1.02y Partial fractions: decompose rational functions

Express \(\frac{3x}{(2-x)(4+x^2)}\) in partial fractions. [5]

\includegraphics{figure_2} Figure 2 shows a hemispherical bowl of radius 5 cm. The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time \(t\) minutes, the depth of water is \(h\) cm and the volume of water in the bowl is \(V\) cm³, where $$V = \frac{1}{3}\pi h^2(15 - h).$$ In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to \(V\).

1. Show that $$\frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)},$$ where \(k\) is a positive constant. [5]
2. Express \(\frac{3(10-h)}{h(15-h)}\) in partial fractions. [3]

Given that when \(t = 0\), \(h = 5\),

1. show that $$h^2(15-h) = 250e^{-kt}.$$ [6]

Given also that when \(t = 2\), \(h = 4\),

1. find the value of \(k\) to 3 significant figures. [3]

$$\text{f}(x) = \frac{x(3x-7)}{(1-x)(1-3x)}, \quad |x| < \frac{1}{3}.$$

1. Find the values of the constants \(A\), \(B\) and \(C\) such that $$\text{f}(x) = A + \frac{B}{1-x} + \frac{C}{1-3x}.$$ [4]
2. Evaluate $$\int_0^{\frac{1}{4}} \text{f}(x) \, dx,$$ giving your answer in the form \(p + \ln q\), where \(p\) and \(q\) are rational. [5]
3. Find the series expansion of f(x) in ascending powers of \(x\) up to and including the term in \(x^3\), simplifying each coefficient. [5]

$$f(x) = \frac{5 - 8x}{(1 + 2x)(1 - x)^2}.$$

1. Express \(f(x)\) in partial fractions. [5]
2. Find the series expansion of \(f(x)\) in ascending powers of \(x\) up to and including the term in \(x^2\), simplifying each coefficient. [6]
3. State the set of values of \(x\) for which your expansion is valid. [1]

$$f(x) = \frac{15-17x}{(2+x)(1-3x)^2}, \quad x \neq -2, \quad x \neq \frac{1}{3}.$$

1. Find the values of the constants \(A\), \(B\) and \(C\) such that $$f(x) = \frac{A}{2+x} + \frac{B}{1-3x} + \frac{C}{(1-3x)^2}.$$ [5]
2. Find the value of $$\int_{-1}^{0} f(x) \, dx,$$ giving your answer in the form \(p + \ln q\), where \(p\) and \(q\) are integers. [5]

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]

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]

Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0\), \(x = 0.5\), and as \(t\) increases, \(x\) approaches 1. \includegraphics{figure_7} One model for this situation is given by the differential equation $$\frac{dx}{dt} = x(1-x).$$

1. Verify that \(x = \frac{1}{1+e^{-t}}\) satisfies this differential equation, including the initial condition. [6]
2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [3]

An alternative model for this situation is given by the differential equation $$\frac{dx}{dt} = x^2(1-x),$$ with \(x = 0.5\) when \(t = 0\) as before.

1. Find constants \(A\), \(B\) and \(C\) such that \(\frac{1}{x^2(1-x)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{1-x}\). [4]
2. Hence show that \(t = 2 + \ln\left(\frac{x}{1-x}\right) - \frac{1}{x}\). [5]
3. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [2]

Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced. The population \(x\), in thousands, of red squirrels is modelled by the equation $$x = \frac{a}{1 + kt},$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0\), \(x = 2.5\).

1. Show that \(\frac{dx}{dt} = -\frac{kx^2}{a}\). [3]
2. Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate \(a\) and \(k\). [3]
3. What is the long-term population of red squirrels predicted by this model? [1]

The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac{dy}{dt} = 2y - y^2.$$ When \(t = 0\), \(y = 1\).

1. Express \(\frac{1}{2y - y^2}\) in partial fractions. [4]
2. Hence show by integration that \(\ln\left(\frac{y}{2-y}\right) = 2t\). Show that \(y = \frac{2}{1 + e^{-2t}}\). [7]
3. What is the long-term population of grey squirrels predicted by this model? [1]

Using partial fractions, find \(\int \frac{x}{(x+1)(2x+1)} dx\). [7]

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

1. Find \(\int \frac{1}{1 + \frac{t}{2}} dt\) [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1 + \frac{t}{2})} = \frac{A}{t} + \frac{Bt + C}{1 + \frac{t}{2}}$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M \sim \frac{K}{.1 + \frac{t}{2}}$$ 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 2m. Which model fits this information better? [3]

A curve has equation $$y = \frac{4x - 3a}{2(x^2 + a^2)},$$ where \(a\) is a positive constant.

1. Explain why the curve has no asymptotes parallel to the \(y\)-axis. [2]
2. Find, in terms of \(a\), the set of values of \(y\) for which there are no points on the curve. [5]
3. Find the exact value of \(\int_a^{2a} \frac{4x - 3a}{2(x^2 + a^2)} dx\), showing that it is independent of \(a\). [5]

Express \(\frac{2x^3 + x + 12}{(2x - 1)(x^2 + 4)}\) in partial fractions. [7]

Simplify fully \(\frac{2x^3 + x^2 - 7x - 6}{x^2 - x - 2}\). [4]

In this question you must show detailed reasoning.

1. Prove that \((\cot \theta + \cosec \theta)^2 = \frac{1 + \cos \theta}{1 - \cos \theta}\). [4]
2. Hence solve, for \(0 < \theta < 2\pi\), \(3(\cot \theta + \cosec \theta)^2 = 2 \sec \theta\). [5]

Find the value of \(\int_1^2 \frac{6x + 1}{6x^2 - 7x + 2} dx\), expressing your answer in the form \(m\ln 2 + n\ln 3\), where \(m\) and \(n\) are integers. [8 marks]

Which of the straight lines given below is an asymptote to the curve $$y = \frac{ax^2}{x-1}$$ where \(a\) is a non-zero constant? Circle your answer. [1 mark] \(y = ax + a\) \quad\quad \(y = ax\) \quad\quad \(y = ax - a\) \quad\quad \(y = a\)