# A straight forward related rates problem with a little twist

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 dont 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.

I$$$$ll be really thankful.

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

• Dr. Richard, I am studying for my GRE as well and this question comes from: math.uchicago.edu/~min/GRE/files/week6.pdf If you look at question 12, this question, none of the answers are equivalent to your derivation when $10$ is substituted for $h$, why is that? Commented Oct 22, 2019 at 12:02

Starting at time $$t,$$ if you wait for $$\delta$$ hours, where $$\delta$$ is a very small positive number, the part of the pyramid that melts will be a pyramidal shell with the exact same outer surface as the pyramid at time $$t$$ and with a thickness of $$\delta$$ cm.

It should be clear from this that the rate of loss of volume is proportional to the surface area. In fact, if the surface area at time $$t$$ is $$A(t)$$, measured in cm$$^2$$, the rate of loss of volume at time $$t$$ is $$A(t)$$, measured in cm$$^3$$/hour.

The linear dimensions of the pyramid also shrink at a constant rate which you can also determine, but you don't really need to know for this question.

This may not be a satisfactory solution for your grader and you may want to develop a more explicit approach. But you can still use this method to check any other method you attempt.

A more explicit method could involve working out the rate at which the linear dimensions decrease. For this, note that after $$y$$ hours, the four base vertices will be $$y$$ cm above the original base after $$y$$ hours and also $$y$$ cm from the nearest original slant faces of the pyramid, and the apex vertex will be $$y$$ cm from all four of the original slant faces of the pyramid. You can also assume the new apex is directly below the original one on the perpendicular line to the base, so you only need to consider one of the slant faces to work out the exact location of the apex.