A spotlight on the ground is shining on a wall 16m away

Question

A spotlight on the ground is shining on a wall 16m away. If a woman 2m tall walks from the spotlight toward the building at a speed of 0.6m/s, how fast is the length of her shadow on the building decreasing when she is 4m from the building?

Answer 1

Let x be the distance of the woman from the spotlight on the ground. Let y be the height of the shadow on the wall.

This forms two similar triangles with bases of x and 16 and heights of 2 and y.

The height/base ratios of the similar triangle are equivalent:

2/x = y/16

Taking the derivative with respect to time of both sides of this equation:

(-2/x²) dx/dt = (1/16) dy/dt

dy/dt = (-32/x²) dx/dt

when x = 16-4 = 12 and dx/dt = 0.6

dy/dt = (-32/144)(0.6) = -0.133 m/s or decreasing at 0.133 m/s since the sign is negative.

You can also use xy = 32 by multiplying the means and extremes and then use the product rule to implicitly take the derivative.

ydx/dt + xdy/dt = 0

dy/dt = -(y/x)dx/dt

when x = 12, y = 32/12 = 8/3

dy/dt = -((8/3)/12)0.6 = -0.133 m/s

Answer 2