Problems

By considering the graphs of \(y=kx\) and \(y=\sin x,\) show that the equation \(kx=\sin x,\) where \(k>0,\) may have \(0,1,2\) or \(3\) roots in the interval \((4n+1)\frac{\pi}{2} < x < (4n+5)\frac{\pi}{2},\) where \(n\) is a positive integer. For a certain given value of \(n\), the equation has exactly one root in this interval. Show that \(k\) lies in an interval which may be written \(\sin\delta < k < \dfrac{2}{(4n+1)\pi},\) where \(0 < \delta < \frac{1}{2}\pi\) and \[ \cos\delta=\left((4n+5)\frac{\pi}{2}-\delta\right)\sin\delta. \] Show that, if \(n\) is large, then \(\delta\approx\dfrac{2}{(4n+5)\pi}\) and obtain a second, improved, approximation.

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Solution:

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Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:

• be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
• not touch the second gradient

The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.

The points \(O,A,B\) and \(C\) are the vertices of a uniform square lamina of mass \(M.\) The lamina can turn freely under gravity about a horizontal axis perpendicular to the plane of the lamina through \(O\). The sides of the lamina are of length \(2a.\) When the lamina is haning at rest with the diagonal \(OB\) vertically downwards it is struck at the midpoint of \(OC\) by a particle of mass \(6M\) moving horizontally in the plane of the lamina with speed \(V\). The particle adheres to the lamina. Find, in terms of \(a,M\) and \(g\), the value which \(V^{2}\) must exceed for the lamina and particle to make complete revolutions about the axis.

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Solution:

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Consider the moment of inertia of the lamina. The MoI about the centre of mass is \(\frac1{12}M((2a)^2 + (2a)^2) = \frac23Ma^2\). //el axis theorem, tells us the moment of inertia about \(O\) is \(I_O = I_G + Md^2_{OG} = \frac23Ma^2 + M2a^2 = \frac83Ma^2\) Moment of inertia of particle is \(6Ma^2\) Total moment of inertial is: \(\frac{26}{3}Ma^2\). Conservation of angular momentum states that \(6M \frac{\sqrt{2}}2Va = \frac{26}{3}Ma^2 \omega \Rightarrow \omega = \frac{9\sqrt{2}V}{26a}\) Consider the centre of mass (in the frame drawn) \begin{array}{c|c|c} \text{Shape} & \text{Mass} & \text{COM} \\ \hline \text{Square} & M & (0,-\sqrt{2}a) \\ \text{Particle} & 6M & (-\frac{\sqrt{2}}2a, -\frac{\sqrt{2}}{2}a) \\ \text{combined} & 7M & \left ( \frac{-3\sqrt{2}}{7} a, -\frac{4\sqrt{2}}{7}a \right) \end{array} The lamina/particle system will complete full circles if it still has positive angular velocity at the peak, ie: \begin{align*} && \underbrace{\frac12 I \omega^2}_{\text{initial rotational energy}} + mgh_{start} &\geq mgh_{top} \\ && \frac 12 \frac{26}{3} Ma^2 \frac{9^2 \cdot 2 V^2}{26^2 a^2} - (7M)g\frac{4\sqrt{2}}{7}a &\geq (7M)g\frac{5\sqrt{2}}{7}a \\ \Rightarrow && \frac{V^2 \cdot 27}{26} &\geq 9\sqrt{2}ga \\ \Rightarrow && V^2 & \geq \frac{26\sqrt{2}}{3}ga \end{align*}

A uniform smooth wedge of mass \(m\) has congruent triangular end faces \(A_{1}B_{1}C_{1}\) and \(A_{2}B_{2}C_{2},\) and \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) are perpendicular to these faces. The points \(A,B\) and \(C\) are the midpoints of \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) respectively. The sides of the triangle \(ABC\) have lengths \(AB=AC=5a\) and \(BC=6a.\) The wedge is placed with \(BC\) on a smooth horizontal table, a particle of mass \(2m\) is placed at \(A\) on \(AC,\) and the system is released from rest. The particle slides down \(AC,\) strikes the table, bounces perfectly elastically and lands again on the table at \(D\). At this time the point \(C\) of the wedge has reached the point \(E\). Show that \(DE=\frac{192}{19}a.\)

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Solution: Conservation of energy, tells us that \(2m \cdot g \cdot 4a = 8amg\) is equal to \(\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2\). Conservation of momentum (horizontally) tells us that \(m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}\).

