Question

A person 6 feet tall is walking away from a lamppost that is 15 ft tall at a rate of 6 ft/sec. At what rate is the end of the person's shadow moving away from the lamppost?

I got 10 ft/sec. Is this correct?

Answer 1

The tip of the shadow is moving at the rate of `=4(ft)/sec`

Explanation:

Let the distance of the person from the bottom of the light post be `=x ft`

And the length of his shadow is `=y ft`

Form the similar triangles

`(x+y)/(15)=y/6`

`6(x+y)=15y`

`6x+6y=15y`

`9y=6x`

`y=2/3x`

Differentiating wrt `t`

`dy/dt=2/3dx/dt`

We know that `dx/dt=6(ft)/sec`

Therefore,

`dy/dt=2/3*6=12/3=4(ft)/sec`

Did you perform the calculations in the same way as you I did?