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The base of a $13-ft$ ladder that is leaning against a wall begins to slide away from the wall. When the base is 12 ft from the wall and moving at the rate of $3 ft/sec$, how fast is the top of the ladder sliding down the wall? Here's an illustration. I'd like to know what process/steps to take to solve any similar problems. Thanks in advance!

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The diagram is incomplete, the picture of me close to the top of the ladder has been left out. I should have known better than to use a $13$ foot ladder.

Let $x=x(t)$ be the distance of the foot of the ladder from the foot of the wall. Let $y=y(t)$ be the distance of the top of the ladder from the ground.

By the Pythagorean Theorem, we have $x^2+y^2=13^2$. Differentiate with respect to time $t$. we get $$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.\tag{1}$$ Now freeze the situation at the instant that $x=12$. At that instant, $y=5$ and $\frac{dx}{dt}=3$. From (1) we can find $\frac{dy}{dt}$ at that instant.

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    $\begingroup$ Haha! Hopefully next time you'll be in the picture. Thanks for the help! $\endgroup$ – Kenshin Jun 30 '14 at 0:49
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    $\begingroup$ You are welcome. This falls under a standard category of problem, Related Rates. We know how fast some quantity $x$ is changing, and want to find how fast some related quantity $y$ is changing. We find a relationship linking $x$ and $y$, and differentiate. $\endgroup$ – André Nicolas Jun 30 '14 at 1:27
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You know that $x^2+y^2=13^3.$ Now, if the ladder moves, we $x$ and $y$ are functions of time. Taking derivatives in the equality above we have $2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}=0,$ or, equivalently $x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}=0.$ If you parametrize $x(t)=12+3t$ ($12$ is the initial distance from the wall and $3$ the velocity) you get that

$$(12+3t)3+y(t)\frac{dy}{dt}=0\implies 36+9t+y(t)\frac{dy}{dt}=0.$$ Since $y(t)=\sqrt{169-(12+3t)^2}$ you get

$$\frac{dy}{dt}=-\frac{36+9t}{\sqrt{169-(12+3t)^2}}=-\frac{3t+12}{\sqrt{6-8t-t^2}}$$

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  • $\begingroup$ Thanks a lot for the help! $\endgroup$ – Kenshin Jun 30 '14 at 0:49
  • $\begingroup$ You're welcome. $\endgroup$ – mfl Jun 30 '14 at 0:50

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