related rates problem

dimon

New member
Joined
Jul 9, 2006
Messages
15
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
Joined
Jan 28, 2005
Messages
5,587
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:
                        |
                        *
                     *  |
              13  *     |
               *        | y
            *           |
         *              |
    --*-----------------*--
               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]: \(\displaystyle \:\frac{dy}{dt}\:=\;-\frac{12\text{ ft}}{5\text{ ft}}\)\(\displaystyle \left(5\text{ ft/sec}\right) \:=\:-12\text{ ft/sec}\)
 
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