Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 3 PROBLEMS PLUS ¤ 141
(b) By the Law of Cosines, |AP | 2 = |OA| 2 + |OP| 2 − 2 |OA||OP| cos θ
⇒
120 2 =40 2 + |OP| 2 − 2 · 40 |OP| cos θ ⇒ |OP| 2 − (80 cos θ) |OP| − 12,800 = 0 ⇒
|OP| = 1 2
80 cos θ ±
√ 6400 cos2 θ +51,200 =40cosθ ± 40 √ cos 2 θ +8=40 cos θ + √ 8+cos 2 θ cm
[since |OP| > 0]. As a check, note that |OP| =160cm when θ =0and |OP| =80 √ 2 cm when θ = π 2 .
(c) By part (b), the x-coordinate of P is given by x =40 cos θ + √ 8+cos 2 θ ,so
dx
dt = dx
dθ
dθ
dt =40 − sin θ −
2cosθ sin θ
2 √ · 12π = −480π sin θ 1+
8+cos 2 θ
In particular, dx/dt =0cm/swhenθ =0and dx/dt = −480π cm/swhenθ = π 2 .
13. Consider the statement that
d
dx (eax sin bx) =ae ax sin bx + be ax cos bx,and
d n
dx n (eax sin bx) =r n e ax sin(bx + nθ). Forn =1,
cos θ
√ cm/s.
8+cos2 θ
re ax sin(bx + θ) =re ax [sin bx cos θ +cosbx sin θ] =re ax a
r sin bx + b r cos bx
= ae ax sin bx + be ax cos bx
since tan θ = b a
⇒ sin θ = b r and cos θ = a . So the statement is true for n =1.
r
But
Assume it is true for n = k. Then
d k+1
dx k+1 (eax sin bx)= d
r k e ax sin(bx + kθ) = r k ae ax sin(bx + kθ)+r k e ax b cos(bx + kθ)
dx
= r k e ax [a sin(bx + kθ)+b cos(bx + kθ)]
sin[bx +(k +1)θ] =sin[(bx + kθ)+θ] =sin(bx + kθ)cosθ +sinθ cos(bx + kθ) = a sin(bx + kθ)+ b cos(bx + kθ).
r r
Hence, a sin(bx + kθ)+b cos(bx + kθ) =r sin[bx +(k +1)θ].So
d k+1
dx k+1 (eax sin bx) =r k e ax [a sin(bx+kθ)+b cos(bx+kθ)] = r k e ax [r sin(bx+(k +1)θ)] = r k+1 e ax [sin(bx+(k +1)θ)].
Therefore, the statement is true for all n by mathematical induction.
15. It seems from the figure that as P approaches the point (0, 2) from the right, x T →∞and y T → 2 + .AsP approaches the
point (3, 0) from the left, it appears that x T → 3 + and y T →∞.Soweguessthatx T ∈ (3, ∞) and y T ∈ (2, ∞). Itis
more difficult to estimate the range of values for x N and y N. We might perhaps guess that x N ∈ (0, 3),
and y N ∈ (−∞, 0) or (−2, 0).
In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to find the equation of the
tangent line: x2
9 + y2
4 =1 ⇒ 2x 9 + 2y 4 y0 =0,soy 0 = − 4 9
x
. So at the point (x0,y0) on the ellipse, an equation of the
y