To begin this problem, let's define some constants:

\(a\) = the rate at which sand is being dumped on the pile, and will have units of volume per time.

\(b\) = the ratio of the base radius to the height of the pile.

And our variables:

\(V\) = the volume of the pile at time \(t\).

\(r\) = the base radius of the pile.

\(h\) = the height of the pile.

And so immediately, we may write:

\(\displaystyle \d{V}{t}=a\)

Now, the volume \(V\) of a cone is given by:

\(\displaystyle V=\frac{\pi}{3}r^2h\)

We know:

\(\displaystyle b=\frac{r}{h}\implies r=bh\)

Hence:

\(\displaystyle V=\frac{\pi}{3}(bh)^2h=\frac{\pi b^2}{3}h^3\)

Implicitly differentiating with respect to \(t\), we obtain:

\(\displaystyle \d{V}{t}=\pi b^2h^2\d{h}{t}\)

And so we have:

\(\displaystyle \frac{a}{\pi b^2}=h^2\d{h}{t}\)

Letting \(h_0\) be the initial height of the pile, and exchanging dummy variables of integration, we may integrate as follows:

\(\displaystyle \frac{a}{\pi b^2}\int_0^t \,du=\int_{h_0}^{h} v^2\,dv\)

Applying the FTOC, there results:

\(\displaystyle \frac{a}{\pi b^2}t=\frac{1}{3}\left(h^3-h_0^3\right)\)

And so, we find:

\(\displaystyle h(t)=\sqrt[3]{\frac{3a}{\pi b^2}t+h_0^3}\)