CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
3.8 LINEARIZATION AND DIFFERENTIALS
s 30 13 and ( 16) 8.875 ft/sec Ê œ È ds œ 60 œ 32 ¸
dt 30È13 È13
(b) cos )"œ 90s Ê sin )" ddt)" œ 90s# †dsdt Ê ddt)" œs sin #90)" †dsdt œ90sx †dsdt. Therefore, xœ60 and sœ30È13 rad/sec; sin cos
Ê ddt œ 90 32 œ 658 œ90s Ê ddt œ 90s dsdt Ê ddt œs cos 90 dsdt
30 13 (60) 13
) ) )
" # # )
# # #
Š È ‹ †ŠÈ ‹ )# )# † †
. Therefore, x 60 and s 30 13 rad/sec.
œ sx90†dsdt œ œ È Ê ddt)# œ 658
(c) ddt)" s sin #90) dtds s90 xs dxdt 90s# dxdt x# 908100 dxdt lim ddt)"
" #
œ † œ ˆ †x‰† † œ œ Ê
s
ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰ ˆ ‰
xÄ !
lim ( 15) rad/sec;
œ œ œ œ œ
xÄ !ˆx# 908100‰ 6 ddt# s cos# 90 dsdt Šs#90‹ˆ ‰ ˆ ‰xs dxdt ˆ s90# ‰ ˆ ‰dxdt
" ) #
) † †x
s
lim rad/sec œˆx#908100‰dxdt Ê ddt# œ"6
xÄ ! )
38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance between the ships. By the Law of Cosines, D#œa#b#2ab cos 120°
2a 2b a b . When a 5, 14, b 3, and 21, then
Ê dDdt œ#"D dadt dbdt dbdt dadt‘ œ dadt œ œ dbdt œ dDdt œ 4132D
where Dœ7. The ships are moving dDdt œ29.5 knots apart.
11. f(x)œsin x Ê f (x)w œcos x
(a) L(x)œf (0)(xw 0) f(0)œ1(x 0) 0 L(x) x at x 0
Ê œ œ
(b) L(x)œf ( )(xw 1 1)f( )1 œ ( 1)(x1)0 L(x) x at x
Ê œ 1 œ1
12. f(x)œcos x Ê f (x)w œ sin x
(a) L(x)œf (0)(xw 0) f(0)œ0(x 0) 1 L(x) 1 at x 0
Ê œ œ
(b) L(x)œfwˆ1#‰ ˆx#1‰ fˆ 1#‰ ( 1) xœ ˆ 1#‰ Ê0 L(x)œ x 12 at xœ 1#
13. f(x)œsec x Ê f (x)w œsec x tan x
(a) L(x)œf (0)(xw 0) f(0)œ0(x 0) 1 L(x) 1 at x 0
Ê œ œ
(b) L(x)œfwˆ13‰ ˆx13‰ fˆ 13‰
2œ È3 xˆ 13‰ Ê2 L(x)œ 2 2È3 xˆ 13‰ at xœ 13
14. f(x)œtan x Ê f (x)w œsec x#
(a) L(x)œf (0)(xw 0) f(0)œ1(x œ0) 0 x L(x) x at x 0
Ê œ œ
(b) L(x)œfwˆ ‰ ˆ14 x14‰fˆ ‰14 œ2 xˆ 14‰1 L(x) 1 2 x at x
Ê œ ˆ 14‰ œ14
15. f xwa bœ " ka xbk". We have fa b! œ " and fwa b! œk. L xa bœ ! !fa b fwa bax ! œ " b k xa ! œ " b kx 16. (a) f xa bœ " a xb'œ " a xb‘'¸ " ' œ " 'a xb x
(b) f xa bœ " #x œ # " a xb‘"¸ # " " a ba xb‘œ # #x (c) f xa bœ " a xb"Î#¸ " ˆ "#‰xœ " #x
(d) f xa bœ " È x# œ # " È Š x##‹"Î#¸ # " È Š # #"x#‹œ # " È Š x%#‹ (e) f xa bœ % $a xb"Î$œ %"Î$ˆ" $%x‰"Î$¸ %"Î$ˆ" " $$ %x‰œ %"Î$ˆ" %x‰
(f) f xa bœ " ˆ # "x‰2Î$œ " ’ ˆ # "x‰“2Î$¸ " #$ˆ # "x‰œ " '$#x 17. (a) (1.0002)&!œ(10.0002)&!¸ 1 50(0.0002)œ 1 .01œ1.01 (b) 1.009$È œ(10.009)"Î$¸ 1 ˆ ‰"3 (0.009)œ 1 0.003œ1.003
