The members of a family of curves in the \(x,y\) plane satisfy the differential equation \begin{equation*} y\frac{dy}{dx} - y^2 = x^2 - x. \end{equation*} By multiplying this equation by a suitable function of \(x\) and integrating, or otherwise, obtain the curve which passes through the point \((0, 1)\). Show that this curve also passes through the point \((-a, 0)\) where \(a > 0\) and \(a = -\ln a\).
Solution: \begin{align*} && y \frac{\d y}{\d x} - y^2 &= x^2 - x \\ \Rightarrow && e^{-2x}2y \frac{\d y}{\d x} - 2e^{-2 x}y^2 &= 2e^{-2x}(x^2 - x) \\ \Rightarrow && \frac{\d }{\d x} \left (y^2 e^{-2x} \right)&= 2e^{-2x}(x^2-x) \\ \Rightarrow && y^2e^{-2x} &= \int 2e^{-2x}(x^2-x) \d x \\ &&y^2e^{-2x} &= -e^{-2x}x^2 + C \\ \Rightarrow && y^2 &= -x^2 +Ce^{2x} \\ x = 0, y =1: && 1 &= C \\ \Rightarrow && y^2 &= -x^2 + e^{2x} \\ \Rightarrow && y &= \sqrt{-x^2+e^{2x}} \end{align*} This will be \(0\) when \(x^2 = e^{2x} \Rightarrow |x| = e^{x} \Rightarrow \ln |x| = x\), ie exactly the point described
A player deals cards from a pack of 52 in sets of four. The first set of four consists of cards of different suits. What is the probability that the last set of four consists of cards of different suits? Had the first set of four consisted of cards of the same suit, what would the probability have been that the last set of four were also of one suit?
In San Theodoros execution is by firing squad at dusk. Executions take place at any time between 6 and 7 pm with equal probability, and as darkness falls the aim of the soldiers worsens at a steady rate, so that at 6 pm their aim is perfectly true, at 6.30 they miss their target with probability \(\frac{1}{2}\) and by 7 pm they always miss. General Tapioca, ruler of San Theodoros, a liberal, has ordered on humanitarian grounds that on exactly half the executions the firing squad shall use blank rounds. Tintin, a reporter, is sentenced to die by firing squad but survives. What is the probability that he faced live rounds?
Solution: \begin{align*} && \mathbb{P}(\text{live rounds} | \text{survives}) &= \frac{\mathbb{P}(\text{live rounds and survives})}{\mathbb{P}(\text{survives})} \\ &&&= \frac{\mathbb{P}(\text{choose live rounds})\mathbb{P}(\text{missed}|\text{live rounds})}{\mathbb{P}(\text{choose live rounds})\mathbb{P}(\text{missed}|\text{live rounds})+\mathbb{P}(\text{choose blank rounds})\mathbb{P}(\text{missed}|\text{blank rounds})} \\ &&&= \frac{\frac12 \cdot \frac12 }{\frac12 \cdot \frac12 + \frac12 \cdot 1} \\ &&&= \frac{1}{3} \end{align*} The probability that he is missed is \(\frac12\) since it is equally likely they shoot with \(p\) or \(1-p\) for each \(p\) over the hour.
A shell is fired from a gun with a muzzle velocity \(V\) and an elevation of \(45^{\circ}\) to the horizontal. At the top of its flight the shell splits into two equal fragments which separate with a relative velocity of magnitude \(\sqrt{2}V\) and elevation \(\alpha\) in the plane of the trajectory. Show that the range of one fragment is \begin{equation*} \frac{V^2}{2g}[1 + (1 + \cos\alpha)(\sin\alpha + \sqrt{\sin^2\alpha + 1})], \end{equation*} and find the range of the other.
A particle of mass \(m\) moves along a straight line in a resistive medium. It experiences a retarding force of magnitude \(\lambda v^3 + kv\), where \(v\) is its velocity and \(\lambda\) and \(k\) are positive constants. Given that the initial velocity of the particle is \(w\), find \(v\) as a function of time. Find \(v\) as a function of \(s\), the distance travelled, and show that \(s\) never exceeds \begin{equation*} \frac{m}{\sqrt{\lambda k}}\tan^{-1}\left(w\sqrt{\frac{\lambda}{k}}\right) \end{equation*}
A uniform rod of mass \(m\) and length \(4a\) can rotate freely in a smooth horizontal plane about its midpoint. Initially the rod is at rest. A particle of mass \(m\) travelling in the plane with velocity \(u\) at right angles to the rod collides perfectly elastically with the rod at a distance \(a\) from the centre. Find the velocity of the particle and angular velocity of the rod after collision. Do the particle and the rod undergo a subsequent collision?