C, the length of the wire is approximately 17:99829 cm.
32. At a certain moment, the temperature in a snake cage satisfies d T =dt D 0:008ı C/s. Estimate the rise in temperature over the
next 10 seconds.
SOLUTION
Using the Linear Approximation, the rise in temperature over the next 10 seconds will be
dT
!t D 0:008.10/ D 0:08ı C:
dt
33. The atmospheric pressure at altitude h (kilometers) for 11 $ h $ 25 is approximately
!T "
P .h/ D 128e !0:157h kilopascals.
(a) Estimate !P at h D 20 when !h D 0:5.
(b) Compute the actual change, and compute the percentage error in the Linear Approximation.
SOLUTION
(a) Let P .h/ D 128e !0:157h . Then P 0 .h/ D !20:096e !0:157h . Using the Linear Approximation,
!P " P 0 .h/!h D P 0 .20/.0:5/ D !0:434906 kilopascals:
(b) The actual change in pressure is
P .20:5/ ! P .20/ D !0:418274 kilopascals:
The percentage error in the Linear Approximation is
ˇ
ˇ
ˇ !0:434906 ! .!0:418274/ ˇ
ˇ # 100% " 3:98%:
ˇ
ˇ
ˇ
!0:418274
360
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
34. Theˇresistance R of a copper wire at temperature T D 20ı C is R D 15 $. Estimate the resistance at T D 22ı C, assuming that
dR=d T ˇT D20 D 0:06 $/ı C.
SOLUTION
!T D 2ı C. The Linear Approximation gives us:
ˇ
ˇ
R.22/ ! R.20/ " dR=d T ˇˇ
!T D 0:06 $=ı C.2ı C/ D 0:12 $:
T D20
Therefore, R.22/ " 15 $ C 0:12 $ D 15:12 $.
35. Newton’s Law of Gravitation shows that if a person weighs w pounds on the surface of the earth, then his or her weight at
distance x from the center of the earth is
W .x/ D
wR2
x2
.for x % R/
where R D 3960 miles is the radius of the earth (Figure 2).
(a) Show that the weight lost at altitude h miles above the earth’s surface is approximately !W " !.0:0005w/h. Hint: Use the
Linear Approximation with dx D h.
(b) Estimate the weight lost by a 200-lb football player flying in a jet at an altitude of 7 miles.
h
0
396
FIGURE 2 The distance to the center of the earth is 3960 C h miles.
SOLUTION
(a) Using the Linear Approximation
!W " W 0 .R/!x D !
2wR2
2wh
hD!
" !0:0005wh:
R
R3
(b) Substitute w D 200 and h D 7 into the result from part (a) to obtain
!W " !0:0005.200/.7/ D !0:7 pounds:
36. Using Exercise 35(a), estimate the altitude at which a 130-lb pilot would weigh 129.5 lb.
SOLUTION From Exercise 35(a), the weight loss !W at altitude h (in miles) for a person weighing w at the surface of the earth
is approximately
!W " !0:0005wh
If w D 130 pounds, then !W " !0:065h. Accordingly, the pilot loses approximately 0.065 pounds per mile of altitude gained.
The pilot will weigh 129.5 pounds at the altitude h such that !0:065h D !0:5, or h D 0:5=0:065 " 7:7 miles.
37.
(a)
(b)
(c)
A stone tossed vertically into the air with initial velocity v cm/s reaches a maximum height of h D v 2 =1960 cm.
Estimate !h if v D 700 cm/s and !v D 1 cm/s.
Estimate !h if v D 1,000 cm/s and !v D 1 cm/s.
In general, does a 1 cm/s increase in v lead to a greater change in h at low or high initial velocities? Explain.
SOLUTION A
h0 .v/ D v=980.
stone tossed vertically with initial velocity v cm=s attains a maximum height of h.v/ D v 2 =1960 cm. Thus,
1
980 .700/.1/ " 0:71 cm.