18. f(x)œÈx 1 sin xœ(x1)"Î#sin x Ê f (x)w œˆ ‰"# (x1)"Î#cos x Ê L (x)f œf (0)(xw 0) f(0) (x 0) 1 L (x) x 1, the linearization of f(x); g(x) x 1 (x 1) g (x)
œ 3# Ê f œ 3# œÈ œ "Î# Ê w
(x 1) L (x) g (0)(x 0) g(0) (x 0) 1 L (x) x 1, the linearization of g(x);
œˆ ‰"# "Î# Ê g œ w œ"# Ê g œ "#
h(x)œsin x Ê h (x)w œcos x Ê L (x)h œh (0)(xw 0) h(0)œ(1)(x Ê0) 0 L (x)h œx, the linearization of h(x). L (x)f œL (x)g L (x) implies that the linearization of a sum is equal to the sum of the linearizations.h
19. yœx$3Èxœx$3x"Î# Ê dyœˆ3x##3x"Î#‰ dx Ê dyœŠ3x#2È3x‹ dx
20. yœxÈ1x# œx 1a x#b"Î# Ê dyœ’(1) 1a x#b"Î#(x)ˆ ‰"# a1x#b"Î#( 2x) dx “
1 x 1 x x dx dx
œa #b"Î#ca #b #d œaÈ12xb
1 x
#
#
21. yœ1 x2x# Ê dyœ (2) 11xx(2x)(2x) dxœ 212xx dx
#
## ##
Š a a b b ‹ a #b
22. yœ3 12 xx œ3 12xx Ê dyœ x 3 1 x 2x x dxœ 3x 3 3 dx
9 1 x 9 1 x
È
ˆ È ‰ a b
ˆ ˆ ‰‰ ˆ ‰
a b a b
"Î# "Î#
"Î#
"Î# "Î# "Î# "Î#
#
"Î## "Î##
Š 3 ‹
dy dx
Ê œ "
3Èx 1ˆ Èx‰#
23. 2y$Î#xy œx 0 Ê 3y"Î#dyy dxx dydxœ0 Ê 3yˆ "Î#x dy‰ œ(1y) dx Ê dyœ 3È1 yy x dx 24. xy#4x$Î# œy 0 Ê y dx# 2xy dy6x"Î#dxdyœ0 Ê (2xy1) dyœˆ6x"Î#y dx#‰
dy dx
Ê œ 62xyÈx1y#
25. yœsin 5ˆ Èx‰œsin 5xˆ "Î#‰ Ê dyœˆcos 5xˆ "Î#‰‰ ˆ#5x"Î#‰ dx Ê dyœ 5 cos 5 x dx
2 x ˆ ȉ È
26. yœcos x a b# Ê dyœ c sin xa b# d(2x) dxœ 2x sin x dxa b#
27. yœ4 tanŠ ‹x3$ Ê dyœ4 secŠ #Š ‹ a bx3$ ‹ x dx # Ê dyœ4x sec# #Š ‹x3$ dx
28. yœsec xa #1 b Ê dyœcsec xa #1 tan xb a #1 (2x) dxbd œ2x sec xc a #1 tan xb a #1 dxbd 29. yœ3 csc 1ˆ 2Èx‰œ3 csc 1ˆ 2x"Î#‰ Ê dyœ3ˆcsc 1ˆ 2x"Î#‰‰ cot 1ˆ 2x"Î#‰ ˆx"Î#‰ dx
dy csc 1 2 x cot 1 2 x dx Ê œ È3x ˆ È ‰ ˆ È ‰
30. yœ2 cotŠÈ" ‹œ2 cot xˆ "Î#‰ Ê dyœ 2 csc#ˆx"Î#‰ ˆ"#‰ ˆx$Î#‰ dx Ê dyœ È" csc#ŠÈ" ‹ dx
x x$ x
31. f(x)œx#2x, x!œ1, dxœ0.1 Ê f (x)w œ2x2
(a) f? œf(x!dx)f(x )! œf(1.1)f(1)œ3.41 œ3 0.41 (b) dfœf (x ) dxw ! œ[2(1)2](0.1)œ0.4
(c) fk? dfkœk0.410.4kœ0.01
32. f(x)œ2x#4x3, x! œ 1, dxœ0.1 Ê f (x)w œ4x4 (a) f? œf(x!dx)f(x )! œ f( .9) œf( 1) .02 (b) dfœf (x ) dxw ! œ[4( 1) 4](.1)œ0
(c) fk? dfkœk.02 œ0k .02
33. f(x)œx$x, x!œ1, dxœ0.1 Ê f (x)w œ3x#1 (a) f? œf(x!dx)f(x )! œf(1.1)f(1)œ.231 (b) dfœf (x ) dxw ! œ[3(1)#1](.1)œ.2
(c) fk? dfkœk.231.2kœ.031 34. f(x)œx , x% !œ1, dxœ0.1 Ê f (x)w œ4x$
(a) f? œf(x!dx)f(x )! œf(1.1)f(1)œ.4641 (b) dfœf (x ) dxw ! œ4(1) (.1)$ œ.4
(c) fk? dfkœk.4641.4kœ.0641
35. f(x)œx", x!œ0.5, dxœ0.1 Ê f (x)w œ x# (a) f? œf(x!dx)f(x )! œf(.6)f(.5)œ "
