A particle whose horizontal and upward vertical co-ordinates are \(x\) and \(y\), respectively, moves under gravity in a resisting medium in which the retardation always acts in a direction opposite to the velocity. Show that at time \(t\) $$\frac{d^2y}{dx^2} = -g \left( \frac{dx}{dt} \right)^2,$$ where \(g\) is the acceleration due to gravity. Show also that \(\psi\), the inclination to the horizontal of the tangent to the path when the particle is at a height \(y_0\), is given by $$\tan \psi = \int_{y'}^{y_0} 2g \left( \frac{dx}{dt} \right)^2 dy,$$ where \(y'\) is the maximum height attained. Deduce that the angle at which the particle strikes the ground exceeds the angle of projection whatever the form of the retardation as a function of velocity.
A particle moves in a circle of radius \(a\) about a centre of force which exerts an attraction of magnitude \(\mu r^n\), \(r\) being the distance from the centre. By considering first-order equations for the time variation of the quantity \(\epsilon\), where \(r = a(1 + \epsilon)\) and \(\epsilon\) is considered small, discuss the stability of this motion when it is disturbed
A bead moves on a rough wire which is in the shape of the cycloid whose intrinsic equation is $$s = 4a \sin \psi.$$ The wire is in a vertical plane and its cusps point upwards, \(s\) is measured from the lowest point, and \(\psi\) is the angle between tangent and horizontal. Show that if the particle is released from rest at one of the cusps it just comes to rest again at the bottom of the wire if the coefficient of friction \(\mu\) satisfies the equation $$\mu^2 e^{4\pi} = 1.$$
\(Ox\) and \(Oy\) are two perpendicular horizontal axes through the centre \(O\) of a uniform sphere of radius \(a\). Together with the upward vertical \(Oz\) they form a right-handed set. The sphere is given a spin \(\omega_0\) about \(Ox\) and a velocity \(v_0\) in the direction \(Oy\) and placed on a rough horizontal table. Show that the sphere's motion is such that under certain conditions it may return subsequently to its initial position. Show in particular that it will just begin to roll when it has returned to its initial position if $$a\omega_0 = 6v_0.$$
(i) Evaluate the determinant $$\begin{vmatrix} a^3 & 3a^2 & 3a & 1 \\ a^2 & a^2+2a & 2a+1 & 1 \\ a & 2a+1 & a+2 & 1 \\ 1 & 3 & 3 & 1 \end{vmatrix}.$$ (ii) Show that if $$\begin{vmatrix} a & a^3 & a^4-1 \\ b & b^3 & b^4-1 \\ c & c^3 & c^4-1 \end{vmatrix} = 0,$$ and \(a\), \(b\), \(c\) are all different, then $$bc + ca + ab = \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab}.$$
If in the equation $$x^{3-\lambda} = a^3$$ the number \(\lambda\) is very small, show that an approximate root is given by $$x = a(1 + \frac{1}{3}\lambda\log a).$$ Continue the approximation process and find the root correct to order \(\lambda^2\).
A rectangular American city consists of \(p\) streets running east--west and \(q\) avenues running north--south. Find the number of different routes by which a man could travel from the south-west corner to the north-east corner of the city, it being supposed that he always proceeds either north or east.
Show that $$1^2 - 2^2 + 3^2 - \ldots + (-)^{n-1}n^2 = (-)^{n-1}(n^2 + n)/2.$$ Find also the sum of $$1^3 - 2^3 + 3^3 - \ldots + (-)^{n-1}n^3,$$ distinguishing between the cases when \(n\) is odd and even.
Explain Newton's method for approximation to the real roots of an equation, namely, that in \emph{certain circumstances} if \(a\) is a first approximation to a root of the equation \(f(x) = 0\), then $$a - \frac{f(a)}{f'(a)}$$ is a better one. Apply this to the equation \(\sin x = \lambda x\), where \(\lambda\) is a small positive quantity, and show that \(\pi[1 - \lambda + \lambda^2 - \lambda^3(1 + \frac{1}{6}\pi^2)]\) is a better approximation to the root near \(\pi\) than \(\pi\) itself.
Two functions \(P(x)\) and \(Q(x)\) have the following properties: $$P(0) = 1, \quad P'(x) = Q(x),$$ $$Q(0) = 0, \quad Q'(x) = P(x).$$ Deduce the following properties: \begin{align} P(x)^2 - Q^2(x) &= 1, \quad P(x)P(x+a) - Q(x)Q(x+a) = P(a), \\ P(x)Q(x+a) - Q(x)P(x+a) &= Q(a), \\ (P(x) + Q(x))(P(y) + Q(y)) &= P(x+y) + Q(x+y), \\ P(-x) &= P(x), \quad Q(-x) = -Q(x). \end{align}