Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 14.5 THE CHAIN RULE 0 207
39. First we find :~ implicitly by taking partial derivatives of both sides with respect to R1 :
oR . R 2
oR
1
= Ri. Then by symmetry,
oR R 2 2
oR = R2 . When RI = 25, R2 = 40 and Ra = 50, -R 1 = 2 1 0 7 0 ¢::} R = 2 1°7° n. Since the possible error
oR2 - ~· oR3 R 3
for each Ri is 0.5%, the maximum error of R is attained by setting t:J.R, = 0.005R.. So
oR oR oR 2 ( 1 1 1 ) 1
!:l.R ~=:::: dR = oR !:l.R1
1
+ oR
2
!:l.R2 + oR
3
!:l.R3 = (0.005)R R
1
+ R 2
+ Ra = (0.005)R = l7 ~=:::: 0.059 st.
~ l!:l.wl l!:l.hl . . .
41 . The errors in measurement are at ~ost 27o, so w $ 0 .. 02 and T $ 0.02. The relative error m the calculated surface
area is
!:l.S dS 0.1091(0.425w 0·425 - 1 )h 0·725 dw + 0.1091w 0 .4 25 (0.725h 0·725 - 1 } dh dw dh
s 1':::: s= 0.1091w0.426h0.725 . = 0.425-:;:; + 0.725/i:
To estimate the maximum relative error, we use : = I ~w I = 0.02 and ~~i = I ~hI = 0.02 =>
dS = 0.425 (0.02) + 0. 725 (0.02} = 0.023. Thus the maximum percentage error is approximately 2.3%.
s
43. !:l.z = f(a + !:l.x, b + !:l.y) - f(a, b) =(a+ !:l.x? + (b + !:l.y) 2 - (a 2 + b 2 )
= a 2 + 2a!:l.x + (!:l.x) 2 + b 2 + 2b!:l.y + (!:l.y?- a 2 - b 2 = 2a !:l.x + (!:l.x? + 2b!:l.y + (!:l.y?
But f,(a, b) = 2a and /y(a, b) = 2b and so !:l.z = f x(a, b) !:l.x + /y(a, b) !:l.y + !:l.x !:l.x + !:l.y !:l.y, which is Definition 7
with e:1 = !:l.x and e: 2 = !:l.y. Hence f is differentiable.
45. To show that f is continuous at (a, b) we need to show that lim f(x, y) = f(a, b) or
(z,y)~(a,b)
equivalently lim f(a + !:l.x, b + !:l.y) = f(a, b). Since f is differentiable at (a, b),
(6x,6y)-(O,O )
f(a + !:l.x, b + !:l.y) - f(a, b)= !:l.z = f,(a, b) 6-x + /y(a, b) !:l.y + e:1 !:l.x + e:2 !:l.y, where €1 and €2 ---+ 0 as
(!:l.x, !:l.y} --> (0, 0). Thus f(a + !:l.x, b + !:l.y) = f(a, b)+ f,(a, b) !:l.x + / 11 (a, b) !:l.y + e:1 !:l.x + e:2 !:l.y. Taking the limit of
both sides as (.C:..x, .C:..y) ___.. (0, 0) gives lim f (a + .C:..x, b + .C:..y) = f (a , b). Thus f is continuous at (a, b).
(6 x,61J)- (0 ,0) .
14.5 The Chain Rule
1. z = x? + y 2 + xy, x =sin t, y = et =>
~ &~ &~ t
-d =,.--d +,.--d =(2x+ y}cost+(2y +x)e
t uX t uy t
3. z =· .J1 + x 2 + y 2 , x = In t, y = cost =>
dz oz dx 8z dy 1 2 2 -1/2 1 1 2 2 -1/ 2 . 1 (x . )
- =-- + --= 2 (1+x + y) (2x)·- + 2 (1+x + y) (2y)(-smt) = - - ysmt
dt ox dt 8y dt t · .J1 + x2 + y2 t
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