Problems

The position vectors, relative to an origin \(O\), at time \(t\) of the particles \(P\) and \(Q\) are $$\cos t \; {\bf i} + \sin t\;{\bf j} + 0 \; {\bf k} \text{ and } \cos (t+\tfrac14{\pi})\, \big[{\tfrac32}{\bf i} + { \tfrac {3\sqrt{3}}2} {\bf k}\big] + 3\sin(t+\tfrac14{\pi}) \; {\bf j}\;,$$ respectively, where \(0\le t \le 2\pi\,\).

1. Give a geometrical description of the motion of \(P\) and \(Q\).
2. Let \(\theta\) be the angle \(POQ\) at time \(t\) that satisfies \(0\le\theta\le\pi\,\). Show that \[ \cos\theta = \tfrac{3\surd2}{8} -\tfrac14 \cos( 2t +\tfrac14 \pi)\;. \]
3. Show that the total time for which \(\theta \ge \frac14 \pi\) is \(\tfrac32 \pi\,\).

For \(x \ge 0\) the curve \(C\) is defined by $$ {\frac{\d y} {\d x}} = \frac{x^3y^2}{(1 + x^2)^{5/2}} $$ with \(y = 1\) when \(x=0\,\). Show that \[ \frac 1 y = \frac {2+3x^2}{3(1+x^2)^{3/2}} +\frac13 \] and hence that for large positive \(x\) $$ y \approx 3 - \frac 9 x\;. $$ Draw a sketch of \(C\). On a separate diagram, draw a sketch of the two curves defined for \(x \ge 0\) by $$ \frac {\d z} {\d x} = \frac{x^3z^3}{2(1 + x^2)^{5/2}} $$ with \(z = 1\) at \(x=0\) on one curve, and \(z = -1\) at \(x=0\) on the other.

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Solution: \begin{align*} && {\frac{\d y} {\d x}} &= \frac{x^3y^2}{(1 + x^2)^{5/2}} \\ \Rightarrow &&\int \frac{1}{y^2} \d y &= \int \frac{x^3}{(1+x^2)^{5/2}} \d x \\ \Rightarrow && -\frac1y &= \int \frac{x^3+x-x}{(1+x^2)^{5/2}} \d x \\ &&&= \int \left ( \frac{x}{(1+x^2)^{3/2}}-\frac{x}{(1+x^2)^{5/2}} \right) \d x \\ &&&= \frac{-1}{(1+x^2)^{1/2}} + \frac{1}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{1-3(1+x^2)}{3(1+x^2)^{3/2}} + C \\ &&&= \frac{-3x^2-2}{3(1+x^2)^{3/2}} + C \\ (x,y) = (0,1): &&-1 &= -\frac23 + C \\ \Rightarrow && C &= -\frac13 \\ \Rightarrow && \frac1y &= \frac{3x^2+2}{3(1+x^2)^{3/2}} + \frac13 \end{align*} \begin{align*} y &= \frac{1}{\frac13 +\frac{3x^2+2}{3(1+x^2)^{3/2}} } \\ &= \frac{3}{1+ \frac{3x^2+2}{3(1+x^2)^{3/2}}} \\ &= \frac{3}{1+ \frac{3}{x} + \cdots} \\ &\approx 3 - \frac{9}{x} \end{align*}

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\begin{align*} && \frac {\d z} {\d x} &= \frac{x^3z^3}{2(1 + x^2)^{5/2}} \\ \Rightarrow && \int \frac{1}{z^3} \d z &= \int \frac{x^3}{2(1+x^2)^{5/2}} \\ && -\frac{1}{2z^2} &= -\frac{3x^2+2}{3(1+x^2)^{3/2}} - C \\ (x,z) = (0, \pm 1): && \frac{1}{2} &= \frac{2}{3} + C \\ \Rightarrow && C &= -\frac16 \\ \Rightarrow && \frac{1}{z^2} &= \frac{6x^2+4}{3(1+x^2)^{3/2}} - \frac13 \end{align*} So as \(x \to \infty\) \(z \sim \pm (3 + \frac{2}{x} + \cdots)\) and so:

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Two particles, \(A\) and \(B\), of masses \(m\) and \(2m\), respectively, are placed on a line of greatest slope, \(\ell\), of a rough inclined plane which makes an angle of \(30^{\circ}\) with the horizontal. The coefficient of friction between \(A\) and the plane is \(\frac16\sqrt{3}\) and the coefficient of friction between \(B\) and the plane is \(\frac13 \sqrt{3}\). The particles are at rest with \(B\) higher up \(\ell\) than \(A\) and are connected by a light inextensible string which is taut. A force \(P\) is applied to \(B\).

