# related rates problem

#### dimon

##### New member
I can solve this problem withough calculus, however, I am required to use derivatives. Hopefully someone can help

A 13 feet ladder is leaning agains a house when it's base starts to slide awayu. By the time the base is 12 ft from the house, the base is moving at the rate of 5ft/sec. How fast is the top of the ladder sliding down the wall then?

any help is appreciated

#### soroban

##### Elite Member
Hello, dimon!

A 13-foot ladder is leaning against a house when its base starts to slide away.
By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec.
How fast is the top of the ladder sliding down the wall then?
Code:
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x
We have: $$\displaystyle \,x^2\,+\,y^2\:=\:13^2\;$$ [1]

Differentiate with respect to time: $$\displaystyle \:2x\left(\frac{dx}{dt}\right)\,+\,2y\left(\frac{dy}{dt}\right) \:=\:0$$

Then: $$\displaystyle \:\frac{dy}{dt}\:=\:-\frac{x}{y}\left(\frac{dx}{dt}\right)\;$$ [2]

At that particular instant: $$\displaystyle \,x\,=\,12$$ ft

Substitute into [1]: $$\displaystyle \,12^2\,+\,y^2\:=\:13^2\;\;\Rightarrow\;\;y\,=\,5$$ ft

And we are given: $$\displaystyle \,\frac{dx}{dt} \,=\,5$$ ft/sec

Substitute into [2]: $$\displaystyle \:\frac{dy}{dt}\:=\;-\frac{12\text{ ft}}{5\text{ ft}}$$$$\displaystyle \left(5\text{ ft/sec}\right) \:=\:-12\text{ ft/sec}$$

thanks
that makes sense