This document contains solutions to three related rates problems: 1) Finding the rate at which the bottom of a sliding ladder is moving away from a wall. 2) Finding the rate at which the length of a person's shadow is changing as they walk away from a lamppost. 3) Finding (a) the rate at which the distance from a camera to a rising rocket is changing, and (b) the rate at which the camera's angle of elevation must change to keep the rocket in view. Diagrams and mathematical steps are provided for each solution.

1. Solutions to Worksheet for Section 2.7
Related Rates
V63.0121, Calculus I
Summer 2010
1. A 10 ft ladder leans against the side of a building. If the top of the ladder begins to slide down
the wall at the rate of 2 ft/sec, how fast is the bottom of the ladder sliding away from the wall
when the top of the ladder is 8 ft off the ground?
Solution. Let’s draw a diagram. Let x be the distance from the corner of the floor and wall
to the foot of the ladder, and y the distance from the corner to the head of the ladder.
2 ft/sec
10
ft
y
lad
de
r
x
?
We have
x2 + y 2 = 102
So we can differentiate:
dx dy dx y dy
2x + 2y = 0 =⇒ =−
dt dt dt x dt
dy
When y = 8ft, x = 6 ft. We are also given that = −2 ft/sec. So
dt
dx 8 ft
=− (−2ft/sec) = 8/3ft/sec
dt 6 ft
2. Suppose a 6 ft tall person is 12 ft away from an 18 ft-tall lamppost. If the person is moving
away from the lamppost at a rate of 2 ft/sec, at what rate is the length of the shadow changing?
Solution. Again, we draw a picture:
1

2. A
18 ft
D
6 ft
O C B
x s
The triangles formed by the lamppost and the tip of the man’s shadow (OAB) and the man
and the tip of his shadow (CDB) are similar. And corresponding parts of similar triangles
are in proportion. Hence
s+x s
=
18 6
x
By cross-multiplying and some algebra, we get that s = . Hence
2
ds 1 dx
=
dt 2 dt
dx ds
If = 2 ft/sec, then = 1 ft/sec.
dt dt
3. A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of
elevation of the camera has to change at the correct rate in order to keep the rocket in sight.
Also, the mechanism for focusing the camera has to take into account the increasing distance
from the camera to the rising rocket. Let’s assume the rocket rises vertically and its speed is
600 ft/sec when it has risen 3000 ft.
(a) How fast is the distance from the television camera to the rocket changing at that mo-
ment?
(b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle
of elevation changing at that same moment?
Solution. Here is our picture:
2

3. 600 ft/sec
s
h
4000 ft θ
dh ds dθ
We know at a particular value of h; we want to know and at that time.
dt dt dt
(a) Since h2 + 40002 = s2 , we have
dh ds ds h dh
2h = 2s =⇒ =
dt dt dt s dt
dh
When h = 3000 ft, s = 5000 ft, and since = 600 ft/sec, we have
dt
ds 3000 ft
= × 600 ft/sec = 360 ft/sec
dt 5000 ft
h
(b) Now tan θ = , so
4000
dθ 1 dh
sec2 θ
=
dt 4000 dt
Evaluating when h = 3000 ft, s = 5000 ft gives
2 2
sec2 θ = (s/h) = (5/3) = 25/16
So
dθ 1
= × 600 ft/sec = 0.096 1/sec
dt (4000 ft)(25/16)
3