Question

A man 6.00 ft tall approaches a street light 15.0 ft above the ground at the rate of 4.00 ​ft/s. How fast is the end of the​ man's shadow moving when he is 14.0 ft from the base of the​ light?

Answer 1

Answer:

$\frac{dx}{dt} = 10 ft/s$

Explanation:

As per given figure let say the tip of the shadow is at distance "x" from the base of the lamp

so here we have

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

so we have

$6x = 15 x - 15 y$

$15 y = 9 x$

now we have

$5\frac{dy}{dt} = 2\frac{dx}{dt}$

$5(4) = 2\frac{dx}{dt}$

$\frac{dx}{dt} = 10 ft/s$