Problems

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Solution: \(R(0) = 0\), \(R(1) = \Omega^2 1 = \Omega^2\), \(R(\Omega^2) = \Omega^4 = -\Omega\), \(R(-\Omega) = -\Omega^3 = 1\) \(T(0) = \frac{\Omega^2}1 = \Omega^2\), \(T(1) = \frac{\Omega + \Omega^2}{2\Omega^2+1} = \frac{2\Omega - 1}{2\Omega-1} = 1\) \(T(\Omega^2) = \frac{\Omega^3 + \Omega^2}{2\Omega^4+1} = \Omega \frac{\Omega^2+\Omega}{-2\Omega+1} = \Omega \frac{2\Omega-1}{-2\Omega+1} = - \Omega\) \(T(-\Omega) = \frac{-\Omega^2 + \Omega^2}{-2\Omega^3+1} = \frac{0}{3} = 0\) Thinking of \(R\) and \(S\) as elements of \(S_4\), we have that \(R = (234), S = (134)\), we can also construct \(RS = (14)(23), R^2S = (12)(34), RSR^2S = (13)(24)\). Therefore we have the subgroups \(\{e, (234), (243)\}\) of order \(3\) and the subgroup \(\{e, (12)(34), (13)(24), (14)(23) \}\) of order \(4\). By Lagrange's theorem this means that both \(3\) and \(4\) divide the order of the group, therefore the group has order divisible by \(12\) (and therefore is at least \(12\)). Yes, we cannot produce any odd permutation, for example \((12)\) cannot be produced. (Since all our generators are even permutations).

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Solution: \begin{align*} \begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}^2 &= \begin{pmatrix} 9-8 & 3-3 \\ -24+24 & -8+9 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \mathbf{I} \end{align*} Therefore: \begin{align*} \mathbf{F} &= \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n} \\ &= \mathbf{I} + \sum_{n=1}^{\infty} \frac{1}{(2n)!}t^{2n} \mathbf{I} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}t^{2n+1} \mathbf{A} \\ &= \cosh t \mathbf{I} + \sinh t \mathbf{A} \end{align*} Notice that \begin{align*} \mathbf{F} (\mathbf{I}\cosh t-\mathbf{A}\sinh t) &= (\mathbf{I}\cosh t+\mathbf{A}\sinh t)(\mathbf{I}\cosh t-\mathbf{A}\sinh t) \\ &= \mathbf{I}^2 \cosh^2 t+\mathbf{A}(\sinh t \cosh t - \cosh t \sinh t) - \mathbf{A}^2\sinh^2 t \\ &= \mathbf{I} \cosh^2 t - \mathbf{I} \sinh^2 t \\ &= \mathbf{I} \end{align*} Therefore \(\mathbf{F}^{-1} = \mathbf{I}\cosh t-\mathbf{A}\sinh t\) \begin{align*} \frac{\d \mathbf{F}}{\d t} &= \frac{\d }{\d t} \left [ \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}\right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^n \\ &= \left ( \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^{n-1} \right) \mathbf{A} \\ &= \mathbf{F}\mathbf{A} \end{align*} \begin{align*} && \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}&=\mathbf{0} \\ \Rightarrow && \mathbf{F} \frac{\d \mathbf{r}}{\d t} + \mathbf{FAr} &= \mathbf{0} \\ && \frac{\d }{\d t} \left ( \mathbf{F} \mathbf{r}\right) &= 0 \\ \Rightarrow && \mathbf{Fr} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= \mathbf{F}^{-1}\mathbf{c} \\ &&&= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\mathbf{c} \\ t = 0: && \binom{\alpha}{\beta} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\binom{\alpha}{\beta} \\ &&&= \binom{\alpha \cosh t}{\beta \cosh t} - \binom{-3\alpha-\beta}{8\alpha + 3\beta}\sinh t \\ &&&= \binom{\alpha \cosh t + (3\alpha + \beta) \sinh t}{\beta \cosh t -(8\alpha + 3\beta)\sinh t} \end{align*} as required

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Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)

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Solution:

