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Solução_Calculo_Stewart_6e

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F.<br />

422 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10<br />

67. If f 0 is continuous and f 0 (t) 6= 0for a ≤ t ≤ b, then either f 0 (t) > 0 for all t in [a, b] or f 0 (t) < 0 for all t in [a, b]. Thus,f<br />

is monotonic (in fact, strictly increasing or strictly decreasing) on [a, b]. It follows that f has an inverse. Set F = g ◦ f −1 ,<br />

that is, define F by F (x) =g(f −1 (x)). Thenx = f(t) ⇒ f −1 (x) =t,soy = g(t) =g(f −1 (x)) = F (x).<br />

dy<br />

69. (a) φ =tan −1 ⇒<br />

dx<br />

<br />

d dy<br />

= d ẏ<br />

=<br />

dt dx dt ẋ<br />

t <br />

fact that s =<br />

dx<br />

dt<br />

0<br />

dφ<br />

dt = d <br />

dy<br />

1 d dy<br />

dt tan−1 =<br />

dx 1+(dy/dx) 2 dt dx<br />

ÿẋ − ẍẏ<br />

⇒ dφ<br />

ÿẋ <br />

ẋ 2 dt = 1 − ẍẏ<br />

=<br />

1+(ẏ/ẋ) 2 ẋ 2<br />

<br />

. But dy<br />

dx = dy/dt<br />

dx/dt = ẏ<br />

ẋ<br />

⇒<br />

ẋÿ − ẍẏ<br />

. Using the Chain Rule, and the<br />

ẋ 2 + ẏ2 2 <br />

+<br />

dy<br />

<br />

<br />

2<br />

dt dt ⇒<br />

ds = dx<br />

2 <br />

dt dt + dy<br />

2<br />

dt = ẋ2 + ẏ 21/2 ,wehavethat<br />

dφ<br />

ds = dφ/dt <br />

ẋÿ − ẍẏ<br />

ds/dt = 1 ẋÿ − ẍẏ <br />

=<br />

ẋ 2 + ẏ 2 (ẋ 2 + ẏ 2 )<br />

1/2<br />

(ẋ 2 + ẏ 2 ) .Soκ = dφ<br />

<br />

3/2 ds = ẋÿ − ẍẏ |ẋÿ − ẍẏ|<br />

=<br />

(ẋ 2 + ẏ 2 ) 3/2 (ẋ 2 + ẏ 2 ) . 3/2<br />

(b) x = x and y = f(x) ⇒ ẋ =1, ẍ =0and ẏ = dy<br />

dx , ÿ = d2 y<br />

dx . 2<br />

<br />

1 · (d 2 y/dx 2 ) − 0 · (dy/dx) <br />

d 2 y/dx 2 <br />

So κ =<br />

=<br />

[1 + (dy/dx) 2 ] 3/2 [1 + (dy/dx) 2 ] . 3/2<br />

71. x = θ − sin θ ⇒ ẋ =1− cos θ ⇒ ẍ =sinθ,andy =1− cos θ ⇒ ẏ =sinθ ⇒ ÿ =cosθ. Therefore,<br />

<br />

cos θ − cos 2 θ − sin 2 θ κ =<br />

[(1 − cos θ) 2 +sin 2 θ] = cos θ − (cos 2 θ +sin 2 θ) |cos θ − 1|<br />

3/2 (1 − 2cosθ +cos 2 θ +sin 2 = . The top of the arch is<br />

θ)<br />

3/2 (2 − 2cosθ)<br />

3/2<br />

characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when θ =(2n − 1)π,<br />

so take n =1and substitute θ = π into the expression for κ: κ =<br />

73. The coordinates of T are (r cos θ, r sin θ). SinceTP was unwound from<br />

arc TA, TP has length rθ. Also∠PTQ = ∠PTR− ∠QT R = 1 2 π − θ,<br />

so P has coordinates x = r cos θ + rθ cos 1<br />

2 π − θ = r(cos θ + θ sin θ),<br />

y = r sin θ − rθ sin 1<br />

2 π − θ = r(sin θ − θ cos θ).<br />

|cos π − 1|<br />

(2 − 2cosπ) 3/2 = |−1 − 1|<br />

[2 − 2(−1)] 3/2 = 1 4 .<br />

10.3 Polar Coordinates<br />

1. (a) 2, π 3<br />

<br />

By adding 2π to π 3 , we obtain the point 2, 7π 3<br />

<br />

. The direction<br />

opposite π 3 is 4π 3 ,so −2, 4π 3<br />

requirement.<br />

<br />

is a point that satisfies the r

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