Question

 

A spotlight on the ground shines on a wall of a building 12 ft away.

If a 6-foot tall man walks from the spotlight toward the wall at a speed of of

1.5 ft/s, how fast is the length of his shadow on the building decreasing when

he is 4 ft from the building?

Answer 1

Let x be the horizontal distance the man has walked away from the spotlight to the wall and y be the length of his shadow. Using similar triangles we can write x/6=12/y, because the ratio of the distance walked to the man’s height is the same as the ratio of the distance between the spotlight and the wall and the height of the shadow. So y=72/x. Differentiating with respect to time t, dy/dt=-(72/x²)dx/dt, where dy/dt is the rate of increase of the shadow’s length and dx/dt is the man’s walking speed=1.5ft/s.

Therefore when x=8 ft, the man is 4 ft away from the building, and y=72/8=9 ft.

So, dy/dt=-(72/64)(1.5)=-(9/8)(3/2)=-27/16=-1.6875 ft/s. The shadow is decreasing at the rate of 1.6885 ft/s when the man is 4 ft from the wall.