Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. x =
dx
dt
t
1
cos u
u
du, y = t
1
sin u
u
dx
du,sobyFTC1,wehave
dt = cos t and dy
t
dt = sin t . Vertical tangent lines occur when
t
=0 ⇔ cos t =0. The parameter value corresponding to (x, y) =(0, 0) is t =1, so the nearest vertical tangent
occurs when t = π . Therefore, the arc length between these points is
2
L =
π/2
1
2 dx
+
dt
dy
dt
2
dt =
π/2
1
cos 2
t
+ sin2 t
π/2
dt
dt =
t 2 t 2 1 t = ln t π/2
=ln π 1
2
3. In terms of x and y, wehavex = r cos θ =(1+c sin θ)cosθ =cosθ + c sin θ cos θ =cosθ + 1 c sin 2θ and
2
y = r sin θ =(1+c sin θ)sinθ =sinθ + c sin 2 θ.Now−1 ≤ sin θ ≤ 1 ⇒ −1 ≤ sin θ + c sin 2 θ ≤ 1+c ≤ 2,so
−1 ≤ y ≤ 2. Furthermore,y =2when c =1and θ = π ,whiley = −1 for c =0and θ = 3π 2 2
. Therefore, we need a viewing
rectangle with −1 ≤ y ≤ 2.
To find the x-values, look at the equation x =cosθ + 1 2 c sin 2θ and use the fact that sin 2θ ≥ 0 for 0 ≤ θ ≤ π 2 and
sin 2θ ≤ 0 for − π 2
≤ θ ≤ 0. [Because r =1+c sin θ is symmetric about the y-axis, we only need to consider
− π 2 ≤ θ ≤ π 2 .] So for − π 2
≤ θ ≤ 0, x has a maximum value when c =0and then x =cosθ has a maximum value
of 1 at θ =0. Thus, the maximum value of x must occur on 0, π 2
with c =1.Thenx =cosθ +
1
sin 2θ ⇒
2
dx
= − sin θ +cos2θ = − sin θ +1− dθ 2sin2 θ ⇒ dx = −(2 sin θ − 1)(sin θ +1)=0when sin θ = −1 or 1 dθ 2
[but sin θ 6=−1 for 0 ≤ θ ≤ π ]. If sin θ = 1 ,thenθ = π and
2 2 6
x =cos π + 1 sin π = √ √ 3
6 2 3 4 3. Thus, the maximum value of x is
3
4 3, and,
by symmetry, the minimum value is − 3 4
√
3. Therefore, the smallest
viewing rectangle that contains every member of the family of polar curves
r =1+c sin θ,where0 ≤ c ≤ 1,is √ √
− 3 4 3,
3
4 3 × [−1, 2].
3t 1
5. (a) If (a, b) lies on the curve, then there is some parameter value t 1 such that
1+t 3 1
= a and 3t2 1
= b. Ift
1+t 3 1 =0,
1
the point is (0, 0), which lies on the line y = x. Ift 1 6=0, then the point corresponding to t = 1 t 1
is given by
x = 3(1/t 1)
1+(1/t 1 ) = 3t2 1
3 t 3 1 +1 = b, y = 3(1/t 1) 2
1+(1/t 1 ) = 3t 1
= a. So(b, a) also lies on the curve. [Another way to see
3 +1
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when
t 3 1
3t
1+t 3 = 3t2
1+t 3 ⇒
t = t 2 ⇒ t =0or 1, so the points are (0, 0) and 3
, 3
2 2 .
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