Related rates

4 A watermelon is assumed to be spherical in shape while it is growing. Its mass, \(M \mathrm {~kg}\), and radius, \(r \mathrm {~cm}\), are related by the formula \(M = k r ^ { 3 }\), where \(k\) is a constant. It is also assumed that the radius is increasing at a constant rate of 0.1 centimetres per day. On a particular day the radius is 10 cm and the mass is 3.2 kg . Find the value of \(k\) and the rate at which the mass is increasing on this day.

14. The volume of a spherical balloon of radius \(r \mathrm {~cm}\) is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) The volume of the balloon increases with time \(t\) seconds according to the formula $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 9000 \pi } { ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$
2. Using the chain rule, or otherwise, show that $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$ where \(k\) and \(n\) are constants to be found. Initially, the radius of the balloon is 3 cm .
3. Using the values of \(k\) and \(n\) found in part (b), solve the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { k } { r ^ { n } ( t + 81 ) ^ { \frac { 5 } { 4 } } } \quad t \geqslant 0$$ to obtain a formula for \(r\) in terms of \(t\).
4. Hence find the radius of the balloon when \(t = 175\), giving your answer to 3 significant figures.
   (1)
5. Find the rate of increase of the radius of the balloon when \(t = 175\). Give your answer to 3 significant figures.

   END

13. The volume of a spherical balloon of radius \(r \mathrm {~m}\) is \(V \mathrm {~m} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) Given that the volume of the balloon increases with time \(t\) seconds according to the formula $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 20 } { V ( 0.05 t + 1 ) ^ { 3 } } , \quad t \geqslant 0$$
2. find an expression in terms of \(r\) and \(t\) for \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) Given that \(V = 1\) when \(t = 0\)
3. solve the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 20 } { V ( 0.05 t + 1 ) ^ { 3 } }$$ giving your answer in the form \(V ^ { 2 } = \mathrm { f } ( t )\).
4. Hence find the radius of the balloon at time \(t = 20\), giving your answer to 3 significant figures.
   \includegraphics[max width=\textwidth, alt={}]{c6bde466-61ec-437d-a3b4-84511a98d788-48_2632_1828_121_121}

1. The volume \(V \mathrm {~cm} ^ { 3 }\) of a spherical balloon with radius \(r \mathrm {~cm}\) is given by the formula

$$V = \frac { 4 } { 3 } \pi r ^ { 3 }$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) giving your answer in simplest form. At time \(t\) seconds, the volume of the balloon is increasing according to the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 900 } { ( 2 t + 3 ) ^ { 2 } } \quad t \geqslant 0$$ Given that \(V = 0\) when \(t = 0\)
2. 1. solve this differential equation to show that $$V = \frac { 300 t } { 2 t + 3 }$$
   2. Hence find the upper limit to the volume of the balloon.
3. Find the radius of the balloon at \(t = 3\), giving your answer in cm to 3 significant figures.
4. Find the rate of increase of the radius of the balloon at \(t = 3\), giving your answer to 2 significant figures. Show your working and state the units of your answer.

7. At time \(t\) seconds the length of the side of a cube is \(x \mathrm {~cm}\), the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\), and the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\). The surface area of the cube is increasing at a constant rate of \(8 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\).
Show that

1. \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\), where \(k\) is a constant to be found,
2. \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 2 V ^ { \frac { 1 } { 3 } }\). Given that \(V = 8\) when \(t = 0\),
3. solve the differential equation in part (b), and find the value of \(t\) when \(V = 16 \sqrt { } 2\).

8. A circular stain grows in such a way that the rate of increase of its radius is inversely proportional to the square of the radius. Given that the area of the stain at time \(t\) seconds is \(A \mathrm {~cm} ^ { 2 }\),

1. show that \(\frac { \mathrm { d } A } { \mathrm {~d} t } \propto \frac { 1 } { \sqrt { A } }\).
   (6) Another stain, which is growing more quickly, has area \(S \mathrm {~cm} ^ { 2 }\) at time \(t\) seconds. It is given that $$\frac { \mathrm { d } S } { \mathrm {~d} t } = \frac { 2 \mathrm { e } ^ { 2 t } } { \sqrt { S } }$$ Given that, for this second stain, \(S = 9\) at time \(t = 0\),
2. solve the differential equation to find the time at which \(S = 16\). Give your answer to 2 significant figures. \section*{END}

7 A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is \(r \mathrm {~cm}\) after \(t\) seconds and when \(t = 3\) the radius of the circle is increasing at the rate of 0.5 centimetres per second.
One observer believes that the radius increases at a rate which is proportional to \(\frac { 1 } { ( t + 1 ) }\).

