219 problems found
Three particles \(P_1\), \(P_2\) and \(P_3\) of masses \(m_{1}\), \(m_{2}\) and \(m_{3}\) respectively lie at rest in a straight line on a smooth horizontal table. \(P_1\) is projected with speed \(v\) towards \(P_2\) and brought to rest by the collision. After \(P_2\) collides with \(P_3\), the latter moves forward with speed \(v\). The coefficients of restitution in the first and second collisions are \(e\) and \(e'\), respectively. Show that \[ e'= \frac{m_{2}+m_{3}-m_{1}}{m_{1}}. \] Show that \(2m_1\ge m_2 +m_3\ge m_1\) for such collisions to be possible. If \(m_1\), \(m_3\) and \(v\) are fixed, find, in terms of \(m_1\), \(m_3\) and \(v\), the largest and smallest possible values for the final energy of the system.
The lengths of the sides \(BC\), \(CA\), \(AB\) of the triangle \(ABC\) are denoted by \(a\), \(b\), \(c\), respectively. Given that $$ b = 8+{\epsilon}_1, \, c=3+{\epsilon}_2,\, A=\tfrac{1}{3}\pi + {\epsilon}_3, $$ where \({\epsilon}_1\), \({\epsilon}_2\), and \( {\epsilon}_3\) are small, show that \(a \approx 7 + {\eta}\), where ${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2 + 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$. Given now that $$ {\vert {\epsilon}_1} \vert \le 2 \times 10^{-3}, \ \ \ {\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \ {\vert {\epsilon}_3} \vert \le \sqrt3 \times 10^{-3}, $$ find the range of possible values of \({\eta}\).
Solution: The cosine rule states that: \(a^2 = b^2 + c^2 - 2bc \cos (A)\) Therefore \begin{align*} a^2 &= (8 + \epsilon_1)^2 + (3 + \epsilon_2)^2 - 2(8 + \epsilon_1) (3 + \epsilon_2)\cos \l \frac{\pi}{3} + \epsilon_3 \r \\ &\approx 64 + 16\epsilon_1 + 9 + 6\epsilon_2- 2(24 + 3\epsilon_1+8\epsilon_2) \cos \l \frac{\pi}{3} + \epsilon_3 \r \\ &= 73 + 16\epsilon_1+ 6\epsilon_2 - 2(24 + 3\epsilon_1+8\epsilon_2) \l \cos \l \frac{\pi}{3} \r \cos \epsilon_3 - \sin \l \frac{\pi}{3} \r \sin \epsilon_3 \r \\ &\approx 73 + 16\epsilon_1+ 6\epsilon_2 - (24 + 3 \epsilon_1+8\epsilon_2) + 24\sqrt{3}\epsilon_3 \\ &= 49 + 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3 \\ &= 7^2 + 2 \cdot 7 \cdot \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \\ &\approx \l 7 + \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \r^2 \end{align*} In this approximation, we are ignoring all terms of order \(2\), and using the approximations \(\cos \varepsilon \approx 1, \sin \varepsilon \approx \varepsilon\) Therefore \(a \approx 7 + \frac{ 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14}\). \(\eta\) is maximised if \(\epsilon_1, \epsilon_3\) are and \(\epsilon_2\) is minimized, ie: \begin{align*} \eta &\leq \frac{13 \cdot 2 \cdot 10^{-3} - 2 \cdot 4.9 \cdot 10^{-2} + 24 \sqrt{3} \cdot \sqrt{3} \cdot 10^{-3}}{14} \\ &= 10^{-3} \cdot \frac{26 - 98 + 74}{14} \\ &= 10^{-3} \cdot \frac{1}{7}\end{align*} Similarly, it is maximised when signs are reversed, ie: \(| \eta | \leq 10^{-3} \cdot \frac{1}{7}\)
