Problems

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Solution: \(V_h = \frac43 \pi r_h^3, S_h = 4 \pi r_h^2\) \(\frac{\d V_h}{\d t} = -k4\pi r_h^2 \Rightarrow 4\pi r_h^2 \frac{\d r_h}{\d t} = -k 4\pi r_h^2 \Rightarrow \frac{\d r_h}{\d t} = -k\) Therefore \(r_h = 2R - kt, r_b = 3R - kt\). The height will halve when \(2kt = \frac{5}{2}R \Rightarrow kt = \frac{5}{4}R\) and the two sections will have radii \(\frac{3}{4}R\) and \(\frac{7}{4}R\) and the ratio of the volumes will be: \begin{align*} \frac{\frac{3^3}{4^3}+\frac{7^3}{4^3}}{2^3+3^3} = \frac{37}{224} \end{align*} \begin{align*} && \frac{\d V}{\d t} &= -4\pi k(r_h^2+r_b^2) \\ && \frac{\d S}{\d t} &= -8\pi k (r_h+r_b) \\ \Rightarrow && \frac{\d V}{\d S} &= \frac{r_h^2 + r_b^2}{2(r_h+r_b)} \\ &&&= \frac{(2R-kt)^2+(3R-kt)^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rkt+2k^2t^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rs + 2s^2}{2(5R-2s)} \end{align*} Where \(s = kt\) and \(0 \leq s \leq 2R\). We can maximise this but differentiating wrt to \(s\). \begin{align*} \Rightarrow && &= \frac{(-10R+4s)(10R-4s)+4(13R^2-10Rs+2s^2)}{4(5R-2s)^2} \\ &&&= \frac{-48R^2+40Rs-8s^2}{4(5R-2s)^2} \\ &&&= \frac{-8(s-2R)(s-3R)}{4(5R-2s)^2} \\ &&&<0 \end{align*} Therefore it is largest when \(s = 0\), ie \(\frac{13R^2}{10R} = \frac{13}{10}R\)

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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:

1. \(\mathbf{n} \cdot \mathbf{r} = 0\) is the equation of a plane with normal \(\mathbf{n}\). \(\mathbf{n} \cdot (\mathbf{r}-\mathbf{a}) = 0\) is the equation of a plane through \(\mathbf{a}\) with normal \(\mathbf{n}\). Our expression is: \begin{align*} && \left(\mathbf{a-b}\right) \cdot \mathbf{r}&=\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) \\ &&&=\frac{1}{2}(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}+\mathbf{b}) \\ \Leftrightarrow && \left(\mathbf{a-b}\right) \cdot \left ( \mathbf{r} - \frac12 (\mathbf{a}+\mathbf{b}) \right) &= 0 \end{align*} So this is a plane through \(\frac12 (\mathbf{a}+\mathbf{b})\) perpendicular to \(\mathbf{a}-\mathbf{b}\). ie the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\) perpendicular to the line between them.
2. \begin{align*} && 0 &= \left(\mathbf{a-r}\right)\cdot\left(\mathbf{b-r}\right) \\ &&&= \mathbf{a} \cdot \mathbf{b} - \mathbf{r} \cdot (\mathbf{a}+\mathbf{b}) + \mathbf{r}\cdot\mathbf{r} \\ &&&= \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) \cdot \left ( \mathbf{r}- \frac12(\mathbf{a}+\mathbf{b}) \right) - \frac14 \left (\mathbf{a}\cdot\mathbf{a}+2\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{b} \right) +\mathbf{a}\cdot\mathbf{b} \\ &&&= \left | \mathbf{r} - \frac12 \left (\mathbf{a}+\mathbf{b} \right) \right|^2 - \left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|^2 \end{align*} Therefore this is a sphere, centre \(\frac12 \left (\mathbf{a}+\mathbf{b} \right)\) radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\)
3. This is a sphere centre \(\mathbf{a}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)
4. This is a sphere centre \(\mathbf{b}\) radius \(\frac1{\sqrt{2}} \left|\mathbf{a-b}\right|\)

