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\includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-2_435_472_932_794} The diagram shows a container in the form of a right circular cone. The angle between the axis and the slant height is \(\alpha\), where \(\alpha = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)\). Initially the container is empty, and then liquid is added at the rate of \(14 \mathrm {~cm} ^ { 3 }\) per minute. The depth of liquid in the container at time \(t\) minutes is \(x \mathrm {~cm}\).

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container when the depth is \(x \mathrm {~cm}\) is given by $$V = \frac { 1 } { 12 } \pi x ^ { 3 } .$$ [The volume of a cone is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
2. Find the rate at which the depth of the liquid is increasing at the instant when the depth is 8 cm . Give your answer in cm per minute correct to 2 decimal places.