Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.5 THE CHAIN RULE ET SECTION 14.5 ¤ 187
37. TheareaoftherectangleisA = xy,and∆A ≈ dA is an estimate of the area of paint in the stripe. Here dA = ydx+ xdy,
so with dx = dy = 3+3
12 = 1 2 , ∆A ≈ dA = (100) 1
2
+ (200)
1
2
= 150 ft 2 . Thus there are approximately 150 ft 2 of paint
in the stripe.
39. First we find ∂R implicitly by taking partial derivatives of both sides with respect to R 1 :
∂R 1
∂ 1
= ∂ [(1/R 1)+(1/R 2 )+(1/R 3 )]
⇒ −R −2 ∂R
= −R −2
1 ⇒ ∂R = R2
.Thenbysymmetry,
∂R 1 R
∂R 1 ∂R 1 ∂R 1
∂R
= R2
,
∂R 2
R 2 2
∂R
= R2
.WhenR 1 =25, R 2 =40and R 3 =50,
∂R 3
R 2 3
1
R = 17
200
R 2 1
⇔ R = 200
17 Ω.
Since the possible error for each R i is 0.5%, the maximum error of R is attained by setting ∆R i =0.005R i .So
∆R ≈ dR = ∂R ∆R 1 + ∂R ∆R 2 + ∂R
1
∆R 3 =(0.005)R 2 + 1 + 1
=(0.005)R = 1 ≈ 0.059 Ω.
17
∂R 1 ∂R 2 ∂R 3 R 1 R 2 R 3 41. The errors in measurement are at most 2%,so
∆w
w ≤ 0.02 and ∆h
h ≤ 0.02. The relative error in the calculated surface
area is
∆S
S
≈ dS S = 0.1091(0.425w0.425−1 )h 0.725 dw +0.1091w 0.425 (0.725h 0.725−1 ) dh
=0.425 dw 0.1091w 0.425 h 0.725 w +0.725 dh h
To estimate the maximum relative error, we use dw
w = ∆w
w =0.02 and dh
h = ∆h
h =0.02 ⇒
dS
S
=0.425 (0.02) + 0.725 (0.02) = 0.023. Thus the maximum percentage error is approximately 2.3%.
43. ∆z = f(a + ∆x, b + ∆y) − f(a, b) =(a + ∆x) 2 +(b + ∆y) 2 − (a 2 + b 2 )
= a 2 +2a ∆x +(∆x) 2 + b 2 +2b ∆y +(∆y) 2 − a 2 − b 2 =2a ∆x +(∆x) 2 +2b ∆y +(∆y) 2
But f x (a, b) =2a and f y (a, b) =2b and so ∆z = f x (a, b) ∆x + f y (a, b) ∆y + ∆x ∆x + ∆y ∆y,whichisDefinition 7
with ε 1 = ∆x and ε 2 = ∆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
(x,y)→(a,b)
equivalently
lim
(∆x,∆y)→(0,0)
f(a + ∆x, b + ∆y) =f(a, b). Sincef is differentiable at (a, b),
f(a + ∆x, b + ∆y) − f(a, b) =∆z = f x (a, b) ∆x + f y (a, b) ∆y + ε 1 ∆x + ε 2 ∆y, where 1 and 2 → 0 as
(∆x, ∆y) → (0, 0). Thusf(a + ∆x, b + ∆y) =f(a, b)+f x(a, b) ∆x + f y(a, b) ∆y + ε 1 ∆x + ε 2 ∆y. Taking the limit of
both sides as (∆x, ∆y) → (0, 0) gives
lim
(∆x,∆y)→(0,0)
f(a + ∆x, b + ∆y) =f(a, b). Thus f is continuous at (a, b).
15.5 The Chain Rule ET 14.5
1. z = x 2 + y 2 + xy, x =sint, y = e t ⇒ dz
dt = ∂z dx
∂x dt + ∂z dy
=(2x + y)cost +(2y + x)et
∂y dt
3. z = 1+x 2 + y 2 , x =lnt, y =cost ⇒
dz
dt = ∂z dx
∂x dt + ∂z dy
∂y dt = 1 (1 + 2 x2 + y 2 ) −1/2 (2x) · 1
t + 1 (1 + 2 x2 + y 2 ) −1/2 1
x
(2y)(− sin t) =
1+x2 + y 2 t − y sin t