Related rates of change

2 A spherical balloon of radius \(r \mathrm {~cm}\) has volume \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\). The balloon is inflated at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of \(r\) when \(r = 8\).

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 .

6 When the gas in a balloon is kept at a constant temperature, the pressure \(P\) in atmospheres and the volume \(V \mathrm {~m} ^ { 3 }\) are related by the equation $$P = \frac { k } { V } ,$$ where \(k\) is a constant. [This is known as Boyle's Law.]
When the volume is \(100 \mathrm {~m} ^ { 3 }\), the pressure is 5 atmospheres, and the volume is increasing at a rate of \(10 \mathrm {~m} ^ { 3 }\) per second.

1. Show that \(k = 500\).
2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} V }\) in terms of \(V\).
3. Find the rate at which the pressure is decreasing when \(V = 100\).

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 A giant spherical balloon is being inflated in a theme park. The radius of the balloon is increasing at a rate of 12 cm per hour. Find the rate at which the surface area of the balloon is increasing at the instant when the radius is 150 cm . Give your answer in \(\mathrm { cm } ^ { 2 }\) per hour correct to 2 significant figures.
[0pt] [Surface area of sphere \(= 4 \pi r ^ { 2 }\).]

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.

3 The volume, \(V\) cubic metres, of water in a reservoir is given by $$V = 3 ( 2 + \sqrt { h } ) ^ { 6 } - 192 ,$$ where \(h\) metres is the depth of the water. Water is flowing into the reservoir at a constant rate of 150 cubic metres per hour. Find the rate at which the depth of water is increasing at the instant when the depth is 1.4 metres.

3 The area of a circular stain is growing at a rate of \(1 \mathrm {~mm} ^ { 2 }\) per second. Find the rate of increase of its radius at an instant when its radius is 2 mm .

6 Oil is leaking into the sea from a pipeline, creating a circular oil slick. The radius \(r\) metres of the oil slick \(t\) hours after the start of the leak is modelled by the equation $$r = 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) .$$

1. Find the radius of the slick when \(t = 2\), and the rate at which the radius is increasing at this time.
2. Find the rate at which the area of the slick is increasing when \(t = 2\).

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 .

5 A spherical balloon of radius \(r \mathrm {~cm}\) has volume \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\). The balloon is inflated at a constant rate of \(10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of \(r\) when \(r = 8\).

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 \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].