OCR MEI C4 (Core Mathematics 4)

1 In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) } .$$

1. Find \(\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t\).
2. Find constants \(A , B\) and \(C\) such that $$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } }$$
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$ where \(K\) is a constant.
4. When \(t = 1 , M = 25\). Calculate \(K\). What is the mass of the chemical in the long term?

2 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } }$$
4. What does this solution indicate about the long-term height of the tree?
5. After a year, the tree has grown to a height of 2 m . Which model fits this information better? \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5699ea45-6a4d-4681-a044-4a337e30588c-3_273_461_190_830} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure} Fig. 9 shows a hemispherical bowl, of radius 10 cm , filled with water to a depth of \(x \mathrm {~cm}\). It can be shown that the volume of water, \(V \mathrm {~cm} ^ { 3 }\), is given by $$V = \pi \left( 10 x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 } \right) .$$ Water is poured into a leaking hemispherical bowl of radius 10 cm . Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in \(\mathrm { cm } ^ { 3 } \mathrm {~s} ^ { - 1 }\), given by the equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = k ( 20 - x )$$ where \(k\) is a constant.
6. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\), and hence show that \(\pi x \frac { \mathrm {~d} x } { \mathrm {~d} t } = k\).
7. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where $$T = \frac { 50 \pi } { k }$$ Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(k x \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
8. Show that, \(t\) seconds later, \(\pi ( 20 - x ) \frac { \mathrm { d } x } { \mathrm {~d} t } = - k\).
9. Solve this differential equation. Hence show that the bowl empties in \(3 T\) seconds.

4 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
5. Which of the two models fits the data better?