OCR MEI C4 (Core Mathematics 4)

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

1. Find \(\int \frac{t}{1+t^2} dt\). [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1+t^2)} = \frac{A}{t} + \frac{Bt+C}{1+t^2}.$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac{Kt}{\sqrt{1+t^2}},$$ where \(K\) is a constant. [6]
4. When \(t = 1\), \(M = 25\). Calculate \(K\). What is the mass of the chemical in the long term? [4]

The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h,$$ 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 \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]

The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).

1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]
2. What does this solution indicate about the long-term height of the tree? [1]
3. After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]

\includegraphics{figure_3} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ 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 cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.

1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]

Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(kx\) cm\(^3\) s\(^{-1}\).

1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
2. Solve this differential equation. Hence show that the bowl empties in 37 seconds. [6]

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

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. [3]
2. Verify that \(v\) satisfies the differential equation \(\frac{dv}{dt} = 10 - 2v\). [3]

In a second model, \(v\) satisfies the differential equation $$\frac{dv}{dt} = 10 - 0.4v^2.$$ As before, when \(t = 0\), \(v = 0\).

1. Show that this differential equation may be written as $$\frac{10}{(5-v)(5+v)} \frac{dv}{dt} = 4.$$ Using partial fractions, solve this differential equation to show that $$t = \frac{1}{4} \ln\left(\frac{5+v}{5-v}\right).$$ [8] This can be re-arranged to give \(v = \frac{5(1-e^{-4t})}{1+e^{-4t}}\). [You are not required to show this result.]
2. Verify that this model also gives a terminal velocity of 5 m s\(^{-1}\). Calculate the velocity after 0.5 seconds as given by this model. [3]

The velocity of the particle after 0.5 seconds is measured as 3 m s\(^{-1}\).

1. Which of the two models fits the data better? [1]