Question

A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per secon

Answer 1

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

$\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}$

Solving for Y given  $X=2m/s$ we get

$\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s$