Related rates with cones, hemispheres, and bowls (variable depth)

3 \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Fig. 1 shows an open tank in the shape of a triangular prism. The vertical ends \(A B E\) and \(D C F\) are identical isosceles triangles. Angle \(A B E =\) angle \(B A E = 30 ^ { \circ }\). The length of \(A D\) is 40 cm . The tank is fixed in position with the open top \(A B C D\) horizontal. Water is poured into the tank at a constant rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). The depth of water, \(t\) seconds after filling starts, is \(h \mathrm {~cm}\) (see Fig. 2).

1. Show that, when the depth of water in the tank is \(h \mathrm {~cm}\), the volume, \(V \mathrm {~cm} ^ { 3 }\), of water in the tank is given by \(V = ( 40 \sqrt { } 3 ) h ^ { 2 }\).
2. Find the rate at which \(h\) is increasing when \(h = 5\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a hemispherical bowl.
Water is flowing into the bowl at a constant rate of \(180 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
When the height of the water is \(h \mathrm {~cm}\), the volume of water \(V \mathrm {~cm} ^ { 3 }\) is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 90 - h ) , \quad 0 \leqslant h \leqslant 30$$ Find the rate of change of the height of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 15\) Give your answer to 2 significant figures.

5. \begin{figure}[h] \end{figure} A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm , as shown in Figure 2. Water is flowing into the container. When the height of water is \(h \mathrm {~cm}\), the surface of the water has radius \(r \mathrm {~cm}\) and the volume of water is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(V = \frac { 4 \pi h ^ { 3 } } { 27 }\).
   [0pt] [The volume \(V\) of a right circular cone with vertical height \(h\) and base radius \(r\) is given by the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).] Water flows into the container at a rate of \(8 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find, in terms of \(\pi\), the rate of change of \(h\) when \(h = 12\). \section*{Question 5 continued} \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

3. \begin{figure}[h] \end{figure} A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is \(h \mathrm {~m}\), the volume \(V \mathrm {~m} ^ { 3 }\) is given by $$V = \frac { 1 } { 12 } \pi \cdot h ^ { 2 } ( 3 - 4 h ) , \quad 0 \leqslant h \leqslant 0.25$$

1. Find, in terms of \(\pi , \frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 0.1\) Water flows into the bowl at a rate of \(\frac { \pi } { 800 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find the rate of change of \(h\), in \(\mathrm { m } \mathrm { s } ^ { - 1 }\), when \(h = 0.1\)

4. \begin{figure}[h] \end{figure} 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 \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(V = \frac { 1 } { 9 } \pi h ^ { 3 }\) [0pt] [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 \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\),
2. 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 \(\pi\).

3. \begin{figure}[h] \end{figure} A bowl with circular cross section and height 20 cm is shown in Figure 2.
The bowl is initially empty and water starts flowing into the bowl.
When the depth of water is \(h \mathrm {~cm}\), the volume of water in the bowl, \(V \mathrm {~cm} ^ { 3 }\), is modelled by the equation $$V = \frac { 1 } { 3 } h ^ { 2 } ( h + 4 ) \quad 0 \leqslant h \leqslant 20$$ Given that the water flows into the bowl at a constant rate of \(160 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), find, according to the model,

1. the time taken to fill the bowl,
2. the rate of change of the depth of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 5\)

6. \includegraphics[max width=\textwidth, alt={}, center]{14a2477a-c40e-4b4b-bc39-7100d1df9b4d-2_397_488_1299_632} The diagram shows a vertical cross-section through a vase.
The inside of the vase is in the shape of a right-circular cone with the angle between the sides in the cross-section being \(60 ^ { \circ }\). When the depth of water in the vase is \(h \mathrm {~cm}\), the volume of water in the vase is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(V = \frac { 1 } { 9 } \pi h ^ { 3 }\). The vase is initially empty and water is poured in at a constant rate of \(120 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find, to 2 decimal places, the rate at which \(h\) is increasing
   1. when \(h = 6\),
   2. after water has been poured in for 8 seconds.

4 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
2. Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
   [0pt] [You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
3. Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).

1 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is \(45 ^ { \circ }\). Water is poured into the cone at a constant rate of \(5 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the height of the water surface above the vertex O of the cone is \(h \mathrm {~cm}\), and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure} Find \(V\) in terms of \(h\). Hence find the rate at which the height of water is increasing when the height is 10 cm .
[0pt] [You are given that the volume \(V\) of a cone of height \(h\) and radius \(r\) is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].

4 Water flows into a bowl at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) (see Fig. 4). \begin{figure}[h] \end{figure} When the depth of water in the bowl is \(h \mathrm {~cm}\), the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \pi h ^ { 2 }\). Find the rate at which the depth is increasing at the instant in time when the depth is 5 cm .

7 Fig. 4 shows a cone. The angle between the axis and the slant edge is \(30 ^ { \circ }\). Water is poured into the cone at a constant rate of \(2 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the radius of the water surface is \(r \mathrm {~cm}\) and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Write down the value of \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
2. Show that \(V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }\), and find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
   [0pt] [You may assume that the volume of a cone of height \(h\) and radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]
3. Use the results of parts (i) and (ii) to find the value of \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when \(r = 2\).

3 \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.

4 Water flows into a bowl at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) (see Fig. 4). \begin{figure}[h] \end{figure} When the depth of water in the bowl is \(h \mathrm {~cm}\), the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \pi h ^ { 2 }\). Find the rate at which the depth is increasing at the instant in time when the depth is 5 cm .

4 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is \(45 ^ { \circ }\). Water is poured into the cone at a constant rate of \(5 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the height of the water surface above the vertex O of the cone is \(h \mathrm {~cm}\), and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure} Find \(V\) in terms of \(h\).
Hence find the rate at which the height of water is increasing when the height is 10 cm .
[0pt] [You are given that the volume \(V\) of a cone of height \(h\) and radius \(r\) is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].