1
D 980
.1000/.1/ D 1:02 cm.
(a) If v D 700 and !v D 1, then !h " h0 .v/!v D
(b) If v D 1000 and !v D 1, then !h " h0 .v/!v
(c) A one centimeter per second increase in initial velocity v increases the maximum height by approximately v=980 cm. Accordingly, there is a bigger effect at higher velocities.
38. The side s of a square carpet is measured at 6 m. Estimate the maximum error in the area A of the carpet if s is accurate to
within 2 centimeters.
Linear Approximation and Applications
S E C T I O N 4.1
361
Let s be the length in meters of the side of the square carpet. Then A.s/ D s 2 is the area of the carpet. With a D 6
and !s D 0:02 (note that 1 cm equals 0:01 m), an estimate of the size of the error in the area is given by the Linear Approximation:
SOLUTION
!A " A0 .6/!s D 12 .0:02/ D 0:24 m2
In Exercises 39 and 40, use the following fact derived from Newton’s Laws: An object released at an angle # with initial velocity
v ft/s travels a horizontal distance
sD
1 2
v sin 2# ft (Figure 3)
32
y
x
FIGURE 3 Trajectory of an object released at an angle #.
39. A player located 18.1 ft from the basket launches a successful jump shot from a height of 10 ft (level with the rim of the basket),
at an angle # D 34ı and initial velocity v D 25 ft/s.)
(a) Show that !s " 0:255!# ft for a small change of !#.
(b) Is it likely that the shot would have been successful if the angle had been off by 2ı ?
1
SOLUTION Using Newton’s laws and the given initial velocity of v D 25 ft=s, the shot travels s D 32 v 2 sin 2t D
where t is in radians.
(a) If # D 34ı (i.e., t D 17
90 "), then
#
$
#
$
625
17
625
17
"
!s " s 0 .t/!t D
cos
" !t D
cos
" !# &
" 0:255!#:
16
45
16
45
180
625
32
sin 2t ft,
(b) If !# D 2ı , this gives !s " 0:51 ft, in which case the shot would not have been successful, having been off half a foot.
40. Estimate !s if # D 34ı , v D 25 ft/s, and !v D 2.
SOLUTION
Using Newton’s laws and the fixed angle of # D 34ı D
sD
17
90 ",
the shot travels
1 2
17
v sin ":
32
45
With v D 25 ft/s and !v D 2 ft/s, we find
1
17"
.25/ sin
& 2 D 2:897 ft:
16
45
41. The radius of a spherical ball is measured at r D 25 cm. Estimate the maximum error in the volume and surface area if r is
accurate to within 0:5 cm.
!s " s 0 .v/!v D
SOLUTION The volume and surface area of the sphere are given by V D
!r D ˙0:5, then
4
3
3"r
and S D 4" r 2 , respectively. If r D 25 and
!V " V 0 .25/!r D 4".25/2 .0:5/ " 3927 cm3 ;
and
!S " S 0 .25/!r D 8".25/.0:5/ " 314:2 cm2 :
42. The dosage D of diphenhydramine for a dog of body mass w kg is D D 4:7w 2=3 mg. Estimate the maximum allowable error
in w for a cocker spaniel of mass w D 10 kg if the percentage error in D must be less than 3%.
SOLUTION
We have D D kw 2=3 where k D 4:7. The Linear Approximation yields
!D "
2 !1=3
kw
!w;
3
so
!D
"
D
2
!1=3 !w
3 kw
kw 2=3
D
2 !w
&
3 w
If the percentage error in D must be less than 3%, we estimate the maximum allowable error in w to be
!w "
3w !D
3.10/
&
D
.:03/ D 0:45 kg
2
D
2
362
APPLICATIONS OF THE DERIVATIVE
CHAPTER 4
43. The volume (in liters) and pressure P (in atmospheres) of a certain gas satisfy P V D 24. A measurement yields V D 4 with a
possible error of ˙0:3 L. Compute P and estimate the maximum error in this computation.
