A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).

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- Apr 17th 2008, 11:13 AMDINOCALC09Speed
A 20 foot ladder slides down a wall at 5ft/sec. At what speed is the bottom sliding out when the top is 10 feet from the floor (in ft/sec).

- Apr 17th 2008, 11:36 AMTheEmptySet
let y be the height of the ladder on the wall

using the pythagorean theorem we get

$\displaystyle x^2+y^2=400$ note: we also get when y=10 that $\displaystyle x=3\sqrt{10}$

taking the derivative with respect to time we get

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

plugging in all the info gives

$\displaystyle 2(3\sqrt{10})\frac{dx}{dt}+2(10)(5)=0 \iff \frac{dx}{dt}=\frac{50}{3\sqrt{10}}=\frac{5\sqrt{1 0}}{3}$ - Apr 17th 2008, 11:44 AMDINOCALC09
i got 5/sqrt(3)

Taking the derivatives of the pythagorean theorem,

$x\left(\frac {dx}{dt}\right) + y\left(\frac {dy}{dt}\right) = z\left(\frac {dz}{dt}\right)$

But $\left(\frac {dz}{dt}\right) = 0$ so

$x\left(\frac {dx}{dt}\right) = - y\left(\frac {dy}{dt}\right)$

Then from pythagoras, $x^{2} + y^{2} = z^{2}$. We are looking for $x$ at $y = 10$ so $x = \sqrt {400 - 100} = \sqrt {300}$.

Then we have

$\sqrt {300}\left(\frac {dx}{dt}\right) = - 10(5)\Rightarrow\left(\frac {dx}{dt}\right) = \frac {50}{\sqrt {300}} = \frac {5}{\sqrt {3}}$

The negative sign just gives direction so it isn't really important.