posted by jordan .
a man 6 ft tall is walking toward a building at the rate of 4 ft/sec. if there is a light on the ground 40 ft from the building, how fast is the man's shadow on the building growing shorter when he is 30 ft from the building?
differential calculus -
Using similar triangles, if the man is x from the light, and his shadow is h tall,
h/40 = 6/x
1/40 dh/dt = -6/x^2 dx/dt
So, when the man is 10 ft from the light,
1/40 dh/dt = -6/100 * 4
dh/dt = -9.6 ft/s