Problems

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Solution:

Splitting the arc up into strips of width \(\delta \theta\), then we must have \begin{align*} && \sum r\cos \theta (r \delta \theta) &= \bar{x}\sum (r \delta \theta) \\ \lim_{\delta \theta \to 0}: && \int_{-\theta}^{\theta} r^2 \cos \theta \d \theta &= \bar{x}2 \theta r \\ \Rightarrow && 2r^2 \sin \theta &= \bar{x} 2 \theta r \\ \Rightarrow && \bar{x} &= \frac{r\sin \theta}{\theta} \end{align*} The centre of mass will lie on the line of symmetry. It also must lie at the center of the base of the semi-circle (see diagram). Using a coordinate frame where that point is the origin we must have \begin{align*} && 0 &= -2r \cdot 2h - 4h \cdot h + \pi r \frac{r}{\frac{\pi}{2}} \\ &&&= -4rh-4h^2+2r^2\\ \Rightarrow && 0 &= r^2-2rh-h^2 \\ \Rightarrow && \frac{r}{h} &= 1 \pm \sqrt{3} \\ \Rightarrow && r &= (1+\sqrt{3})h \\ && h & = \frac12 (\sqrt{3}-1) r \end{align*}

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Solution: Suppose she applies a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) backwards, then \begin{align*} && M v \frac{dv}{dx} &= -\frac{1}{2}F(V^{2}+v^{2})/V^{2} \\ \Rightarrow && M\int_{V}^0 \frac{2v}{V^2+ v^2} \d v &= - \frac{F}{V^2} x \\ \Rightarrow && M \left [ -\log(V^2+v^2) \right]_0^V &= -\frac{Fx}{V^2} \\ \Rightarrow && -M \ln 2&= -\frac{Fx}{V^2} \end{align*} Therefore she will stop quickly enough if \(d > \frac{V^2M \ln 2}{F}\) If she attempts to use the right-angled method, then she will travel a distance at most \(r\) where \(r\) is the radius of her circle: \begin{align*} && F &= M \frac{V^2}{r} \\ \Rightarrow && r &= \frac{MV^2}{F} \end{align*} Therefore she can always avoid the wall if \(d > \frac{MV^2}{F}\). There are two potential issues with being a particle. Firstly we would need to account for any variation in the distance to the wall (which could be accounted for by changing \(d\)). Secondly when she enters circular motion she will rotate and therefore we might need to consider her inertia as well as just her velocity when modelling.

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Solution: In the initial position, since the system is in equilibrium the tension in the two ropes must be \(500g\). Therefore since \(T = \frac{\lambda x}{l} \Rightarrow x = \frac{10 \cdot 500 g}{490\, 000} = \frac1{10}\) so the initial extension is \(\frac1{10}\) By conservation of momentum, if the initial speed of the plank + Nellie is \(V\), we must have \(4000 \cdot 5 = 5000 V \Rightarrow V = 4\) \begin{array}{ccc} & \text{GPE} & \text{EPE} & \text{KE} \\ \hline \text{Initially} & 5000gh & 2 \cdot \frac12 \frac{\lambda}{l} \frac{1}{100} & \frac12 \cdot 5000 \cdot 4^2 \\ & 49\,000h & 490 & 40\,000 \\ \text{Finally} & 0 & 2 \cdot \frac12 \frac{\lambda}{l} (h + \frac1{10})^2 & 0 \\ & 0 & 49\,000 (h+\frac1{10})^2 & 0 \end{array} By conservation of energy, we can set up a quadratic: \begin{align*} && 49\,000 (h+\frac1{10})^2 &= 49\,000h + 40\,490 \\ \Rightarrow && 49\,000(h + \frac1{10})^2 &= 49\,000(h + \frac1{10})+35\, 590 \\ \Rightarrow&& h + \frac1{10} &= 1.488092\cdots \\ \Rightarrow && h &= 1.49 \,\, (3\text{ s.f.}) \end{align*} When she gets off the plank, it will move according to: \begin{align*} \text{N2}(\uparrow): && \frac{\lambda x}{l} -1000g &= -1000 \ddot{x} \\ && 49 x-g &= -\ddot{x} \\ \Rightarrow && x &= A \sin 7t + B \cos 7t + 0.2 \\ && x(0) = 1.49, &x'(0) = 0 \\ \Rightarrow && B = -1.69, & A=0 \end{align*} If we continued under this motion the string would definitely reach a point \(0.1\) above \(0\), and therefore the ropes would go slack.

