Problems

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Solution:

1. \(\,\) \begin{align*} \bf{COE}: && \frac12 m \cdot 5ag &= mg\cdot 2a + \frac12 m v^2 \\ \Rightarrow && v^2 &= ag \\ && v &= \sqrt{ag} \end{align*} If it just clears the second wall, we must have: \begin{align*} && 0 &= \sqrt{ag} \sin \theta t - \frac12 gt^2 \\ \Rightarrow && t &= \frac{2\sqrt{ag}\sin \theta}{g} \\ && a &= \sqrt{ag} \cos \theta t \\ &&&=\sqrt{ag} \cos \theta \frac{2\sqrt{ag}\sin \theta}{g} \\ &&&= a \sin 2 \theta \\ \Rightarrow && \theta &= 45^{\circ} \end{align*} Imagine firing the particle backwards from the top of the wall at \(45^\circ\) then \begin{align*} && -2a &= \sqrt{ag}\cdot \left ( -\frac1{\sqrt{2}} \right) t - \frac12 g t^2 \\ \Rightarrow && 0 &= gt^2+\sqrt{2ag} t -4a \\ &&&= (\sqrt{g}t -\sqrt{2} \sqrt{a})(\sqrt{g}t +2\sqrt{2} \sqrt{a}) \\ \Rightarrow && t &= \sqrt{\frac{2a}{g}} \\ \Rightarrow && s &= \left ( -\frac1{\sqrt{2}} \right) \sqrt{ag} \sqrt{\frac{2a}{g}} \\ &&&= -a \end{align*} Therefore the \(A\) is \(a\) from the wall.
2. When it passes over the first wall, \begin{align*} \bf{COE}: && \frac52amg &= (2a+h)mg + \frac12 m v^2 \\ \Rightarrow && v^2 &= (a-2h)g \end{align*} Now imagine firing a particle with this speed in any direction. The question is asking whether we can ever travel \(2a\) without descending more than \(h\). \begin{align*} && a &= \sqrt{(a-2h)g} \cos \beta t \\ \Rightarrow && t &= \frac{a}{\sqrt{(a-2h)g} \cos \beta}\\ && -h &= \sqrt{(a-2h)g} \sin \beta t - \frac12 g t^2 \\ &&&= a \tan \beta - \frac12 \frac{a^2}{(a-2h)} \sec^2 \beta \\ &&&= a \tan \beta - \frac{a^2}{2(a-2h)}(1+ \tan^2 \beta )\\ \Rightarrow && 0 &= \frac{a^2}{2(a-2h)} \tan^2 \beta-a \tan \beta + \frac{a^2-2ah+4h^2}{2(a-2h)} \\ && \Delta &= a^2 - \frac{a^2}{a-2h} \frac{a^2-2ah+4h^2}{a-2h} \\ &&&= \frac{a^2}{(a-2h)^2}\left ( a^2-4ah+4h^2-a^2+2ah-4h^2\right) \\ &&&= \frac{a^2}{(a-2h)^2}\left ( -2ah\right) < 0 \\ \end{align*} So there are no solutions if \(h > 0\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X+Y=r) &= \sum_{k=0}^r \mathbb{P}(X = k, Y = r-k) \\ &&&= \sum_{k=0}^r \mathbb{P}(X = k)\mathbb{P}( Y = r-k) \\ &&&= \sum_{k=0}^r \frac{e^{-\lambda T} (\lambda T)^k}{k!}\frac{e^{-\mu T} (\mu T)^{r-k}}{(r-k)!}\\ &&&= \frac{e^{-(\mu+\lambda)T}}{r!}\sum_{k=0}^r \binom{r}{k}(\lambda T)^k (\mu T)^{r-k}\\ &&&= \frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^r}{r!} \end{align*} Therefore \(X+Y \sim Po \left ( (\mu+\lambda)T \right)\)
2. \(\,\) \begin{align*} && \mathbb{P}(X = r | X+Y = k) &= \frac{\mathbb{P}(X=r, Y = k-r)}{\mathbb{P}(X+Y=k)} \\ &&&= \frac{\frac{e^{-\lambda T} (\lambda T)^r}{r!}\frac{e^{-\mu T} (\mu T)^{k-r}}{(k-r)!}}{\frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^k}{k!}} \\ &&&= \binom{k}{r} \left ( \frac{\lambda}{\lambda + \mu} \right)^r \left ( \frac{\mu}{\lambda + \mu} \right)^{k-r} \end{align*} Therefore \(X|X+Y=k \sim B(k, \frac{\lambda}{\lambda + \mu})\)
3. \(P(X=1|X+Y = 1) = \frac{\lambda}{\lambda + \mu}\)
4. Let \(X_1, Y_1\) be the time to the first fish are caught by Adam and Eve, then \begin{align*} && \mathbb{P}(X_1, Y_1 > t) &= \mathbb{P}(X_1> t) \mathbb{P}( Y_1 > t) \\ &&&= e^{-\lambda t}e^{-\mu t} \\ &&&= e^{-(\lambda+\mu)t} \\ \Rightarrow && f_{\max(X_1,Y_1)}(t) &= (\lambda+\mu)e^{-(\lambda+\mu)} \end{align*} Therefore the expected time is \(\frac1{\mu+\lambda}\)

