Originally Posted by

**scherz0** Hello everyone,

I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?

Thank you.

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1. A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ feet.

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Let: $\displaystyle y $ = height of the ladder on wall

$\displaystyle x $ = distance of the base of the ladder to the wall

Therefore:

$\displaystyle \frac{dy}{dt} = 4 $ $\displaystyle \frac{dx}{dt} = ?$

$\displaystyle x^2 + y^2 = 25 $

$\displaystyle \frac{d}{dx}(x^2 + y^2 = 25) $

$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 $

$\displaystyle 2x\frac{dx}{dt} + 8y = 0 $

$\displaystyle \frac{dx}{dt} = \frac{-4y}{x} $

I am not sure what to substitute as x and y here.