I am currently reading How to Solve It by G. Polya. I am confused by the conclusion in one of his examples. You can find it here in this pdf on page 29.
The problems states: "Water is flowing into a conical vessel at the rate r. The vessel has the shape of a right circular cone, with horizontal base, the vertex pointing downwards; the radius of the base is a, the altitude of the cone b. Find the rate at which the surface is rising when the depth of the water is y. Finally obtain the numerical value of the unknown supposing that a = 4 ft., b = 3 ft., r = 2 cu ft. per minute, and y = 1 ft."
The ultimate solution comes out to be: $$\frac{dV}{dt}=\frac{\pi a^2y^2}{b^2}\frac{dy}{dt}$$
I am confused why the denominator is b^2? I get that this is mostly about find the volume as a function of y. I am just glitching on how this equation fulfills the solution.
We have that $V=\frac{1}{3}\pi r^2h$. By similar triangles, $\frac{h}{r}=\frac{b}{a}$, so this is just $V=\frac{1}{3}\pi\frac{a^2h^3}{b^2}$. Differentiating, $\frac{dV}{dt}=\frac{\pi a^2h^2}{b^2}\frac{dh}{dt}$, as desired.