Problems

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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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Solution: The path can be broken down into \(5\) sections. 1. The section from \((-R,0)\) to \((-a,0)\) which will have distance \(R-a\) and is unchangeable. 2. The distance from \((-a,0)\) to \((-b, \frac{a^2-b^2}{2a})\) whose distance we will calculate shortly. 3. The distance from \((-b, \frac{a^2-b^2}{2a})\) to \((b, \frac{a^2-b^2}{2a})\) which will have distance \(\pi b\). 4. The distance from \((b, \frac{a^2-b^2}{2a})\) to \((a,0)\) which will have the same distance as 2. 5. The distance from \((a,0)\) to \((R,0)\) which will have distance \(R-a\) and we have no control over. \begin{align*} \text{distance 2.} &= \int_b^a \sqrt{1 + \left ( \frac{x}{a}\right)^2 } \d x \end{align*} We want to minimize the total, by varying \(b\), so it makes sense to differentiate and set to zero. \begin{align*} &&0&= -2\sqrt{1+\frac{b^2}{a^2}} + \pi \\ \Rightarrow && \frac{\pi^2}{4} &= 1 + \frac{b^2}{a^2} \\ \Rightarrow && b &= a \sqrt{\frac{\pi^2}{4}-1} \end{align*} Since \(\pi \approx 3\) this point is outside our range \(0 \leq b \leq a\), and our derivative is always positive. Therefore the distance is always increasing and the king would be better off going around the hill as soon as he arrives at it.

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Solution: After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days. Therefore the outstanding balance will be: \begin{align*} c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\ &= \frac{c[(1+k)^n-1]}{k} \end{align*} At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\). We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is: \begin{align*} (1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k} \end{align*} Therefore the net position, \(S_{T+m}\) will be: \begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\ &= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k} \end{align*} Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)

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Solution: \begin{align*} && f(x) &= \sin 2x \cos x \\ &&&= \frac12 \l \sin 3x + \sin x \r \\ \Rightarrow && f^{(1988)}(x) &= \frac12 \l 3^{1988} (-1)^{994} \sin 3x+ (-1)^{994} \sin x \r \\ &&&= \boxed{\frac12 \left (3^{1998} \sin 3x + \sin x \right)} \\ \\ f^{(1988)}(x) = 0: && 0 &= 3^{1988} \sin 3x + \sin x \\ \Rightarrow && 0 &= 3^{1988} ( 3\sin x-4\sin^3 x) + \sin x \\ \Rightarrow && 0 &= \sin x \left (1+3^{1989}-4\cdot 3^{1988}\sin^{2} x \right) \end{align*} Since \(\sin x\) will first contribute a zero when \(x = \frac{\pi}{2}\) we focus on the second bracket, in particular, we need: \begin{align*} && \sin^2 x &= \frac{3}{4} \left ( 1 + \frac{1}{3^{1988}} \right) \\ \Rightarrow && \sin x &= \frac{\sqrt{3}}2 \left (1 + \frac{1}{2 \cdot 3^{1988}} + \cdots \right ) \end{align*} Since near \(\frac{\pi}{3}\), \begin{align*} \sin (\frac{\pi}{3} + \epsilon) &= \sin \frac{\pi}{3} \cos \epsilon + \cos \frac{\pi}{3} \sin \epsilon \\ &\approx \frac{\sqrt{3}}{2} (1-\epsilon^2 + \cdots ) + \frac{1}{2}(\epsilon + \cdots) \\ &= \frac{\sqrt{3}}2 + \frac12 \epsilon + \cdots \end{align*} Therefore by comparison we can see that \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} 3^{-1988}\) will be a very good approximation for the root.

