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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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  • $\begingroup$ Doh! I tried that earlier but screwed up the arithmetic. Thanks. $\endgroup$ – Matt Munson Dec 14 '11 at 9:32
  • $\begingroup$ @Gerry Myerson If you give me the answer to this question in a step by step manner, I would understand it very clearly. $\endgroup$ – Dhamnekar Winod Jun 3 '17 at 14:22
  • $\begingroup$ I did give it in a step-by-step manner. Which step(s) don't you understand? $\endgroup$ – Gerry Myerson Jun 4 '17 at 6:46

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