related rates problem

[dimon]

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]

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?

                        |
                        *
                     *  |
              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}$

 

[dimon]

thanks
that makes sense