# Related Rates

• Feb 19th 2007, 04:58 AM
Related Rates
A man 6 ft tall is walking toward a building at the rate of 5ft/sec. If there is a light on the ground 50ft from the building, how fast is the man's shadow on the building growing shorter when he is 30 ft from the building?
:confused:
• Feb 19th 2007, 11:43 AM
Soroban
i
Quote:

A man 6 ft tall is walking toward a building at the rate of 5ft/sec.
If there is a light on the ground 50ft from the building, how fast is the man's shadow
on the building growing shorter when he is 30 ft from the building?

First, make a sketch . . .
Code:

```      |     D *       |  *       |      *  B       |          *     s |          |  *       |          |      *       |          |6          *       |          |              *       |    x    |      50-x        *     E * - - - - - + - - - - - - - - - - - * A                   C       : - - - - - - - 50  - - - - - - - - :```

The light is at A; the man is: BC = 6.

The wall is at the far left; the length of his shadow is: s = DE.

Let x = EC, then CA = 50 - x.
. . We are given: dx/dt = -5 ft/sec

Right triangle DEA is similar to right triangle BCA.
. . . . . . . . . . . . .s . . . . . .6
. . so we have: . --- . = . -------
. . . . . . . . . . . . 50 . . . .50 - x

. . which simplifies to: .s .= .300 / (50 - x)

Differentiate with respect to time: .ds/dt .= .300(dx/dt) / (50 - x)²

When x = 30 and dx/dt = -5: .ds/dt .= .300(-5) / (50- 30)² .= .-15/4

Therefore, the length of the shadow is decreasing at 3¾ feet/sec.