SOLUTION Given P V
P 0 D !24V !2 and
D 24 and V D 4, it follows that P D 6 atmospheres. Solving P V D 24 for P yields P D 24V !1 . Thus,
!P " P 0 .4/!V D !24.4/!2 .˙0:3/ D ˙0:45 atmospheres:
44. In the notation of Exercise 43, assume that a measurement yields V D 4. Estimate the maximum allowable error in V if P
must have an error of less than 0.2 atm.
SOLUTION
From Exercise 43, with V D 4, we have
3
!P " ! !V
2
or
2
!V D ! !P:
3
If we require j!P j $ 0:2, then we must have
j!V j $
2
.0:2/ D 0:133333 L:
3
In Exercises 45–54, find the linearization at x D a.
45. f .x/ D x 4 ,
SOLUTION
46. f .x/ D
SOLUTION
aD1
Let f .x/ D x 4 . Then f 0 .x/ D 4x 3 . The linearization at a D 1 is
1
,
x
L.x/ D f 0 .a/.x ! a/ C f .a/ D 4.x ! 1/ C 1 D 4x ! 3:
aD2
1
x
Let f .x/ D
D x !1 . Then f 0 .x/ D !x !2 . The linearization at a D 2 is
1
1
1
L.x/ D f 0 .a/.x ! a/ C f .a/ D ! .x ! 2/ C D ! x C 1:
4
2
4
47. f .#/ D sin2 #,
SOLUTION
48. g.x/ D
SOLUTION
aD
!
4
Let f .#/ D sin2 #. Then f 0 .#/ D 2 sin # cos # D sin 2#. The linearization at a D
is
!
"" 1
"
1
L.#/ D f 0 .a/.# ! a/ C f .a/ D 1 # !
C D#! C :
4
2
4
2
x2
, aD4
x!3
Let g.x/ D
!
4
x2
x!3 .
Then
g 0 .x/ D
.x ! 3/.2x/ ! x 2
x 2 ! 6x
D
:
.x ! 3/2
.x ! 3/2
The linearization at a D 4 is
49. y D .1 C x/!1=2 , a D 0
SOLUTION
L.x/ D g 0 .a/.x ! a/ C g.a/ D !8.x ! 4/ C 16 D !8x C 48:
Let f .x/ D .1 C x/!1=2 . Then f 0 .x/ D ! 12 .1 C x/!3=2 . The linearization at a D 0 is
1
L.x/ D f 0 .a/.x ! a/ C f .a/ D ! x C 1:
2
50. y D .1 C x/!1=2 , a D 3
Let f .x/ D .1 C x/!1=2 . Then f 0 .x/ D ! 12 .1 C x/!3=2 , f .a/ D 4!1=2 D
so the linearization at a D 3 is
SOLUTION
L.x/ D f 0 .a/.x ! a/ C f .a/ D !
51. y D .1 C x 2 /!1=2 ,
SOLUTION
1
2,
1
and f 0 .a/ D ! 12 .4!3=2 / D ! 16
,
1
1
1
11
.x ! 3/ C D ! x C :
16
2
16
16
aD0
Let f .x/ D .1 C x 2 /!1=2 . Then f 0 .x/ D !x.1 C x 2 /!3=2 , f .a/ D 1 and f 0 .a/ D 0, so the linearization at a is
L.x/ D f 0 .a/.x ! a/ C f .a/ D 1:
S E C T I O N 4.1
Linear Approximation and Applications
363
52. y D tan!1 x, a D 1
SOLUTION
Let f .x/ D tan!1 x. Then
f 0 .x/ D
1
"
1
; f .a/ D ; and f 0 .a/ D ;
4
2
1 C x2
so the linearization of f .x/ at a is
53. y D e
p
SOLUTION
L.x/ D f 0 .a/.x ! a/ C f .a/ D
x,
aD1
Let f .x/ D e
p
x.
