Solução_Calculo_Stewart_6e

F.
186 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
(3, 2, 6) is given by
f(x, y, z) ≈ f(3, 2, 6) + f x(3, 2, 6)(x − 3) + f y(3, 2, 6)(y − 2) + f z(3, 2, 6)(z − 6)
=7+ 3 7 (x − 3) + 2 7 (y − 2) + 6 7 (z − 6) = 3 7 x + 2 7 y + 6 7 z
Thus (3.02) 2 +(1.97) 2 +(5.99) 2 = f(3.02, 1.97, 5.99) ≈ 3 (3.02) + 2 (1.97) + 6 7 7 7
(5.99) ≈ 6.9914.
23. From the table, f(94, 80) = 127. Toestimatef T (94, 80) and f H (94, 80) we follow the procedure used in Section 15.3
f(94 + h, 80) − f(94, 80)
[ET 14.3]. Since f T (94, 80) = lim
, we approximate this quantity with h = ±2 and use the
h→0 h
values given in the table:
f T (94, 80) ≈
f(96, 80) − f(94, 80)
2
=
135 − 127
2
=4, f T (94, 80) ≈
f(92, 80) − f(94, 80)
−2
=
119 − 127
−2
f(94, 80 + h) − f(94, 80)
Averaging these values gives f T (94, 80) ≈ 4. Similarly, f H (94, 80) = lim
,soweuseh = ±5:
h→0 h
f H(94, 80) ≈
f(94, 85) − f(94, 80)
5
=
132 − 127
5
=1, f H(94, 80) ≈
Averaging these values gives f H (94, 80) ≈ 1. The linear approximation, then, is
f(94, 75) − f(94, 80)
−5
f(T,H) ≈ f(94, 80) + f T (94, 80)(T − 94) + f H (94, 80)(H − 80)
≈ 127 + 4(T − 94) + 1(H − 80) [or 4T + H − 329]
=
122 − 127
−5
Thus when T =95and H =78, f(95, 78) ≈ 127 + 4(95 − 94) + 1(78 − 80) = 129, so we estimate the heat index to be
approximately 129 ◦ F.
25. z = x 3 ln(y 2 ) ⇒ dz = ∂z ∂z
dx +
∂x ∂y dy =3x2 ln(y 2 ) dx + x 3 1 ·
y 2(2y) dy =3x2 ln(y 2 ) dx + 2x3
y dy
27. m = p 5 q 3 ⇒ dm = ∂m ∂m
dp +
∂p ∂q dq =5p4 q 3 dp +3p 5 q 2 dq
29. R = αβ 2 cos γ ⇒ dR = ∂R ∂R ∂R
dα + dβ +
∂α ∂β ∂γ dγ = β2 cos γdα+2αβ cos γdβ− αβ 2 sin γdγ
31. dx = ∆x =0.05, dy = ∆y =0.1, z =5x 2 + y 2 , z x =10x, z y =2y. Thus when x =1and y =2,
dz = z x (1, 2) dx + z y (1, 2) dy = (10)(0.05) + (4)(0.1) = 0.9 while
∆z = f(1.05, 2.1) − f(1, 2) = 5(1.05) 2 +(2.1) 2 − 5 − 4=0.9225.
33. dA = ∂A ∂A
dx + dy = ydx+ xdyand |∆x| ≤ 0.1, |∆y| ≤ 0.1. Weusedx =0.1, dy =0.1 with x =30, y =24;
∂x ∂y
then the maximum error in the area is about dA =24(0.1) + 30(0.1) = 5.4 cm 2 .
35. ThevolumeofacanisV = πr 2 h and ∆V ≈ dV is an estimate of the amount of tin. Here dV =2πrh dr + πr 2 dh,soput
dr =0.04, dh =0.08 (0.04 on top, 0.04 on bottom) and then ∆V ≈ dV =2π(48)(0.04) + π(16)(0.08) ≈ 16.08 cm 3 .
Thus the amount of tin is about 16 cm 3 .
=4
=1