Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 3 PROBLEMS PLUS ¤ 143
19. Since ∠ROQ = ∠OQP = θ, the triangle QOR is isosceles, so
|QR| = |RO| = x. By the Law of Cosines, x 2 = x 2 + r 2 − 2rx cos θ. Hence,
2rx cos θ = r 2 ,sox =
sin θ = y/r), and hence x →
r2
2r cos θ = r
2cosθ .Notethatasy → 0+ , θ → 0 + (since
r
2cos0 = r .Thus,asP is taken closer and closer
2
to the x-axis, the point R approaches the midpoint of the radius AO.
21. lim
x→0
sin(a +2x) − 2sin(a + x)+sina
x 2
= lim
x→0
sin a cos 2x +cosa sin 2x − 2sina cos x − 2cosa sin x +sina
x 2
= lim
x→0
sin a (cos 2x − 2cosx +1)+cosa (sin 2x − 2sinx)
x 2
= lim
x→0
sin a (2 cos 2 x − 1 − 2cosx +1)+cosa (2 sin x cos x − 2sinx)
x 2
= lim
x→0
sin a (2 cos x)(cos x − 1) + cos a (2 sin x)(cos x − 1)
x 2
2(cos x − 1)[sin a cos x +cosa sin x](cos x +1)
= lim
x→0 x 2 (cos x +1)
−2sin 2 2
x [sin(a + x)]
sin x sin(a + x)
= lim
= −2 lim ·
x→0 x 2 (cos x +1)
x→0 x cos x +1 = sin(a +0)
−2(1)2 cos 0 + 1 = − sin a
23. Let f(x) =e 2x and g(x) =k √ x [k >0]. From the graphs of f and g,
So we must have k √ a =
k
4 √ a
k =2e 1/2 =2 √ e ≈ 3.297.
25. y =
⇒
we see that f will intersect g exactly once when f and g share a tangent
line. Thus, we must have f = g and f 0 = g 0 at x = a.
f(a) =g(a) ⇒ e 2a = k √ a ()
and f 0 (a) =g 0 (a) ⇒ 2e 2a = k
2 √ ⇒ e 2a = k
a
4 √ a .
√ 2 k a = ⇒ a = 1
4k
.From(), 4 e2(1/4) = k 1/4 ⇒
x
√
a2 − 1 − 2
√
a2 − 1 arctan sin x
a + √ a 2 − 1+cosx .Letk = a + √ a 2 − 1. Then
y 0 =
=
1
√
a2 − 1 − 2
√
a2 − 1 ·
1
1+sin 2 x/(k +cosx) · cos x(k +cosx)+sin2 x
2 (k +cosx) 2
1
√
a2 − 1 − 2
√
a2 − 1 · k cos x +cos2 x +sin 2 x
(k +cosx) 2 +sin 2 x = 1
√
a2 − 1 − 2
√
a2 − 1 · k cos x +1
k 2 +2k cos x +1
= k2 +2k cos x +1− 2k cos x − 2
√
a2 − 1(k 2 +2k cos x +1)
=
k 2 − 1
√
a2 − 1(k 2 +2k cos x +1)
But k 2 =2a 2 +2a √ a 2 − 1 − 1=2a a + √ a 2 − 1 − 1=2ak − 1,sok 2 +1=2ak,andk 2 − 1=2(ak − 1).
[continued]