4.
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\caption{Figure 1}
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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 \(V \mathrm {~cm} ^ { 3 }\).
- Show that \(V = \frac { 1 } { 9 } \pi h ^ { 3 }\)
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[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 \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), - 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 \(\pi\).