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?


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?



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 $h(t)$.


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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.]


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