3
(b) dfœf (x ) dxw ! œ ( 4)ˆ ‰10" œ 52 (c) fk? dfkœ ¸ "3 52¸œ 15"
36. f(x)œx$2x3, x!œ2, dxœ0.1 Ê f (x)w œ3x#2 (a) f? œf(x!dx)f(x )! œf(2.1)f(2)œ1.061 (b) dfœf (x ) dxw ! œ(10)(0.10)œ1
(c) fk? dfkœk1.061 œ1k .061
37. Vœ 431r $ Ê dVœ4 r dr 1#! 38. Vœx $ Ê dVœ3x dx#! 39. Sœ6x # Ê dSœ12x dx!
40. Sœ1rÈr#h#œ1r ra#h#b"Î#, h constant Ê dSdr œ1ar#h#b"Î#1r r r† a#h#b"Î#
dS dr, h constant Ê dSdr œ r h r Ê œ
r h
2r h
r h
1a b 1 1
È
a b
É
# # #
# #
# #
!
#! #
41. Vœ1r h, height constant # Ê dVœ2 r h dr 1! 42. Sœ2 rh 1 Ê dSœ2 r dh1 43. Given rœ2 m, drœ.02 m
(a) Aœ1r # Ê dAœ2 r dr1 œ2 (2)(.02)1 œ.08 m1 # (b) ˆ.08411‰(100%)œ2%
44. Cœ2 r and dC1 œ2 in. Ê dCœ2 dr 1 Ê drœ "1 Ê the diameter grew about in.; A12 œ1r # Ê dAœ2 r dr1 2 (5)œ 1 ˆ ‰"1 œ10 in.#
45. The volume of a cylinder is Vœ1r h. When h is held fixed, we have # dVdr œ #1rh, and so dVœ #1rh dr. For hœ $! in., rœ ' in., and drœ !Þ& in., the volume of the material in the shell is approximately dVœ #1rh drœ # ' $! !Þ&1a ba ba b in .œ ")! ¸ &'&Þ&1 $
46. Let )œangle of elevation and hœheight of building. Then hœ $!tan , so dh) œ $!sec d . We want dh#) ) l l !Þ!%h, which gives: l$!sec d#) )l !Þ!%l$!tan)l Ê cos"#)l l d) !Þ!%cos sin )) Ê l l !Þ!%d) sin cos ) )Ê l l !Þ!%d) sin &"#1 cos &"#1
radian. The angle should be measured with an error of less than radian (or approximatley degrees), œ !Þ!" !Þ!" !Þ&(
which is a percentage error of approximately !Þ('%.
47. Vœ1h $ Ê dVœ3 h dh; recall that V1 # ? ¸dV. Then k?VkŸ(1%)(V)œ(1)100a1h$b Ê dVk kŸ (1)100a1h$b 3 h dh dh h % h. Therefore the greatest tolerated error in the measurement Ê k 1 # kŸ (1)100a1h$b Ê k kŸ 300" œˆ"3 ‰
of h is %."3
48. (a) Let D represent the inside diameter. Then Vi œ1r h# œ1ˆ ‰D#i #hœ 1D h4i# and hœ10 Ê Vœ 5 D1#i# Ê dVœ5 D dD . Recall that V1 i i ? ¸dV. We want k?VkŸ(1%)(V) Ê dVk kŸˆ100" ‰ Š5 D1##i‹œ140D#i
5 D dD 200. The inside diameter must be measured to within 0.5%.
Ê 1 i i Ÿ1D Ê Ÿ
40 D
# dD
i i
i
(b) Let D represent the exterior diameter, h the height and S the area of the painted surface. Se œ1D he ÊdSœ1hdDe
. Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter Ê dSS œ dDDe
e
is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to within 5%, the tanks's exterior diameter must be measured to within 5%.
