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?

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- February 19th 2007, 04:58 AM^_^Engineer_Adam^_^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: - February 19th 2007, 11:43 AMSoroban
^_^Engineer_Adam^_^!

iQuote:

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.