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The question is a ladder is 10 ft long and leans against a wall. The top of the ladder slides down at 3 m/s and the bottom slides away from the wall at 4 m/s. Find how high the top of the ladder is up the wall.

I know the general solution to these problems starts with using Pythagorean's theorem but I'm confused how I start this question when I'm not given x (distance from the wall). I've tried to re-arrange Pythagorean's theorem as follows:

l^2 - x^2 = y^2
Plug into Pythagorean's theorem:
l^2 = x^2 + l^2 - x^2

But that doesn't get me anywhere. I know how to do implicit differientation, so could someone explain or give a hint as to how I find x?

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Hint: from $x^2+y^2=10$ take implicit derivative and divide by 2, get $$x\cdot x'+y \cdot y'=0.$$ You know $x'=4,\ y'=-3$ so you have $4x-3y=0$. Now solve this along with $x^2+y^2=10.$

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  • $\begingroup$ Thank you, your explanation also makes a lot of sense. I guess I was sort of circling around it but didn't quite get it, although I now understand it! $\endgroup$ – Cucko Oooo Mar 19 '13 at 21:39
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Start with $l^2-x^2=y^2$ and do implicit differentiation. Note that $l$ isn't changing, so that gives you $$-2x\frac{dx}{dt}=2y\frac{dy}{dt}.\tag{1}$$ You also know from Pythagorean Theorem that $$x=\sqrt{l^2-y^2},$$ since $x$ is simply the horizontal distance from the wall to the bottom of the ladder, so we can rewrite $(1)$ in terms of $y$ and known quantities. After plugging in our values for $l$, $\frac{dx}{dt}$ and $\frac{dy}{dt}$, we then solve for $y.$ Don't forget that $\frac{dy}{dt}$ will be negative. (Why?)

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    $\begingroup$ Thank you so much, your explanation makes a lot of sense. dy/dt would be negative because the ladder is sliding down the wall at the calculated rate. $\endgroup$ – Cucko Oooo Mar 19 '13 at 21:36
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In addition to the two hints already given, check your units. I hope your calculus teacher doesn't assign problems with ladders in feet and speed in meters/second.

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  • $\begingroup$ Oops, that was my fault when copying down the question, thank you for catching that! $\endgroup$ – Cucko Oooo Mar 19 '13 at 21:40

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