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Solution:

1. The probability they pull the key out on the \(k\)th attempt will be \(\left ( \frac{n-1}{n} \right)^{k-1} \frac1n\), so we want: \begin{align*} \E[G_1] &= \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \frac1n \\ &= \frac{1}n \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \\ &= \frac1n \frac{1}{\left (1 - \frac{n-1}{n} \right)^2} \\ &= \frac{1}{n} \frac{n^2}{1^2} = n \end{align*}
2. In version 2, the probability the correct key comes out at the \(k\)th attempt is \(\frac1n\) (assume we take out all the keys, then the correct key is equally likely to appear in all of the space). Therefore \(\E[G_2] = \frac1n (1 + 2 + \cdots + n) = \frac{n+1}{2}\)
3. The probability the key comes out on the correct attempt is: \begin{align*} && \mathbb{P}(G_3 = k) &= \frac{n-1}{n} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \cdots \frac{n+k-3}{n+k-2} \cdot \frac{1}{n+k-1} \\ &&&= \frac{n-1}{(n+k-2)(n+k-1)} \\ \\ &&k \cdot \mathbb{P}(G_3 = k) &= \frac{k(n-1)}{(n+k-2)(n+k-1)} \\ &&&= \frac{(n-1)(2-n)}{n+k-2} + \frac{(n-1)^2}{n+k-1} \\ &&&= \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n-k+2} \\ \Rightarrow && \E[G_3] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(G_3 = k) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n+k-2} \right) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} \right) +\underbrace{\sum_{k=1}^{\infty} \frac{n-1}{n-k+2}}_{\to \infty} \\ \end{align*}

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Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since \(\frac{1}{^{N+r}\C_{r}} \to 0\)} \\ &= \frac{r+1}{r} \end{align*} When \(r = 2\), we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since \(k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4\), this only makes sense if \(n \geq 3\) \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if \(n \geq 3\)} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

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Solution:

1. We can map \(a \mapsto 0\), rotate the whole plane, then shift the plane back to \(a\), so \(z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})\)
2. \(z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})\) \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle \(\phi + \theta\) and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If \(\phi + \theta = 2\pi\), then \(z \mapsto z + (b-a)(1 - e^{i\phi})\) which is a translation.
3. If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If \(\phi + \theta \neq 2 \pi\) then \(RS = ST\) if \(b = a\) or \(e^{i\phi} = e^{i \theta}\) ie rotation through the same angle.

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Solution: \begin{align*} A &= -(\alpha \beta + \gamma\delta + \alpha\gamma+\beta\delta+\alpha \delta + \beta\gamma) \\ &= -q \end{align*}

1. The corresponding cubic equation is: \begin{align*} && 0 &= y^3 - 3y^2-40y+(120-36) \\ &&&= y^3 -3y^2 - 40y + 84 \\ &&&= (y-7)(y-2)(y+6) \end{align*} Therefore \(\alpha\beta + \gamma \delta = 7\)
2. \begin{align*}(\alpha+\beta)(\gamma+\delta) &= \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\ &= 3 -(\alpha\beta + \gamma\delta) \\ &=3-7 = -4 \end{align*} Let \(\alpha\beta\) and \(\gamma\delta\) be the roots of a quadratic; then the quadratic will be \(t^2-7t+10 = 0 \Rightarrow t = 2,5\) so \(\alpha\beta = 5\)
3. \(\alpha\beta = 5, \gamma\delta = 2\) Consider the quadratic with roots \(\alpha+\beta\) and \(\gamma+\delta\), then \(t^2-4 = 0 \Rightarrow t = \pm 2\). Suppose \(\alpha+\beta = 2, \gamma+\delta=-2\) then \(\alpha, \beta = 1 \pm 2i, \gamma,\delta = -1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = -6 \neq 6\) Suppose \(\alpha+\beta = -2, \gamma+\delta=2\) then \(\alpha, \beta = -1 \pm 2i, \gamma,\delta = 1 \pm i\) \(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = 6\), therefore these are there roots. (In some order): \(1 \pm i, -1 \pm 2i\)

