Question

A conical water tank with vertex down has a radius of 10 feet at the top and is 28 feet high. If water flows into the tank at a rate of 20 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 15 feet deep?

Answer 1

I think i have a solution although this is a little rough and ready.

Answer 2

Ok so the transient equation you are given as:\[\frac{ d(V) }{ dt }\]
Now we need to derive an expression for the change in volume with height.
So first we need to get the change in volume for a cone. Start with the simple area for a circle.
and integrate it for a given height i.e:\[V = \int\limits_{0}^{h} \pi r^{2}.dh\]
In the case of say a cylinder r is independent of height but for a cone this is not true.
because as h changes so does the height.

We therefore need to create an expression for how the radius changes with height.
we can consider the following:

because of the simple trigonometry for a right angled triangle the following must be true:\[\tan(a) = \frac{ r }{ h }=\frac{ R }{ H }\]
therefore we can create the following expression to relate r to h:
\[r = h \frac{ R }{ H}\]