It is required to approximate a given function \(\f(x)\), over the interval \(0 \le x \le 1\), by the linear function \(\lambda x\), where \(\lambda\) is chosen to minimise \[ \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x . \] Show that \[ \lambda = 3 \int_0^1 x\f(x)\,\d x. \] The residual error, \(R\), of this approximation process is such that \[ R^2 = \int_0^1 \big(\f(x)-\lambda x \big)^{\!2}\,\d x. \] Show that \[ R^2 = \int_0^1 \big(\f(x)\big)^{\!2}\,\d x -\tfrac{1}{3} \lambda ^2. \] Given now that \(\f(x)= \sin (\pi x/n)\), show that (i) for large \(n\), \(\lambda \approx \pi/n\) and (ii) \(\lim_{n \to \infty}R = 0.\) Explain why, prior to any calculation, these results are to be expected. [You may assume that, when \(\theta\) is small, $\sin \theta \approx \theta-\frac{1}{6}\theta^3$ and \(\cos \theta \approx 1 - \frac{1}{2}\theta^2.\)]
Solution: \begin{align*} && g(\lambda) &= \int_0^1 \big(\f(x)-\lambda x \big)^{\!2} \,\d x \\ &&&= \int_0^1 \left ( f(x)^2 -2\lambda xf(x) + \lambda^2 x^2\right) \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ \end{align*} Differentiating (or completing the square) it is clear the minimum occurs when \(\displaystyle \lambda = 3 \int_0^1 xf(x) \d x\) \begin{align*} && R^2 &= \int_0^1 (f(x) - \lambda x )^2 \d x \\ &&&= \frac13\lambda^2 - 2\lambda \int_0^1 x f(x) \d x + \int_0^1 f(x)^2 \d x \\ &&&= \frac13 \left (\lambda -3\int_0^1 xf(x) \d x \right)^2 -\frac13 \left ( 3\int_0^1 xf(x) \d x \right)^2+\int_0^1 f(x)^2 \d x \\ \end{align*} When \(\lambda = 3\int_0^1 xf(x) \d x \) clearly this is the desired result. \begin{align*} && \lambda &= 3\int_0^1 xf(x) \d x \\ &&&= 3\int_0^1 x \sin(\pi x /n) \d x \\ &&&= 3 \left [-x \frac{n}{\pi} \cos (\pi x /n) \right]_0^1 + \frac{3n}{\pi} \int_0^1 \cos(\pi x /n) \d x \\ &&&= -\frac{3n}{\pi}\cos(\pi/n) + \frac{3n}{\pi} \left [ \frac{n}{\pi} \sin(\pi x /n)\right]_0^1 \\ &&&= -\frac{3n}{\pi} \cos(\pi/n) + \frac{3n^2}{\pi^2} \sin(\pi /n) \\ \text{for large }n: &&&\approx -\frac{3n}{\pi}\left ( 1 - \frac12\frac{\pi^2}{n^2} + o(1/n^4)\right) + \frac{3n^2}{\pi^2} \left (\frac{\pi}{n} - \frac16 \frac{\pi^3}{n^3} +o(1/n^5) \right) \\ &&&= \left (\frac32 -\frac12\right)\frac{\pi}{n} + o(1/n^3) \\ &&&= \frac{\pi}{n} + o(1/n^2) \end{align*} Therefore for large \(n\), \(\lambda \approx \frac{\pi}n\) \begin{align*} && \int_0^1 \sin^2(\pi x/n) \d x &= \frac12\int_0^1(1- \cos(2\pi x/n)) \d x\\ &&&= \frac12\left ( 1 - \frac{n}{2\pi}\left[\sin(2\pi x/n) \right]_0^1 \right) \\ &&&= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) \\ \\ && R^2 &= \frac12 -\frac{n}{4\pi}\sin(2\pi /n) - \frac13 \left ( \frac{\pi}{n}+o(1/n^2)\right)^2 \\ &&&= \frac12 - \left ( \frac{1}{2} -\frac16\frac{\pi}{n}+o(1/n^3) \right) - o(1/n^2) \\ &&& = \frac16 \frac{\pi}{n} + o(1/n^2) \\ &&&\to 0 \text{ as } n \to \infty \end{align*} We should expect these results as for \(n\) very large \(\sin(\pi x/n) \approx \frac{\pi }{n}x\) so the best linear approximation is likely to be \(\lambda = \frac{\pi}{n}\) and we should expect it to improve to the point that we cannot tell the difference, ie \(R^2 \to 0\)
A group of biologists attempts to estimate the magnitude, \(N\), of an island population of voles ({\it Microtus agrestis}). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, \(p_N\), of this occurrence is given by $$ p_N = k{{{\big((N-200)!\big)}^2} \over {N!(N-389)!}}, $$ where \(k\) is independent of \(N\). The biologists then estimate \(N\) by calculating the value of \(N\) for which \(p_N\) is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Write down the probability that this sample contains exactly \(j\) marked voles, leaving your answer in terms of binomial coefficients. Deduce that $$ \sum_{j=0}^{200}{389 \choose j}{3247 \choose {200-j}} = {3636 \choose 200}. $$
