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

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.)

$$x^2+y^2=L^2,$$ 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.