Solução_Calculo_Stewart_6e

F.
190 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
31. x 2 + y 2 + z 2 =3xyz, soletF (x, y, z) =x 2 + y 2 + z 2 − 3xyz =0. Then by Equations 7
∂z
∂x = −Fx F z
2x − 3yz 3yz − 2x
= − =
2z − 3xy 2z − 3xy
and
∂z
∂y = −Fy F z
33. x − z =arctan(yz),soletF (x, y, z) =x − z − arctan(yz) =0.Then
∂z
∂x = −Fx F z
∂z
∂y = −F y
F z
2y − 3xz 3xz − 2y
= − =
2z − 3xy 2z − 3xy .
1
= −
= 1+y2 z 2
1
−1 −
1+(yz) (y) 1+y + y 2 z 2
2
1
−
1+(yz) (z)
z
= −
2 1+y
= −
2 z 2
z
1
−1 −
1+(yz) (y) 1+y 2 z 2 = −
+ y 1+y + y 2 z 2
2 1+y 2 z 2
35. Since x and y are each functions of t, T (x, y) is a function of t,sobytheChainRule, dT
dt = ∂T dx
∂x dt + ∂T dy
∂y dt .After
3 seconds, x = √ 1+t = √ 1+3=2, y =2+ 1 t dx
3 =2+1 (3) = 3,
3
dt = 1
2 √ 1+t = 1
2 √ 1+3 = 1 dy
,and
4 dt = 1 3 .
Then dT
dt
dx
dy
= Tx(2, 3) + Ty(2, 3)
dt dt =4
1
4 +3 1
3 =2. Thus the temperature is rising at a rate of 2 ◦ C/s.
37. C =1449.2+4.6T − 0.055T 2 +0.00029T 3 +0.016D, so ∂C
∂T =4.6 − 0.11T +0.00087T 2 and ∂C
∂D =0.016.
According to the graph, the diver is experiencing a temperature of approximately 12.5 ◦ C at t =20minutes, so
∂C
∂T =4.6 − 0.11(12.5) + 0.00087(12.5)2 ≈ 3.36. By sketching tangent lines at t =20to the graphs given, we estimate
dD
dt ≈ 1 dT
and
2 dt ≈−1 dC
. Then, by the Chain Rule,
10 dt = ∂C dT
∂T dt + ∂C dD
∂D dt ≈ (3.36)
− 1 10 +(0.016) 1
2 ≈−0.33.
Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 0.33 m/sperminute.
39. (a) V = wh,sobytheChainRule,
dV
dt = ∂V d
∂ dt + ∂V dw
∂w dt + ∂V dh d dw dh
= wh + h + w
∂h dt dt dt dt =2· 2 · 2+1· 2 · 2+1· 2 · (−3) = 6 m3 /s.
(b) S =2(w + h + wh),sobytheChainRule,
dS
dt = ∂S d
∂ dt + ∂S dw
∂w dt + ∂S dh
d
dw
dh
=2(w + h) +2( + h) +2( + w)
∂h dt dt dt dt
=2(2+2)2+2(1+2)2+2(1+2)(−3) = 10 m 2 /s
(c) L 2 = 2 + w 2 + h 2 ⇒ 2L dL
dt
dL/dt =0m/s.
d dw dh
=2 +2w +2h = 2(1)(2) + 2(2)(2) + 2(2)(−3) = 0
dt dt dt ⇒
41.
dP dT
=0.05,
dt dt =0.15, V =8.31 T dV
and
P dt = 8.31 dT
P dt − 8.31 T dP
. Thus whenP =20and T = 320,
P 2 dt
dV 0.15
dt =8.31 20 − (0.05)(320)
≈−0.27 L/s.
400
43. Let x be the length of the first side of the triangle and y the length of the second side. The area A of the triangle is given by
A = 1 xy sin θ where θ is the angle between the two sides. Thus A is a function of x, y,andθ,andx, y,andθ are each in
2