The points \(A\), \(B\) and \(C\) in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers \(a\), \(b\) and \(c\) representing \(A\), \(B\) and \(C\) satisfy \[2c= (a+b) +\mathrm{i}\sqrt3(b-a).\] Find a similar relation in the case that \(A\), \(B\) and \(C\) are the vertices of an equilateral triangle described clockwise.
A particle of mass \(m\) is initially at rest on a rough horizontal surface. The particle experiences a force \(mg\sin \pi t\), where \(t\) is time, acting in a fixed horizontal direction. The coefficient of friction between the particle and the surface is \(\mu\). Given that the particle starts to move first at \(t=T_0\), state the relation between \(T_0\) and \(\mu\).
A long string consists of \(n\) short light strings joined together, each of natural length \(\ell\) and modulus of elasticity \(\lambda\). It hangs vertically at rest, suspended from one end. Each of the short strings has a particle of mass \(m\) attached to its lower end. The short strings are numbered \(1\) to \(n\), the \(n\)th short string being at the top. By considering the tension in the \(r\)th short string, determine the length of the long string. Find also the elastic energy stored in the long string. A uniform heavy rope of mass \(M\) and natural length \(L_0\) has modulus of elasticity \(\lambda\). The rope hangs vertically at rest, suspended from one end. Show that the length, \(L\), of the rope is given by \[ L=L_0\biggl(1+ \frac{Mg}{2\lambda}\biggr), \] and find an expression in terms of \(L\), \(L_0\) and \(\lambda\) for the elastic energy stored in the rope.
A circular wheel of radius \(r\) has moment of inertia \(I\) about its axle, which is fixed in a horizontal position. A light string is wrapped around the circumference of the wheel and a particle of mass \(m\) hangs from the free end. The system is released from rest and the particle descends. The string does not slip on the wheel. As the particle descends, the wheel turns through \(n_1\) revolutions, and the string then detaches from the wheel. At this moment, the angular speed of the wheel is \(\omega_0\). The wheel then turns through a further \(n_2\) revolutions, in time \(T\), before coming to rest. The couple on the wheel due to resistance is constant. Show that \[ \frac12 \omega_0 T = 2 \pi n_2\] and \[ I =\dfrac {mgrn_1T^2 -4\pi mr^2n_2^2}{4\pi n_2(n_1+n_2)}\;. \]
Let \(X\) be a random variable with a Laplace distribution, so that its probability density function is given by \[ \f(x) = \frac12 \e^{-\vert x \vert }\;, \text{ \ \ \ \(-\infty < x < \infty \)\ }. \tag{\(*\)} \] Sketch \(\f(x)\). Show that its moment generating function \({\rm M}_X(\theta)\) is given by ${\rm M}_X(\theta) = (1-\theta^2)^{-1}\( and hence find the variance of \)X$. A frog is jumping up and down, attempting to land on the same spot each time. In fact, in each of \(n\) successive jumps he always lands on a fixed straight line but when he lands from the \(i\)th jump (\(i=1\,,2\,,\ldots\,,n\)) his displacement from the point from which he jumped is \(X_i\,\)cm, where \(X_i\) has the distribution \((*)\). His displacement from his starting point after \(n\)~jumps is \(Y\,\)cm (so that \(Y=\sum\limits_{i=1}^n X_i\)). Each jump is independent of the others. Obtain the moment generating function for \(Y/ \sqrt {2n}\) and, by considering its logarithm, show that this moment generating function tends to \(\exp(\frac12\theta^2)\) as \(n\to\infty\). Given that \(\exp(\frac12\theta^2)\) is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a \(5\%\) chance that the frog lands 25~cm or more from his starting point.
A box contains \(n\) pieces of string, each of which has two ends. I select two string ends at random and tie them together. This creates either a ring (if the two ends are from the same string) or a longer piece of string. I repeat the process of tying together string ends chosen at random until there are none left. Find the expected number of rings created at the first step and hence obtain an expression for the expected number of rings created by the end of the process. Find also an expression for the variance of the number of rings created. Given that \(\ln 20 \approx 3\) and that \(1+ \frac12 + \cdots + \frac 1n \approx \ln n\) for large \(n\), determine approximately the expected number of rings created in the case \(n=40\,000\).
A positive integer with \(2n\) digits (the first of which must not be \(0\)) is called a balanced number if the sum of the first \(n\) digits equals the sum of the last \(n\) digits. For example, \(1634\) is a \(4\)-digit balanced number, but \(123401\) is not a balanced number.
Solution:
Solution:
Prove the identities \(\cos^4\theta -\sin^4\theta \equiv \cos 2\theta\) and $\cos^4 \theta + \sin^4 \theta \equiv 1 - {\frac12} \sin^2 2 \theta$. Hence or otherwise evaluate \[ \int_0^{\frac{1}{2}\pi} \cos^4 \theta \; \d \theta \;\;\;\; \mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^4 \theta \; \d \theta \,. \] Evaluate also \[ \int_0^{\frac{1}{2}\pi} \cos^6 \theta \; \d \theta \;\;\;\; \mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^6 \theta \; \d \theta \,. \]
Solution: \begin{align*} && \cos^4 \theta - \sin^4 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \\ &&&= \cos^2 \theta - \sin^2 \theta \\ &&&= \cos 2 \theta \\ \\ && 1&= (\cos^2 \theta + \sin^2 \theta)^2 \\ &&&= \cos^4 \theta + \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta \\ &&&= \cos^4 \theta + \sin^4 \theta + \frac12 ( \sin^2 2 \theta) \\ \Rightarrow && \cos^4 \theta + \sin^4 \theta &= 1 - \tfrac12 \sin^2 2 \theta \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^4 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^4 \theta \d \theta \\ && I-J &= \int_0^{\pi/2} \cos 2 \theta \d \theta = 0 \\ && I+J &= \int_0^{\pi/2} (1- \frac12 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac14 \int_0^{\pi} \sin^2 \theta \d \theta \\ &&&= \frac{\pi}{2} - \frac{\pi}{8} \\ &&&= \frac{3\pi}{8} \\ \Rightarrow && I=J &= \frac{3\pi}{16} \end{align*} \begin{align*} && \cos^6 \theta + \sin^6 \theta &= (\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ &&&= 1-\tfrac12 \sin^2 2\theta - \tfrac14 \sin^2 2 \theta \\ &&&= 1 - \tfrac34 \sin^2 2 \theta \\ %&& \cos^6 \theta - \sin^6 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac12 \sin^2 2 \theta + \tfrac14 \sin^2 2 \theta) \\ %&&&= \cos 2 \theta (1 - \tfrac14 \sin^2 2 \theta) \\ \end{align*} \begin{align*} && I &= \int_{0}^{\pi/2} \cos^6 \theta \d \theta \\ && J &= \int_0^{\pi/2} \sin^6 \theta \d \theta \\ && I-J &= 0 \\ && I+J &= \int_0^{\pi/2} (1 - \tfrac34 \sin^2 2 \theta) \d \theta \\ &&&= \frac{\pi}{2} - \frac{3\pi}{16} = \frac{5\pi}{16} \\ \Rightarrow && I = J &= \frac{5\pi}{32} \end{align*}