Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

202 0 CHAPTER 14 PARTIAL DERIVATIVES
. 83. By the Chain Rule, taking the partial derivative of both sides with respect to R 1 gives
8R- 1 8R 8[(1/Rl) + (1/R2) + (1/R3)] - 2 8R _2 8R R 2
8R 8R1 = 8R1 or -R 8R 1
= - R1 . Thus 8R 1
= _Rf'
85. If we fix K = Ko , P.(L,-Ko) is a function of a single variable L, and ~~ =a I is a separable differentia; equation. Then
dP dL J dP J dL
p = aL =>- p = a L =>- In IPI = a In 1-41 + C (Ko), where G_(Ko) can depend on K 0 • Then
- .
IPI = e lniLI-f;C(Ko), andsinceP > 0 andL >0, we have P = ec:dnLeC(Ko) = eC(T- f xv(x, y) = 4 and fv(x, y) = 3x- y =>- fvx(x, y) = 3. Since f xy and f vx are continuous
everywhere but f:x:v(x, y) f- f 11
.,(x, y), Clairaut's Theorem implies that such a function f(x, y) does not exist.
95. By the geometry of partial derivatives, the slope of the tangent line is fx(l , 2). By implicit differentiation of
4x 2 + 2y 2 + z 2 = 16, we get8x + 2z (8zj8x) = 0 => 8zf8x = - 4x/z, so when x = 1 and z = 2 we have ·
8zj8x = -2. So the slope is fx(1, 2) = -2. Thus the tangent line is given by z - 2 = -2(x - 1), y = 2. Taking the
parameter to bet = x - 1, we can write parametric equations for this line: x = 1 + t, y = 2, z = 2 - 2t.
97. By Clairaut's Theorem, f xvv = (f:x:y) 11
= (/ 11 x}y = fvxv = Uv).,'ll = Uv)yx = fvvx·
99. Let g(x) = f(x, 0) = x(x 2 )- 3 1 2 e 0 = ·x ix J'- 3 • But we are using the point (1, 0), so near (1, 0), g(x) = x- 2 • Then
g'(x) = - 2x- 3 and g'(1) = -2, so using (l) we have fx(l, 0) = g'(l) = - 2.
101. (a) (b) For (x, y) f- (0, 0),
(3x2y- y3)(x2 + y2) - (x3y- xy3)(2x)
f x(x,y) =
(x2+y2)2
x4y + 4 x2y3 _ ys
(x2 + y2)2
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