Considering the pulley system with the lift (now of mass \(M-m\)) and the counterweight of mass \(M\). Once they start moving, since they are connected by a light inextensible string they must move with the same speed (and by extension the same acceleration). (Up to sign) \begin{align*} \text{N2(lift,}\uparrow):&&(M-m)a &= T-(M-m)g \\ \text{N2(couterweight,}\downarrow):&&Ma &= Mg - T \\ \Rightarrow && (2M-m)a &= mg \\ \Rightarrow && a &= \frac{mg}{2M-m} \end{align*} We could treat the situation as the tile travelling a distance of \(h\) with acceleration \(\displaystyle g \left ( 1 + \frac{m}{2M-m} \right) = g \frac{2M}{2M-m}\). \begin{align*} t &= \sqrt{\frac{2h}{g \frac{2M}{2M-m}}} \\ &= \sqrt{\frac{(2M-m)h}{Mg}} \\ \end{align*} Since the collision between the lift and tile is perfectly inelastic, they immediately coalesce. There is also an impulse in the pulley system, which goes over the pulley, which means there is an impulse acting vertically on the lift and the counterweight. Assume afterwards the lift (and tile) is travelling upwards with speed \(V\) and the counterweight is travelling downwards with speed \(V\) (ie velocity \(-V\)). \begin{align*} \text{for the lift/tile}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= MV - ((M-m)at +m(-g)t) \\ &&&= MV - Mat + m(a-g)t \\ \text{for the counterweight}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= M(-V) - (M(-a)t) \\ &&&= -MV +Mat \\ \Rightarrow && 2MV &= m(g-a)t + 2Mat \\ &&&= t (2Ma -ma+mg) \\ &&&= 0 \\ \Rightarrow && V &= 0 \end{align*} Therefore, the lift ends up stationary. The energy lost in the collision is: \begin{align*} && E &= KE_{before} - KE_{after} \\ &&&= \underbrace{\frac12 (M-m)a^2 t^2}_{lift} + \underbrace{\frac12 mg^2 t^2}_{tile} + \underbrace{\frac12 Ma^2 t^2}_{counterweight} - \underbrace{0}_{\text{everything is at rest after}} \\ &&&= \frac12 \l (M-m)a^2 + mg^2 + Ma^2 \r t^2 \\ &&&= \frac12 \l 2Ma^2-ma^2 + mg^2 \r t^2 \\ &&&= \frac12 \l (2M-m)a^2 + mg^2 \r t^2 \\ &&&= \frac12 \l mga + mg^2 \r t^2 \\ &&&= \frac12 mg (a + g)t^2 \\ &&&= \frac12 mg \left ( \frac{mg}{2M-m} + g\right ) \frac{(2M-m)h}{Mg} \\ &&&= \frac12 mg \left ( \frac{mg +2Mg - mg}{2M-m} \right) \frac{(2M-m)h}{Mg} \\ &&&= mgh \end{align*} as required.

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Solution:

