related rates...calc?

Question

A man 6 ft tall walks at a rate of 5 ft/s away from a lamppost that is 19ft high. At what rate is the lenght of his shadow changing when he is 35 ft away from the lamppost?

Answer 1

We first need to find a formula that relates the distance of the man from the lamppost with the length of the shadow.
I’m sorry I can’t draw this for you.

I’m going to form a triangle from the bottom of the lamp post to the top of the lamppost to the end of the man’s shadow. The man is within the triangle with his head touching the hypotenuse. I’m going to use x to represent the distance from the man to the lamppost, and y to represent the length of the shadow.

From geometry, we know that the ratio of the lamppost height to the length from the lamppost to the end of the shadow is the same as the height of the man to the length of the shadow.

19 / (x + y) = 6 / y
19y = 6x + 6y
13y = 6x
Take d/dt on both sides
13(dy/dt) = 6(dx/dt)
You know that dx/dt = 5 ft/s
13(dy / dt) = 6(5 ft/s)
dy / dt = 30/13 ft/s

So no matter what distance he is away from the lamppost, as long as he is constantly walking at 5 ft/s his shadow will always grow at 30/13 ft/s.

Answer 2

Starting with Michael's

13y = 6x

we get

y = 6x / 13

We are asked for dy/dt.

dy/dt = (dx/dt) * (dy/dx)
= 5 * (6 / 13) = 30 / 13