Hello, qu2010!

1. A spotlight is on the ground 100 feet from a vertical building.

A 6-foot man starts at the spotlight and walks towards the building at 5 ft/sec. Code:

C *
| * A
| *
y | | *
| |6 *
| | *
* - - - * - - - - - - - * - -
D x B 100-x L
: - - - - 100 - - - - - :

The spotlight is at $\displaystyle L.$

The man is: .$\displaystyle AB = 6$

His shadow is: .$\displaystyle y = CD$

His distance from the building is: .$\displaystyle x \,=\,DB$

. . Hence: .$\displaystyle BL \:=\:100-x$

We are given: .$\displaystyle \frac{dx}{dt} = -5$ ft/sec.

From similar right triangles $\displaystyle CDL\text{ and }ABL\text{ we have: }\:\frac{y}{100} \:=\:\frac{6}{100-x}$

. . Then: .$\displaystyle y \:=\:600(100-x)^{-1}$

Differentiate with respect to time: .$\displaystyle \frac{dy}{dt} \:=\:-600(100-x)^{-2}\left(-\frac{dx}{dt}\right)$

. . and we have: .$\displaystyle \boxed{\frac{dy}{dt} \;=\;\frac{600}{(100-x)^2}\,\frac{dx}{dt}}$

a) How fast is the top of the man's shadow moving down the building

when he is 50 feet from the building? When $\displaystyle x = 50$ . . .

. . $\displaystyle \frac{dy}{dt} \;=\;\frac{600}{(100-50)^2}\,(-5) \;=\;\boxed{-\frac{6}{5}\text{ ft/sec}}$

b) How fast is the top of the man's shadow moving down the building

when he is 25 feet from the building? When $\displaystyle x = 25$ . . .

. . $\displaystyle \frac{dy}{dt} \;=\;\frac{600}{(100-25)^2}\,(-5) \;=\;\boxed{-\frac{8}{15}\text{ ft/sec}}$