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A $13$ foot ladder is leaning against a wall. If the top slips down the wall at a rate of $4 ft/s$, how fast will the foot be moving away from the wall when the top is $10$ feet above the ground?

I got $\frac{-8\sqrt{69}}{20}$ ft/s but this doesn't seem to be the answer, does anyone know why?

I have $169=x^2+y^2$ and $\frac{dx}{dt}$=4ft/s

By differentiating, $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ => $8x+2y\frac{dy}{dt}=0$

If $y=10$, $x=\sqrt{69}$ => $-8(\sqrt{69})=2(10)\frac{dy}{dt}$

Which gave me $\frac{8\sqrt{69}}{20}$ ft/s

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Two issues: $\frac{dx}{dt}$ should be negative, since the ladder is moving down the wall, not being pushed up the wall, and you seem to have mixed up $x$ and $y$. If $\frac{dx}{dt}$ is the rate at which the top of the ladder is moving down the wall, then we should have $x=10$ instead of $y=10$, since $x$ is the height of the top of the ladder.

We have $\frac{dx}{dt}=-4$, $x=10$, and $y=\sqrt{69}$, which leads to

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$ $$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$ $$10(-4)+\sqrt{69}\frac{dy}{dt}=0$$ $$\sqrt{69}\frac{dy}{dt}=40$$ $$\frac{dy}{dt}=\frac{40}{\sqrt{69}}$$

Therefore, the foot of the ladder is moving away from the wall at $\frac{40}{\sqrt{69}}\approx4.82\text{ ft/s}$. (Brief sanity check: our answer is positive, which confirms that the foot of the ladder is indeed moving away from the wall.)

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$x^2+y^2=L^2,$ where $y$ is vertical wall intercept and $x$ is ground intercept, ladder length being $L$. Usual sign convention for $(x,y) $ directions. We have $ x= \sqrt{69}, y=10$ at the given instant.

So differentiating $ \, x \dot x+y \dot y=0,$

$$ \sqrt{69} \dot x + 10(-4) \rightarrow \dot x= \dfrac{40}{\sqrt{69} }$$ where $y$ speed downwards is taken negative, the lower end slides to right, as it should.

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