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 . . .
| * B
s | | *
| | *
| |6 *
| | *
| x | 50-x *
E * - - - - - + - - - - - - - - - - - * A
: - - - - - - - 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.