Conical geometry differential equations

10 \includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-4_335_875_262_635} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semivertical angle is \(60 ^ { \circ }\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt { } h\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).

1. Show that \(h\) and \(t\) satisfy a differential equation of the form $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A h ^ { - \frac { 3 } { 2 } } ,$$ where \(A\) is a positive constant.
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\).
3. Find the time at which the depth reaches \(\frac { 1 } { 2 } H\).
   [0pt] [The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

10. \begin{figure}[h] \end{figure} Diagram not drawn to scale Figure 3 shows a container in the shape of an inverted right circular cone which contains some water. The cone has an internal radius of 3 m and a vertical height of 5 m as shown in Figure 3. At time \(t\) seconds,the height of the water is \(h\) metres,the volume of the water is \(V \mathrm {~m} ^ { 3 }\) and water is leaking from a hole in the bottom of the container at a constant rate of \(0.02 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\) [The volume of a cone of radius \(r\) and height \(h\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) .]

1. Show that,while the water is leaking, $$h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 1 } { \mathrm { k } \pi }$$ where \(k\) is a constant to be found. Given that the container is initially full of water,
2. express \(h\) in terms of \(t\) .
3. Find the time taken for the container to empty,giving your answer to the nearest minute.

10 A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of \(0.01 \mathrm {~m} ^ { 3 }\) per second. At time \(t\) seconds the fertiliser remaining in the container forms an inverted cone of height \(h\) metres.
[0pt] [The volume of a cone is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

1. Show that \(h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 9 } { 400 \pi }\).
2. Express \(h\) in terms of \(t\).
3. Find the time it takes to empty the container, giving your answer to the nearest minute.

8 \includegraphics[max width=\textwidth, alt={}, center]{cb83836f-753f-4b3a-99e8-a18aff0f49ff-07_360_489_1027_788} The diagram shows a water tank which is shaped as an inverted cone with semi-vertical angle \(30 ^ { \circ }\) and height 50 cm . Initially the tank is full, and the depth of the water is 50 cm . Water flows out of a small hole at the bottom of the tank. The rate at which the water flows out is modelled by \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - 2 h\), where \(V \mathrm {~cm} ^ { 3 }\) is the volume of water remaining and \(h \mathrm {~cm}\) is the depth of water in the tank \(t\) seconds after the water begins to flow out. Determine the time taken for the tank to become empty.
[0pt] [For a cone with base radius \(r\) and height \(h\) the volume \(V\) is given by \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]

\includegraphics{figure_10} A tank containing water is in the form of a cone with vertex \(C\). The axis is vertical and the semi-vertical angle is \(60°\), as shown in the diagram. At time \(t = 0\), the tank is full and the depth of water is \(H\). At this instant, a tap at \(C\) is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where \(h\) is the depth of water at time \(t\). The tank becomes empty when \(t = 60\).

1. Show that \(h\) and \(t\) satisfy a differential equation of the form $$\frac{dh}{dt} = -Ah^{-\frac{1}{2}},$$ where \(A\) is a positive constant. [4]
2. Solve the differential equation given in part (i) and obtain an expression for \(t\) in terms of \(h\) and \(H\). [6]
3. Find the time at which the depth reaches \(\frac{1}{2}H\). [1]

[The volume \(V\) of a cone of vertical height \(h\) and base radius \(r\) is given by \(V = \frac{1}{3}\pi r^2 h\).]