We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

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Solution: This follows from the parallel axis theorem. The moment of inertia of both the system and the body will be equal to the moment of inertia about the axis through the centre of mass plus the distance from the axis. Suppose we have an axis through the centre of the rod, then consider the coordinate frame with the axis and as the \(y\) axis and the intersection between rod and axis at the origin. Suppose the angle between the rod and the \(x\) axis is \(\theta\) Then the moment of inertia for the rod will be: \begin{align*} \int_{-a}^a\frac{M}{2a} x^2 \cos^2 \theta \d x &= \frac{M}{2a}\frac23 a^3 \cos^2 \theta \\ &= \frac13 M \cos^2 \theta a^2 \end{align*} Suppose we put a weights of mass \(\frac16\) at where the end of the rod would be, and a weight of mass \(\frac23\) at the centre, then the moment of inertial would be: \begin{align*} I &= \frac23M \cdot 0^2 + \frac16M (a \cos \theta)^2+\frac16 (a \cos \theta)^2 \\ &= \frac13 M \cos^2 \theta a^2 \end{align*} Therefore it has the same mass (\(M\)), centre of mass (\(O\)) and moment of inertia for any axis through the COM so the two systems are dynamically equivalent. A uniform lamina can be broken down into a system with a rod of mass \(\frac23 M\) through the middle parallel to one side and rods of mass \(\frac16 M\) on each of those parallel sides. Those rods are then equivalent to a particle at the centre mass \(\frac23 \cdot \frac23 M = \frac49 M\) a mass at the centre of those sides of mass \(\frac23 \cdot \frac 16 M = \frac19 M\), a mass at the centre of the parallel sides of mass \(\frac16 \cdot \frac23 M = \frac 19 M\) and masses at the corners of mass \(\frac16 \cdot \frac16 M = \frac1{36} M\) The moment of inertia of a square lamina side length \(2a\) mass \(M\) about a diagonal through the centre will be: \begin{align*} I &= \sum_{\text{points}} md^2 \\ &= 2 \cdot \frac1{36}M \cdot \frac12 (2a)^2 \\ &= \frac19 Ma^2 \end{align*} Suppose the rod is on \([0, 1]\), then we must have: \(\displaystyle \int_0^1 \rho x \d x = m \Rightarrow \rho = 2m\). The centre of mass will be at: \begin{align*} \overline{x} &= \frac1m \int_0^1 2m x^2 \d x \\ &= \frac23 \end{align*} The moment of inertial for a line through \((\frac23, 0)\) with angle \(\theta\) will be: \begin{align*} I &= \int_0^12mx \l\frac23 - x \r^2\cos^2 \theta \d x \\ &= 2m\cos^2 \theta \cdot \frac1{36} \\ &= \frac1{18}m \cos^2 \theta \end{align*} Therefore if the particles have mass \(m_0, m_{1/2}, m_1\) we must have: \begin{align*} &&m &= m_0 + m_{1/2} + m_1 \\ &&\frac23m &= \frac12 m_{1/2} + m_1 \\ &&\frac1{18}m \cos^2 \theta &= m_0\frac49 \cos^2 \theta + m_{1/2}\frac1{36} \cos^2 \theta + m_1 \frac1{9} \cos^2 \theta \\ \Rightarrow && m &= 8 m_0 +\frac12 m_{1/2}+2m_1 \\ \Rightarrow && m_0 &= 0 \\ && m_{1/2} &= \frac23 \\ && m_1 &= \frac13 \end{align*} Since \(m_0 = 0\) the particle at the "thin" end of the rod could be ignored.

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Solution:

\begin{align*} \text{N2}(\nwarrow): && T - mg \sin \theta &= m \left ( \frac{v^2}{r}\right) \\ &&&= \frac{m v^2}{a} \\ \text{COE}:&& \underbrace{0}_{\text{assume initial GPE level is }0} &= \frac12 m v^2 - mga\sin \theta \\ \Rightarrow && v^2 &= 2ag \sin \theta \\ \Rightarrow && T &= \frac{m}{a} \cdot 2 ag \sin \theta + mg \sin \theta \\ &&&= 3mg \sin\theta \end{align*} Considering the force on the edge of the table will be: \begin{align*} && \mathbf{R} &= \binom{-T}{0} + \binom{T \cos \theta}{-T \sin \theta} \\ &&&= \binom{T(1-\cos \theta)}{-T \sin \theta} \\ &&&= 3mg \sin \theta \binom{1-\cos \theta}{-\sin \theta} \\ \Rightarrow && |\mathbf{R}| &= 3mg \sin \theta \sqrt{(1-\cos \theta)^2 + \sin ^2 \theta} \\ &&&= 3mg \sin \theta \sqrt{2 - 2 \cos \theta} \\ &&&= 3mg \sin \theta\sqrt{4 \sin^2 \tfrac{\theta} {2}} \\ &&&= 6mg \sin \theta |\sin \tfrac{\theta} {2} | \\ s = \sin \tfrac \theta2:&&&= 12mg s^2 \sqrt{1-s^2} \end{align*} We can maximise \(V = x\sqrt{1-x}\) by differentiating: \begin{align*} && \frac{\d V}{\d x} &= \sqrt{1-x} - \frac{x}{2\sqrt{1-x}} \\ &&&= \sqrt{1-x} \left ( 1 - \frac{x}{2-2x}\right) \\ &&&= \sqrt{1-x} \frac{2-3x}{2-2x} \\ \Rightarrow && x &= \frac23 \end{align*} Therefore the maximum for will be: \begin{align*} |\mathbf{R}| &= 12 mg \frac 23 \sqrt{\frac13} \\ &= 8mg/\sqrt{3} \end{align*} as required.

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Solution:

1. 

