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I've been scratching my head over this problem for at least an hour.

"The top of a ladder slides down a vertical wall at a rate of 0.15 m/s. At the moment when the bottom of the later is 3 m from the wall, it slides away from the wall at a rate of 0.2 m/s. How long is the ladder."

I have,

a'(t)=-0.15,

$a(t)=c-0.15t$,

$b(t)=3.0$, when $t =$ now

b'(t)=0.2, when $t =$ now

$c=$ length $= \sqrt{a^2 +b^2} = \sqrt{a^2 + 9}$

I'm aware that $c=5$ and that $a=4$, and I'm also aware that when b'>a' then $a > b$ (and vice versa). In fact, I think I can even infer from the answer that \frac {a'}{b'} = \frac{b}{a} (is this correct?) and use that to solve the problem a posteriori. Unfortunately, I can't figure out how to solve this as a related rates problem.

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$a^2+b^2$ is constant, so its derivative is zero. But its derivative is 2aa'+2bb'. So if you know three of the quantities a,a',b,b', you can find the fourth; and if you know both $a$ and $b$, then you can get the length.

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    I did give it in a step-by-step manner. Which step(s) don't you understand?2017-06-04