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Solution: Let \(X \sim N(0,1)\), and $\displaystyle X_{M}=\begin{cases} X & \text{if }X < M,\\ M & \text{if }X\geqslant M. \end{cases} $. Then we can calculate: \begin{align*} \mathbb{E}[X_M] &= \int_{-\infty}^M xf_X(x)\,dx + M\mathbb{P}(X \geq M) \\ &= \int_{-\infty}^M x \frac1{\sqrt{2\pi}}e^{-\frac12x^2}\,dx + M\mathbb{P}(X \geq M) \\ &= \left [ -\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2} \right ]_{-\infty}^M + M (1-\mathbb{P}(X < M)) \\ &= -\phi(M) + M(1-\Phi(M)) \end{align*} Let \(B \sim B\left (1024, \frac12 \right)\) be the number of potential bus passengers. Then \(B \approx N(512, 256) = N(512, 16^2)\) which is a good approximation since both \(np\) and \(nq\) are large. The question is asking us, how much additional profit would the bus company get if they ran an additional bus. Currently each week they is (on average) \(512\) passengers worth of demand, but they can only supply \(496\) seats, so we should expect that there is demand for another bus. The question is how much that demand is worth. Using the first part of the question, we can see that their profit is something like a `capped normal', \(X_M\), except we are scaled and with a different cap. So we are interested in $\displaystyle Y_{M}=\begin{cases} B & \mbox{if }B< M,\\ M & \mbox{if }B\geqslant M. \end{cases}\(, but since \)B \approx N\left (512,16^2\right)$ this is similar to \begin{align*} Y_{M}&=\begin{cases} 16X+512 & \mbox{if }16X+512< M,\\ M & \mbox{if }16X+512\geq M. \end{cases} \\ &= \begin{cases} 16X+512 & \mbox{if }X< \frac{M-512}{16},\\ M & \mbox{if }X \geq \frac{M-512}{16}. \end{cases} \\ &= 16X_{\frac{M-512}{16}} + 512\end{align*} We are interested in \(\mathbb{E}[Y_{16\times31}]\) and \(\mathbb{E}[Y_{17\times31}]\), which are \(16\mathbb{E}[X_{-1}]+512\) and \(16\mathbb{E}[Y_{\frac{15}{16}}]+512\) Since \(\frac{15}{16} \approx 1\), lets look at \(16(\mathbb{E}[X_1] - \mathbb{E}[X_{-1}])\) \begin{align*} \mathbb{E}[X_1] - \mathbb{E}[X_{-1}] &= \left ( -\phi(1) + 1-\Phi(1)\right) - \left ( - \phi(-1) -(1 - \Phi(-1)) \right ) \\ &= -\phi(1) + \phi(-1) + 1-\Phi(1) + 1 - \Phi(-1) \\ &= 1 - \Phi(1) + \Phi(1) \\ &= 1 \end{align*} Therefore the extra \(31\) will fill roughly \(16\) of them. (This is a slight overestimate, which is worth bearing in mind). A better approximation might be that \(\mathbb{E}[X_t] - \mathbb{E}[X_{-1}] = \frac{t +1}{2}\) for \(t \approx 1\), (since we want something increasing). This would give us an approximation of \(15.5\), which is very close to the `true' answer. Therefore, over \(50\) bus runs, we should earn roughly \(800\) vloska extra from an additional bus. (Again an overestimate, and with an uncertain pay-off, they should consider offering maybe \(600\)). Since this is the future, we can quite easily calculate the exact values using the binomial distribution on a computer. This gives the true value as \(15.833\), and so they should pay up to \(791\)