1
"
.x ! 1/ C :
2
4
Then
1 p
1
f 0 .x/ D p e x ; f .a/ D e; and f 0 .a/ D e;
2
2 x
so the linearization of f .x/ at a is
L.x/ D f 0 .a/.x ! a/ C f .a/ D
1
1
e.x ! 1/ C e D e.x C 1/:
2
2
54. y D e x ln x, a D 1
SOLUTION
Let f .x/ D e x ln x. Then
f 0 .x/ D
ex
C e x ln x; f .a/ D 0; and f 0 .a/ D e;
x
so the linearization of f .x/ at a is
L.x/ D f 0 .a/.x ! a/ C f .a/ D e.x ! 1/:
55. What is f .2/ if the linearization of f .x/ at a D 2 is L.x/ D 2x C 4?
SOLUTION
f .2/ D L.2/ D 2.2/ C 4 D 8.
56. Compute the linearization of f .x/ D 3x ! 4 at a D 0 and a D 2. Prove more generally that a linear function coincides with
its linearization at x D a for all a.
SOLUTION
is
Let f .x/ D 3x ! 4. Then f 0 .x/ D 3. With a D 0, f .a/ D !4 and f 0 .a/ D 3, so the linearization of f .x/ at a D 0
L.x/ D !4 C 3.x ! 0/ D 3x ! 4 D f .x/:
With a D 2, f .a/ D 2 and f 0 .a/ D 3, so the linearization of f .x/ at a D 2 is
L.x/ D 2 C 3.x ! 2/ D 2 C 3x ! 6 D 3x ! 4 D f .x/:
More generally, let g.x/ D bx C c be any linear function. The linearization L.x/ of g.x/ at x D a is
L.x/ D g 0 .a/.x ! a/ C g.a/ D b.x ! a/ C ba C c D bx C c D g.x/I
i.e., L.x/ D g.x/.
p
p
57. Estimate 16:2 using the linearization L.x/ of f .x/ D x at a D 16. Plot f .x/ and L.x/ on the same set of axes and
determine whether the estimate is too large or too small.
SOLUTION
f .x/ is
Let f .x/ D x 1=2 , a D 16, and !x D 0:2. Then f 0 .x/ D
L.x/ D f 0 .a/.x ! a/ C f .a/ D
1 !1=2
2x
and f 0 .a/ D f 0 .16/ D
1
8.
The linearization to
1
1
.x ! 16/ C 4 D x C 2:
8
8
p
Thus, we have 16:2 " L.16:2/ D 4:025. Graphs of f .x/ and L.x/ are shown below. Because the graph of L.x/ lies above the
graph of f .x/, we expect that the estimate from the Linear Approximation is too large.
y
5
4
3
2
1
L(x)
0
5
f (x)
10
15
20
25
x
364
APPLICATIONS OF THE DERIVATIVE
CHAPTER 4
p
p
58.
Estimate 1= 15 using a suitable linearization of f .x/ D 1= x. Plot f .x/ and L.x/ on the same set of axes and
determine whether the estimate is too large or too small. Use a calculator to compute the percentage error.
SOLUTION
The nearest perfect square to 15 is 16. Let f .x/ D
p1
x
1
! 128
. The linearization is
and a D 16. Then f 0 .x/ D ! 12 x !3=2 and f 0 .a/ D f 0 .16/ D
L.x/ D f 0 .a/.x ! a/ C f .a/ D !
1
1
.x ! 16/ C :
128
4
Then
1
1
1
33
p " L.15/ D !
.!1/ C D
D 0:257813:
128
4
128
15
Graphs of f .x/ and L.x/ are shown below. Because the graph of L.x/ lies below the graph of f .x/, we expect that the estimate
from the Linear Approximation is too small. The percentage error in the estimate is
ˇ
ˇ
ˇ p1 ! 0:257813 ˇ
ˇ 15
ˇ
ˇ
ˇ # 100% " 0:15%
ˇ
ˇ
1
p
ˇ
ˇ
15
y
0.8
0.6
f (x)
0.4
0.2
L(x)
0
5
10
15
20
25
x
In Exercises 59–67, approximate using linearization and use a calculator to compute the percentage error.
