A paraboloidal bucket is formed by rotating the curve \(ay = x^2\) (\(0 \leq y \leq a\)) about the \(y\)-axis which is vertical. Water runs out of the bucket, initially full, through a small hole at \(y = 0\). The volume of water issuing per unit time is proportional to \(h^\alpha\), where \(h\) is the depth of the water remaining in the bucket at time \(t\), and \(\alpha\) is a constant (\(0 < \alpha < 2\)). At time \(t_1\) the bucket is half-empty (in terms of volume); it becomes totally empty at time \(t_2\). Find \(t_1/t_2\), showing that it depends on \(\alpha\) only.
By repeated integration by parts, or otherwise, show that \[\frac{1}{n!} \int_0^1 (1-t)^n e^t dt = e - \sum_{r=0}^n \frac{1}{r!}.\] Prove that \(0 \leq \int_0^1 (1-t)^n e^t dt \leq 1\) (\(n \geq 1\)). Deduce that \[\left|e - \sum_{r=0}^n \frac{1}{r!}\right| \leq \frac{1}{n!}.\] [By convention, \(0! = 1\).]
Sketch the `\(2m\)-rose' defined in polar coordinates by \(r = |\sin m\theta|\), for \(m = 1, 2, 3\). Show that for all integers \(m > 0\) the total area of the petals is independent of \(m\), and evaluate this area.
A measuring device has an indicator whose position satisfies the equation \[\frac{d^2x}{dt^2} + x = -2k\frac{dx}{dt}.\] Initially, \[x(0) = 1, \left.\frac{dx}{dt}\right|_{t=0} = -k.\] Find the solution \(x(t)\) when \(k > 0\), \(k \neq 1\). Sketch the graph of \(x(t)\) in the two cases \(k = 1\), \(k = 2\). Show that, for those values of \(k\) between 0 and 1 for which \(|x(2)| < 10^{-3}\), we have \(|x(3)| \geq 10^{-3}\).
A curve in the Cartesian plane goes through the origin, touching the \(x\)-axis there; at any point the product of its radius of curvature \(R\) and its arc-length \(s\) (measured from \(O\)) is a constant, \(a^2\). Obtain the intrinsic equation of the curve and deduce that it may be parametrized thus: \[\begin{cases} dx = a(2\psi)^{-\frac{1}{2}}\cos\psi d\psi,\\ dy = a(2\psi)^{-\frac{1}{2}}\sin\psi d\psi. \end{cases}\] Draw a rough sketch of the curve. [You may assume if you wish that \[\int_0^\infty \frac{\cos\psi}{\sqrt{\psi}}d\psi = \int_0^\infty \frac{\sin\psi}{\sqrt{\psi}}d\psi = \sqrt{\frac{\pi}{2}}.\]]
\(ABC\) is an acute angled triangle and \(P\) is the foot of the perpendicular from \(A\) to \(BC\). \(X\) is a variable point on the line \(BC\) and is equally likely to be anywhere between \(B\) and \(C\). Let \(u\), \(v\) and \(h\) denote the lengths of \(BP\), \(CP\) and \(AP\) respectively. Find expressions for the expected value and the variance of the area of the triangle \(APX\) in terms of the parameters \(u\), \(v\) and \(h\).
A box contains \(b\) black and \(r\) red balls. Balls are drawn from it at random one at a time. After each draw the drawn ball is replaced and \(c\) balls of its colour are added to the box. Prove by induction or otherwise that the probability \(p(n)\) that a black ball is drawn on the \(n\)th occasion is \(b/(r+b)\). What is the expected number of black balls in the box immediately before the \((n+1)\)th draw?
Two independent random variables \(X\) and \(Y\) are each uniformly distributed between 0 and 2. Find the probability that \(X^m Y^n \leq 1\) in the cases (i) \(m = n = 1\), (ii) \(m = 2\), \(n = -1\).
In a sample of 50 male undergraduates at Cambridge in 1900 the mean height was found to be 68.93 in. In a sample of 25 male undergraduates at Cambridge in 1974 the mean height was 70.66 in. It may be assumed that the heights of male undergraduates are always normally distributed with a standard deviation of 2.5 in. Is it reasonable to suppose that there has been an increase in the average height of male undergraduates at Cambridge over the past 74 years? Explain carefully the reasoning you use.
A simple pendulum has length \(l\) and is deflected through an angle \(\theta(t)\) from the vertical. Without making any approximations, write down the equation of motion and deduce the equation of energy if \(\alpha\) is the greatest value of \(\theta\) reached. Show that the period is given by \[2\left(\frac{l}{g}\right)^\frac{1}{2} \int_0^\alpha (\sin^2 \frac{1}{2}\alpha - \sin^2 \frac{1}{2}\theta)^{-\frac{1}{2}} d\theta.\] By making the substitution \(\sin \frac{1}{2}\theta = \sin \frac{1}{2}\alpha \sin \psi\) and expanding the integrand appropriately, show that, for small values of \(\alpha\), the period is approximately \[2\pi\left(\frac{l}{g}\right)^\frac{1}{2} \left(1 + \frac{1}{16}\alpha^2\right).\]