49. Vœ1r h, h is constant # Ê dVœ2 rh dr; recall that V1 ? ¸dV. We want k?VkŸ 1000" V Ê dVk kŸ10001r h# 2 rh dr dr (.05%)r a .05% variation in the radius can be tolerated.
Ê k 1 kŸ 10001r h# Ê k kŸ#000r œ Ê 50. Volumeœ(x?x)$œx$3x ( x)# ? 3x( x)? #( x)? $
51. Wœ œ a bg a bg" Ê dWœ bg# dgœ b dgg# Ê dWdWmoon œ œ 5.232 #œ37.87, so a change of
earth
Š ‹
Š ‹
b dg (5.2) b dg (32)
#
#
ˆ ‰
gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
52. (a) Tœ21Š ‹Lg "Î# Ê dTœ21ÈLˆ"#g$Î#‰ dgœ 1ÈL g$Î# dg
(b) If g increases, then dg0 Ê dT0. The period T decreases and the clock ticks more frequently. Both the pendulum speed and clock speed increase.
(c) 0.001œ 1È100 980ˆ $Î#‰ dg Ê dg¸ 0.977 cm/sec # Ê the new g¸979 cm/sec#
53. The error in measurement dxœ(1%)(10)œ0.1 cm; Vœx $ Ê dVœ3x dx# œ3(10) (0.1)# œ30 cm $ Ê the percentage error in the volume calculation is ˆ100030 ‰(100%)œ3%
54. Aœs # Ê dAœ2s ds; recall that A? ¸dA. Then k?AkŸ(2%)Aœ 1002s# œ 50s# Ê dAk kŸ50s# Ê 2s dsk kŸ 50s# ds (1%) s the error must be no more than 1% of the true value.
Ê k kŸ(2s)(50)s# œ 100s œ Ê
55. Given Dœ100 cm, dDœ1 cm, Vœ 431ˆ ‰D# $œ 1D6$ Ê dVœ #1D dD# œ #1(100) (1)# œ 10#%1. Then dVV (100%) 10 %œ”1010 •a bœ”1010 •%œ3%
6 6
% '
# #
' '
1 1
1 # 1
56. Vœ 431r$œ 431ˆ ‰D# $ œ1D6$ Ê dVœ 1D## dD; recall that V? ¸dV. Then k?VkŸ(3%)Vœˆ1003 ‰Š1D6$‹
dV dD dD (1%) D the allowable percentage error in
œ 1200D$ Ê k kŸ 1200D$ Ê ¹1D## ¹Ÿ1#D00$ Ê k kŸ100D œ Ê measuring the diameter is 1%.
57. A 5% error in measuring t Ê dtœ(5%)tœ 20t . Then sœ16t # Ê dsœ32t dtœ32tˆ ‰20t œ32t20# œ 16t10# œˆ ‰10" s (10%)s a 10% error in the calculation of s.
œ Ê
58. From Example 8 we have dVV œ4 drr . An increase of 12.5% in r will give a 50% increase in V.
59. lim 1 60. lim lim (1)(1) 1
xÄ0 xÄ0 xÄ0
È1 x È1 0
1 1 x x cos x
tan x sin x
x "
# œ 0# œ œ ˆ ‰ ˆ ‰œ œ
61. E(x)œf(x)g(x) Ê E(x)œf(x)m(x a) c. Then E(a)œ0 Ê f(a)m(a œa) c 0 Ê œ c f(a). Next we calculate m: lim xÄa xE(x)aœ0 Ê lim xÄa f(x)m(xxaa) c œ0 Ê lim xÄa ’f(x)xf(a)a m“œ0 (since cœf(a))
f (a) m 0 m f (a). Therefore, g(x) m(x a) c f (a)(x a) f(a) is the linear approximation, Ê w œ Ê œ w œ œ w
as claimed.
