Question

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 1.3 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer 1

Let y = length of shadow on the building
We can form similar triangles here so that we have,
y/2 = 12/x, where x = distance of man from the spotlight
or y = 2(12)/x
y = 24/x

taking the derivative of y with respect to time t:
dy/dt = - 24(dx/dt)/x^2

From the given conditions of the problem, dx/dt = 1.3 m/s and
x = 12m - 4m = 8m.

Therefore,
dy/dt = -24(1.3 m/s)/(8 m)^2
dy/dt = -0.49 m/s answer

Hope this helps.