Solução_Calculo_Stewart_6e

F.
140 ¤ CHAPTER 3 PROBLEMS PLUS
TX.10
d k
Now assume S k is true, that is, sin 4 x +cos 4 x =4 k−1 cos
4x + k π
dx k 2 .Then
d k+1
dx k+1 (sin4 x +cos 4 x)= d
d
k
dx dx k (sin4 x +cos 4 x) = d 4 k−1 cos
4x + k π 2
dx
which shows that S k+1 is true.
Therefore,
Another proof: First write
= −4 k−1 sin
4x + k π d
2 ·
dx
=4 k sin −4x − k π 2
4x + k
π
2 = −4 k sin
4x + k π 2
=4 k cos π
2 − −4x − k π 2
=4 k cos 4x +(k +1) π 2
d n
dx n (sin4 x +cos 4 x)=4 n−1 cos
4x + n π 2 for every positive integer n, by mathematical induction.
sin 4 x +cos 4 x =(sin 2 x +cos 2 x) 2 − 2sin 2 x cos 2 x =1− 1 2 sin2 2x =1− 1 (1 − cos 4x) = 3 + 1 cos 4x
4 4 4
Then we have dn
dx n (sin4 x +cos 4 x)= dn 3
dx n 4 + 1
4 cos 4x = 1
4 · 4n cos 4x + n π
=4 n−1 cos 4x + n π
.
2
2
9. We must find a value x 0 such that the normal lines to the parabola y = x 2 at x = ±x 0 intersectatapointoneunitfromthe
points
±x 0 ,x 2 0 . The normals to y = x 2 at x = ±x 0 have slopes − 1 and pass through
±x 0 ,x 2 0 respectively, so the
±2x 0
normals have the equations y − x 2 0 = − 1
2x 0
(x − x 0) and y − x 2 0 = 1
2x 0
(x + x 0). The common y-intercept is x 2 0 + 1 2 .
We want to find the value of x 0 for which the distance from 0,x 2 0 + 1 2
to
x0,x 2 0
equals 1. The square of the distance is
(x 0 − 0) 2 + x 2 0 − x 2 0 + 1 2
the center of the circle is at 0, 5 4
.
2
= x 2 0 + 1 4 =1 ⇔ x 0 = ± √ 3
2 .Forthesevaluesofx 0,they-intercept is x 2 0 + 1 2 = 5 4 ,so
Another solution: Let the center of the circle be (0,a). Then the equation of the circle is x 2 +(y − a) 2 =1.
Solving with the equation of the parabola, y = x 2 ,wegetx 2 +(x 2 − a) 2 =1 ⇔ x 2 + x 4 − 2ax 2 + a 2 =1 ⇔
x 4 +(1− 2a)x 2 + a 2 − 1=0. The parabola and the circle will be tangent to each other when this quadratic equation in x 2
has equal roots; that is, when the discriminant is 0. Thus,(1 − 2a) 2 − 4(a 2 − 1) = 0
⇔
1 − 4a +4a 2 − 4a 2 +4=0 ⇔ 4a =5,soa = 5 4 . The center of the circle is 0, 5 4
.
11. We can assume without loss of generality that θ =0at time t =0,sothatθ =12πt rad. [The angular velocity of the wheel
is 360 rpm =360· (2π rad)/(60 s) =12π rad/s.] Then the position of A as a function of time is
A =(40cosθ, 40 sin θ) = (40 cos 12πt, 40 sin 12πt),sosin α =
y 40 sin θ
= = sin θ = 1 sin 12πt.
1.2 m 120 3 3
(a) Differentiating the expression for sin α,wegetcos α · dα
dt = 1 3 · 12π · cos 12πt =4π cos θ. Whenθ = π 3 ,wehave
sin α = 1 √
√ 2
3
3 11
3 sin θ = 6 ,socos α = dα
1 − = and
6 12 dt = 4π cos π 3
cos α = 2π
= 4π √ 3
√ ≈ 6.56 rad/s.
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