Question

A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.

Answer 1

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light $=15 feet$

height of man$=6 feet$

$\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s$

From diagram

$\frac{15}{y}=\frac{6}{y-x}$

$5(y-x)=2y$

$3y=5x$

differentiate both sides

$3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Tip of shadow is moving at the rate of

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s$

(b)rate at which length of his shadow  is changing

Length of shadow is $y-x$

differentiating w.r.t time

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s$