1
59. p
17
SOLUTION
to f .x/ is
1
Let f .x/ D x !1=2 , a D 16, and !x D 1. Then f 0 .x/ D ! 12 x !3=2 , f 0 .a/ D f 0 .16/ D ! 128
and the linearization
L.x/ D f 0 .a/.x ! a/ C f .a/ D !
Thus, we have
60.
p1
17
" L.17/ " 0:24219. The percentage error in this estimate is
ˇ
ˇ
ˇ p1 ! 0:24219 ˇ
ˇ 17
ˇ
ˇ
ˇ # 100% " 0:14%
ˇ
ˇ
1
p
ˇ
ˇ
17
1
101
SOLUTION
to f .x/ is
1
1
1
3
.x ! 16/ C D !
xC :
128
4
128
8
Let f .x/ D x !1 , a D 100 and !x D 1. Then f 0 .x/ D !x !2 , f 0 .a/ D f 0 .100/ D !0:0001 and the linearization
L.x/ D f 0 .a/.x ! a/ C f .a/ D !0:0001.x ! 100/ C 0:01 D !0:0001x C 0:02:
Thus, we have
1
" L.101/ D !0:0001.101/ C 0:02 D 0:0099:
101
The percentage error in this estimate is
ˇ
ˇ
ˇ 1 ! 0:0099 ˇ
ˇ 101
ˇ
ˇ
ˇ # 100% " 0:01%
1
ˇ
ˇ
101
1
61.
.10:03/2
SOLUTION
to f .x/ is
Let f .x/ D x !2 , a D 10 and !x D 0:03. Then f 0 .x/ D !2x !3 , f 0 .a/ D f 0 .10/ D !0:002 and the linearization
L.x/ D f 0 .a/.x ! a/ C f .a/ D !0:002.x ! 10/ C 0:01 D !0:002x C 0:03:
Linear Approximation and Applications
S E C T I O N 4.1
365
Thus, we have
1
" L.10:03/ D !0:002.10:03/ C 0:03 D 0:00994:
.10:03/2
The percentage error in this estimate is
ˇ
ˇ
ˇ
1
ˇ
ˇ .10:03/2 ! 0:00994 ˇ
ˇ
ˇ # 100% " 0:0027%
ˇ
ˇ
1
ˇ
ˇ
.10:03/2
62. .17/1=4
SOLUTION
f .x/ is
Let f .x/ D x 1=4 , a D 16, and !x D 1. Then f 0 .x/ D
L.x/ D f 0 .a/.x ! a/ C f .a/ D
1 !3=4
,
4x
f 0 .a/ D f 0 .16/ D
1
32
and the linearization to
1
48
and the linearization to
1
1
3
.x ! 16/ C 2 D
xC :
32
32
2
Thus, we have .17/1=4 " L.17/ D 2:03125. The percentage error in this estimate is
ˇ
ˇ
ˇ .17/1=4 ! 2:03125 ˇ
ˇ
ˇ
ˇ
ˇ # 100% " 0:035%
ˇ
ˇ
.17/1=4
63. .64:1/1=3
SOLUTION
f .x/ is
Let f .x/ D x 1=3 , a D 64, and !x D 0:1. Then f 0 .x/ D
L.x/ D f 0 .a/.x ! a/ C f .a/ D
1 !2=3
,
3x
f 0 .a/ D f 0 .64/ D
1
1
8
.x ! 64/ C 4 D
xC :
48
48
3
Thus, we have .64:1/1=3 " L.64:1/ " 4:002083. The percentage error in this estimate is
ˇ
ˇ
ˇ .64:1/1=3 ! 4:002083 ˇ
ˇ
ˇ
ˇ
ˇ # 100% " 0:000019%
ˇ
ˇ
.64:1/1=3
64. .1:2/5=3
SOLUTION
is
Let f .x/ D .1 C x/5=3 and a D 0. Then f 0 .x/ D
5
3 .1 C
x/2=3 , f 0 .a/ D f 0 .0/ D
L.x/ D f 0 .a/.x ! a/ C f .a/ D
Thus, we have .1:2/5=3 " L.0:2/ D
65. cos!1 .0:52/
SOLUTION
and the linearization to f .x/
5
x C 1:
3
5
3 .0:2/ C 1
D 1:3333. The percentage error in this estimate is
ˇ
ˇ
ˇ .1:2/5=3 ! 1:3333 ˇ
ˇ
ˇ
ˇ
ˇ # 100% " 1:61%
ˇ
ˇ
.1:2/5=3
Let f .x/ D cos!1 x and a D 0:5. Then
f 0 .x/ D ! p
and the linearization to f .x/ is
5
3
1
1 ! x2
; f 0 .a/ D f 0 .0/ D !