1. Write down a differential equation for this situation, using \(k\) as a constant of proportionality.
2. Show that \(k = 2\).
3. Calculate the radius of the circle after 10 seconds according to this model. Another observer believes that the rate of increase of the radius of the circle is proportional to \(\frac { 1 } { ( t + 1 ) ( t + 2 ) }\).
4. Write down a new differential equation for this new situation. Using the same initial conditions as before, find the value of the new constant of proportionality.
5. Hence solve the differential equation.
6. Calculate the radius of the circle after 10 seconds according to this model.

7 When a stone is dropped into still water, ripples move outwards forming a circle of rippled water. At time \(t\) seconds after the stone hits the water the radius of the circle of ripples is increasing at a rate that is inversely proportional to the radius When the radius is 200 cm the rate of increase of the radius is 5 cm per second. Write down the differential equation that represents this situation.

8 At time \(t\) seconds, the radius of a spherical balloon is \(r \mathrm {~cm}\). The balloon is being inflated so that the rate of increase of its radius is inversely proportional to the square root of its radius. When \(t = 5 , r = 9\) and, at this instant, the radius is increasing at \(1.08 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).

1. Write down a differential equation to model this situation, and solve it to express \(r\) in terms of \(t\).
2. How much air is in the balloon initially?
   [0pt] [The volume of a sphere is \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]

8 In the year 2000 the population density, \(P\), of a village was 100 people per \(\mathrm { km } ^ { 2 }\), and was increasing at the rate of 1 person per \(\mathrm { km } ^ { 2 }\) per year. The rate of increase of the population density is thought to be inversely proportional to the size of the population density. The time in years after the year 2000 is denoted by \(t\).

1. Write down a differential equation to model this situation, and solve it to express \(P\) in terms of \(t\).
2. In 2008 the population density of the village was 108 people per \(\mathrm { km } ^ { 2 }\) and in 2013 it was 128 people per \(\mathrm { km } ^ { 2 }\). Determine how well the model fits these figures.

1. A circular stain is growing.

The rate of increase of its radius is inversely proportional to the square of the radius. At time \(t\) seconds the circular stain has radius \(r \mathrm {~cm}\) and area \(A \mathrm {~cm} ^ { 2 }\).
a. Show that \(\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { k } { \sqrt { A } }\). Given that

• the initial area of the circular stain is \(0.09 \mathrm {~cm} ^ { 2 }\).
• after 10 seconds the area of the circular stain is \(0.36 \mathrm {~cm} ^ { 2 }\).
  b. Solve the differential equation to find a complete equation linking \(A\) and \(t\).

7 The height of the tide in a certain harbour is \(h\) metres at time \(t\) hours. Successive high tides occur every 12 hours. The rate of change of the height of the tide can be modelled by a function of the form \(a \cos ( k t )\), where \(a\) and \(k\) are constants. The largest value of this rate of change is 1.3 metres per hour. Write down a differential equation in the variables \(h\) and \(t\). State the values of the constants \(a\) and \(k\).

\includegraphics{figure_9} A container in the shape of a cuboid has a square base of side \(x\) and a height of \((10 - x)\). It is given that \(x\) varies with time, \(t\), where \(t > 0\). The container decreases in volume at a rate which is inversely proportional to \(t\). When \(t = \frac{1}{10}\), \(x = \frac{1}{2}\) and the rate of decrease of \(x\) is \(\frac{20}{37}\).

1. Show that \(x\) and \(t\) satisfy the differential equation $$\frac{dx}{dt} = \frac{-1}{2t(20x - 3x^2)}$$ [5]
2. Solve the differential equation, obtaining an expression for \(t\) in terms of \(x\). [6]