^_^Engineer_Adam^_^!
i 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  50x *
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.