Problems

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Solution: The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\) The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\) But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin. \(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so: \begin{align*} && \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\ \Rightarrow && k(a^2-ap)-kap &= a^2 \\ \Rightarrow && k(a^2-2ap) &= a^2 \\ \Rightarrow && k &= \frac{a^2}{a^2-2ap} \end{align*} Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\) The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\). \begin{align*} && s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2 \end{align*} The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\) Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required. Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\) The area of triangle \(PQT\) is : \begin{align*} \frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\ &= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\ &= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\ &= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\ &= \frac{ap^2}{a^2-ap} \end{align*}

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Solution:

As the string wraps around, the total length in contact will be \(a \theta\). The end contact point will be at \((a\cos \theta, a\sin \theta)\) and the string will be tangential to that. The tangent (unit) vector will be \(\binom{-\sin \theta}{\cos \theta}\), and so the particle will be at \(\binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta}\). The velocity will be: \begin{align*} \frac{\d}{\d t} \binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta} &= \binom{-a \sin \theta \cdot \dot{\theta} -(L-a \theta) \cos \theta \cdot \dot{\theta} + a \sin \theta \cdot \dot{\theta} }{a \cos \theta \cdot \dot{\theta} + (L-a \theta) \sin \theta \cdot \dot{\theta} - a \cos \theta \cdot \dot{\theta}} \\ &= \binom{-(L-a \theta) \cos \theta \cdot \dot{\theta} }{ (L-a \theta) \sin \theta \cdot \dot{\theta}} \\ \end{align*} Therefore the speed is \((L-a\theta) \dot{\theta}\). By conservation of energy, we must have that speed is constant, ie: \begin{align*} && (L - a \theta)\dot{\theta} &= v \\ \Rightarrow && \int_0^{L/a} (L - a \theta)\d \theta &= \int_0^T v \d t \\ \Rightarrow && vT &= \frac{L^2}{a} - a\frac{L^2}{2a^2} \\ &&&= \frac{L^2}{2a} \\ \Rightarrow && T &= \frac{L^2}{2av} \end{align*} as requried

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Solution: Consider some point once the string is moving, there will be \(x\) above the table and \(l - x\) hanging in the air. For the hanging string we must have \((l-x)mg - T = -(l-x)m\ddot{x}\). For the string on the table we must have that \(T = -xm \ddot{x}\). Eliminating T, we have \((l-x)g = -l \ddot{x}\) Solving the differential equation, we must have \(x = A \cosh \sqrt \frac{g}{l}t+B \sinh\sqrt \frac{g}{l}t+l\), Since \(x(0) = d, \dot{x}(0) = 0 \Rightarrow B = 0, A = (-d)\). Therefore \(x = l-(l-d) \cosh \sqrt \frac{g}{l} t \Rightarrow t =\sqrt \frac{l}{g} \cosh^{-1} \l \frac{l-x}{l-d} \r\) and we go over the edge when \(x = 0\), ie \(\sqrt \frac{l}{g} \cosh^{-1} \l \frac{l}{l-d} \r\)

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Solution:

The force acting vertically on each of the outer books must be (by symmetry) \(\frac{nW}{2}\). The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say \(Q\) since increasing this force doesn't make any task less achievable. Looking at an end book, it will have force \(\frac{nW}{2}\) acting on one side, but it this force needs to not slip, ie \(\frac{nW}{2} \leq \mu Q\) \begin{align*} && \frac{nW}{2} &\leq \mu Q \\ \Rightarrow && n &\leq \frac{2\mu Q}{W} \\ && \frac{nW}{2} & \leq P \\ && n & \leq \frac{2P}{W} \\ \Rightarrow && n &\leq \frac2{W}\min \left (P, \mu Q \right) \end{align*}

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Solution:

The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.

