HOMEWORK 5 FOR MATH 2250 - MODEL SOLUTION
(1) (4 pts) A rock thrown vertically upward from the surface of the moon at a velocity of 24 m/sec (about 86 km/h) reaches a height of s = 24t − 0.8t2 m in t
sec.
(a) Find the rock’s velocity and acceleration at time t.
velocity: v(t) =
ds
= 24 − 1.6t (m/s).
dt
dv
= −1.6 (m/s2 ).
dt
(b) How long does it take the rock to reach its highest point?
At highest point, v(t) = 0.
acceleration: a(t) =
24
= 15.
1.6
It takes 15 seconds to reach the highest point.
(c) How high does the rock go?
v(t) = 24 − 1.6t = 0 ⇒ t =
s(15) = 24 · 15 − 0.8(15)2 = 180.
Answer: 180 m.
(d) How long does it take the rock to reach half its maximum height?
The highest point is 180, so the half of it is 90.
s(t) = 24t − 0.8t2 = 90 ⇒ 0.8t2 − 24t + 90 = 0 ⇒ 4t2 − 120t + 450 = 0.
If you solve this equation by using quadratic formula,
√
√
√
120 ± 1202 − 4 · 4 · 450
120 ± 7200
30 ± 15 2
t=
=
=
.
8
8
2
Or if you use a calculator,
√
√
30 − 15 2
30 + 15 2
≈ 4.39,
≈ 25.61.
2
2
Answer: 4.39, 25.61 sec.
(e) How long is the rock aloft?
It is aloft until s(t) = 0 again.
s(t) = 0 ⇒ 24t − 0.8t2 = 0 ⇒ t(24 − 0.8t) = 0 ⇒ t = 0 or t =
Answer: 30 sec.
Date: February 18, 2012.
1
24
= 30.
0.8
2
HOMEWORK 5 FOR MATH 2250 - MODEL SOLUTION
(2) (2 pts) Find
dy
of
dx
y=
cos x
.
1 + sin x
d
d
(1 + sin x) dx
cos x − cos x dx
(1 + sin x)
dy
=
2
dx
(1 + sin x)
(1 + sin x)(− sin x) − cos x cos x
− sin x − sin2 x − cos2 x
=
=
(1 + sin x)2
(1 + sin x)2
− sin x − 1
−(1 + sin x)
1
.
=
=
=
−
(1 + sin x)2
(1 + sin x)2
1 + sin x
(3) (2 pts) Find
dp
of
dq
p = (1 + csc q) cos q.
dp
= − sin q − csc2 q.
p = cos q + cot q ⇒
dq
Or
dp
d
d
= (1 + csc q) cos q + cos q (1 + csc q)
dq
dq
dq
= (1 + csc q)(− sin q) + cos q(− csc q cot q)
1 cos q
1
)(− sin q) + cos q(−
)
= (1 +
sin q
sin q sin q
cos2 q
= − sin q − 1 −
= −1 − sin q − cot2 q.
sin2 q
Or, you can rewrite it as
1
− sin q − 1 − cot2 q = − sin q −
= − sin q − csc2 q.
2
sin q