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Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

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Solution: \((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

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Solution:

1. \(\,\) \begin{align*} && T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\ &&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\ u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\ &&&= T_\infty - T(x^{-1}) \end{align*}
2. Let \(v = \frac{u+a}{1-au}\) then \begin{align*} && \frac{\d v}{\d u} &= \frac{(1-au) \cdot 1 - (u+a)\cdot(-a)}{(1-au)^2} \\ &&&= \frac{1-au+au+a^2}{(1-au)^2} \\ &&&= \frac{1+a^2}{(1-au)^2} \\ \\ && \frac{1+v^2}{1+u^2} &= \frac{1 + \left ( \frac{u+a}{1-ua} \right)^2}{1+u^2} \\ &&&= \frac{(1-ua)^2+(u+a)^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+u^2a^2+u^2+a^2}{(1-ua)^2(1+u^2)} \\ &&&= \frac{(1+u^2)(1+a^2)}{(1-ua)^2(1+u^2)} \\ &&&= \frac{1+a^2}{(1-ua)^2} \end{align*} if \(a > 0, x < \frac1a\) then \begin{align*} && T(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_{v=a}^{v=\frac{a+x}{1-ax}} \frac{1}{1+u^2} \frac{1+u^2}{1+v^2} \d v \\ &&&= T\left ( \frac{x+a}{1-ax} \right) - T(a) \\ \\ \Rightarrow && T(x^{-1}) &= T_\infty - T(x) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T(a) \\ &&&= T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) + T_\infty-T(a^{-1}) \\ &&&= 2T_\infty - 2T\left ( \frac{x+a}{1-ax} \right) -T(a^{-1}) \end{align*} \(b > 0, y > \frac1b\) then \(y> 0, b > \frac1y\) (same as letting \(x = \frac1y, a = \frac1b\) \begin{align*} && T(y) &= 2T_\infty - 2T \left ( \frac{\frac1y+\frac1b}{1-\frac1{by}} \right) + T(b) \\ \Rightarrow && T(y) &= 2T_\infty - 2T \left ( \frac{b+y}{by-1} \right) + T(b) \\ \end{align*}
3. Letting \(y = b = \sqrt{3}\) in the final equation \begin{align*} && T(\sqrt{3}) &= 2T_{\infty} - T \left ( \frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}\sqrt{3}-1} \right) -T (\sqrt{3}) \\ &&&= 2T_\infty - 2T(\sqrt{3}) \\ \Rightarrow && T(\sqrt{3}) &= \tfrac23 T_\infty \end{align*} Let \(x = \sqrt2 - 1, a = 1\) so, \begin{align*} && T(\sqrt2 -1) &= T \left ( \frac{\sqrt2-1+1}{1-\sqrt2+1} \right)-T(1) \\ &&&= T \left ( \frac{\sqrt{2}}{2-\sqrt{2}} \right) - T(1) \\ &&&= T(\frac{\sqrt{2}(2+\sqrt{2})}{2}) - T(1) \\ &&&= T(\sqrt{2}+1) - T(1) \\ &&&= T_\infty - T(\sqrt2-1)-T(1) \\ \Rightarrow && T(\sqrt{2}-1) &= \frac12T_\infty-\frac12T(1) \\ && T(1) &= T_\infty - T(1) \\ \Rightarrow && T(1) &= \frac12 T_\infty \\ \Rightarrow && T(\sqrt2-1) &= \frac12T_\infty - \frac14T_\infty \\ &&&= \frac14 T_\infty \end{align*}

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Solution:

1. The tangent has equation: \begin{align*} && 0 &= \frac{Xx}{a^2} + \frac{Yy}{b^2} -1 \\ \Rightarrow &&&= \frac{Xa(1-t^2)}{a^2(1+t^2)} + \frac{Y2bt}{b^2(1+t^2)} - 1 \\ \Rightarrow &&0&= Xb(1-t^2) + Y2at - ab(1+t^2)\\ &&&= -(b(a+X)t^2 -2aYt +b(a-X)) \\ \Rightarrow && 0 &= (a+X)bt^2-2aYt+b(a-X) \\ \\ && 0 <\Delta &= 4a^2Y^2 - 4(a+X)b(a-X)b \\ &&&= 4(a^2Y^2-b^2(a^2-X^2)) \\ \Leftrightarrow && a^2Y^2 &> (a^2-X^2)b^2 \end{align*} Therefore there are two roots to the quadratic, ie two values of the parameter \(t\) which works. The condition is equivalent to \(\frac{X^2}{a^2} + \frac{Y^2}{b^2} > 1\). ie from any point outside the ellipse there are two tangent lies.
   