Solution: There will be \(200\) marked vols out of \(N\), and we are finding \(11\) of them. There are \(\binom{200}{11}\) ways to chose the \(11\) marked voles and \(\binom{N - 200}{200-11}\) ways to choose the unmarked voles. The total number of ways to choose \(200\) voles is \(\binom{N}{200}\). Therefore the probability is \begin{align*} p_N &= \frac{\binom{200}{11} \cdot \binom{N - 200}{200-11}}{\binom{N}{200}} \\ &= \binom{200}{11} \cdot \frac{ \frac{(N-200)!}{(189)!(N - 389)!} }{\frac{N!}{(N-200)!(200)!}} \\ &= \binom{200}{11} \frac{200!}{189!} \frac{\big((N-200)!\big)^2}{N!(N-389)!} \end{align*} As required and \(k = \binom{200}{11} \frac{200!}{189!}\). We want to maximise \(\frac{(N-200)!^2}{N!(N-389)!}\), we will do this by comparing consecutive \(p_N\). \begin{align*} \frac{p_{N+1}}{p_N} &= \frac{\frac{(N+1-200)!^2}{(N+1)!(N+1-389)!}}{\frac{(N-200)!^2}{N!(N-389)!}} \\ &= \frac{(N-199)!^2 \cdot N! \cdot (N-389)!}{(N+1)!(N-388)!(N-200)!^2} \\ &= \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} \\ \end{align*} \begin{align*} && \frac{p_{N+1}}{p_N} &> 1 \\ \Leftrightarrow && \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} & > 1 \\ \Leftrightarrow && (N-199)^2 & > (N+1) \cdot (N-388) \\ \Leftrightarrow && N^2-2\cdot199N+199^2 & > N^2 - 387N -388 \\ \Leftrightarrow && -398N+199^2 & > - 387N -388 \\ \Leftrightarrow && 199^2+388 & > 11N\\ \Leftrightarrow && \frac{199^2+388}{11} & > N\\ \Leftrightarrow && 3635\frac{4}{11} & > N\\ \end{align*} Therefore \(p_N\) is increasing if \(N \leq 3635\), so we should take \(N = 3636\). \[ \P(\text{exactly } j \text{ marked voles}) = \frac{\binom{389}{j} \cdot \binom{3636 - 389}{200-j}}{\binom{3636}{200}}\] Since \begin{align*} && 1 &= \sum_{j=0}^{200} \P(\text{exactly } j \text{ marked voles}) \\ && &= \sum_{j=0}^{200} \frac{\binom{389}{j} \cdot \binom{3247}{200-j}}{\binom{3636}{200}} \\ \Leftrightarrow&& \binom{3636}{200} &= \sum_{j=0}^{200} \binom{389}{j} \cdot \binom{3247}{200-j} \end{align*}
For this question, you may use the following approximations, valid if \(\theta \) is small: \ \(\sin\theta \approx \theta\) and \(\cos\theta \approx 1-\theta^2/2\,\). A satellite \(X\) is directly above the point \(Y\) on the Earth's surface and can just be seen (on the horizon) from another point \(Z\) on the Earth's surface. The radius of the Earth is \(R\) and the height of the satellite above the Earth is \(h\).
Solution:
A tortoise and a hare have a race to the vegetable patch, a distance \(X\) kilometres from the starting post, and back. The tortoise sets off immediately, at a steady \(v\) kilometers per hour. The hare goes to sleep for half an hour and then sets off at a steady speed \(V\) kilometres per hour. The hare overtakes the tortoise half a kilometre from the starting post, and continues on to the vegetable patch, where she has another half an hour's sleep before setting off for the return journey at her previous pace. One and quarter kilometres from the vegetable patch, she passes the tortoise, still plodding gallantly and steadily towards the vegetable patch. Show that \[ V= \frac{10}{4X-9} \] and find \(v\) in terms of \(X\). Find \(X\) if the hare arrives back at the starting post one and a half hours after the start of the race.