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We know that the particle must remain on the slope, and so \(v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}\). In conclusion, we have: \begin{align*} && 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\ &&&= \frac{19}{4}mv_{wedge}^2 \\ \Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag \end{align*}. To calculate the time the ball bounces for, note that: \(s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}\). During this time, the wedge (and ball) who horizontally are moving apart with speed \(\frac32 v_{wedge}\) we have they move apart by: \begin{align*} && DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\ &&&= \frac{6}{g} v_{wedge}^2 \\ &&&= \frac{6}{g}\frac{32}{19}ag \\ &&&= \frac{192}{19}a \end{align*}

A particle \(P\) is projected, from the lowest point, along the smooth inside surface of a fixed sphere with centre \(O\). It leaves the surface when \(OP\) makes an angle \(\theta\) with the upward vertical. Find the smallest angle that must be exceeded by \(\theta\) to ensure that \(P\) will strike the surface below the level of \(O\). You may find it helpful to find the time at which the particle strikes the sphere.

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Solution:

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\begin{align*} %\text{COE}: && \frac12 m u^2 - mga &= \frac12mv^2 + mga\cos \theta \\ \text{N2}(\swarrow): && R+mg\cos\theta &= \frac{m v^2}{a} \\ R = 0: && v^2 &= ag\cos \theta \\ \end{align*} So the particle will become a projectile moving tangent to the circle with \(v^2 = ag \cos \theta\). Therefore the velocity will be \(\displaystyle \sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta}\). We have: \begin{align*} && \mathbf{s} &= a\binom{\sin \theta}{\cos \theta}+\sqrt{ag \cos \theta}\binom{-\cos \theta}{\sin \theta} t + \frac12 \binom{0}{-g} t^2 \\ \Rightarrow && a^2 &= \mathbf{s} \cdot \mathbf{s} \\ &&&= a^2 + ag\cos \theta t^2 + \frac1{4} g^2t^4 -ag \cos \theta t^2 - \sqrt{ag \cos \theta} \sin \theta g t^3 \\ \Rightarrow && 0 &= \frac14 g t - \sqrt{ag \cos \theta} \sin \theta \\ \Rightarrow && t &= \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \end{align*} At this time, the vertical position will be: \begin{align*} && s_y &= a \cos \theta + \sqrt{ag \cos \theta} \sin \theta \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} - \frac12 g \left ( \frac{4\sqrt{a g \cos \theta} \sin \theta}{g} \right)^2 \\ &&&= a \cos \theta + 4a\cos \theta \sin^2 \theta - 8a\cos \theta \sin^2 \theta \\ &&&= a \cos \theta - 4 a \cos \theta \sin^2 \theta \\ &&&= a \cos \theta (1-4 \sin^2 \theta) \\ \underbrace{\Rightarrow}_{s_y < 0} && 0 &> 1 - 4 \sin^2 \theta \\ \Rightarrow && \sin\theta &> \frac12 \\ \Rightarrow && \theta & > \frac{\pi}{6} \end{align*}

The edges \(OA,OB,OC\) of a rigid cube are taken as coordinate axes and \(O',A',B',C'\) are the vertices diagonally opposite \(O,A,B,C,\) respectively. The four forces acting on the cube are \[ \begin{pmatrix}\alpha\\ \beta\\ \gamma \end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\ 0\\ 1 \end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix}\mbox{ at }B'\ (a,0,a). \] The moment of the system about \(O\) is zero: find \(\lambda,\mu\) and \(\nu\).

1. Given that \(\alpha=\beta=\gamma=0,\) find the system consisting of a single force at \(B\) together with a couple which is equivalent to the given system.
2. Given that \(\alpha=2,\beta=3\) and \(\gamma=2,\) find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.

An unbiased twelve-sided die has its faces marked \(A,A,A,B,B,B,B,B,B,B,B,B.\) In a series of throws of the die the first \(M\) throws show \(A,\) the next \(N\) throws show \(B\) and the \((M+N+1)\)th throw shows \(A\). Write down the probability that \(M=m\) and \(N=n\), where \(m\geqslant0\) and \(n\geqslant1.\) Find

1. the marginal distributions of \(M\) and \(N\),
2. the mean values of \(M\) and \(N\).

1. A rod of unit length is cut into pieces of length \(X\) and \(1-X\); the latter is then cut in half. The random variable \(X\) is uniformly distributed over \([0,1].\) For some values of \(X\) a triangle can be formed from the three pieces of the rod. Show that the conditional probability that, if a triangle can be formed, it will be obtuse-angled is \(3-2\sqrt{2.}\)
2. The bivariate distribution of the random variables \(X\) and \(Y\) is uniform over the triangle with vertices \((1,0),(1,1)\) and \((0,1).\) A pair of values \(x,y\) is chosen at random from this distribution and a (perhaps degenerate) triangle \(ABC\) is constructed with \(BC=x\) and \(CA=y\) and \(AB=2-x-y.\) Show that the construction is always possible and that \(\angle ABC\) is obtuse if and only if \[ y>\frac{x^{2}-2x+2}{2-x}. \] Deduce that the probability that \(\angle ABC\) is obtuse is \(3-4\ln2.\)

