Hello, anfieldgirl!
Sounds like "Home Alone" . . .
Carlo is perched precariously on top of a 10foot ladder leaning
against the back wall of an apartment building (spying on an enemy of his)
when it starts to slide down the wall at a rate of 4 ft per minute.
Carlo's accomplice, Alex, is standing on the ground 6 ft away from the wall.
How fast is the base of the ladder moving when it hits Alex? Code:

*
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y  \ 10
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*    *   *
x A
We are given: .$\displaystyle \frac{dy}{dt} \:=\:\text{}4$ ft/min.
We know that: .$\displaystyle x^2 + y^2 \:=\:10^2$
Differentiate with respect to time: .$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \:=\:0 \quad\Rightarrow\quad \frac{dx}{dt} \:=\:\frac{y}{x}\,\frac{dy}{dt}$
When $\displaystyle x = 6,\:y = 8$
So we have: .$\displaystyle \frac{dx}{dt} \;=\;\frac{8}{6}(4) \;=\;\frac{16}{3}$
The base of the ladder hits Alex at $\displaystyle 5\tfrac{1}{3}$ ft/min.