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

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236 0 CHAPTER 14 PARTIAL DERIVATIVES<br />

(c) The gradient vector of a function points in the direction of maximum rate of increase of the function. On a graph of the<br />

function, the gradient points in the direction of steepest ascent.<br />

15. (a) f has a local maximum at {a, b) if f(x, y) :5 f(a, b) when {x, y) is near {a, b).<br />

(b) f has an absolute maximum at {a, b) if f(x, y) :5 f(a, b) tor all points (x, y) in the domain of f .<br />

(c) f has a local minimum at (a, b) if f(x, y) ~ f (a, b) when (x, y) is near (a, b).<br />

(d) f has an absolute minimum at (a, b) if f(x, y) ~ f(a, b) for aU points (x, y) in the domain of f.<br />

(e) f has a saddle point at (a, b) if f(a, b) is a local maximum in one direction but a local minimum in another.<br />

16. (a) By Theorem 14.7.2, iff has a local maximum at (a, b) and the first-order partial derivatives of j exist there, then<br />

/ :r(a, b) = 0 and / 11 (a, b)= 0.<br />

(b) A critical point off is a point (a, b) such that [.(a, b) = 0 and j 11 (a, b) = 0 or one of these partial derivatives does<br />

not exist.<br />

17. See (3) in Section 14.7.<br />

18. (a) See Figure II and the accompanying discussion in Section 14.7.<br />

(b) See Theorem 14.7.8.<br />

(c) See the proc<strong>ed</strong>ure outlin<strong>ed</strong> in (9) in Section 14.7.<br />

19. See the discussion beginning on page 981 [ET 957]; see "Two Constraints" on page 985 [ET 961].<br />

TRUE-FALSE QUIZ<br />

1. True. fv(a, b) = lim f(a, b +h) - f(a, b) from Equation 14.3.3. Let h = ·y- b. Ash~ 0, y --+ b. Then by substituting,<br />

h~o h ,<br />

f(a,y)-f(a,b)<br />

we get f 11 ( a, b) = li m b .<br />

u~ b y-<br />

5. False. See Example 14.2.3.<br />

7. Tru~. Iff has a local minimum and f is differentiable at (a, b) then by Theorem 14.7.2, f x(a, b) = 0 and fu (a, b) = 0, so<br />

V f(a,b) = (f,(a,b),fv(a,b)) = (0,0} = 0.<br />

9. False. V f(x, y) = (0, 1/ y).<br />

11. True. V f = (cosx, cosy), so IV / I = v'cos 2 x + cos 2 y . But lcosBI :5 1, so IV !I :5 .J?.. Now<br />

Du f(x, y) = V f · u = IV f l l.u l cos{:}, but u is a unit vector, so IDu f (x, y)l :5 .J2 · 1 · 1 = ./2.<br />

@ 2012 c~nsab~ Learning. All Rights Rcscr\'cd. Mlly not be scann<strong>ed</strong>. copi<strong>ed</strong>, orduplicntcd. or post<strong>ed</strong> to a publicly accessible website. in whole or in p:u1.

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