A spotlight on the ground is shining on a wall 12m12m away.

Question

A spotlight on the ground is shining on a wall 12m12m away. If a woman 2m2m tall walks from the spotlight toward the building at a speed of 0.6m/s,0.6m/s,how fast is the length of her shadow on the building decreasing when she is 2m2m from the building?

Answer 1

First find given rates and unknown rates:

Given rate: Speed the woman walks (We'll call this dx/dt)

Unknown rate: Speed the shadow decreases (We'll call this ds/dt)

We want a way to compare dx/dt and ds/dt

First we need an equation to compare x (distance the woman is from the spotlight and s (the length of the shadow.

It's best to draw a diagram here. You should get similar triangles with:

x/2 = 12/s (multiply both sides by 2

x = 24/s

Now take derivatives with respect to t on both sides.

dx/dt = (-24/s²) ds/dt

We're give dx/dt. We need s when she's 2 meters from the building. x = 12 - 2 = 10

Solve for s: (10) = 24/s

Therefore, s = 2.4

Plug in and solve for ds/dt

0.6 = -24/(2.4)²(ds/dt)