L.x/ D f 0 .a/.x ! a/ C f .a/ D !
p
2 3
;
3
p
2 3
"
.x ! 0:5/ C :
3
3
Thus, we have cos!1 .0:52/ " L.0:02/ D 1:024104. The percentage error in this estimate is
ˇ
ˇ
ˇ cos!1 .0:52/ ! 1:024104 ˇ
ˇ
ˇ
ˇ
ˇ # 100% " 0:015%:
ˇ
ˇ
cos!1 .0:52/
66. ln 1:07
SOLUTION
Let f .x/ D ln.1 C x/ and a D 0. Then f 0 .x/ D
1
1Cx ,
f 0 .a/ D f 0 .0/ D 1 and the linearization to f .x/ is
L.x/ D f 0 .a/.x ! a/ C f .a/ D x:
Thus, we have ln 1:07 " L.0:07/ D 0:07. The percentage error in this estimate is
ˇ
ˇ
ˇ ln 1:07 ! 0:07 ˇ
ˇ
ˇ # 100% " 3:46%:
ˇ
ˇ
ln 1:07
366
CHAPTER 4
APPLICATIONS OF THE DERIVATIVE
67. e !0:012
SOLUTION
Let f .x/ D e x and a D 0. Then f 0 .x/ D e x , f 0 .a/ D f 0 .0/ D 1 and the linearization to f .x/ is
L.x/ D f 0 .a/.x ! a/ C f .a/ D 1.x ! 0/ C 1 D x C 1:
Thus, we have e !0:012 " L.!0:012/ D 1 ! 0:012 D 0:988. The percentage error in this estimate is
ˇ
ˇ
ˇ e !0:012 ! 0:988 ˇ
ˇ
ˇ
ˇ
ˇ # 100% " 0:0073%:
ˇ
ˇ
e !0:012
68.
Compute the linearization L.x/ of f .x/ D x 2 ! x 3=2 at a D 4. Then plot f .x/ ! L.x/ and find an interval I around
a D 4 such that jf .x/ ! L.x/j $ 0:1 for x 2 I .
SOLUTION
Let f .x/ D x 2 ! x 3=2 and a D 4. Then f 0 .x/ D 2x ! 32 x 1=2 , f 0 .4/ D 5 and
L.x/ D f .a/ C f 0 .a/.x ! a/ D 8 C 5.x ! 4/ D 5x ! 12:
The graph of y D f .x/ ! L.x/ is shown below at the left, and portions of the graphs of y D f .x/ ! L.x/ and y D 0:1 are shown
below at the right. From the graph on the right, we see that jf .x/ ! L.x/j < 0:1 roughly for 3:6 < x < 4:4.
y
12
10
8
6
4
2
x
2
4
x
6
3.6
3.8
4.0
4.2
4.4
p
p
69. Show that the Linear Approximation to f .x/ D x at x D 9 yields the estimate 9 C h ! 3 " 16 h. Set K D 0:01 and show
that jf 00 .x/j $ K for x % 9. Then verify numerically that the error E satisfies Eq. (5) for h D 10!n , for 1 $ n $ 4.
