7.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0e7cab3d-c1e6-4420-93b4-eca5af704432-07_591_730_294_735}
\captionsetup{labelformat=empty}
\caption{Diagram not drawn to scale}
\end{figure}
Figure 4
[0pt]
[ The volume of a cone of base radius \(r\) and height \(h\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) ]
Figure 4 shows a container in the shape of an inverted right circular cone which contains some water.
The cone has an internal base radius of 2.5 m and a vertical height of 4 m .
At time \(t\) seconds
- the height of the water is \(h \mathrm {~m}\)
- the volume of the water is \(V \mathrm {~m} ^ { 3 }\)
- the water is modelled as leaking from a hole at the bottom of the container at a rate of
$$\left( \frac { \pi } { 512 } \sqrt { h } \right) m ^ { 3 } s ^ { - 1 }$$
- Show that, while the water is leaking
$$h ^ { \frac { 3 } { 2 } } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 1 } { 200 }$$
Given that the container was initially full of water
- find an equation, in terms of \(h\) and \(t\), to model this situation.