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

246 0 CHAPTER 14 PROBLEMS PLUS
Since x > 0, we must have x = ~w, in which case cosO=~. soB = f, sinO=~ .
k = '{!w, b1 = ~w,ln = ~w.
and A= {!:w 2 • As in Example 14.7.6, we can argue from the physical nature of this problem that we have found a local
maximum of A. Now checking the boundary of A, let
g( B) = A( w /2, B) = ~ w 2 sin B - ~w 2 sin B + iw 2 sin B cos 6 = i w 2 sin 26, 0 < B ::; ~. Clearly g is maximized when
sin2B = 1 in which case A= iw 2 . Also along the line B = ~. let h(x) = A(x, ~) = wx- 2x 2 , 0 < x < ~w . =>
h'(x) = w- 4x = 0
*? x = i w, and h(~w) = w(~w) - 2(tw) 2 = iw 2 . Since -kw 2 < "If w 2 , we conclude that
the local maximum found earlier was an absolute maximum.
2 2
(b) If the metal were bent into a semi-circular gutter of radius r, we would have w = 1rr and A= ~11'r 2 = ~1r(~) = ~7r.
S . ../3 w2 w 2 mce - > ---, 1t • wou ld b e b etter to be n d th e meta l · mto a gutter WI 'th a sem1c1rcu · · I ar cross-section. .
211' 1 2
5. Letg(x,y) = xf(~) · Theng,.(:,y) = !(~) +xf'(~) ( - : 2 ) = !(~ ) - ~ f'(~;) and
gy ( x, y) = xf' ( ~) (;) = f' ( ~ ). Thus the tange?t plane at ( x0 , y 0 , zo) on the surface has equation
[t ( ;~ ) - y0 x0 1 !' ( ~~ )] x + [r ( ~~ )] y - z = 0. But any plane whose equation is of the form ax+ by+ cz = 0
passes through the origin. Thus the origin is the ~ommon point of intersection.
7. Since we are minimizing the area of the ellipse, and the circle lies above the x -axis,
y
the ellipse will intersect the circle for only one value ofy. This y-value must
satisfY both the equation of the circle and the equation of the ellipse. Now
2 2 2
x 2
+ Y 2
= 1 => x 2 = a 2
(b 2 -y 2 ). Substitutingintotheequationofthe
a b b
2
circle gives ~2 (b 2 - y 2 ) + y 2 - 2y = 0 =>
In order for there to be orlly one solution to this quadratic equation, the discrirninant must be 0, so 4- 4a 2 2 2
b ~ a
b 2 - a 2 b 2 + a 4 = 0. The area of the ellipse is A(a, b) = 1rab, and we minimize this function subject to the constraint ·
g(a, b) = b 2 - a 2 b 2 + a 4 = 0.
= 0 =>
-rrb
2a(2a 2 -
2) (1),
b
.>. = 2 b(l7r~ a 2 ) (2), b2 - a 2 b 2 + a 4 = 0 (3). Comparing (1) and (2) gives 2
a( 2
; b _ b2).
21rb 2 = 47ra 4 -:=:> a 2 = ~ b. Substitute this into (3) to get b = ~ => a = jf.
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