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Solution: \begin{align*} && y_0(x) &= e^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_0(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_0(x) &= \lim_{x \to \pm \infty} 2x\lambda e^{-\lambda x^2} \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= 4x^2 \lambda^2 e^{-\lambda x^2} + 2\lambda e^{-\lambda x^2} \\ \\ && y''_0 - \omega^2 x^2 y_0+(2\cdot 0 + 1) \omega y_0 &= e^{-\lambda x^2} \l 4x^2 \lambda^2 + 2 \lambda - \omega^2 x^2 + \omega\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_0\) satisfies if \(\lambda = \frac{\omega}{2}\) Similarly for \(y_1\), \begin{align*} && y_1(x) &= xe^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_1(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_1(x) &= \lim_{x \to \pm \infty} \l -2x^2 \lambda e^{-\lambda x^2} + e^{-\lambda x^2} \r \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= e^{-\lambda x^2} \l 4x^3 \lambda^2-4x\lambda - 2x\lambda \r \\ &&&= e^{-\lambda x^2} \l 4x^3 \lambda^2-6x\lambda \r \\ && y''_1 - \omega^2 x^2 y_1+(2\cdot 1 + 1) \omega y_1 &= e^{-\lambda x^2} \l 4x^3\lambda^2-6x\lambda-\omega^2x^3+3\omega x\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore \(y_1\) satisfies if \(\lambda = \frac{\omega}{2}\) \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) &= y'_my'_n+y_my''_n - y'_ny'_m-y_ny''_m \\ &= y_my''_n - y_ny''_m \\ &= y_m(\omega^2 x^2 y_n - (2n+1)\omega y_n) - y_n(\omega^2 x^2 y_m - (2m+1)\omega y_m) \\ &= y_my_n (2m-2n)\omega \\ &= 2(m-n) \omega y_my_n \end{align*} Therefore: \begin{align*} \int_{-\infty}^{\infty} y_m(x)y_n(x) \d x &= \int_{-\infty}^{\infty} \frac{1}{2(m-n)} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) \d x \\ &= \frac{1}{2(m-n)} \left [ y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right]_{-\infty}^{\infty} \\ &\to 0 \end{align*} This condition is known as Orthogonality. In fact this question is talking about a Sturm-Liouville orthogonality condition, in particular for the quantum harmonic oscillator, and the eigenfunctions are related to Hermite polynomials.

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Solution: \(\frac{dy}{dt} = 2a, \frac{dx}{dt} = 2at \Rightarrow \frac{dy}{dx} = \frac{1}{t}\). Therefore the equation of the tangent will be \(\frac{y - 2at}{x-at^2} = \frac{1}{t} \Rightarrow y = \frac1tx +at\) and normal will be \(\frac{y-2at}{x-at^2} = -t \Rightarrow y = t(at^2-x+2a)\). The tangents will meet when: \begin{align*} && \begin{cases} t_iy -x &= at_i^2 \\ t_j y - x &= at_j^2 \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)y &= a(t_i-t_j)(t_i+t_j) \\ \Rightarrow && y &= a(t_i+t_j) \\ && x &= at_it_j \end{align*} The normals will meet when: \begin{align*} && \begin{cases} y +t_i x &= at_i^3+2at_i \\ y +t_j x &= at_j^3+2at_j \\ \end{cases} \\ \Rightarrow &&(t_i - t_j)x &= a(t_i-t_j)(t_i^2+t_it_j+t_j^2+2) \\ \Rightarrow && x&= a(t_i^2+t_it_j+t_j^2+2) \\ && y &= -at_it_j(t_i+t_j) \end{align*} Therefore the area of our triangles will be: \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}at_1t_2 & a(t_1+t_2) & 1\\ at_2t_3 & a(t_2+t_3) & 1\\ at_3t_1 & a(t_3+t_1) & 1 \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2t_3 & (t_2+t_3) & 1\\ t_3t_1 & (t_3+t_1) & 1 \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}t_1t_2 & (t_1+t_2) & 1\\ t_2(t_3-t_1) & (t_3-t_1) & 0\\ t_1(t_3-t_2) & (t_3-t_2) & 0 \end{pmatrix} \\ &= \frac{a^2}{2}|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} and \begin{align*} \tfrac{1}{2}\det\begin{pmatrix}a(t_1^2+t_1t_2+t_2^2+2) & -at_1t_2(t_1+t_2) & 1\\ a(t_2^2+t_2t_3+t_3^2+2) & -at_2t_3(t_2+t_3) & 1\\ a(t_3^2+t_3t_1+t_1^2+2) & -at_3t_1(t_3+t_1) & 1\\ \end{pmatrix} &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_2^2+t_2t_3+t_3^2+2) & -t_2t_3(t_2+t_3) & 1\\ (t_3^2+t_3t_1+t_1^2+2) & -t_3t_1(t_3+t_1) & 1\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ t_3^2-t_1^2+t_2(t_3-t_1) & t_2(t_1^2+t_1t_2-t_2t_3-t_3^2) & 0\\ t_3^2-t_2^2+t_1(t_3-t_2) & t_1(t_2^2+t_2t_1-t_1t_3-t_3^2) & 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}\det\begin{pmatrix}(t_1^2+t_1t_2+t_2^2+2) & -t_1t_2(t_1+t_2) & 1\\ (t_3-t_1)(t_3+t_2+t_1) & t_2(t_1-t_3)(t_1+t_3+t_2) & 0\\ (t_3-t_2)(t_3+t_2+t_1) & t_1(t_2-t_3)(t_1+t_2+t_3)& 0\\ \end{pmatrix} \\ &= \frac{a^2}{2}(t_1+t_2+t_3)^2|(t_2-t_1)(t_3-t_2)(t_1-t_3)| \end{align*} as required. The normals will be concurrent iff the area of their triangle is \(0\). This is certainly true if \(t_1+t_2+t_3 = 0\). In fact the only if is also true, since no \(3\) tangents can be concurrent.