1. Show that the least magnitude of \(P\) for which the two particles move upwards along \(\ell\) is \(\frac{11}8 \sqrt{3}\, mg\) and give, in this case, the direction in which \(P\) acts.
2. Find the least magnitude of \(P\) for which the particles do not slip downwards along~\(\ell\).

The points \(A\) and \(B\) are \(180\) metres apart and lie on horizontal ground. A missile is launched from \(A\) at speed of \(100\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(AB\) of \(\arcsin \frac35\). A time \(T\) seconds later, an anti-missile missile is launched from \(B\), at speed of \(200\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(BA\) of \(\arcsin \frac45\). The motion of both missiles takes place in the vertical plane containing \(A\) and \(B\), and the missiles collide. Taking \(g =10\,\)m\,s\(^{-2}\) and ignoring air resistance, find \(T\). \noindent [Note that \(\arcsin \frac35\) is another notation for \(\sin^{-1} \frac35\,\).]

A plane is inclined at an angle \(\arctan \frac34\) to the horizontal and a small, smooth, light pulley~\(P\) is fixed to the top of the plane. A string, \(APB\), passes over the pulley. A particle of mass~\(m_1\) is attached to the string at \(A\) and rests on the inclined plane with \(AP\) parallel to a line of greatest slope in the plane. A particle of mass \(m_2\), where \(m_2>m_1\), is attached to the string at \(B\) and hangs freely with \(BP\) vertical. The coefficient of friction between the particle at \(A\) and the plane is \( \frac{1}{2}\). The system is released from rest with the string taut. Show that the acceleration of the particles is \(\ds \frac{m_2-m_1}{m_2+m_1}g\). At a time \(T\) after release, the string breaks. Given that the particle at \(A\) does not reach the pulley at any point in its motion, find an expression in terms of \(T\) for the time after release at which the particle at \(A\) reaches its maximum height. It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at \(A\) to descend from its point of maximum height to the point at which it was released. Find the ratio \(m_1 : m_2\). \noindent [Note that \(\arctan \frac34\) is another notation for \(\tan^{-1} \frac34\,\).]

The twins Anna and Bella share a computer and never sign their e-mails. When I e-mail them, only the twin currently online responds. The probability that it is Anna who is online is \(p\) and she answers each question I ask her truthfully with probability \(a\), independently of all her other answers, even if a question is repeated. The probability that it is Bella who is online is~\(q\), where \(q=1-p\), and she answers each question truthfully with probability \(b\), independently of all her other answers, even if a question is repeated.

1. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. Show that the probability that the coin did come down heads is \(\frac{1}{2}\) if and only if \(2(ap+bq)=1\).
2. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. I then send the e-mail: `Did the coin come down heads?' and I receive the answer `no'. Show that the probability (taking into account these answers) that the coin did come down heads is \(\frac{1}{2}\,\).
3. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. I then send the e-mail: `Did the coin come down heads?' and I receive the answer `yes'. Show that, if \(2(ap+bq)=1\), the probability (taking into account these answers) that the coin did come down heads is \(\frac{1}{2}\,\).

