A square pyramid, of side length 100 cm and height 100 cm, of ice is melting at a consistent rate such that all of the ice less than y cm from the surface melts after y hours (bottom is also melting). What is the rate of change of the volume, when the height is 10cm?
It looked very straight forward, I went with $V = s^2.h/3$ and then $s=h$, after that $\frac{dV}{dt} = h^2.\frac{dh}{dt}$. I thought I was pretty close and basically done. But I don`t know, but somehow, I'm not really able to figure out a way ahead. I know that I'm supposed to find how V changes from "consistent rate such that all of the ice less than y cm from the surface melts after y hours", but I'm not really able to decipher it.
Maybe it's the way this question is presented, or I'm just not thinking in that sense at the moment, or perhaps I'm just stupid.
Can anyone please help me with this?
I`ll be really thankful.
Best Answer
One thing you are probably missing is that you rather have to consider $$\frac{dV}{dt}(h)$$ And then you also are implicitely provided by the rate $$\frac{dh}{dt}$$ from the statement "y cm from the surface melts after y hours". Thus you simply have to insert the latter into your formula for the former and evaluate that term at $h=10\ cm$.
Thus the only thing remaining so far is the actual value of $\frac{dh}{dt}$. So a little of geometry has to be applied. Consider a vertical plane intersecting your pyramid, running through the tip and parallel to the base sides, reducing the 3D problem to a 2D one. Then you have a isocele triangle of base size $h$ and height $h$, i.e. the lacing edge size is $\frac{\sqrt5}2h$. For that triangle you have to look for the inradius $\rho$, which provides the depth (and thus time) to melt down to nothing. That one calculates as $$\rho=\frac{(\sqrt5-1)}4h$$
Thus I would say you have $$\frac{dV}{dt}(h)=\frac{(\sqrt5-1)}4h^2$$
--- rk