Melting Ice Cream

Let's start with the relationship between the radius and height of the cone. We know that the radius of the cone is 3cm while the height of the cone is 10cm. Due to similar triangles , the radius and height of the liquid ice cream are bound by the relationship $10r = 3h$ throughout the cone. We solve for $r$ so as to eliminate it from the volume formula for the cone. $r = \frac{3h}{10} \longrightarrow V = \frac{1}{3}\pi r^2 h \longrightarrow V = \frac{1}{3}\pi \left(\frac{3h}{10}\right)^2h = \frac{3\pi h^3}{100}$

We're asked to examine the moment in time at this quantity is $\frac{15}{4}\pi$ , so we know this occurs when $h = 5$ . Differentiating both sides of the volume formula leads us to:

$\frac{dV}{dt} = \frac{9\pi h^2}{100}\frac{dh}{dt}$

Plugging in what we know about $\frac{dV}{dt}$ and $h$ , we find that $\frac{dh}{dt} = \frac{8}{9\pi}$ .