Here is a related rates y'all may like to tackle.

"a circular tank with height 2 feet, radius on the top of R and radius on the bottom of r where R>r.

If R=6 ft and r=4.5ft, and it's filling at a rate of 15 ft^3/min, how fast is the height of the water changing when the water's radius is 5?

Of course, this forms a frustum of a cone. But, there is a trick that makes this quite easy to find dh/dt.

Since it is a frustum, extend the sides on down to make a cone, then treat it as such.

If we extend the sides down to the apex, it is at (0,-6).

That means the cone is 8 inches high overall.

By using similar triangles: \frac{r}{h}=\frac{6}{8}\Rightarrow r=\frac{3h}{4}

When r=5, then h=20/3.

V=\frac{\pi}{3}(\frac{3h}{4})^{2}h=\frac{3{\pi}h^{ 3}}{16}

\frac{dV}{dt}=\frac{9{\pi}h^{2}}{16}\cdot\frac{dh} {dt}

We have to find dh/dt, when h=20/3.

15=\frac{9{\pi}(\frac{20}{3})^{2}}{16}\cdot\frac{d h}{dt}

Solving for dh/dt, we get \frac{dh}{dt}=\frac{3}{5\pi}


Now, the easy way I mentioned was by noting that \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}

Where A(t) is the area of the water surface at some time t. In this case, we want to know the surface of the water area when r=5.

That is easy enough, 25{\pi}

Now, all we do is \frac{15}{25\pi}=\frac{3}{5\pi}. Same as above only much quicker.