The current in a straight river of constant width \(h\) flows at uniform speed \(\alpha v\) parallel to the river banks, where \(0<\alpha<1\). A boat has to cross from a point \(A\) on one bank to a point \(B\) on the other bank directly opposite to \(A\). The boat moves at constant speed \(v\) relative to the water. When the position of the boat is \((x,y)\), where \(x\) is the perpendicular distance from the opposite bank and \(y\) is the distance downstream from \(AB\), the boat is pointing in a direction which makes an angle \(\theta\) with \(AB\). Determine the velocity vector of the boat in terms of \(v,\theta\) and \(\alpha.\) The pilot of the boat steers in such a way that the boat always points exactly towards \(B\). Show that the velocity vector of the boat is \[ \begin{pmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\ \tan\theta\dfrac{\mathrm{d}x}{\mathrm{d}t}+x\sec^{2}\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t} \end{pmatrix}. \] By comparing this with your previous expression deduce that \[ \alpha\frac{\mathrm{d}x}{\mathrm{d}\theta}=-x\sec\theta \] and hence show that \[ (x/h)^{\alpha}=(\sec\theta+\tan\theta)^{-1}. \] Let \(s(t)\) be a new variable defined by \(\tan\theta=\sinh(\alpha s).\) Show that \(x=h\mathrm{e}^{-s},\) and that \[ h\mathrm{e}^{-s}\cosh(\alpha s)\frac{\mathrm{d}s}{\mathrm{d}t}=v. \] Hence show that the time of crossing is \(hv^{-1}(1-\alpha^{2})^{-1}.\)
Integers \(n_{1},n_{2},\ldots,n_{r}\) (possibly the same) are chosen independently at random from the integers \(1,2,3,\ldots,m\). Show that the probability that \(\left|n_{1}-n_{2}\right|=k\), where \(1\leqslant k\leqslant m-1\), is \(2(m-k)/m^{2}\) and show that the expectation of \(\left|n_{1}-n_{2}\right|\) is \((m^{2}-1)/(3m)\). Verify, for the case \(m=2\), the result that the expection of \(\left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|\) is \(2(m^{2}-1)/(3m).\) Write down the expectation, for general \(m\), of \[ \left|n_{1}-n_{2}\right|+\left|n_{2}-n_{3}\right|+\cdots+\left|n_{r-1}-n_{r}\right|. \] Desks in an examination hall are placed a distance \(d\) apart in straight lines. Each invigilator looks after one line of \(m\) desks. When called by a candidate, the invigilator walks to that candidate's desk, and stays there until called again. He or she is equally likely to be called by any of the \(m\) candidates in the line but candidates never call simultaneously or while the invigilator is attending to another call. At the beginning of the examination the invigilator stands by the first desk. Show that the expected distance walked by the invigilator in dealing with \(N+1\) calls is \[ \frac{d(m-1)}{6m}[2N(m+1)+3m]. \]
Each time it rains over the Cabbibo dam, a volume \(V\) of water is deposited, almost instanetaneously, in the reservoir. Each day (midnight to midnight) water flows from the reservoir at a constant rate \(u\) units of volume per day. An engineer, if present, may choose to alter the value of \(u\) at any midnight.
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The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]
The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.
Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).
The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).
The curve \(C\) has the differential equation in polar coordinates \[ \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5}, \] and, when \(\theta=\dfrac{\pi}{2},\) \(r=1\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.\) Show that \(C\) forms a closed loop and that the area of the region enclosed by \(C\) is \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]
Solution: First we seek the complementary function. \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\ \Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta \end{align*} Next we seek a particular integral, of the form \(r = C \sin 3 \theta\). \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\ \Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\ \Rightarrow && C &= -1 \\ \end{align*} So our general solution is \(A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta\). Plugging in boundary conditions we obtain: \begin{align*} \theta = \frac{\pi}{2}, r = 1: &&1 &= -B +1 \\ \Rightarrow && B &= 0 \\ \theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\ \Rightarrow && A &= 1 \end{align*} So the general solution is \(r = \sin 2 \theta - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)\) First notice that for \(\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]\) this is positive, and it is zero on the end points, therefore we are tracing out a a loop. The area of the loop will be: \begin{align*} A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta \d \theta \\ &= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta - \sin \theta \right]_{\pi/5}^{3\pi/5} \\ &= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right] \end{align*}
The transformation \(T\) from \(\binom{x}{y}\) to \(\binom{x'}{y'}\) in two-dimensional space is given by \[ \begin{pmatrix}x'\\ y' \end{pmatrix}=\begin{pmatrix}\cosh u & \sinh u\\ \sinh u & \cosh u \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}, \] where \(u\) is a positive real constant. Show that the curve with equation \(x^{2}-y^{2}=1\) is transformed into itself. Find the equations of two straight lines through the origin which transform into themselves. A line, not necessary through the origin, which has gradient \(\tanh v\) transforms under \(T\) into a line with gradient \(\tanh v'\). Show that \(v'=v+u\). The lines \(\ell_{1}\) and \(\ell_{2}\) with gradients \(\tanh v_{1}\) and \(\tanh v_{2}\) transform under \(T\) into lines with gradients \(\tanh v_{1}'\) and \(\tanh v_{2}'\) respectively. Find the relation satisfied by \(v_{1}\) and \(v_{2}\) that is the necessary and sufficient for \(\ell_{1}\) and \(\ell_{2}\) to intersect at the same angle as their transforms. In the case when \(\ell_{1}\) and \(\ell_{2}\) meet at the origin, illustrate in a diagram the relation between \(\ell_{1}\), \(\ell_{2}\) and their transforms.
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