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Solution: In order to receive \(n\) marks over the two papers, where \(m < n \leq 2m\) the student must receive \(k\) and \(n-k\) marks in each paper. Since \(n > m\), \(n-k\) is a valid mark when \(n-k \leq m\) ie when \(n-m\leq k\), therefore the probability is: \begin{align*} \sum_{k = n-m}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k=n-m}^m \frac{1}{(m+1)^2} \\ &= \frac{m-(n-m-1)}{(m+1)^2} \\ &= \frac{2m-n+1}{(m+1)^2} \end{align*} If \(0 \leq n \leq m\) then we need \(n-k\) marks in the second paper to be positive, ie \(n-k \geq 0 \Rightarrow n \geq k\), so \begin{align*} \sum_{k = 0}^n \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k = 0}^n \frac{1}{(m+1)^2} \\ &= \frac{n+1}{(m+1)^2} \end{align*} On the first paper, they can score any number of marks, since \(n > m\), so we must have: \begin{align*} \sum_{k=0}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \frac{1}{m+1} \sum_{k=0}^m \mathbb{P}(\text{scores }n-k\text{ on second papers}) \\ &= \frac{1}{m+1}\l \sum_{k=0}^{n-m} \frac{2m-(n-k)+1}{(m+1)^2} +\sum_{k=n-m+1}^m \frac{n-k+1}{(m+1)^2}\r \end{align*}

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Solution: The roots are \(X_1, X_2 = -B \pm \sqrt{B^2-1}\)

1. The smallest root will be \(-B - \sqrt{B^2-1}\). For this to be larger than \(\frac15\) we must have, \begin{align*} && -B -\sqrt{B^2-1} &\geq \frac15 \\ \Rightarrow && -B - \frac15 &\geq \sqrt{B^2 - 1} \\ \Rightarrow && B^2 + \frac25 B + \frac1{25} &\geq B^2 - 1 \\ \Rightarrow && \frac25 B \geq -\frac{26}{25} \\ \Rightarrow && B \geq -\frac{13}{5} \end{align*} Therefore \(-\frac{13}5 \leq B \leq -1\). Therefore we want: \begin{align*} \frac{\P(-\frac{13}5 \leq B \leq -1)}{\P(B < -1) + \P(B > 1)} &= \frac{\Phi(-1) - \Phi(-\frac{13}{5})}{\Phi(-1)+1-\Phi(1)} \\ &= \frac{0.1586\ldots - 0.0046\ldots}{0.1586\ldots + 1- 0.8413\ldots} \\ &= 0.4853\ldots \\ &= 0.485 \,\,(3 \text{ s.f.}) \end{align*}
2. \(X_1 + X_2 = -2B\). Therefore we want: \begin{align*} \mathbb{E}(|X_1 + X_2| &= \mathbb{E}(|2B|) \\ &= 2 \l\frac{1}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B+\frac{1}{2\Phi(-1)} \int_{-\infty}^{-1} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \r \\ &= \frac{4}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \\ &=\frac{4}{2\sqrt{2 \pi} \Phi(-1)} \left [ -e^{-\frac12 B^2}\right]_1^{\infty} \\ &= \frac{4}{2\sqrt{2 \pi} \Phi(-1) \sqrt{e}} \\ &= 3.0502\ldots \\ &= 3.05\,\, (3\text{ s.f.}) \end{align*}

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Solution:

First notice the integrand is always positive over the range we are integrating, so the integral is greater than \(0\). Since \(\frac{x}{1+x} \leq 1\) for \(x \geq 0\) we can note that: \begin{align*} \int_0^{\infty} \frac{x^2e^{-x}}{1+x} \d x &=\int_0^{\infty} \frac{x}{1+x}xe^{-x} \d x \\ &< \int_0^\infty xe^{-x} \d x \\ &= \left [ -xe^{-x} \right]_0^{\infty} + \int_0^{\infty} e^{-x} \d x \\ &= 0 + 1 \\ &= 1 \end{align*} and so we are done.