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Solution: \begin{array}{c|c|c|c} k & \P(k \text{ goals}) & \P(\geq 2k+1 \text{ fouls}) & \P(k \text{ goals and } \geq 2k+1 \text{ fouls}) \\ \hline 0 & \frac{3-|2|}{9} = \frac19 & \frac{9}{10} & \frac{9}{90}\\ 1 & \frac{3-|2-1|}{9} = \frac29 & \frac{7}{10} & \frac{14}{90} \\ 2 & \frac{3-|2-2|}{9} = \frac39 & \frac{5}{10} & \frac{15}{90} \\ 3 & \frac{3-|2-3|}{9} = \frac29 & \frac{3}{10} & \frac{6}{90} \\ 4 & \frac{3-|2-4|}{9} = \frac19 & \frac{1}{10} & \frac{1}{90} \\ \hline &&& \frac{9+14+15+6+1}{90} = \frac12 \end{array} The probability a team scores more than half its points from fouls is \(\frac12\). Letting \(X\) be the number of times a team committed \(9\) fouls, then \(X \sim B(300, p)\). Consider two hypotheses: \(H_0: p = \frac1{10}\) \(H_1: p < \frac1{10}\) Under \(H_0\), we are interested in \(\P(X \leq 9)\). Since \(300 \frac{1}{10} > 5\) it is appropriate to use a normal approximation, \(N(30, 27)\). Therefore, \begin{align*} && \P(X \leq 9) &\approx \P(3\sqrt{3}Z + 30 \leq 9.5) \\ &&&= \P( Z \leq \frac{9.5-30}{3\sqrt{3}}) \\ &&&= \P(Z \leq \frac{-20.5}{3\sqrt{3}}) \\ &&&< \P(Z \leq -\frac{7}{2}) \end{align*} Which is very small. Therefore there is good evidence to believe there has been a change in the number of fouls.

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Solution: Given the probability density after \(3\) years is proportional to \(\frac{t^2}{(1+t^2)^2}\) then we must have that: \begin{align*} && 1 &= A \int_0^{\infty} \frac{t^2}{(1+t^2)^2} \, \d t \\ &&&= A \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{\infty} + \frac{A}2 \int_0^{\infty} \frac{1}{1+t^2} \d t \\ &&&= \frac{A}{2} \frac{\pi}{2} \\ \Rightarrow && A &= \frac{4}{\pi} \end{align*} In order to fail a test on the fifth anniversary, there are two possibilities for when we went faulty. We could have gone faulty before \(4\) years, got lucky once and then failed the second test, or gone faulty in the next year and then failed the first test. \begin{align*} \P(\text{rusty before } 4 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{1} + \frac{2}{\pi} \int_0^{1} \frac{1}{1+t^2} \d t \\ &= -\frac{1}{\pi} + \frac{2}{\pi} \frac{\pi}{4} \\ &= \frac12 - \frac{1}{\pi} \\ &\approx 0.181690\cdots \\ \\ \P(\text{rusty before } 5 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{2} + \frac{2}{\pi} \int_0^{2} \frac{1}{1+t^2} \d t \\ &= -\frac{4}{5\pi} + \frac{2}{\pi} \tan^{-1} 2 \\ &\approx 0.450184\cdots \\ \end{align*} Therefore: \begin{align*} \P(\text{fails 5th anniversary}) &= \P(\text{rusty before } 4 \text{ years}) \P(\text{pass one, fail other}) + \\ & \quad \quad + \P(\text{rusty between 4 and 5 years}) \P(\text{fail}) \\ &= 0.181690\cdots \cdot \frac{1}{4} + \frac{1}{2} ( 0.450184\cdots- 0.181690\cdots) \\ &= \frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots \\ &= 0.1796688\cdots \\ &= 18.0\%\,\, (3\text{ s.f.}) \end{align*} We also must have that: \begin{align*} \P(\text{rusty at 4 years}|\text{destroyed at 5}) &= \frac{\P(\text{rusty at 4 years and destroyed at 5})}{\P(\text{destroyed at 5})} \\ &= \frac{0.181690\cdots \cdot \frac{1}{4}}{\frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots} \\ &= 0.252811\cdots \\ &= 25.3\%\,\,(3\text{ s.f.}) \end{align*}