   + - ↺
   \begin{align*} \mathbb{E}(A \mid \text{exactly one point on diameter}) &= \int_{-1}^1\int_0^\pi \frac12 (x-0)\cdot 1 \cdot \sin(\pi - \theta) \frac{1}{\pi} \d \theta \frac{1}{2} \d x \\ &= \int_{-1}^1\frac1{2\pi} x \d x \cdot \left [ -\cos \theta \right]_0^\pi \\ &= \frac{1}{2\pi} \end{align*}
2. 

   + - ↺
   \begin{align*} \mathbb{E}(A \mid \text{no point on diameter}) &= \int_0^{\pi} \frac12 \cdot 1 \cdot 1 \cdot \sin x \cdot 2(\pi - x)/\pi^2 \d x \\ &= \frac1{\pi^2} \int_0^\pi \sin x (\pi - x) \d x \\ &= \frac1{\pi^2} \int_0^\pi x\sin x \d x \\ &= \frac1{\pi^2} \left [ \sin x - x \cos x \right]_0^{\pi} \\ &= \frac{1}{\pi} \end{align*}

If both points lie on the diameter the area of the triangle is \(0\). Therefore: \begin{align*} \mathbb{E}(A) &= \frac{1}{2\pi} \mathbb{P}(\text{exactly one point on diameter}) + \frac{1}{\pi}\mathbb{P}(\text{no points on diameter}) \\ &= \frac1{2\pi} \cdot \left (2 \cdot \frac{2}{2+\pi} \cdot \frac{\pi}{2+\pi} \right) + \frac{1}{\pi} \cdot \left ( \frac{\pi}{2+\pi} \cdot \frac{\pi}{2+\pi}\right) \\ &= \frac{1}{\pi} \frac{2\pi + \pi^2}{(2+\pi)^2} \\ &= \frac{1}{2+\pi} \end{align*}

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Solution: \begin{align*} \mathbb{P}(\text{exactly }k\text{ faults after test}) &= \mathbb{P}(k\text{ faults in non-tested, 0 in batch})+\mathbb{P}(k-1\text{ faults in non-tested, 1 in batch}) \\ &=\binom{N}{k}(1-p)^{N-k}p^k\binom{n}{0}(1-p)^n+\binom{N}{k-1}(1-p)^{N-k+1}p^{k-1}\binom{n}{1}(1-p)^{n-1}p \\ &= (1-p)^{N-k+n}p^k \cdot \left ( \binom{N}{k}+n\binom{N}{k-1} \right) \\ &= (1-p)^{N-k+n}p^k \cdot \left (\frac{N!}{k!(N-k)!}+\frac{N!n}{(k-1)!(N-k+1)!}\right) \\ &= (1-p)^{N-k+n}p^k \frac{N!}{k!(N-k+1)!} \cdot \left ((N-k+1)+nk \right) \\ &= \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \end{align*} When \(k = N+1\) we get: \begin{align*} \frac{(N+1)n N!}{(N+1)!} p^{N+1}(1-p)^{N+n-k} &= np^{N+1}(1-p)^{N+n-k} \end{align*} and the probability is: \begin{align*} \mathbb{P}(\text{exactly }N+1\text{ faults after test}) &= \mathbb{P}(N\text{ faults in non-tested, 1 in batch}) \\ &= \binom{N}{N}p^N \cdot \binom{n}{1}p(1-p)^{N-1} \\ &= np^{N+1}(1-p)^{N+n-k} \end{align*} So the formula does work for \(k = N+1\). \begin{align*} \mathbb{E}(faults) &= \sum_{k=0}^{N+1} k \cdot \mathbb{P}(\text{exactly }k\text{ faults after test}) \\ &= \sum_{k=0}^{N+1} k \cdot \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!(k-1)!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \left[N+1+k\left(n-1\right)\right] p(1-p)^{n-1}\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \\ &= p(1-p)^{n-1} \cdot \left ( (N+1+n-1)\sum_{k=1}^{N+1} \binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1}+ (n-1)\sum_{k=1}^{N+1} (k-1)\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \right) \\ &= p(1-p)^{n-1} \left ((N+1+n-1) + (n-1)pN \right) \\ &= \left[N+n+pN(n-1)\right]p(1-p)^{n-1} \end{align*}

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Solution:

We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*} We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}