I've got this question: ) A ladder 15ft long leans against a vertical wall. If the top slides down at 2ft per second, how fast along the ground is the base moving when it is 5ft from the wall?
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I'm not entirely sure how to approach this. the only thing I can work out is that since the top is sliding at 2ft per second, then the bottom must be too. any solutions?
thanks
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By Pythagoras's theorem, $l^2 = h^2 + b^2$
Differentiating with respect to t, we get $2l\frac{dl}{dt}=2h\frac{dh}{dt} + 2b\frac{db}{dt}$
Since the length of the ladder is constant. Hence, $\frac{dl}{dt}= 0.$ Also h = 12 feets.
Hence, $\frac{db}{dt} = -\frac{2*h*\frac{dh}{dt}}{\frac{db}{dt}} = \frac{2*12*2}{2*5} = -4.8 ft/s $
[The minus sign denotes that the end of the ladder touching the surface is sliding away from the wall.]
Let $h(t)$ be the height of the top at time $t$ and $b(t)$ be the position of the base at time $t$. At the beginning we have $h(0) = 15,\ b(0) = 0$. Also $t$ is in second. Moreover you know that $h'(t) = -2$.
Clearly for any $t$, we must have $b(t)^2 + h(t)^2 = 15^2$. Talking the derivative (w.r.t $t$) of this expression gives: $$ 2b'(t)b(t) + 2h'(t)h(t) = 0 $$ Now you can rearrange it to extract $b'(t)$ knowing that $b(t) = 5$ and using pythagorian relation for $h(t)$.