Suppose the first two cases are true, then by symmetry it suffices to show that we can prove either of the second cases are true. (Since everything is symmetric in \(\mathbf{a}\) and \(\mathbf{b}\)). It's useful to note that \(\mathbf{r}\cdot \mathbf{r} = \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b}\) from the second condition. \begin{align*} \left|\mathbf{r-a}\right|^{2} &= \mathbf{r} \cdot \mathbf{r}-2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r}\cdot \mathbf{b} + \mathbf{r}\cdot \mathbf{a} -\mathbf{a}\cdot\mathbf{b} - 2\mathbf{a}\cdot \mathbf{r} + \mathbf{a}\cdot \mathbf{a} \\ &= \mathbf{r} \cdot ( \mathbf{b} - \mathbf{a}) + \mathbf{a} \cdot (\mathbf{a}-\mathbf{b}) \\ &= -\frac{1}{2}(\left|\mathbf{a}\right|^{2}-\left|\mathbf{b}\right|^{2}) + |\mathbf{a}|^2- \mathbf{a}\cdot\mathbf{b} \\ &= \frac{1}{2} |\mathbf{a}-\mathbf{b}|^2 \end{align*} as required. To show the other direction, consider Geometrically, these cases are equivalent, because together they both describe a circle of radius \(\left |\frac12 \left ( \mathbf{a} - \mathbf{b}\right) \right|\) in the plane halfway between \(\mathbf{a}\) and \(\mathbf{b}\)

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Solution: Suppose we divide a sphere of radius \(r\) up into slices of thickness \(\delta x\). Then the force acting on \(P\) will be: \begin{align*} F &= \sum_{\text{slices}} 2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right) \\ &= \sum_{i=-r/\delta x}^{r/\delta x} 2\pi mk\frac{M}{\frac43 \pi r^3}\delta x\left(1-\frac{i \delta x}{((1-(i\delta x)^2)+(i \delta x)^{2})^{\frac{1}{2}}}\right) \\ &\to \int_{-r}^r \frac{1}{2} \frac{mkM}{r^3}(1-t) \d t \\ &=\frac{mkM}{r^2} \end{align*} We can see that the particle will have a force attracting it towards the centre, with magnitude \(\frac{kmM}{r^2}\), therefore and since \(\frac{\d v}{\d t} = \frac{\d v}{\d r} \frac{\d r}{\d t}\) we must have: \(v \frac{\d v}{\d r}m = - \frac{kmM}{r^2}\) and dividing by \(m\) we get exactly the result we seek. \begin{align*} && v \frac{\d v}{\d r} &= \frac{-kM}{r^2} \\ \Rightarrow && \frac{v^2}{2}+C &= \frac{kM}{r} \\ r = R, v =0: && C &= \frac{kM}{R} \\ \Rightarrow && v^2&= 2kM\left ( \frac1r - \frac1R\right ) \\ \Rightarrow && \frac{\d r}{\d t} &= -\sqrt{2kM\left ( \frac1r - \frac1R\right )} \\ \Rightarrow && -\sqrt{2kM}T &= \int_{r=R}^{r=R\cos^2 \alpha} \frac{1}{\sqrt{\frac1r-\frac1R}} \d r \\ r = R\cos^2 \theta: && -\sqrt{2kM}T &= \int_{\theta = 0}^{\theta = \alpha} \frac{\sqrt{R}}{\sqrt{\sec^2 \theta - 1}} \cdot R \cdot 2 \cdot (-\cos \theta) \cdot \sin \theta \d \theta \\ \Rightarrow && T &= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\sqrt{\sec^2 \theta - 1}} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha \frac{2 \cos \theta \sin \theta}{\tan \theta} \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 2\cos^2 \theta \d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \int_0^\alpha 1 + \cos 2 \theta\d \theta \\ &&&= \sqrt{\frac{R^3}{2kM}} \left [1 + \frac12 \sin 2 \theta \right]_0^\alpha \\ &&&= \sqrt{\frac{R^3}{2kM}} \left (\alpha + \frac12 \sin 2 \alpha \right) \\ \end{align*}