The number of printing errors on any page of a large book of \(N\) pages is modelled by a Poisson variate with parameter \(\lambda\) and is statistically independent of the number of printing errors on any other page. The number of pages in a random sample of \(n\) pages (where \(n\) is much smaller than \(N\) and \(n\ge2\)) which contain fewer than two errors is denoted by \(Y\). Show that \(\P(Y=k) = \binom n k p^kq^{n-k}\) where \(p=(1+\lambda)e^{-\lambda}\) and \(q=1-p\,\). Show also that, if \(\lambda\) is sufficiently small,

1. \(q\approx \frac12 \lambda^2\,\);
2. the largest value of \(n\) for which \(\P(Y=n)\ge 1-\lambda\) is approximately \(2/\lambda\,\);
3. $ \P(Y>1 \;\vert\; Y>0) \approx 1-n(\lambda^2/2)^{n-1}\;.$

The probability density function \(\f(x)\) of the random variable \(X\) is given by $$ \f(x) = k\left[{\phi}(x) + {\lambda}\g(x)\right],\,\,\,\, $$ where \({\phi}(x)\) is the probability density function of a normal variate with mean~0 and variance~1, \(\lambda \) is a positive constant, and \(\g(x)\) is a probability density function defined by \[ \g(x)= \begin{cases} 1/\lambda & \mbox{for \(0 \le x \le {\lambda}\)}\,;\\ 0& \mbox{otherwise} . \end{cases} \] Find \(\mu\), the mean of \(X\), in terms of \(\lambda\), and prove that \(\sigma\), the standard deviation of \(X\), satisfies. $$ \sigma^2 = \frac{\lambda^4 +4{\lambda}^3+12{\lambda}+12} {12(1 + \lambda )^2}\;. $$ In the case \(\lambda=2\):

1. draw a sketch of the curve \(y=\f(x)\);
2. express the cumulative distribution function of \(X\) in terms of \(\Phi(x)\), the cumulative distribution function corresponding to \(\phi(x)\);
3. evaluate \(\P(0

Show that \(\sin A = \cos B\) if and only if \(A = (4n+1)\frac{\pi}{2}\pm B\) for some integer \(n\). Show also that \(\big\vert\sin x \pm \cos x \big\vert \le \sqrt{2}\) for all values of \(x\) and deduce that there are no solutions to the equation \(\sin\left( \sin x \right) = \cos \left( \cos x \right)\). Sketch, on the same axes, the graphs of \(y= \sin \left( \sin x \right)\) and \(y = \cos \left( \cos x \right)\). Sketch, not on the previous axes, the graph of \(y= \sin \left(2 \sin x \right)\).

Show Solution

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Solution: \begin{align*} && \sin A &= \cos B \\ \Leftrightarrow && 0 &= \sin A - \cos B \\ &&&= \sin A - \sin ( \frac{\pi}{2} - B) \\ &&&= 2 \sin \left ( \frac{A + B - \frac{\pi}{2}}{2} \right) \cos \left (\frac{A - B + \frac\pi2}{2} \right) \\ \Leftrightarrow && n \pi &= \frac{A+B - \frac{\pi}{2}}{2}, n\pi + \frac{\pi}{2} = \frac{A-B+\frac{\pi}{2}}{2} \\ \Leftrightarrow && A \pm B &= 2n\pi + \frac{\pi}{2} \\ &&&= (4n+1) \frac{\pi}{2} \end{align*} \begin{align*} |\sin x \pm \cos x| &= | \sqrt{2} \sin(x \pm \frac{\pi}{4} )| \\ & \leq \sqrt{2} \end{align*} Therefore if \(\sin(\sin x) = \cos (\cos x)\) we must have that \(|\sin x \pm \cos x| = |(4n+1) \frac{\pi}{2}| \geq \frac{\pi}{2} > 1.5 > \sqrt{2}\) contradiction.

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Find the general solution of the differential equation \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;\), where \(a\ne0\,\), and show that it can be written in the form \(\displaystyle y^2(x^2+a^2)= c^2\,\), where \(c\) is an arbitrary constant. Sketch this curve. Find an expression for \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)\) and show that \[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;. \]

1. Show that, if \(0 < c < a^2\), the points on the curve whose distance from the origin is least are \(\displaystyle \l 0\,,\;\pm \frac{c}{a}\r\).
2. If \(c > a^2\), determine the points on the curve whose distance from the origin is least.

Show Solution

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Solution: \begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).

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\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}

1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)

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