Calculus Related Rates

Question

A spotlight on the ground is shining on a wall 24m away. If a woman 2m tall walks from the spotlight toward the building at a speed of 0.8m/s, how fast is the length of her shadow on the building decreasing when she is 2m from the building?

Answer 1

Draw a picture of the situation.

Let

S be the point where the spotlight is located,

B be the base of the building on the ground,

SB = 24 m is the distance of the spotlight from the building,

W be the spot where the woman is at some instant of time t,

H be the top of her head,

WH = 2 m is the height of the woman,

v = 0.8 m/s be the speed of the woman,

x = SW be the distance of the woman from the spotlight,

Y be the projection of the point H on the building,

y = BY be the height of her shadow on the building.

We need to calculate dy/dt.

The triangles SWH and SBY are congruent, and therefore

BY/SB = WH/SW

Substituting

y/24 = 2/x

Simplifying

y = 48/x

Differentiting

dy/dt = -48/x² dx/dt

Substituting dx/dt = v

dy/dt = -48×0.8/x² = -38.4/x²

When the woman is 2m from the building then x = 22. At that point

dy/dt = -38.4/22² = -0.08 m/s

So the answer is that when the woman is 2 m from the building the shadow is decreasing

8 cm per second.