Problems

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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*}

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Solution:

The initial velocity is \(\begin{pmatrix} v \cos \theta \\ v \sin \theta \end{pmatrix}\). The trajectory will be: \(\begin{pmatrix} x_0 + (v \cos \theta) t \\ (v \sin \theta)t -\frac12 g t^2 \end{pmatrix}\) we must have that for some time \(t\), this is equal to \(\begin{pmatrix} 0 \\ h \end{pmatrix}\) So \(t = -\frac{x_0}{v \cos \theta}\) and so \begin{align*} &&h &= (v \sin \theta)t -\frac12 g t^2 \\ &&&= -x_0\tan \theta - \frac12 g \frac{x_0^2}{v^2 \cos^2 \theta} \\ &&&= -x_0\tan \theta - \frac{g}{2v^2 \cos^2 \theta}x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} \sec^2 \theta x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} (1+\tan^2 \theta )x_0^2 \\ &&&= -\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2+\frac{v^2}{2g}-\frac{g}{2v^2}x_0^2 \\ \Rightarrow && \frac{g}{2v^2}x_0^2 &= \frac{v^2}{2g}-h-\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2 \\ \Rightarrow && x_0^2 &= \frac{v^2(v^2-2gh)}{g^2}-K^2 \end{align*} Therefore \(\displaystyle |x_0| \leq \frac{v}{g}\sqrt{v^2-2gh}\)

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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: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

Where the line makes a length longer than \(\frac{4}{\pi}\) it will make an angle at the origin of \(2\sin^{-1} \frac{2}{\pi}\). Therefore the probability of being larger than this is \(\frac{2\pi - 2 \times 2\sin^{-1} \frac{2}{\pi}}{2 \pi} = 1 - \frac{2}{\pi} \sin^{-1} \frac{2}{\pi} \approx 0.560\)