Suppose \(a > b > 0\). Show that the circle of curvature of the ellipse \begin{align*} x^2/a^2 + y^2/b^2 = 1 \end{align*} at the point \((0, -b)\) is \begin{align*} b^2x^2 + (by + b^2 - a^2)^2 = a^4. \end{align*} The tangent to the ellipse at \((a \cos \theta, b \sin \theta)\) meets the circle at \(P\) and \(Q\), and the lines from \((0, -b)\) to \(P\) and \(Q\) meet the \(x\)-axis at \(X_1\) and \(X_2\). Show that the distance from \(X_1\) to \(X_2\) is equal to the distance between the foci of the ellipse.
A sequence of numbers \(x_0, x_1, \ldots\) is defined by \begin{align*} x_0 &= 0,\\ x_{n+1} &= x_n + \frac{1}{2k}(x^{2k} - x_n^{2k}), \end{align*} where \(-1 \leq x \leq 1\) and \(k\) is a positive integer. Show that \begin{align*} x_0 \leq x_1 \leq \ldots \leq x_n \leq x_{n+1} \leq \ldots \leq |x| \end{align*} and find the limit of the sequence \(x_0, x_1, x_2, \ldots\).
Show that \(e^{x}/x \to \infty\) as \(x \to \infty\). Sketch the graph of the function \begin{align*} f(x) = x \log_e(x) \quad (x > 0). \end{align*} Solve the equation \begin{align*} \int_0^x f(t) dt = 0. \end{align*}
A computer prints out a list of \(M\) integers. Each integer has been chosen independently and at random from the range 1 to \(N\), with equal probability assigned to each of the \(N\) possible values. What is the probability
A machine produces boiled sweets in 100 kg batches, at the rate of 10 tonnes per day. When the machine is working properly, the chance of a given batch being bad is known to be \(\frac{1}{10}\), different batches being independent: however, the machine is prone to develop a fault which increases the chance of bad batches. At the end of a day in which 2 tonnes of bad sweets were produced, the quality control officer reasoned as follows: 'The machine has produced 8 tonnes of good sweets and 2 tonnes of bad, as against the expected tonnages of 9 and 1 respectively. Hence \begin{align*} \chi^2 = (9-8)^2/9 + (2-1)^2/1 = 10/9, \end{align*} which is not significant against \(\chi_1^2\).' Is his argument satisfactory, and is there evidence for the machine being faulty? Does it matter that the same value of \(\chi^2\) is obtained on a day when all the sweets produced are good?
Solution: His argument is erroneous. He essentially is losing information about the size of the sample, we should be using: \begin{align*} && \chi^2 &= \frac{(90-80)^2}{90} + \frac{(20-10)^2}{10} \\ &&&= \frac{100}{9} = 11.11 \end{align*} which is significant at \(p=0.001\), so there is a very strong evidence the machine is faulty. When all the sweets are good the issue is Chi-squared is a two-sided test, so we are still learning that the machine is acting well out of it's distribution which is not good.
A yacht sails North with speed \(V\) into a wind of speed \(W\) coming from \(\theta^\circ\) East of North. Relative to the yacht, what is the speed and direction of the wind? By correctly setting the sail, it is possible to obtain a wind force on the yacht proportional to the square of the relative speed of the wind (with a constant of proportionality \(C_L\)) in a direction perpendicular to the relative direction of the wind. The sideways component of this wind force is absorbed by a fin under the yacht and causes negligible sideways drift of the yacht. When the yacht is travelling at a constant velocity the forwards component of the wind force balances the drag of the water on the hull, which is proportional to the square of the speed of the yacht with a constant of proportionality \(C_D\). Show that in the steady motion \begin{align*} C_D V^2 = C_L(W^2 + 2WV\cos\theta + V^2)W\sin\theta. \end{align*} By sketching a graph of the right-hand side of the above equation as a function of \(V\), or otherwise, show that there is just one solution for \(V\). Find its approximate value when \(C_D \gg C_L\) and when \(C_D \ll C_L\).
A vibrating carbon dioxide molecule can be thought of as three particles constrained to move along a line, the outer two particles each of mass 16 units being joined to the central particle of mass 12 units by identical springs. If the displacements of the three particles from their equilibrium positions are \(x_1\), \(x_2\) and \(x_3\) (\(x_2\) referring to the central particle), write down the equation of motion for each particle. Show that these equations can be satisfied by two modes of vibration \begin{align*} \text{(I)} \quad &x_1 = \cos \omega t, \quad x_2 = 0, \quad x_3 = -\cos \omega t\\ \text{and} \quad \text{(II)} \quad &x_1 = \cos \Omega t, \quad x_2 = -A \cos \Omega t, \quad x_3 = \cos \Omega t \end{align*} with suitable choices of \(\omega\), \(\Omega\) and \(A\). Show that the ratio of the frequencies of the two modes is \(\sqrt{\frac{4}{3}}\).
Two perfectly elastic balls collide without loss of energy. Show that the relative speed of the balls is the same before and after the collision. A child has a collection of perfectly elastic balls of various sizes. He arranges three of them separated by tiny gaps in a vertical line, the lightest being at the top, at a height much greater than any diameter and drops them simultaneously on to a rigid floor. Show that if the ratio of the masses is 1 : 2 : 6 the lightest ball rises to a height of \(9h\), the other balls being at rest. Show further that with a different choice of sizes the upper ball can move to a height of almost \(49h\).
When a soap film is punctured, a circular hole grows rapidly under the action of surface tension. It is observed that the mass of the film from the hole is concentrated on the rim of the hole and is spread evenly around the rim. Let the soap film, before being punctured, have a thickness \(h\) and a density \(\rho\), and let the radius of the hole at time \(t\) be \(r(t)\). How much mass is there in that segment of the rim which subtends a small angle \(\delta\theta\) at the centre of the hole? Write down Newton's equation of motion for this small segment given that the surface tension gives rise to a net outwards force on the segment of \(2T \cdot r\delta\theta\). Thence show that \begin{align*} r^4\ddot{r} = \frac{2T}{\rho h}r^4 + \text{constant}, \end{align*} and conclude that when the hole is large it grows like \begin{align*} r = t\left(\frac{2T}{\rho h}\right)^{\frac{1}{2}} + \text{constant}. \end{align*}