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Solution: First notice that \(u_n = \alpha^n\) and \(u_n = \beta^n\) both satisfy the recurrence, since: \begin{align*} && a \alpha^2 + b \alpha + c &= 0 \\ \Rightarrow && a \alpha^{n+2} + b \alpha^{n+1} + c \alpha^n &= 0 \\ \Rightarrow && a u_{n+2} + bu_{n+1} + cu_n &=0 \end{align*} Notice also that if \(u_n\) and \(v_n\) both satisfy the recurrence, then any linear combination of them will satisfy the recurrence: \begin{align*} && \begin{cases} au_{n+2} + bu_{n+1} + cu_n &= 0 \\ av_{n+2} + bv_{n+1} + cv_n &= 0 \\ \end{cases} \\ \Rightarrow && a (\lambda u_{n+2}+ \mu v_{n+2}) + b (\lambda u_{n+1}+ \mu v_{n+1}) + c (\lambda u_{n}+ \mu v_{n}) &= 0 \end{align*} by adding a linear combination of the top two equations. Therefore it suffices to check that the constants \(A\) and \(B\) are such that we match \(u_0\) and \(u_1\). \(\frac{u_1 - \beta u_0}{\alpha - \beta} + \frac{u_1 - \alpha u_0}{\beta - \alpha} = u_0\) and \(\frac{u_1 - \beta u_0}{\alpha - \beta}\alpha + \frac{u_1 - \alpha u_0}{\beta - \alpha}\beta = u_1\). So we are done. Suppose we have another sequence, then we first notice that the first and second terms must be identical to each other. Suppose the first \(k\) terms are identical, then since the \(k+1\)th term depends only on the \(k\) and \(k-1\)th terms (both of which are equal) the \(k+1\)th term is the same. Therefore, by the principle of mathematical induction, all terms are the same. First notice that if you put \(v_n = (n+1)w_n\) we have \begin{align*} && 8(n+3)(n+2)(n+1)w_{n+2} - 2(n+3)(n+2)(n+1)w_{n+1} - (n+3)(n+2)(n+1)w_n &= 0 \\ \Rightarrow && 8w_{n+2}-2w_{n+1}-w_n &= 0 \end{align*} This has characteristic equation \(8\lambda^2 - 2\lambda - 1 = 0 \Rightarrow \lambda = \frac12, -\frac14\). Therefore the general solution is \(w_n = A \l \frac12 \r^n + B \l -\frac14\r^n\) and \(v_n = (n+1)\l A \l \frac12 \r^n + B \l -\frac14\r^n \r\). When \(n = 0\) we have \(A+B = 0 \Rightarrow B =-A\). When \(n=1\) we have \(1 = 2 \l \frac{A}{2} + \frac{A}{4} \r \Rightarrow A = \frac{4}{3}\), therefore \[ v_n = \frac{4}{3}(n+1) \l \frac{1}{2^n} + \l -\frac14\r^n \r\]

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Solution: There are many possible parametric forms, for example \(z = i + 2e^{it}, z = 2\ cos \theta + (1 + 2\sin \theta)i\) etc. It is a circle radius \(2\) about the point \(i\).

1. \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{2i + 2e^{it}}{2e^{it}} \\ &= 2 + ie^{-it} \end{align*} This is obvious a circle radius \(1\) about the point \(2\).
   

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2. If \(z\) is real, then \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(z+i)^2}{z^2+1} \\ &= \frac{z^2-1 + 2zi}{z^2+1} \end{align*} We can quickly notice this describes a circle radius \(1\) about \(0\). Alternatively, \(|z+i| = |z-i| \Rightarrow |\frac{z+i}{z-i}| = 1\) so we must be talking about points on the unit circle. Since this is a Mobius transform we know it maps lines and circles to lines and circles, therefore it must map to the unit circle;
3. If \(z\) is purely imaginary, say \(it\) then: \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(it+i)(i-it)}{(-1+t)^2} \\ &= \frac{t^2-1}{(t-1)^2} \end{align*} Which is purely real, and can take all real values.
   

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