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 6m from the building?

Question

The spotlight is 20 metres away.

We're going through derivatives right now so so far I've tried variations of this:

dx/dy = 40/((2-x)^2)

@ x =6

40/ (14^2) - 0.6

But I can't get the right answer

Answer 1

Imagine a large triangle formed by the spotlight at the origin with a base of 20 and the shadow height on the wall y. A similar triangle inside the large triangle is x meters from the spotlight and 2m high, the height of the woman.

By similar triangles:

ShadowHeight/DistanceLightToWall = WomanHeight/DistanceLightToWoman

Let y be the shadow height. Let x = distance from light to the woman.

y/20 = 2/x

or

xy = 40

Take the derivative of both sides:

xdy/dt +ydx/dt = 0 By the product rule.

Solve for dy/dy which is what we want:

dy/dt = -(y/x)dx/dt dx/dt is given as 0.6 m/s

When the woman is 6 m from the wall, she is 20 - 6 meters from the light, or x = 14 m.

Remember from the similar triangles

y/20 = 2/x = 2/14 --> y = 40/14 = 20/7

Therefore dy/dt = -(20/7)(1/14)(0.6) = -0.12245 m/s

The minus sign indicates that the rate is decreasing because y is getting shorter.