62. (a) i. Q aa bœf a implies that ba b !œf a .a b
ii. Since Q xwa bœb" #b x#a a , Q ab wa bœf a implies that bwa b "œf a .wa b iii. Since Q xwwa bœ #b , Q a# wwa bœfwwa ba implies that b"œ fww#a ba .
In summary, b!œf a , ba b "œf a , and bwa b "œ fww#a ba. (b) f xa bœ " a xb"
f xwa bœ " " a xb a# " œ " b a xb# fwwa bx œ # " a xb a$ " œ # " b a xb$
Since fa b! œ ", fwa b! œ ", and fwwa b! œ #, the coefficients are b!œ ", b"œ ", b# œ œ "## . The quadratic approximation is Q xa bœ " x x .#
(c) As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) g xa bœx"
g xwa bœ "x# g xwwa bœ #x$
Since ga b" œ ", gwa b" œ ", and gwwa b" œ # , the coefficients are b! œ ", b"œ ", b#œ œ "## . The quadratic
approximation is Q xa bœ " ax " b ax "b#.
As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.
(e) h xa bœ " a xb"Î#
h xwa bœ " "#a xb"Î#
h xwwa bœ " "%a xb$Î#
Since ha b! œ ", hwa b! œ "#, and hwwa b! œ "% , the coefficients are b!œ ", b"œ"#, b# œ"% œ ". The quadratic
2 8
approximation is Q xa bœ " x# x8#.
As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function u x at xa b œa is L xa bœu aa bu a xwa ba abœb!b x"a a , whereb b and b are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization! "
for f x at xa b œ ! " is x; the linearization for g x at xa b œ " " is ax "b or # x; and the linearization for h x ata b xœ ! " is .x#
63. (a) xœ1
(b) xœ1; mœ2.5, e1¸2.7 xœ0; mœ1, e0œ1 xœ 1; mœ0.3, e1¸0.4
64. If f has a horizontal tangent at xœa, then f (a)w œ0 and the linearization of f at xœa is L(x)œf(a)f (a)(xw œa) f(a)0 (x† œa) f(a). The linearization is a constant.
65. Find v when ml l œ "Þ!"m . m! œ Ê m " œm!Ê " œ Ê " œ Êv œc "
"
# #
m v v m v
c c m c m m
m m
! ! ! !
##
# #
# # # # #
# # #
É vc É É Š ‹
v c dv c dm, dm m dv . m m ,
Ê l l œ É" mm Ê œ † Š" mm ‹ Šmm ‹ œ !Þ!" Ê œ c m ˆ ‰ œ
m
! ! ! ! m!
# # # #
# # $
$
" "!"
# "Î# # "!! "!!
! !
"
Ê m
m!#
#
dvœ c m† m œ 1 ¸0.69c. Body at restÊv œ ! and vœv dv
" "!! !!!
"!" " ! !
! !
#
"!"$ $
"!!$ ! "!"#!#
"!!# !#
"!!#
"!"#
m3 m
m ÍÍ Í Ì
Ê
ˆ ‰
vÊ œ0.69c.
66. (a) The successive square roots of 2 appear to converge to the number 1. For tenth roots the convergence is more rapid.
(b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge
to 1.
A graph indicates what is going on:
Starting on the line yœx, the succesSive square roots are found by moving to the graph of yœÈx and then across to the line yœx again. From any positive starting value x, the iterates converge to 1.
67-70. Example CAS commands:
: Maple
with(plots):
a:= 1: f:=x -> x• • 3 x 2 2*x;
plot(f(x), x= 1..2);
diff(f(x),x);
fp := unapply ( ,x);ww
L:=x -> f(a)fp(a)*(xa);
plot({f(x), L(x)}, x= 1..2);
err:=x -> abs(f(x)L(x));
plot(err(x), x= 1..2, title = absolute error function ); # #
err( 1);
: (function, x1, x2, and a may vary):
Mathematica
Clear[f, x]
{x1, x2} = { 1, 2}; a = 1;
f[x_]:=x3x 2 2x Plot[f[x], {x, x1, x2}]
lin[x_]=f[a]f'[a](xa) Plot[{f[x], lin[x]}, {x, x1, x2}]
err[x_]=Abs[f[x]lin[x]]
Plot[err[x], {x, x1,x 2}]
err//N
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) eps = 0.5; del = 0.4
Plot[{err[x], eps},{x, adel, adel}]