Solution: If \(T_1\) is the time they meet after \(\frac12\)km from the starting points and \(T_2\) is the time they meet a second time, then \begin{align*} && \frac12 &= vT_1 \\ &&&= V(T_1-\tfrac12) \\ && X - \frac54 &= vT_2 \\ && X + \frac54 &= V(T_2 - 1) \\ && \frac{T_2}{T_1} &= \frac{4X-5}{2} \\ && X + \frac54 + V &= VT_2 \\ && \frac12 + \frac12 V &= VT_1 \\ \Rightarrow && \frac{T_2}{T_1} &= \frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && \frac{4X-5}{2}&=\frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && V(4X-9) &= 10 \\ \Rightarrow && V &= \frac{10}{4X-9} \\ \\ && T_1 &= \frac{1}{2V} + \frac12 \\ &&&= \frac{4X+1}{20} \\ && v &= \frac{1}{2T_1} \\ &&&= \frac{10}{4X+1} \end{align*} \begin{align*} && 2X &= \frac12 V \\ \Rightarrow && 2X(4X-9) &= 5 \\ \Rightarrow && 0 &= 8X^2-18X-5 \\ &&&= (4X+1)(2X-5) \\ \Rightarrow && X &= -\frac14, \frac52 \end{align*} Since \(X\) is positive, we must have \(X = \frac52\)km
Solution:
In the \(Z\)--universe, a star of mass \(M\) suddenly blows up, and the fragments, with various initial speeds, start to move away from the centre of mass \(G\) which may be regarded as a fixed point. In the subsequent motion the acceleration of each fragment is directed towards \(G\). Moreover, in accordance with the laws of physics of the \(Z\)--universe, there are positive constants \(k_1\), \(k_2\) and \(R\) such that when a fragment is at a distance \(x\) from \(G\), the magnitude of its acceleration is \(k_1x^3\) if \(x < R\) and is \(k_2x^{-4}\) if \(x \ge R\). The initial speed of a fragment is denoted by \(u\).
An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius \(R\) in a horizontal plane at a constant angular speed \(\omega\). A shell is fired from \(O\), the centre of this circle, with initial speed \(V\) and angle of elevation \(\alpha\). Show that if \(V^2 < gR\), then no matter what the value of \(\alpha\), or what vertical plane the shell is fired in, the shell cannot hit the target. Assume now that \(V^2 > gR\) and that the shell hits the target, and let \(\beta\) be the angle through which the target rotates between the time at which the shell is fired and the time of impact. Show that \(\beta\) satisfies the equation $$ g^2{{\beta}^4} - 4{{\omega}^2}{V^2}{{\beta}^2} +4{R^2}{{\omega}^4}=0. $$ Deduce that there are exactly two possible values of \(\beta\). Let \(\beta_1\) and \(\beta_2\) be the possible values of \(\beta\) and let \(P_1\) and \(P_2\) be the corresponding points of impact. By considering the quantities \((\beta_1^2 +\beta_2^2) \) and \(\beta_1^2\beta_2^2\,\), or otherwise, show that the linear distance between \(P_1\) and \(P_2\) is \[ 2R \sin\Big( \frac\omega g \sqrt{V^2-Rg}\Big) \;. \]
Solution: \begin{align*} && 0 &= V\sin \alpha t-\frac12 gt^2 \\ \Rightarrow && t &= \frac{2V \sin \alpha}{g} \\ && R &= V \cos \alpha \, t \\ &&&= \frac{2V^2 \sin \alpha \cos \alpha}{g} \\ &&&= \frac{V^2 \sin 2 \alpha}{g} \end{align*} Therefore the max distance is \(\frac{V^2}{g}\), therefore we cannot hit the target if \(R > \frac{V^2}{g} \Rightarrow gR > V^2\). We have \(\beta = \omega t \Rightarrow t = \frac{\beta}{\omega}\) \begin{align*} && \sin \alpha &= \frac{gt}{2V} \\ && \cos \alpha &= \frac{R}{Vt} \\ \Rightarrow && 1 &= \left (\frac{gt}{2V} \right)^2 + \left ( \frac{R}{Vt} \right)^2 \\ &&&= \left (\frac{g\beta}{2V \omega} \right)^2 + \left ( \frac{R\omega}{V\beta} \right)^2 \\ &&&= \frac{g^2 \beta^2}{4 V^2 \omega^2} + \frac{R^2 \omega^2}{V^2 \beta ^2} \\ \Rightarrow && 4V^2 \omega^2 \beta^2 &= g^2 \beta^4 + 4R^2 \omega^4 \\ \Rightarrow && 0 &= g^2 \beta^4 - 4\omega^2 V^2 \beta^2+4R^2\omega^4 \end{align*} This (quadratic) equation in terms of \(\beta^2\) has two solution if \(\Delta = 16\omega^4V^4-16g^2R^2\omega^4 =16\omega^4(V^4-g^2R^2) > 0\) since \(V^2 > gR\). Since \(\beta > 0\) there are exactly two solutions, once we have values for \(\beta\). First notice, \begin{align*} && \beta_1^2 + \beta_2^2 &= \frac{4\omega^2V^2}{g^2} \\ && \beta_1^2\beta_2^2 &= \frac{4R^2\omega^4}{g^2} \end{align*} Then notice the positions of \(P_1\) and \(P_2\) are \((R\cos \beta_1 , R\sin \beta_1)\) and \((R\cos \beta_2, R\sin \beta_2)\). \begin{align*} && d^2 &= R^2\left ( \cos \beta_1 - \cos \beta_2 \right)^2 + R^2 \left ( \sin \beta_1 - \sin \beta_2 \right)^2 \\ &&&= 2R^2 - 2R^2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2) \\ &&&= 2R^2-2R^2\cos(\beta_1 - \beta_2) \\ &&&= 2R^2 \left (1-\cos(\sqrt{(\beta_1-\beta_2)^2} \right ) \\ &&&= 2R^2 \left (1 - \cos\left ( \sqrt{\frac{4\omega^2 V^2}{g^2} - \frac{4R\omega^2}{g}} \right) \right) \\ &&&= 2R^2 \left (1 - \cos\left (\frac{2\omega}{g} \sqrt{V^2 - Rg} \right) \right) \\ &&&= 4 R^2 \sin^2 \left (\frac{\omega}{g} \sqrt{V^2 - Rg} \right) \end{align*} which gives us the required result.
It is known that there are three manufacturers \(A, B, C,\) who can produce micro chip MB666. The probability that a randomly selected MB666 is produced by \(A\) is \(2p\), and the corresponding probabilities for \(B\) and \(C\) are \(p\) and \(1 - 3p\), respectively, where \({{0} \le p \le {1 \over 3}}.\) It is also known that \(70\%\) of MB666 micro chips from \(A\) are sound and that the corresponding percentages for \(B\) and \(C\) are \(80\%\) and \(90\%\), respectively. Find in terms of \(p\), the conditional probability, \(\P(A {\vert} S)\), that if a randomly selected MB666 chip is found to be sound then it came from \(A\), and also the conditional probability, \(\P(C {\vert} S)\), that if it is sound then it came from \(C\). A quality inspector took a random sample of one MB666 micro chip and found it to be sound. She then traced its place of manufacture to be \(A\), and so estimated \(p\) by calculating the value of \(p\) that corresponds to the greatest value of \(\P(A {\vert} S)\). A second quality inspector also a took random sample of one MB666 chip and found it to be sound. Later he traced its place of manufacture to be \(C\) and so estimated \(p\) by applying the procedure of his colleague to \(\P(C {\vert} S)\). Determine the values of the two estimates and comment briefly on the results obtained.
You play the following game. You throw a six-sided fair die repeatedly. You may choose to stop after any throw, except that you must stop if you throw a 1. Your score is the number obtained on your last throw. Determine the strategy that you should adopt in order to maximize your expected score, explaining your reasoning carefully.
