Spherical geometry differential equations

10 In a model of the expansion of a sphere of radius \(r \mathrm {~cm}\), it is assumed that, at time \(t\) seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \(t = 0\), \(r = 5\) and \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 2\).

1. Show that \(r\) satisfies the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$ [The surface area \(A\) and volume \(V\) of a sphere of radius \(r\) are given by the formulae \(A = 4 \pi r ^ { 2 }\), \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
2. Solve this differential equation, obtaining an expression for \(r\) in terms of \(t\).
3. Deduce from your answer to part (ii) the set of values that \(t\) can take, according to this model.

1. A balloon is being inflated.

In a simple model,

• the balloon is modelled as a sphere
• the rate of increase of the radius of the balloon is inversely proportional to the square root of the radius of the balloon

At time \(t\) seconds, the radius of the balloon is \(r \mathrm {~cm}\).

1. Write down a differential equation to model this situation. At the instant when \(t = 10\)
   • the radius is 16 cm
   • the radius is increasing at a rate of \(0.9 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\)
   • Solve the differential equation to show that
   $$r ^ { \frac { 3 } { 2 } } = 5.4 t + 10$$
2. Hence find the radius of the balloon when \(t = 20\) Give your answer to the nearest millimetre.
3. Suggest a limitation of the model.

1. A large spherical balloon is deflating.

At time \(t\) seconds the balloon has radius \(r \mathrm {~cm}\) and volume \(V \mathrm {~cm} ^ { 3 }\) The volume of the balloon is modelled as decreasing at a constant rate.

1. Using this model, show that $$\frac { \mathrm { d } r } { \mathrm {~d} t } = - \frac { k } { r ^ { 2 } }$$ where \(k\) is a positive constant. Given that
   • the initial radius of the balloon is 40 cm
   • after 5 seconds the radius of the balloon is 20 cm
   • the volume of the balloon continues to decrease at a constant rate until the balloon is empty
   • solve the differential equation to find a complete equation linking \(r\) and \(t\).
   • Find the limitation on the values of \(t\) for which the equation in part (b) is valid.

1. A spherical mint of radius 5 mm is placed in the mouth and sucked. Four minutes later, the radius of the mint is 3 mm .

In a simple model, the rate of decrease of the radius of the mint is inversely proportional to the square of the radius. Using this model and all the information given,

1. find an equation linking the radius of the mint and the time.
   (You should define the variables that you use.)
2. Hence find the total time taken for the mint to completely dissolve. Give your answer in minutes and seconds to the nearest second.
3. Suggest a limitation of the model.

11 A balloon is being inflated. The balloon is modelled as a sphere with radius \(x \mathrm {~cm}\) at time \(t \mathrm {~s}\). The volume \(V \mathrm {~cm} ^ { 3 }\) is given by \(\mathrm { V } = \frac { 4 } { 3 } \pi \mathrm { x } ^ { 3 }\). The rate of increase of volume is inversely proportional to the radius of the balloon. Initially, when \(t = 0\), the radius of the balloon is 5 cm and the volume of the balloon is increasing at a rate of \(21 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).

1. Show that \(x\) satisfies the differential equation \(\frac { \mathrm { dx } } { \mathrm { dt } } = \frac { 105 } { 4 \pi \mathrm { x } ^ { 3 } }\).
2. Find the radius of the balloon after two minutes.
3. Explain why the model may not be suitable for very large values of \(t\).

7 A giant snowball is melting. The snowball can be modelled as a sphere whose surface area is decreasing at a constant rate with respect to time. The surface area of the sphere is \(A \mathrm {~cm} ^ { 2 }\) at time \(t\) days after it begins to melt.

1. Write down a differential equation in terms of the variables \(A\) and \(t\) and a constant \(k\), where \(k > 0\), to model the melting snowball.
2. 1. Initially, the radius of the snowball is 60 cm , and 9 days later, the radius has halved. Show that \(A = 1200 \pi ( 12 - t )\).
      (You may assume that the surface area of a sphere is given by \(A = 4 \pi r ^ { 2 }\), where \(r\) is the radius.)
   2. Use this model to find the number of days that it takes the snowball to melt completely.

A spherical ball of ice of radius 12 cm is placed in a bucket of water. In a model of the situation, • the ball remains spherical as it melts • \(t\) minutes after the ball of ice is placed in the bucket, its radius is \(r\) cm • the rate of decrease of the radius of the ball of ice is inversely proportional to the square of the radius • the radius of the ball of ice is 6 cm after 15 minutes Using the model and the information given,

1. find an equation linking \(r\) and \(t\), [5]
2. find the time taken for the ball of ice to melt completely, [2]
3. On Diagram 1 on page 27, sketch a graph of \(r\) against \(t\). [1]