   + - ↺
   Clearly there are two tangents when \(X = \pm a\) (except \((X,Y) = (\pm a, 0)\).
2. We must have \(p\) and \(q\) are roots of \(0 = (a+X)bt^2-2aYt+b(a-X)\), ie \(pq = \frac{b(a-X)}{(a+X)b} \Rightarrow (a+X)pq = a-X\). Similarly \(p+q = \frac{2aY}{(a+X)b}\) Given that the tangents meet the \(y\)-axis at \((0, y_i)\) we must have \(abt^2-2ay_it + ab = 0\), so \begin{align*} && 0 &= abp^2-2ay_1p + ab \\ && 0 &= abq^2-2ay_2q + ab \\ \Rightarrow && y_1 &= \frac{ab(p^2+1)}{2ap} \\ && y_2 &= \frac{ab(q^2+1)}{2aq} \\ \Rightarrow && 2b &= \frac{ab(p^2+1)}{2ap} +\frac{ab(q^2+1)}{2aq} \\ &&&= \frac{ab(pq(p+q)+p+q)}{2apq} \\ \Rightarrow && 4pq &= pq(p+q)+p+q \\ \Rightarrow && 4 \frac{b(a-X)}{(a+X)b} &= \frac{2aY}{(a+X)b} \left ( \frac{b(a-X)}{(a+X)b} + 1 \right) \\ && &= \frac{2aY}{(a+X)b} \frac{2ab}{(a+X)b} \\ \Rightarrow && 4b^2(a^2-X^2) &= 4a^2bY \\ \Rightarrow && 1 &= \frac{Y}{b} + \frac{X^2}{a^2} \end{align*} as required.

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Solution: \begin{align*} \sum_{m=1}^n a_m(b_{m+1} -b_m) +\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) &= \sum_{m=1}^n \left (a_{m+1}b_{m+1}-a_mb_m \right) \\ &= a_{n+1}b_{n+1} - a_1b_1 \end{align*} And the result follows.

1. Let \(b_m = \sin m x \), \(a_m = \cosec \frac{x}{2}\), so \begin{align*} && \sum_{m=1}^n \cosec \frac{x}{2} \left (\sin (m+1)x - \sin mx \right) &= \sum_{m=1}^n \cosec \frac{x}{2} 2 \cos \left ( \frac{2m+1}{2}x \right) \sin \left ( \frac{(m+1)-m}{2}x \right) \\ &&&=2 \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right)\\ \\ \Rightarrow && \sum_{m=1}^n\cos \left ( (m + \tfrac12)x \right) &= \tfrac12 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \end{align*}
2. \(\,\) \begin{align*} && b_{m+1}-b_m &= \sin m x \sin \tfrac12 x \\ &&&= \frac12 \left ( \cos (m-\tfrac12)x - \cos (m+\tfrac12)x \right)\\ \Rightarrow && b_m &= -\tfrac12 \cos (m - \tfrac12)x\\ && a_m &= m \\ \Rightarrow && \sum_{m=1}^n m \sin m x \sin \tfrac12 x &= (n+1) b_{n+1} - 1 \cdot b_1 - \sum_{m=1}^n b_{m+1} \cdot 1 \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac12\sum_{m=1}^n \cos(m+\tfrac12)x \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+\tfrac12\cos(\tfrac12x) + \tfrac14 \cosec \tfrac{x}{2}\left ( \sin(n+1)x - \sin x \right) \\ &&&= -(n+1) \tfrac12\cos(n+1-\tfrac12)x+ \tfrac14 \cosec \tfrac{x}{2}\sin(n+1)x \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\cos(n+\tfrac12)x\sin\tfrac12x \right) \\ &&&= \tfrac12\cosec\tfrac{x}2 \left (\tfrac12 \sin (n+1)x-(n+1)\tfrac12 \left ( \sin (n+1)x - \sin nx \right) \right) \\ &&&= \tfrac14 \cosec \tfrac{x}{2} \left ( -n \sin (n+1)x +(n+1) \sin n x \right) \end{align*} Therefore \(p = -\frac{n}4, q = \frac{n+1}{4}\)

Notice the connection here to integration by parts.