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Solution: \begin{questionparts} \item Show that, if \(o(g)=n\) and \(g^{N}=e,\) then \(n\) divides \(N.\) Using the division algorithm, write \(N = qn + r\) where \(0 \leq r < n\) to divide \(N\) by \(n\). Then we have \(e = g^N = g^{qn + r} = g^{qn}g^r = (g^{n})^qg^r = e^qg^r = g^r\) therefore \(r\) is a number smaller than \(n\) such that \(g^r = e\). Therefore either \(r = 0\) or \(o(g) = r\), but by definition \(o(g) = n\) therefore \(r = 0\) and \(n \mid N\). \item Let \(g\) and \(h\) be elements of \(G\). Prove that, for any integer \(m,\) \[ gh^{m}g^{-1}=(ghg^{-1})^{m}. \] \((ghg^{-1})^m = \underbrace{(ghg^{-1})(ghg^{-1})\cdots(ghg^{-1})}_{m \text{ times}} = gh(g^{-1}g)h(g^{-1}g)\cdots (g^{-1}g)hg^{-1} = gh^mg^{-1}\) \item Let \(g\) and \(h\) be elements of \(G\), such that \(g^{5}=e,h\neq e\) and \(ghg^{-1}=h^{2}.\) Prove that \(g^{2}hg^{-2}=h^{4}\) and find \(o(h).\) \(g^2hg^{-2} = g(ghg^{-1})g^{-1} = gh^2g^{-1} = (ghg^{-1})^2 = (h^2)^2 = h^4\). \(h = g^{5}hg^{-5} = g^4ghg^{-1}g^{-4} = g^4h^2g^{-4} = g^3(ghg^{-1})^2g^{-3} = g^3h^4g^{-3} = h^32\) Therefore \(e = h^{31}\). Therfore \(o(h) \mid 31 \Rightarrow \boxed{o(h) = 31}\) since \(31\) is prime and \(o(h) \neq 1\)

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Solution:

It's straightforward to see that \(\dot{r} = -\frac{v}{\sqrt{2}}\) and \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v\delta t/\sqrt{2}}{r - v \delta t/\sqrt{2}} = \frac{v}{r\sqrt{2}} \delta t + o(\delta t^2) \Rightarrow \dot{\theta} = \frac{v}{r \sqrt{2}}\) Solving this system, we can see that \(r(t) = \frac{l - vt}{\sqrt{2}}\) and \(\frac{\d r}{\d \theta} = -r \Rightarrow r = \lambda e^{-\theta}\) where \(\lambda = \frac{l}{\sqrt{2}}\) where \(l\) is the initial square side length. By the cosine rule, we can see that \(r(t + \delta t)^2 = r^2+(v\delta t)^2 - 2rv\delta t \cos (\frac12 (\pi - \frac{2\pi}{n}))\), ie: \(\frac{r(t + \delta t)^2 - r^2}{\delta t} = - 2r v \sin(\frac{\pi}{n}) \Rightarrow \dot{r} = - v \sin (\frac{\pi}{n})\). We can also observe that \(\tan (\theta(t + \delta t) - \theta(t)) = \frac{v \delta t \sin (\frac{\pi}{2} - \frac{\pi}{n})}{r - v \delta t \cos (\frac{\pi}{2} - \frac{\pi}{n})} \Rightarrow \dot{\theta} = \frac{v}{r} \cos (\frac{\pi}{n})\). Combining these, we can see that \(\frac{\d r}{\d \theta} = - r \tan (\frac{\pi}{n}) \Rightarrow r = \lambda e^{-\tan(\frac{\pi}{n})t}\) where \(\lambda\) is the initial radius of the circumscribed circle.

