Solução_Calculo_Stewart_6e

F.
182 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
variables, so Definition 4 says that f xy (x, y) = ∂
∂y [f f x(x, y + h) − f x(x, y)
x(x, y)] = lim
h→0 h
f x(3, 2+h) − f x(3, 2)
f xy(3, 2) = lim
.
h→0 h
We can estimate this value using our previous work with h =0.2 and h = −0.2:
f xy (3, 2) ≈ f x(3, 2.2) − f x (3, 2)
0.2
=
16.8 − 12.2
0.2
=23, f xy (3, 2) ≈ f x(3, 1.8) − f x (3, 2)
−0.2
Averaging these values, we estimate f xy (3, 2) to be approximately 23.25.
=
⇒
7.5 − 12.2
−0.2
71. u = e −α2 k 2t sin kx ⇒ u x = ke −α2 k 2t cos kx, u xx = −k 2 e −α2 k 2t sin kx,andu t = −α 2 k 2 e −α2 k 2t sin kx.
Thus α 2 u xx = u t .
73. u =
1
x2 + y 2 + z 2 ⇒ u x = − 1 2
(x 2 + y 2 + z 2 ) −3/2 (2x) =−x(x 2 + y 2 + z 2 ) −3/2 and
u xx = −(x 2 + y 2 + z 2 ) −3/2 − x − 3 2
(x 2 + y 2 + z 2 ) −5/2 (2x) = 2x2 − y 2 − z 2
(x 2 + y 2 + z 2 ) 5/2 .
By symmetry, u yy = 2y2 − x 2 − z 2
(x 2 + y 2 + z 2 ) 5/2 and u zz = 2z2 − x 2 − y 2
(x 2 + y 2 + z 2 ) 5/2 .
Thus u xx + u yy + u zz = 2x2 − y 2 − z 2 +2y 2 − x 2 − z 2 +2z 2 − x 2 − y 2
(x 2 + y 2 + z 2 ) 5/2 =0.
75. Let v = x + at, w = x − at. Thenu t = ∂[f(v)+g(w)]
∂t
u tt = ∂[af 0 (v) − ag 0 (w)]
∂t
=
df (v) ∂v
dv
∂t + dg(w)
dw
=23.5.
∂w
∂t = af 0 (v) − ag 0 (w) and
= a[af 00 (v)+ag 00 (w)] = a 2 [f 00 (v)+g 00 (w)]. Similarly, by using the Chain Rule we have
u x = f 0 (v)+g 0 (w) and u xx = f 00 (v)+g 00 (w). Thus u tt = a 2 u xx .
77. z =ln(e x + e y ) ⇒ ∂z
∂x =
∂ 2 z
∂x 2 = ex (e x + e y ) − e x (e x )
(e x + e y ) 2 =
ex ∂z
and
e x + ey ∂y =
∂ 2 z
∂y = ey (e x + e y ) − e y (e y ) e x+y
=
2 (e x + e y ) 2 (e x + e y ) . Thus 2
∂ 2 z ∂ 2 z ∂ 2 2
∂x 2 ∂y − z
=
2 ∂x∂y
ey ∂z
,so
e x + ey ∂x + ∂z
∂y =
ex
e x + e y +
ey
e x + e y
e x+y
(e x + e y ) 2 , ∂ 2 z
∂x∂y = 0 − ey (e x )
(e x + e y ) 2 = − ex+y
(e x + e y ) 2 , and
e x+y
(e x + e y ) 2 ·
e x+y
(e x + e y ) − 2
= ex + e y
e x + e y =1.
2
−
ex+y
= (ex+y ) 2
(e x + e y ) 2 (e x + e y ) − (ex+y ) 2
4 (e x + e y ) =0 4
79. If we fix K = K 0 ,P(L, K 0 ) is a function of a single variable L,and dP
dL = α P is a separable differential equation. Then
L
dP
dP
P = αdL L ⇒ P = α dL ⇒ ln |P | = α ln |L| + C (K0), whereC(K0) can depend on K0. Then
L
|P | = e α ln|L| + C(K 0) , and since P>0 and L>0,wehaveP = e α ln L e C(K 0) = e C(K 0) e ln Lα = C 1 (K 0 )L α where
C 1(K 0)=e C(K 0) .
81. By the Chain Rule, taking the partial derivative of both sides with respect to R 1 gives
∂R −1
∂R
∂R
∂R 1
= ∂ [(1/R1)+(1/R2)+(1/R3)]
∂R 1
or −R −2 ∂R
= −R −2 ∂R
1 .Thus = R2
.
∂R 1 ∂R 1
R 2 1