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Solution: \begin{align*} \text{N2}(\downarrow): && mg -T &= m\ddot{y} \\ \text{N2}(\rightarrow): && T &= 2m\ddot{x} \\ \Rightarrow && g &= \ddot{y}+2\ddot{x} \\ \Rightarrow && \tfrac12gt^2 &= y + 2x \end{align*} At time \(T = \sqrt{6a/g}\), we have \(y + 2x = 3a\), note also that \(\dot{y}+2\dot{x} = gt\) \begin{array}{ccc} & \text{KE} & \text{GPE} & \text{EPE} \\ \text{Initial} & 0 & 0 & 0 \\ \text{Final} & \frac12m\dot{y}^2 + \frac12(2m)\dot{x}^2 & -mgy & \frac{\lambda (y-x)^2}{2a} \end{array} Also note when we head over the table, \(x = a\) and \(y = a\) \begin{align*} \text{COE}: && 0 &= \frac12m(gT-2\dot{x})^2+m\dot{x}^2-mga+\frac{\lambda(0)^2}{2a} \\ \Rightarrow && 0 &= (gT-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= (\sqrt{6ag}-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= 6\dot{x}^2-4\sqrt{6ag}+4ag \\ \Rightarrow &&&= (\sqrt{6}\dot{x} - 2\sqrt{ag})^2 \\ \Rightarrow && \dot{x} &= \sqrt{2ag/3} \end{align*} as required.

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Solution:

By energy considerations, the initial energy is \(0\). Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{\(\frac{\d}{\d t}\)} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since \(2a > \ell\)} \end{align*} At limiting equilibrium, \(F = \mu R\). \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If \(\ell > 2a\), then the initial reaction force will be \(0\), ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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Solution:

1. \(\,\) \begin{align*} \text{COM}: && 0 &= nm(v-0) - Mu \\ \Rightarrow && u &= \frac{nm}{M}v \\ \\ \Rightarrow && \text{K.E.} &= \underbrace{\tfrac12 nmv^2}_{\text{projectiles}} + \underbrace{\tfrac12Mu^2}_{\text{guns and truck}} \\ &&&= \tfrac12nmv^2 + \tfrac12M \frac{n^2m^2}{M^2}v^2 \\ &&&= \tfrac12 nmv^2 \left (1 + \frac{nm}{M} \right) \end{align*}
2. \(\,\) \begin{align*} \text{COM}: && ((n-r+1)m+M)u_{r-1} &= -m(v-u_{r-1})+ ((n-r)m+M)u_r \\ \Rightarrow && mv &= -((n-r+1)m+M-m)u_{r-1}+((n-r)m+M)u_r \\ \Rightarrow && u_r-u_{r-1} &= \frac{mv}{M+(n-r)m} \\ \\ && u_n &= \frac{mv}{M+(n-1)m} + \frac{mv}{M+(n-2)m} + \cdots + \frac{mv}{M} \\ &&&< \frac{mv}M + \cdots + \frac{mv}{M} \\ &&&= \frac{mn}{M}v = u \end{align*}
3. \(\,\) \begin{align*} && K_r &= \underbrace{K_{r-1}-\frac12(m(n-r+1)+M)u_{r-1}^2}_{\text{energy of already dispersed projectiles}} + \frac12m(v-u_{r-1})^2 + \frac12(m(n-r)+M)u_r^2 \\ \Rightarrow && K_r-K_{r-1} &= \tfrac12(u_r^2-u_{r-1}^2)(M+m(n-r))+\tfrac12mv^2-mvu_{r-1} \\ &&&=\tfrac12mv^2+ \tfrac12(u_r+u_{r-1})mv-mvu_{r-1} \\ &&&= \tfrac12mv^2 + \tfrac12mv(u_r-u_{r-1}) \\ \\ && K_n &= \frac12nmv^2 + \tfrac12mv(u_n - u_0) \\ &&&= \tfrac12nmv^2 + \tfrac12mvu_n \\ &&&< \tfrac12nmv^2 + \tfrac12mvu \\ &&&= \tfrac12nmv^2 + \tfrac12mv \frac{nm}{M}v \\ &&&= \tfrac12nmv^2 \left (1 +\frac{m}{M} \right) \\ &&&\leq K \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\ &&&= \sum_{y=1}^n k(x+y) \\ &&&= nkx + k\frac{n(n+1)}2 \\ \\ && 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\ &&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\ &&&= kn^2(n+1) \\ \Rightarrow && k &= \frac{1}{n^2(n+1)} \\ \Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\ &&&= \frac{n+1+2x}{2n(n+1)} \\ \\ && \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\ &&&\neq \frac{x+y}{n^2(n+1)} \end{align*} Therefore \(X\) and \(Y\) are not independent.
2. \(\,\) \begin{align*} && \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\ &&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\ &&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\ &&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\ &&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\ &&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\ &&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\ &&&= \frac{7n+5}{12} \\ \\ && \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\ &&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\ &&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\ &&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\ &&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\ &&& < 0 \end{align*} since \(\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0\)