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Solution:

1. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-F_r &= 0 \\ \overset{\curvearrowleft}{X}: && lmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg \\ \Rightarrow && R_1 &= 2mg - \frac12mg \\ &&&=\frac32mg \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac32mg &= 0 \\ \Rightarrow && \mu &= \frac{1}{\sqrt{3}} \end{align*}
2. \begin{align*} \text{N2}(\uparrow): && R_1 + R_2\sin(\frac{\pi}{6})-2mg &= 0 \\ \overset{\curvearrowleft}{X}: && \frac12 lmg \cos \tfrac{\pi}{6}+xmg \cos \tfrac{\pi}{6} - l R_2 \cos \tfrac{\pi}{6} &= 0 \\ \\ \Rightarrow && R_2 &= mg(\frac{1}2+\frac{x}{l}) \\ \Rightarrow && R_1 &= 2mg - \frac12mg(\frac{1}2+\frac{x}{l}) \\ &&&=(\frac74 - \frac{x}{2l})mg \\ &&&\geq \frac{5}{4}mg\\ \text{N2}(\rightarrow): && R_2 \cos (\frac{\pi}{6})-\mu R_1& \leq 0 \\ \Rightarrow && \frac{\sqrt{3}}2mg - \mu\frac54mg &\leq 0 \\ \Rightarrow && \mu &\geq \frac{2\sqrt{3}}{5} \end{align*}

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Solution:

Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

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Solution: Applying \(\mathbf{F} = m\mathbf{a} = m \frac{\d \mathbf{v}}{dt}\) we have: \begin{align*} && m \frac{\d \mathbf{v}}{d t} &= m\mathbf{g} - mk(\mathbf{v} - \mathbf{w}) \\ \Rightarrow && \frac{\d \mathbf{v}}{d t} +k \mathbf{v} &= \mathbf{g} + k \mathbf{w} \\ \\ \Rightarrow && e^{k t} \l \frac{\d \mathbf{v}}{d t} +k \mathbf{v} \r &= e^{kt} ( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && \frac{\d}{\d t} \l e^{kt} \mathbf{v} \r &= e^{kt}( \mathbf{g} + k \mathbf{w}) \\ \Rightarrow && e^{kt} \mathbf{v} &= \frac{1}ke^{kt}( \mathbf{g} + k \mathbf{w}) + c \\ \Rightarrow && \mathbf{v}_0 &= \frac{1}{k} ( \mathbf{g} + k \mathbf{w})+c \\ \Rightarrow && \mathbf{v} &= e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g} + \mathbf{w} \\ \Rightarrow && \mathbf{x} &= -\frac{1}{k}e^{-kt} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t+C \\ \Rightarrow && \mathbf{0} &= -\frac{1}{k} \l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + C \\ \Rightarrow && \mathbf{x} &= \frac1{k}\l 1- e^{-kt} \r\l \mathbf{v_0} - \frac{1}{k}\mathbf{g} - \mathbf{w} \r + \frac{1}{k} \mathbf{g}t + \mathbf{w}t \end{align*} Position at time \(t\) is: \begin{align*} && x_x &= \frac1{k} ( 1-e^{-kt})(u_x - w)+wt \\ && x_y &= \frac1{k} ( 1-e^{-kt})(u_x \frac{g}{kw} - \frac{g}{k})+\frac{1}{k}gt \\ &&&= \frac{g}{kw} \left ( ( 1-e^{-kt})(u_x - w)+wt \right) \\ &&&= \frac{g}{kw} x_x \end{align*} Therefore if \(x_x\) is ever \(0\) then \(x_y\) will also be zero. But the ball must eventually hit the ground, and when it does, it will be in the process of scoring an own goal.