p
1
1
SOLUTION Let f .x/ D x. Then f .9/ D 3, f 0 .x/ D 2 x !1=2 and f 0 .9/ D 6 . Therefore, by the Linear Approximation,
f .9 C h/ ! f .9/ D
p
1
h:
6
9Ch!3 "
Moreover, f 00 .x/ D ! 14 x !3=2 , so jf 00 .x/j D 14 x !3=2 . Because this is a decreasing function, it follows that for x % 9,
K D max jf 00 .x/j $ jf 00 .9/j D
1
< 0:01:
108
From the following table, we see that for h D 10!n , 1 $ n $ 4, E $ 12 Kh2 .
p
h
E D j 9 C h ! 3 ! 16 hj
10!1
10!2
10!3
10!4
4:604 # 10!5
4:627 # 10!7
4:629 # 10!9
4:627 # 10!11
1
2
2 Kh
5:00 # 10!5
5:00 # 10!7
5:00 # 10!9
5:00 # 10!11
%
&
70.
The Linear Approximation to f .x/ D tan x at x D !4 yields the estimate tan !4 C h ! 1 " 2h. Set K D 6:2 and
show, using a plot, that jf 00 .x/j $ K for x 2 Π!4 ; !4 C 0:1%. Then verify numerically that the error E satisfies Eq. (5) for h D 10!n ,
for 1 $ n $ 4.
SOLUTION
Let f .x/ D tan x. Then f . !4 / D 1, f 0 .x/ D sec2 x and f 0 . !4 / D 2. Therefore, by the Linear Approximation,
!"
"
!" "
!"
"
f
Ch !f
D tan
C h ! 1 " 2h:
4
4
4
Moreover, f 00 .x/ D 2 sec2 x tan x. The graph of the second derivative over the interval Œ"=4; "=4 C 0:1% is shown below. From
this graph we see that K D max jf 00 .x/j " 6:1 < 6:2.
y
6.1
5.7
5.3
4.9
4.5
4.1
x
0.78 0.80 0.82 0.84
0.86 0.88
Linear Approximation and Applications
S E C T I O N 4.1
367
Finally, from the following table, we see that for h D 10!n , 1 $ n $ 4, E $ 12 Kh2 .
h
10!1
10!2
10!3
10!4
E D j tan. !4 C h/ ! 1 ! 2hj
2:305 # 10!2
2:027 # 10!4
2:003 # 10!6
2:000 # 10!8
1
2
2 Kh
3:10 # 10!2
3:10 # 10!4
3:10 # 10!6
3:10 # 10!8
Further Insights and Challenges
71. Compute dy=dx at the point P D .2; 1/ on the curve y 3 C 3xy D 7 and show that the linearization at P is L.x/ D ! 13 x C 53 .
Use L.x/ to estimate the y-coordinate of the point on the curve where x D 2:1.
SOLUTION
Differentiating both sides of the equation y 3 C 3xy D 7 with respect to x yields
3y 2
dy
dy
C 3x
C 3y D 0;
dx
dx
so
dy
y
D! 2
:
dx
y Cx
Thus,
and the linearization at P D .2; 1/ is
ˇ
dy ˇˇ
1
1
D! 2
D! ;
ˇ
dx .2;1/
3
1 C2
1
1
5
L.x/ D 1 ! .x ! 2/ D ! x C :
3
3
3
Finally, when x D 2:1, we estimate that the y-coordinate of the point on the curve is
1
5
y " L.2:1/ D ! .2:1/ C D 0:967:
3
3
72. Apply the method of Exercise 71 to P D .0:5; 1/ on y 5 C y ! 2x D 1 to estimate the y-coordinate of the point on the curve
where x D 0:55.
