Two identical particles of unit mass move under gravity in a medium for which the magnitude of the retarding force on a particle is \(k\) times its speed. The first particle is allowed to fall from rest at a point \(A\) whilst, at the same time, the second is projected upwards with speed \(u\) from a point \(B\) a positive distance \(d\) vertically above \(A\). Find their distance apart after a time \(t\) and show that this distance tends to the value \[ d+\frac{u}{k} \] as \(t\rightarrow\infty.\)
Solution: Both particles have equations of motion, \(\ddot{x} = -g-k\dot{x}\), so we can note that the distance between them has the equation of motion: \(\ddot{x} = -k \ddot{x} \Rightarrow x = Ae^{-kt} + B\) \begin{align*} && x(0) &= d \\ \Rightarrow && A+B &= d \\ && x'(0) &= u \\ \Rightarrow && -kA &= u \\ \Rightarrow && A &= -\frac{u}{k} \\ \Rightarrow && B &= d+\frac{u}{k} \\ \Rightarrow && x(t) &= -\frac{u}{k}e^{-kt} + d + \frac{u}{k} \to d + \frac{u}{k} \end{align*} as required.
Bread roll throwing duels at the Drones' Club are governed by a strict etiquette. The two duellists throw alternatively until one is hit, when the other is declared the winner. If Percy has probability \(p>0\) of hitting his target and Rodney has probability \(r>0\) of hitting his, show that, if Percy throws first, the probability that he beats Rodney is \[ \frac{p}{p+r-pr}. \] Algernon, Bertie and Cuthbert decide to have a three sided duel in which they throw in order \(\mathrm{A,B,C,A,B,C,}\ldots\) except that anyone who is hit must leave the game. Cuthbert always his target, Bertie hits his target with probability \(3/5\) and Algernon hits his target with probability \(2/5.\) Bertie and Cuthbert will always aim at each other if they are both still in the duel. Otherwise they aim at Algernon. With his first shot Algernon may aim at either Bertie or Cuthbert or deliberately miss both. Faced with only one opponent Algernon will aim at him. What are Algernon's changes of winning if he:
Fly By Night Airlines run jumbo jets which seat \(N\) passengers. From long experience they know that a very small proportion \(\epsilon\) of their passengers fail to turn up. They decide to sell \(N+k\) tickets for each flight. If \(k\) is very small compared with \(N\) explain why they might expect \[ \mathrm{P}(r\mbox{ passengers fail to turn up})=\frac{\lambda^{r}}{r!}\mathrm{e}^{-\lambda} \] approximately, with \(\lambda=N\epsilon.\) For the rest of the question you may assume that the formula holds exactly. Each ticket sold represents \(\pounds A\) profit, but the airline must pay each passenger that it cannot fly \(\pounds B\) where \(B>A>0.\) Explain why, if \(r\) passengers fail to turn up, its profit, in pounds, is \[ A(N+k)-B\max(0,k-r), \] where \(\max(0,k-r)\) is the larger of \(0\) and \(k-r.\) Write down the expected profit \(u_{k}\) when \(k=0,1,2\) and \(3.\) Find \(v_{k}=u_{k+1}-u_{k}\) for general \(k\) and show that \(v_{k}>v_{k+1}.\) Show also that \[ v_{k}\rightarrow A-B \] as \(k\rightarrow\infty.\) Advise Fly By Night on how to choose \(k\) to maximise its expected profit \(u_{k}.\)
Suppose \(X\) is a random variable with probability density \[ \mathrm{f}(x)=Ax^{2}\exp(-x^{2}/2) \] for \(-\infty < x < \infty.\) Find \(A\). You belong to a group of scientists who believe that the outcome of a certain experiment is a random variable with the probability density just given, while other scientists believe that the probability density is the same except with different mean (i.e. the probability density is \(\mathrm{f}(x-\mu)\) with \(\mu\neq0\)). In each of the following two cases decide whether the result given would shake your faith in your hypothesis, and justify your answer.
Solution: Let \(Z \sim N(0,1)\), with a pdf of \(f(x) = \frac{1}{\sqrt{2\pi}} \exp(-x^2/2)\) \begin{align*} && 1 &= \int_{-\infty}^\infty Ax^2 \exp(-x^2/2) \d x \\ &&&= A\sqrt{2\pi} \int_{-\infty}^\infty x^2 \frac{1}{\sqrt{2\pi}} \exp(-x^2/2) \d x \\ &&&= A\sqrt{2\pi} \E[Z^2] = A\sqrt{2\pi} \\ \Rightarrow && A &= \frac{1}{\sqrt{2\pi}} \end{align*}