For \(x>0\) find \(\int x\ln x\,\mathrm{d}x\). By approximating the area corresponding to \(\int_{0}^{1}x\ln(1/x)\, \d x\) by \(n\) rectangles of equal width and with their top right-hand vertices on the curve \(y=x\ln(1/x)\), show that, as \(n\rightarrow\infty\), \[ \frac{1}{2}\left(1+\frac{1}{n}\right)\ln n-\frac{1}{n^{2}}\left[\ln\left(\frac{n!}{0!}\right)+\ln\left(\frac{n!}{1!}\right)+\ln\left(\frac{n!}{2!}\right)+\cdots+\ln\left(\frac{n!}{(n-1)!}\right)\right]\rightarrow\frac{1}{4}. \] {[}You may assume that \(x\ln x\rightarrow0\) as \(x\rightarrow0\).{]}

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Solution: Integrating by parts we obtain: \begin{align*} \int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\ &= \frac12 x^2 \ln x - \frac14 x^2 + C \end{align*}

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We should have: \begin{align*} \int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \left [ -\frac12 x^2 \ln x + \frac14 x^2 \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \frac{1}{4} &=\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\ &= \lim_{n \to \infty} \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n - \sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\ \end{align*}

In the triangle \(OAB,\) \(\overrightarrow{OA}=\mathbf{a},\) \(\overrightarrow{OB}=\mathbf{b}\) and \(OA=OB=1\). Points \(C\) and \(D\) trisect \(AB\) (i.e. \(AC=CD=DB=\frac{1}{3}AB\)). \(X\) and \(Y\) lie on the line-segments \(OA\) and \(OB\) respectively, in such a way that \(CY\) and \(DX\) are perpendicular, and \(OX+OY=1\). Denoting \(OX\) by \(x\), obtain a condition relating \(x\) and \(\mathbf{a\cdot b}\), and prove that \[ \frac{8}{17}\leqslant\mathbf{a\cdot b}\leqslant1. \] If the angle \(AOB\) is as large as possible, determine the distance \(OE,\) where \(E\) is the point of intersection of \(CY\) and \(DX\).

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Solution:

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Denoting \(\overrightarrow{OY}\) by \(\mathbf{y}\) and \(\overrightarrow{OC}\) by \(\mathbf{c}\) etc, we have: \begin{align*} \mathbf{c} &= \frac23 \mathbf{a} + \frac13 \mathbf{b} \\ \mathbf{d} &= \frac13 \mathbf{a} + \frac23 \mathbf{b} \\ \mathbf{x} &= \lambda \mathbf{a} \\ \mathbf{y} &= (1-\lambda) \mathbf{b} \\ 0 &= (\mathbf{d}-\mathbf{x}) \cdot (\mathbf{c} - \mathbf{y}) \\ &=((\frac13 -\lambda)\mathbf{a} + \frac23 \mathbf{b})\cdot(\frac23 \mathbf{a} + (\frac13-1+\lambda) \mathbf{b} ) \\ &= \frac{2}{3} \cdot (\frac13-\lambda) +\frac23 \cdot(\lambda - \frac23)+(\frac{4}{9}+(\frac13-\lambda)(-\frac23+\lambda))\mathbf{a}\cdot\mathbf{b} \\ &= -\frac{2}{9} + (\frac{4}{9} - \frac{2}{9}+\lambda-\lambda^2)\mathbf{a}\cdot \mathbf{b} \\ &= - \frac{2}{9} + (\frac{2}{9} + \lambda - \lambda^2)\mathbf{a}\cdot \mathbf{b} \\ \mathbf{a}\cdot \mathbf{b} &= \frac{2/9}{2/9+\lambda - \lambda^2} \end{align*} Since \(0 \leq \lambda - \lambda^2 \leq \frac14\), \(\frac{\frac29}{\frac29+\frac14} = \frac8{17} \leq \mathbf{a}\cdot\mathbf{b} \leq 1\) If \(\angle AOB\) is as large as possible, \(\mathbf{a}\cdot\mathbf{b}\) is as small as possible, ie \(\lambda = \frac12\) and \(\mathbf{a}\cdot \mathbf{b} = \frac{8}{17}\) First notice that the length \(OM\) to the midpoint of \(AB\) is \(\sqrt{\frac14 (\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})} = \sqrt{\frac14 (2 + 2\mathbf{a}\cdot \mathbf{b})} = \sqrt{\frac12 + \frac4{17}} = \sqrt{\frac{25}{34}} = \frac{5}{\sqrt{34}}\) Notice that \(XYE\) and \(DCE\) are similar triangles, and so the heights satisfy \(\frac{h_1}{h_2} = \frac{\frac12}{\frac13} = \frac32\). Therefore the length \(OE\) is \(\frac12 \frac{5}{\sqrt{34}} + \frac{3}{5} \frac12 \frac{5}{\sqrt{34}} = \frac{8}{10} \frac{5}{\sqrt{34}} = \frac{4}{\sqrt{34}}\)