Question

A man 6 ft tall walks at a rate of 7 ft/sec away from a lamppost that is 14 ft high. At what rate is the length of his shadow changing when he is 60 ft away from the lamppost? g

Answer 1

Let $d$ be the distance between the lamppost and the man, and $s$ the length of the man's shadow cast by light. Then the man and lamppost form a set of similar right triangles (see image) in which

$\dfrac{s+d}{14}=\dfrac s6$

We can solve for $s$:

$\dfrac d{14}=\dfrac{2s}{21}\implies s=\dfrac{3d}4$

Differentiating both sides with respect to time $t$:

$\dfrac{\mathrm ds}{\mathrm dt}=\dfrac34\dfrac{\mathrm dd}{\mathrm dt}$

Since $d$ is changing at a rate of 7 ft/sec, it follows that $s$ is changing at a 25% slow rate, or 7*3/4 = 21/4 = 5.25 ft/sec.