Calculus
posted by bryan on .
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

make a sketch, place the man x m from the light
let the length of the shadow on the wall me y m
by similar triangles : 2/x = y/12
xy = 24
x dy/dt + y dx/dt = 0
dy/dt = y dx/dt / x
when he is 4 m from the wall x = 8 (look at our diagram)
so in xy=24 > 8y = 24 or y = 3
and dx/dt = 1.6
dy/dt = 3(1.6)/8
=  .6 m
at that moment his shadow is decreasing at .6 m /s