2020 ACJC Promo Q13

(A cone of radius \(r\) and height \(h\) has volume \(V=\frac{1}{3}\pi {{r}^{2}}h\).)

1. An ornament consists of a sphere of fixed radius \(r\) cm inscribed in a right circular cone. The sphere is in contact with the base of the cone at the point \(P\), and with the inner surfaces of the cone at the points \(D\) and \(E\), respectively, as shown in the diagram. Each of the lengths \(AB\) and \(AC\) makes an angle of \(\theta \) radians with the downward vertical.
   
   1. Show that the volume of the cone, \(V\)cm\(^{3}\), is given by:
      \(V=\frac{1}{3}\pi {{r}^{3}}{{\left( 1+\mathrm{cosec}\theta  \right)}^{3}}{{\tan }^{2}}\theta \).[2]
   2. Using differentiation, show that the volume of the cone is minimum when \(\sin \theta =\frac{1}{3}\).[5]
2. Another ornament in the shape of an hour-glass is made up of two identical right circular cones of fixed height \(H\)cm and slant height \(\sqrt{6}H\)cm as shown in the diagram below. The hour-glass is filled with some luminous liquid such that when inverted, the liquid in the upper cone flows into the lower cone at a rate of \(4\)cm\(^{3}\)s\(^{-1}\) through a hole of negligible size at \(K\). At time \(t\) seconds, the depth of the liquid in the lower cone is \(h\)cm. Show that \(W\)cm\(^{3}\), the volume of liquid in the lower cone at time \(t\) s, is given by:
   \(W=\frac{5}{3}\pi \left[ {{H}^{3}}-{{\left( H-h \right)}^{3}} \right]\).[2]
   
   Find the rate of increase of the depth of the liquid in the lower cone when the depth of the liquid in the lower cone is \(\frac{1}{25}H\)cm. Leave your answer in exact form in terms of \(\pi \) and \(H\).[3]