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:

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





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



Solving for dh/dt, we get


Now, the easy way I mentioned was by noting that

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,

Now, all we do is . Same as above only much quicker.