4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0c4a3759-ecaa-47c3-a071-ce25fd11159f-12_978_1264_121_411}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of $30 ^ { \circ }$, as shown in Figure 1. The height of the container is 50 cm .

When the depth of the water in the container is $h \mathrm {~cm}$, the surface of the water has radius $r \mathrm {~cm}$ and the volume of water is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac { 1 } { 9 } \pi h ^ { 3 }$\\[0pt]
[You may assume the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ for the volume of a cone.]

Given that the volume of water in the container increases at a constant rate of $200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$,
\item find the rate of change of the depth of the water, in $\mathrm { cm } \mathrm { s } ^ { - 1 }$, when $h = 15$

Give your answer in its simplest form in terms of $\pi$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4 2018 Q4 [6]}}