CALCULUS 2210–001: HOMEWORK 1, STUDY GUIDE
13.1: #2. y = 32 x for all values of x ∈ R .
13.1: #18. x + y = 9 for all values of 0 ≤ x ≤ 9 (why?)
13.2: #5. w = 12 (u + v) = 12 u + 12 v.
13.3: #4.
(a) a · b = −60, |a| = 12, and |b| = 5, so cos(θ) = −1, whence θ = π.
(b) a · b = −50, |a| = 5 and |b| = 10, so cos(θ) = −1 and this means that θ = π.
√
√
√
(c) |a| = 10, |b| = 40, a · b = −20. Therefore, cos(θ) = −20/ 400 = −1, θ = π.
√
√
(d) |a| = 2, |b| = 12, a · b = 4 3. Therefore, cos(θ) = 1, and so θ = 0.
13.3: #34.
(a) pru (u) =
(b)
(c)
(d)
u·u
|u|2 u = u.
pr−u (u) = −u·u
|u|2 u = −u.
pru (−u) = u·(−u)
|u|2 u = −u.
pr−u (−u) = (−u)·(−u)
(−u)
|u|2
= −u.
13.4: #18. We have r(t) = sin(2t)i + cosh(t)j and h(t) = ln(2t − 3), so that
Dt [h(t)r(t)] = h0 (t)r(t) + h(t)r0 (t)
3
hsin(2t), cosh(t)i + ln(3t − 2) h2 cos(2t), sinh(t)i .
=
3t − 2
13.4: #42. We know that a(t) = −32j. Therefore, integrate to get v(t) = −32tj + C
for some constant vector C. Because v(0) = −32j, this means that C = 20i; i.e., v(t) =
20i − 32tj. Integrate again to see that r(t) = 20ti − 16t2 j + Dfor a constant vector D.
Since r(0) = 4j, D = 4j, and we get r(t) = 20ti + 4 − 16t2 j. The time t when the
y-coordinate of r(t) is zero describes the time when the ball hits the floor. This it the
t that solves 4 − 16t2 = 0; i.e., t = 12 . The tangent to the trajectory, at this time, is
v( 12 ) = 20i − 16j. Thus, the angle between the tangent and the line, at time t = 12 is
√
described by cos(θ) = v( 12 ) · (−r( 12 ))/|v( 21 )| |r( 21 )| = −20/ 656, so that θ ≈ 2.466◦ .
13.4: #46. Let us work consistently with miles and seconds as our units. We know
Kepler’s law (see exercise 44):
2
4π
2
r3 ,
T =
gR2
1
where T is the the time it takes to complete one revolution, r is the distance to the moon,
and k = gR2 , where g ≈ 32.17 feet/second2 , and R ≈ 3, 960 miles. Solve for r 3 to obtain
2 2
r 3 = gR4πT2 , so that after taking 13 -rd roots, we have
r=
gR2 T 2
4π 2
1/3
.
(Eq-1)
Now we plug the numerical values in, but we need to ensure that all units are used
consistently. Note that g ≈ 32.17 feet/sec2 . There are 5, 280 feet per mile, so that
g ≈ 32.17/5280 ≈ 0.006 miles per sec2 . Now R ≈ 3, 960 miles. Finally, T ≈ 27.32 days,
but there are 24 × 60 × 60 = 86, 400 seconds per day, so T ≈ 27.32 × 86, 400 ≈ 2.36 × 106
seconds. Plug this all into (Eq-1) to get r ≈ 238, 019 miles.

1/3
R2
T2
z
}|
{
z
}|
{
}| {
 z0.006
× (3, 960)2 × (2.36 × 106 )2 


r≈
≈ 238, 019 miles.

2
4π


g
2