Related rates of change

9 The volume \(V \mathrm {~m} ^ { 3 }\) of a large circular mound of iron ore of radius \(r \mathrm {~m}\) is modelled by the equation \(V = \frac { 3 } { 2 } \left( r - \frac { 1 } { 2 } \right) ^ { 3 } - 1\) for \(r \geqslant 2\). Iron ore is added to the mound at a constant rate of \(1.5 \mathrm {~m} ^ { 3 }\) per second.
[0pt]

1. Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m . [3]
2. Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second.

2 The volume of a spherical balloon is increasing at a constant rate of \(50 \mathrm {~cm} ^ { 3 }\) per second. Find the rate of increase of the radius when the radius is 10 cm . [Volume of a sphere \(= \frac { 4 } { 3 } \pi r ^ { 3 }\).]

7 The point \(P ( x , y )\) is moving along the curve \(y = x ^ { 2 } - \frac { 10 } { 3 } x ^ { \frac { 3 } { 2 } } + 5 x\) in such a way that the rate of change of \(y\) is constant. Find the values of \(x\) at the points at which the rate of change of \(x\) is equal to half the rate of change of \(y\).

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 \(h = 5\).

8. The volume \(V\) of a spherical balloon is increasing at a constant rate of \(250 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate of increase of the radius of the balloon, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), at the instant when the volume of the balloon is \(12000 \mathrm {~cm} ^ { 3 }\).
Give your answer to 2 significant figures.
[0pt] [You may assume that the volume \(V\) of a sphere of radius \(r\) is given by the formula \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]

10. \begin{figure}[h] \end{figure} Figure 4 shows a right circular cylindrical rod which is expanding as it is heated.
At time \(t\) seconds the radius of the rod is \(x \mathrm {~cm}\) and the length of the rod is \(6 x \mathrm {~cm}\).
Given that the cross-sectional area of the rod is increasing at a constant rate of \(\frac { \pi } { 20 } \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\), find the rate of increase of the volume of the rod when \(x = 2\) Write your answer in the form \(k \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) where \(k\) is a rational number.

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.

4. \begin{figure}[h] \end{figure} A regular icosahedron of side length \(x \mathrm {~cm}\), shown in Figure 1, is expanding uniformly. The icosahedron consists of 20 congruent equilateral triangular faces of side length \(x \mathrm {~cm}\).

1. Show that the surface area, \(A \mathrm {~cm} ^ { 2 }\), of the icosahedron is given by $$A = 5 \sqrt { 3 } x ^ { 2 }$$ Given that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the icosahedron is given by $$V = \frac { 5 } { 12 } ( 3 + \sqrt { 5 } ) x ^ { 3 }$$
2. show that \(\frac { \mathrm { d } V } { \mathrm {~d} A } = \frac { ( 3 + \sqrt { 5 } ) x } { 8 \sqrt { 3 } }\) The surface area of the icosahedron is increasing at a constant rate of \(0.025 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\)
3. Find the rate of change of the volume of the icosahedron when \(x = 2\), giving your answer to 2 significant figures.

4. \begin{figure}[h] \end{figure} A cone, shown in Figure 2, has

• fixed height 5 cm
• base radius \(r \mathrm {~cm}\)
• slant height \(l \mathrm {~cm}\)
  1. Find an expression for \(l\) in terms of \(r\)

Given that the base radius is increasing at a constant rate of 3 cm per minute,

find the rate at which the total surface area of the cone is changing when the radius of the cone is 1.5 cm . Give your answer in \(\mathrm { cm } ^ { 2 }\) per minute to one decimal place.
[0pt] [The total surface area, \(S\), of a cone is given by the formula \(S = \pi r ^ { 2 } + \pi r l\) ]

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 \(h = 12\).

6. The area \(A\) of a circle is increasing at a constant rate of \(1.5 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\). Find, to 3 significant figures, the rate at which the radius \(r\) of the circle is increasing when the area of the circle is \(2 \mathrm {~cm} ^ { 2 }\).
(5)

6. Oil is leaking from a storage container onto a flat section of concrete at a rate of \(0.48 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). The leaking oil spreads to form a pool with an increasing circular cross-section. The pool has a constant uniform thickness of 3 mm . Find the rate at which the radius \(r\) of the pool of oil is increasing at the instant when \(r = 5 \mathrm {~cm}\). Give your answer, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-19_104_95_2617_1786}

3. \begin{figure}[h] \end{figure} Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After \(t\) seconds the radius of the rod is \(x \mathrm {~cm}\) and the length of the rod is \(5 x \mathrm {~cm}\). The cross-sectional area of the rod is increasing at the constant rate of \(0.032 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\).

1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) when the radius of the rod is 2 cm , giving your answer to 3 significant figures.
2. Find the rate of increase of the volume of the rod when \(x = 2\).
   \section*{LU}

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 \(h = 0.1\)

2. \begin{figure}[h] \end{figure} Figure 1 shows a metal cube which is expanding uniformly as it is heated. At time \(t\) seconds, the length of each edge of the cube is \(x \mathrm {~cm}\), and the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} x } = 3 x ^ { 2 }\) Given that the volume, \(V \mathrm {~cm} ^ { 3 }\), increases at a constant rate of \(0.048 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\),
2. find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\), when \(x = 8\)
3. find the rate of increase of the total surface area of the cube, in \(\mathrm { cm } ^ { 2 } \mathrm {~s} ^ { - 1 }\), when \(x = 8\)

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 \(h = 5\)

2. \begin{figure}[h] \end{figure} Figure 1 shows a cube which is increasing in size.
At time \(t\) seconds,

• the length of each edge of the cube is \(x \mathrm {~cm}\)
• the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\)
• the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\)

Given that the surface area of the cube is increasing at a constant rate of \(4 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\)

1. show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\) where \(k\) is a constant to be found,
2. show that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = V ^ { p }\) where \(p\) is a constant to be found.

4 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 \(V = 100\).

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.

1. A balloon is filled with air at a constant rate of \(80 \mathrm {~cm} ^ { 3 }\) per second.

Assuming that the balloon is spherical as it is filled, find to 3 significant figures the rate at which its radius is increasing at the instant when its radius is 6 cm .

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 \(r = 2\).

4 A piston can slide inside a tube which is closed at one end and encloses a quantity of gas (see Fig. 4). \begin{figure}[h] \end{figure} The pressure of the gas in atmospheric units is given by \(p = \frac { 100 } { x }\), where \(x \mathrm {~cm}\) is the distance of the piston from the closed end. At a certain moment, \(x = 50\), and the piston is being pulled away from the closed end at 10 cm per minute. At what rate is the pressure changing at that time?

7 An oil slick is circular with radius \(r \mathrm {~km}\) and area \(A \mathrm {~km} ^ { 2 }\). The radius increases with time at a rate given by \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.5\), in kilometres per hour.

1. Show that \(\frac { \mathrm { dA } } { \mathrm { d } t } = \pi r\).
2. Find the rate of increase of the area of the slick at a time when the radius is 6 km .

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