Lesson 19: Related Rates Worksheet Solutions

Solutions to Worksheet for Lesson 19 (Section 4.1) Related Rates Math 1a November 7, 2007 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.

10

2 ft/sec

ft r

de

lad

y

x

?

We have x2 + y 2 = 102 So we can differentiate:

dy dx y dy dx + 2y = 0 =⇒ =− dt dt dt x dt dy When y = 8 ft, x = 6 ft. We are also given that = −2 ft/sec. So dt 2x

dx 8 ft =− (−2 ft/sec) = 8/3 ft/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:

A

18 ft D 6 ft O

x

C

s

B

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 1 dx ds = dt 2 dt If

ds dx = 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 moment? (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:

600 ft/sec

s

h

4000 ft We know

θ

dh ds dθ at a particular value of h; we want to know and at that time. dt dt dt

(a) Since h2 + 40002 = s2 , we have 2h

ds ds h dh dh = 2s =⇒ = dt dt dt s dt

When h = 3000 ft, s = 5000 ft, and since

dh = 600 ft/sec, we have dt

ds 3000 ft = × 600 ft/sec = 360 ft/sec dt 5000 ft (b) Now tan θ =

h , so 4000

1 dh dθ = dt 4000 dt Evaluating when h = 3000 ft, s = 5000 ft gives sec2 θ

2

2

sec2 θ = (s/h) = (5/3) = 25/16 So

1 dθ = × 600 ft/sec = 0.096 1/sec dt (4000 ft)(25/16)