SOLUTION
Differentiating both sides of the equation y 5 C y ! 2x D 1 with respect to x yields
5y 4
dy
dy
C
! 2 D 0;
dx
dx
so
dy
2
D 4
:
dx
5y C 1
Thus,
and the linearization at P D .0:5; 1/ is
ˇ
dy ˇˇ
2
1
D
D ;
dx ˇ.0:5;1/
3
5.1/2 C 1
L.x/ D 1 C
#
$
1
1
1
5
x!
D xC :
3
2
3
6
Finally, when x D 0:55, we estimate that the y-coordinate of the point on the curve is
y " L.0:55/ D
5
1
.0:55/ C D 1:017:
3
6
73. Apply the method of Exercise 71 to P D .!1; 2/ on y 4 C 7xy D 2 to estimate the solution of y 4 ! 7:7y D 2 near y D 2.
368
APPLICATIONS OF THE DERIVATIVE
CHAPTER 4
SOLUTION
Differentiating both sides of the equation y 4 C 7xy D 2 with respect to x yields
4y 3
dy
dy
C 7x
C 7y D 0;
dx
dx
so
dy
7y
D! 3
:
dx
4y C 7x
Thus,
ˇ
dy ˇˇ
7.2/
14
D!
D! ;
ˇ
3
dx .!1;2/
25
4.2/ C 7.!1/
and the linearization at P D .!1; 2/ is
L.x/ D 2 !
14
14
36
.x C 1/ D ! x C :
25
25
25
Finally, the equation y 4 ! 7:7y D 2 corresponds to x D !1:1, so we estimate the solution of this equation near y D 2 is
y " L.!1:1/ D !
14
36
.!1:1/ C
D 2:056:
25
25
74. Show that for any real number k, .1 C !x/k " 1 C k!x for small !x. Estimate .1:02/0:7 and .1:02/!0:3 .
SOLUTION
Let f .x/ D .1 C x/k . Then for small !x, we have
f .!x/ " L.!x/ D f 0 .0/.!x ! 0/ C f .0/ D k.1 C 0/k!1 .!x ! 0/ C 1 D 1 C k!x
"
"
Let k D 0:7 and !x D 0:02. Then L.0:02/ D 1 C .0:7/.0:02/ D 1:014.
Let k D !0:3 and !x D 0:02. Then L.0:02/ D 1 C .!0:3/.0:02/ D 0:994.
75. Let !f D f .5 C h/ ! f .5/, where f .x/ D x 2 . Verify directly that E D j!f ! f 0 .5/hj satisfies (5) with K D 2.
SOLUTION
Let f .x/ D x 2 . Then
!f D f .5 C h/ ! f .5/ D .5 C h/2 ! 52 D h2 C 10h
and
E D j!f ! f 0 .5/hj D jh2 C 10h ! 10hj D h2 D
1
1
.2/h2 D Kh2 :
2
2
76. Let !f D f .1 C h/ ! f .1/ where f .x/ D x !1 . Show directly that E D j!f ! f 0 .1/hj is equal to h2 =.1 C h/. Then prove
that E $ 2h2 if ! 21 $ h $ 12 . Hint: In this case, 21 $ 1 C h $ 32 .
SOLUTION
Let f .x/ D x !1 . Then
!f D f .1 C h/ ! f .1/ D
1
h
!1D!
1Ch
1Ch
ˇ
ˇ
E D j!f ! f 0 .1/hj D ˇˇ!
ˇ
ˇ
h
h2
C hˇˇ D
:
1Ch
1Ch
and
If ! 12 $ h $ 12 , then
1
2
$1Ch $
3
2
and
2
3
$
1
1Ch
$ 2. Thus, E $ 2h2 for ! 21 $ h $ 21 .
4.2 Extreme Values
Preliminary Questions
1. What is the definition of a critical point?
SOLUTION A critical point is a value of the independent variable x in the domain of a function f at which either f 0 .x/ D 0 or
f 0 .x/ does not exist.