$dh/dt$ given $dV/dt$

Question

I have a textbook problem whose answer is different from the one I got.

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The problem goes:

A conical egg timer is letting sand through from top to bottom at a rate of $0.02\,\mathrm{cm^3s^{-1}}$. find an expression for the rate of change of height ${\dfrac{dh}{dt}}$.

(There is a diagram showing the full height of the conical egg timer as $5\,\text{cm}$ and the radius of its base as $2\,\text{cm}$. The value of $h$ is the height of the sand in the conical egg timer.)

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My Attempt:

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What I understand from this question is that:

1. The conical egg timer is an upside-down cone, so its volume is ${V={\frac{1}{3}{\pi}r^{2}h}}$, which is also the same as the volume of the sand within it at the beginning.

2. The rate of change of volume of the sand within the conical egg timer with respect to time is ${\dfrac{dV}{dt}}=-0.02$.

3. The diagram shows the radius at the base of the conical egg timer (or "top" since it is upside-down) to be ${2}$ and its full height from base to "point" to be ${5}$.

4. The sand exits from the "point" so its height decreases from its base (the top).

5. The question is asking me to find the rate of change of the height of the sand within the timer with respect to time, or ${\dfrac{dh}{dt}}$.

Given this information, I attempted to answer the problem:

0. If I am correct, the values given at the start are:

   ${\dfrac{dV}{dt}}=-0.02$

   and (at the very beginning, when the cone is full of sand, assuming it starts full)

   $h=5$ and $r=2$.

1. First, I differentiated the volume ${V}$ with respect to ${h}$, giving ${\dfrac{dV}{dh}={\frac{1}{3}}{\pi}{r^{2}}}$ (using the power rule for differentiation).

   $\to {\dfrac{dV}{dh}={\frac{1}{3}}{\pi}{r^{2}}}$

2. Secondly, I used the relationship of the derivative of a function and the derivative of its inverse to get ${\dfrac{dh}{dV}}={\dfrac{1}{\dfrac{dV}{dh}}}={\dfrac{1}{\frac{1}{3}{\pi}r^{2}}}={\dfrac{3}{{\pi}r^{2}}}$ (using the inverse function theorem (I don't fully understand why derivatives of inverses are reciprocals to one another but that's the rule I used)).

   $\to {\dfrac{dh}{dV}}={\dfrac{3}{{\pi}r^{2}}}$

3. Thirdly, I multiplied ${\dfrac{dV}{dt}}$ and ${\dfrac{dh}{dV}}$ to get ${\dfrac{dh}{dt}}=-0.02 \times{\dfrac{3}{{\pi}r^{2}}}={\dfrac{-0.06}{{\pi}r^{2}}}={\dfrac{-3}{50{\pi}r^{2}}}$ (using the chain rule for differentiation).

   $\to {\dfrac{dh}{dt}}={\dfrac{-3}{50{\pi}r^{2}}}$

4. Then, I used the initial values of the height and radius of the sand in the cone to write ${\dfrac{dh}{dt}}$ in terms of $h$.

   The initial values of $h$ and $r$ are the values found in the diagram:

   $h=5$ and $r=2$

   $\to {\dfrac{dh}{dt}}={\dfrac{-3}{50{\pi}r^{2}}}={\dfrac{-3}{50{\pi}({2})^{2}}}={\dfrac{-3}{200{\pi}}}={\dfrac{-3}{40h{\pi}}}={\dfrac{-3}{8{h^{2}}{\pi}}}={\dfrac{-3}{8{\pi}h^{2}}}$ (replacing $r$ in the equation with $2$ and then replacing every multiple of $5$ in the equation with $h$).

   $\to$ my answer is ${\dfrac{dh}{dt}}={\dfrac{-3}{8{\pi}h^{2}}}$.

C) The answer given in the textbook however, is ${\dfrac{dh}{dt}}={\dfrac{-1}{8{\pi}h^{2}}}$.

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Question:

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I'd like to ask for help with regards to finding where I made my error(s), so I hope my working is easy to follow- I am also unsure about whether this is the correct way to answer the question in the first place - thanks.

Answer 1

Note that in a cone, the radius isn't constant. More specifically, the radius is dependent on the height (you can use similar triangles to find their ratio). When you try to find $\frac{dV}{dh}$, you must consider that $\frac{dr}{dh}\neq 0$.

It would be better to substitute an expression relating $r$ and $h$ and then find $\frac{dV}{dh}$.