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.