On-line Math 21

On-line Math 21

2.4  Applications of the derivative

Example 1 A man, walking at night, is walking directly towards a streetlight. The light is 10 feet off the ground, and the man is 6 feet tall and walking at 4 feet per second. When the man is 8 feet from the streetlight, how fast is the length of his shadow changing?

Solution

This one is unusual, in that it does not involve the Pythagorean Theorem. It is actually simpler than it appears, but confuses a lot of people because you are now expecting to use the Pythagorean Theorem on every one of these problems.

Reducing the question to its basic parts, let s be the length of the shadow, and let x be the distance from the man to the streetlight. You do not want to know the length of that ray of light from the streetlight to the ground; you also don't know anything about it, so leave it out of the problem. The only other facts you have are the height of the man (6 feet) and that of the streetlight (10 feet). That information gives the following drawing:

You know that dx/dt = -4 (walking towards the streetlight), and that the height of the man is 6 , the light is 10 . You want to know what ds/dt is when x = 8 .

Both these triangles are right triangles (presuming that both the streetlight and the man are standing upright, and that the ground is flat), and so both triangles have a right angle. They also both share the angle made by the ground and the ray of light at the tip of the shadow, so all three angles of the triangle are the same; that is, the two triangles are similar. Similar triangles have the same ratios of corresponding sides, so the ratio of base/altitude is the same for the two triangles. For the larger triangle, the altitude is 10, and the base is (x+s) ; for the smaller triangle, the base is 6 and the altitude is s , so
(x+s)
10
= s
6
.
Simplifying,
x
10
=
s
6
- s
10
=
5s-3s
30
=
s
15
.
Differentiating both sides, and substituting dx/dt = -4 (he is walking towards the streetlight).
1
10
dx
dt
=
1
15
ds
dt
,
1
10
(-4)
=
1
15
ds
dt
, or
ds
dt
=
-6,
so the shadow is shrinking at a rate of 6 feet/sec. This is the value for all values of x , the part about x = 8 is irrelevant.

Copyright (c) 2000 by David L. Johnson.


File translated from TEX by TTH, version 2.61.
On 24 Nov 2000, 17:45.