Solution: Once you have thrown, all previous throws are irrelevant so the only thing which can affect your decision is the current throw. Therefore the strategy must consist of a list of states we re-throw from, and a list of states we stick on. It must also be the case that if we stick on \(k\) we stick on \(k+1\) (otherwise we can improve our strategy by switching those two values around). Therefore we can form a table of our expected score: \begin{array}{c|c|c} \text{stop on} & \text{possible outcomes} & \E[\text{score}] \\ \hline \geq 2 & \{1,2,3,4,5,6\} & \frac{21}{6} = 3.5 \\ \geq 3 & \{1,3,4,5,6\} & \frac{19}{5} = 3.8 \\ \geq 4 & \{1,4,5,6\} & \frac{16}{4} = 4 \\ \geq 5 & \{1,5,6\} & \frac{12}{3} = 4 \\ =6 & \{1,6\} & \frac{7}{2} = 3.5 \end{array} Therefore the optimal strategy is to stop on \(4\) or higher. If we cared about variance we might look at the variance of the two best strategies, \(4\) or higher has a variance of \(\frac{1+16+25+36}{4} - 16 = 3.5\) and \(5\) or higher has a variance of \(\frac{1+25+36}3 - 16 = \frac{14}3 > 3.5\) so \(4\) or higher is probably better in most scenarios.
Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square. The result changes if, instead of maximising the sum of lengths of sides of the rectangle, we seek to maximise the sum of \(n\)th powers of the lengths of those sides for \(n\geqslant 2\). What happens if \(n=2\)? What happens if \(n=3\)? Justify your answers.
Solution: We can always rotate the circle so that sides are parallel to the \(x\) and \(y\) axes. Therefore if one corner is \((a,b)\) the other coordinates are \((-a,b), (a,-b), (-a,-b)\) and the perimeter will be \(4(a+b)\). Therefore we wish to maximise \(4(a+b)\) subject to \(a^2+b^2 = \text{some constant}\). Notice that \(\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}\) with equality when \(a = b\), therefore the maximum is a square. If \(n = 2\) then we are looking at \(2((2a)^2+(2b)^2) = 8(a^2+b^2)\) which is constant for all rectangles. If \(n=3\) we are maximising \(16(a^3+b^3) = 16(a^3+(c^2-a^2)^{3/2})\) which is maximised when \(a = 0, c\)
Hank's Gold Mine has a very long vertical shaft of height \(l\). A light chain of length \(l\) passes over a small smooth light fixed pulley at the top of the shaft. To one end of the chain is attached a bucket \(A\) of negligible mass and to the other a bucket \(B\) of mass \(m\). The system is used to raise ore from the mine as follows. When bucket \(A\) is at the top it is filled with mass \(2m\) of water and bucket \(B\) is filled with mass \(\lambda m\) of ore, where \(0<\lambda<1\). The buckets are then released, so that bucket \(A\) descends and bucket \(B\) ascends. When bucket \(B\) reaches the top both buckets are emptied and released, so that bucket \(B\) descends and bucket \(A\) ascends. The time to fill and empty the buckets is negligible. Find the time taken from the moment bucket \(A\) is released at the top until the first time it reaches the top again. This process goes on for a very long time. Show that, if the greatest amount of ore is to be raised in that time, then \(\lambda\) must satisfy the condition \(\mathrm{f}'(\lambda)=0\) where \[\mathrm{f}(\lambda)=\frac{\lambda(1-\lambda)^{1/2}} {(1-\lambda)^{1/2}+(3+\lambda)^{1/2}}.\]
To celebrate the opening of the financial year the finance minister of Genland flings a Slihing, a circular coin of radius \(a\) cm, where \(0 < a < 1\), onto a large board divided into squares by two sets of parallel lines 2 cm apart. If the coin does not cross any line, or if the coin covers an intersection, the tax on yaks remains unchanged. Otherwise the tax is doubled. Show that, in order to raise most tax, the value of \(a\) should be \[\left(1+{\displaystyle \frac{\pi}{4}}\right)^{-1}.\] If, indeed, \(a=\left(1+{\displaystyle \frac{\pi}{4}}\right)^{-1}\) and the tax on yaks is 1 Slihing per yak this year, show that its expected value after \(n\) years will have passed is \[ \left(\frac{8+\pi}{4+\pi}\right)^{n}.\]