Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 15.3 PARTIAL DERIVATIVES ET SECTION 14.3 ¤ 183
83. By Exercise 82, PV = mRT ⇒ P = mRT
V
Since T = PV
∂P
,wehaveT
mR ∂T
∂P
,so
∂T = mR
V
∂V
∂T = PV
mR · mR
V · mR
P
= mR.
.Also,PV = mRT ⇒ V =
mRT
P
and ∂V
∂T = mR
P .
85.
∂K
∂m = 1 2 v2 ,
∂K
∂v = mv, ∂ 2 K
∂v 2
∂K
= m. Thus
∂m · ∂2 K
∂v = 1 2 2 v2 m = K.
87. f x (x, y) =x +4y ⇒ f xy (x, y) =4and f y (x, y) =3x − y ⇒ f yx (x, y) =3.Sincef xy and f yx are continuous
everywhere but f xy(x, y) 6=f yx(x, y), Clairaut’s Theorem implies that such a function f(x, y) does not exist.
89. By the geometry of partial derivatives, the slope of the tangent line is f x(1, 2). By implicit differentiation of
4x 2 +2y 2 + z 2 =16,weget8x +2z (∂z/∂x) =0 ⇒ ∂z/∂x = −4x/z, sowhenx =1and z =2we have
∂z/∂x = −2. Sotheslopeisf x (1, 2) = −2. Thus the tangent line is given by z − 2=−2(x − 1), y =2.Takingthe
parameter to be t = x − 1, we can write parametric equations for this line: x =1+t, y =2, z =2− 2t.
91. By Clairaut’s Theorem, f xyy =(f xy ) y
=(f yx ) y
= f yxy =(f y ) xy
=(f y ) yx
= f yyx .
93. Let g(x) =f(x, 0) = x(x 2 ) −3/2 e 0 = x |x| −3 .Butweareusingthepoint(1, 0), sonear(1, 0), g(x) =x −2 .Then
g 0 (x) =−2x −3 and g 0 (1) = −2,sousing(1)wehavef x (1, 0) = g 0 (1) = −2.
95. (a) (b) For (x, y) 6= (0, 0),
f x(x, y)= (3x2 y − y 3 )(x 2 + y 2 ) − (x 3 y − xy 3 )(2x)
(x 2 + y 2 ) 2
= x4 y +4x 2 y 3 − y 5
(x 2 + y 2 ) 2
andbysymmetryf y(x, y) = x5 − 4x 3 y 2 − xy 4
(x 2 + y 2 ) 2 .
f(h, 0) − f(0, 0) (0/h 2 ) − 0
f(0,h) − f(0, 0)
(c) f x (0, 0) = lim
= lim
=0and f y (0, 0) = lim
=0.
h→0 h
h→0 h
h→0 h
(d) By (3), f xy (0, 0) = ∂f x
∂y =lim f x (0,h) − f x (0, 0) (−h 5 − 0)/h 4
= lim
h→0 h
h→0 h
f yx(0, 0) = ∂f y
∂x =lim f y(h, 0) − f y(0, 0) h 5 /h 4
= lim =1.
h→0 h
h→0 h
(e) For (x, y) 6= (0, 0), we use a CAS to compute
f xy (x, y) = x6 +9x 4 y 2 − 9x 2 y 4 − y 6
(x 2 + y 2 ) 3
Now as (x, y) → (0, 0) along the x-axis, f xy(x, y) → 1 while as
(x, y) → (0, 0) along the y-axis, f xy (x, y) →−1. Thus f xy isn’t
continuous at (0, 0) and Clairaut’s Theorem doesn’t apply, so there is
no contradiction. The graphs of f xy and f yx are identical except at the
origin, where we observe the discontinuity.
= −1 while by (2),