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Solution: \begin{align*} && \f(x) &= \frac1{(x-1)(x-2)} \\ &&&=\frac{1}{x-2} -\frac{1}{x-1} \\ \end{align*} Therefore, for \(|x| < 1\) \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac{1}{1-x} \\ &&&= \sum_{n=0}^{\infty} x^n - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 \end{align*} where both geometric series converge since \(|x| < 1\) and \(|\frac{x}{2}| < 1\) When \(1 < |x|< 2 \Rightarrow |\frac{1}{x}| < 1\), we must have: \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&&= -\frac12 \frac{1}{1-\frac{x}{2}} + \frac1{x}\frac{1}{1-\frac{1}{x}} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \frac{1}{x} \sum_{n=0}^{\infty} x^{-n} \\ &&&= - \frac12 \sum_{n=0}^\infty \l \frac{x}2 \r^2 - \sum_{n=1}^{\infty} x^{-n} \\ \end{align*} Finally, when \(|x| > 2\), ie \(|\frac{2}{x}| < 1\) we have \begin{align*} && \f(x) &=\frac{1}{x-2} -\frac{1}{x-1} \\ &&& =\frac1{x} \frac{1}{1-\frac{2}{x}} - \frac{1}{x}\frac{1}{1-\frac{1}{x}} \\ &&&= \frac1{x} \sum_{n=0}^{\infty} \l \frac{2}{x} \r^n - \sum_{n=1}^{\infty}x^{-n} \\ &&&= \sum_{n=1}^{\infty} 2^{n-1} x^{-n} - \sum_{n=1}^{\infty}x^{-n} \\ \end{align*}

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Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).

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Solution: Suppose \(k\) is a root of our quadratic. There are two possibilities, if \(k^{-1} = k\) then we must have \(k^2 = 1\) so either \(\pm 1\) is a root or we must have \((x-k)(x-k^{-1}) = x^2+bx+c\). In the first case we can have: \begin{align*} x^2+bx +c = (x-1)^2 &\Rightarrow b = -2, c = 1 \\ x^2+bx +c = (x+1)^2 &\Rightarrow b = 2, c = 1 \\ x^2+bx +c = (x-1)(x+1) &\Rightarrow b = 0, c = 1 \\ \end{align*} In the other cases, \(c = 1\) and \(b = k^{-1}+k\). Therefore we must have \(c = 1\) and \(b\) can take any values. The statement "if \(k\) is a root then \(1-k\) is a root" implies these two roots are different, so we must have \(1-k = k^{-1} \Rightarrow k-k^2 = 1 \Rightarrow k^2-k+1 = 0\) so \(b = -1, c = 1\). Suppose \(x^3+px^2+qx+r = 0\) has the first property, then for any root \(k\) we must have: \(k^3 + pk^2 + qk + r = 0\) and \(1 + pk^{-1} + qk^{-2} + rk^{-3} = 0\) therefore \(x^3+px^2+qx+r\) and \(rx^3+qx^2+px+1 = 0\) must have identical roots (since \(x = \pm\) either wont work here since they imply having the roots \(1, 0\) or \(-1, 2, \frac12\) which is exactly our equation. Therefore \(r = 1, p = q\). Suppose \(x^3 + px^2 + px+1 = 0\) has the property that if \(k\) is a root \(1-k\) is a root, therefore: \begin{align*} 0 &= (1-k)^3+p(1-k)^2 + p(1-k) + 1 \\ &= 1 -3k+3k^2-k^3+p-2kp+pk^2+p-pk+1 \\ &= -k^3+(3+p)k^2+(-3-3p)k+(2+2p) \end{align*} Since these roots must be the same as the original roots, we must have \(3+p = -p, -3-3p = -p, 2+2p = -1 \Rightarrow p = -\frac32\)

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Solution:

1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
   

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2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
   

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If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)