Problems

Solution:

1. Let \(k = 1\), then \begin{align*} \dot{x} &= - x - y \\ \dot{y} &= x - y \\ \dot{x}-\dot{y} &= -2x \\ \ddot{x} &= -\dot{x}-\dot{y} \\ &= -\dot{x} - (\dot{x}+2x) \\ &= -2\dot{x}- 2x \\ \dot{x}+\dot{y} &= -2y \\ \ddot{y} &= \dot{x}-\dot{y} \\ &= -2y-2\dot{y} \end{align*} So we have an auxiliary equation for \(x\) and \(y\) which is \(\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i\). Therefore \(x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t\). We also must have that, \(A = 1, C = 0\), so \(x = e^{-t} \cos t + Be^{-t} \sin t\) and \(y = De^{-t} \sin x\). \begin{align*} \dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\ &= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\ \end{align*} therefore \(B = 0, D = 1\) and \(x = e^{-t} \cos t, y = e^{-t} \sin t\)
   

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   \begin{align*} y &= e^{-t} \sin t \\ \dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\ \dot{x} &= e^{-t} \cos t -e^{-t} \sin t \end{align*}
   

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2. \begin{align*} \dot{x} = -x \\ \dot{y} = x-y \end{align*} So \(x = e^{-t}\). \(\dot{y} + y = e^{-t}\) so \(y = (t+B)e^{-t}\) and so \(y =te^{-t}\).
   

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Solution:

1. \(\,\) \begin{align*} && f(0+x) &= f(0)f(x) \\ \Rightarrow && f(0) &= 0, 1\\ &&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\ \Rightarrow && f(0) &= 1 \end{align*} \begin{align*} && f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\ &&&= f(x) \cdot f'(0) \\ &&&= kf(x) \end{align*} Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)
2. Consider \begin{align*} && g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\ \Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\ \Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\ \Rightarrow && g(0) &= -1, 0, 1 \\ \Rightarrow && g(0) &= 0 \tag{\(|g(0)| < 1\)} \end{align*} \begin{align*} && g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\ &&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &&&= (1-(g(x))^2) \cdot g'(0)\\ &&&= k (1-(g(x))^2) \\ \end{align*} Let \(y = g(x)\) so \begin{align*} && y' &= k(1-y^2) \\ \Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\ \Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\ &&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\ x = 0, y = 0: && 0 &= \ln 1 + C \\ \Rightarrow && C &= 0 \\ \Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\ \Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\ \Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\ &&&= \tanh kx \end{align*}

Solution:

1. Suppose \((x,y)\) is on the line of invariant points, then \begin{align*} &&\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\ \Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*} \end{align*} Therefore either \(x = 0, y = 0\) or \((a-1)(d-1)-bc = 0\) \(\Rightarrow ((a-1)(d-1)-bc)xy = 0\). We also know this is true for all values \(x,y\) on the line of invariant points. If there is one where both \(x \neq 0, y \neq 0\) we are done, otherwise the line of invariant points must be one of the axes. ie but then one of \(\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) is true and we'd also be done. If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since \(\mathbf{A}\) is a linear operator) and so every point is fixed. ie \(\mathbf{A} = \mathbf{I}\).
2. Suppose \((a-1)(d-1) -bc = 0\) and \(b \neq 0\) then I claim that \(y = \frac{1-a}{b}x\) is a line of invariant points. It's clear that the first equation will be satisfied in \((*)\) so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied. If \(b = 0\) then \((a-1)(d-1) = 0\), so our matrix must look like \(\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}\) (if \(d \neq 1\))or \(\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}\). In the first case, the line \(y = \frac{c}{1-d}x\) and in the second \(x = 0\) is an invariant line.
3. Suppose the invariant line is \(y = mx+k\) then we must have that \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\ (c+dm)x + dk \end{pmatrix} \end{align*} and \((c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)\) Since this equation must be true for all values of \(x\), and \(k \neq 0\) we can say that \(mb = d-1\) and \(-c+(a-d)m+m^2b = 0\), ie \(-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0\) if \(m \neq 0\) then \((a-1)\frac{(d-1)}{b} - c = 0\) ie our desired relation is true. If \(m = 0\) then we must have that \(y = k\) is an invariant line, ie \(d-1=0\) and \(c=0\) which also satisfies our relation.

Solution:

1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && y &= \frac{cx}{\sqrt{x^2+p}} \\ && \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\ &&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\ && I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\ &&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p} }} \d y \\ &&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\ &&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\ p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x \end{align*} When \(b = \sqrt{3}, c = \sqrt{2}\) \begin{align*} && I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\ &&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\ &&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\ &&&= \frac{\pi}{3\sqrt{3}} \end{align*} \begin{align*} && I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\ &&&= \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\ &&&= I_1 = \frac{\pi}{3\sqrt{3}} \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\ u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2} \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\ v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\ &&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\ &&&= \frac{\pi}{4} \end{align*}

Solution: \begin{align*} && |z-a|^2 &= (z-a)(z-a)^* \\ &&&= (z-a)(z^*-a^*) \\ &&&= zz^*-az^*-a^*z+aa^* \\ &&&= r^2 \end{align*} Therefore the locus of \(P\) is a circle centre \(a\) radius \(r\).

1. \begin{align*} && 0 &= zz^* - az^* - a^*z + aa^* - r^2 \\ &&&= \frac{1}{\omega \omega^{*}} - \frac{a}{\omega^*} - \frac{a^*}{\omega} + aa^*-r^2 \\ \Rightarrow && 0 &= 1-a\omega-a^*\omega^*+(|a|^2-r^2)\omega\omega^* \\ \Rightarrow && 0 &= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)-\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)+ \frac{1}{|a|^2-r^2} \\ &&&= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}-\frac{|a|^2}{(|a|^2-r^2)^2}+ \frac{1}{|a|^2-r^2} \\ &&&=\omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}- \frac{r^2}{(|a|^2-r^2)^2} \end{align*} Therefore \(\displaystyle \left|\omega-\left ( \frac{a^*}{|a|^2-r^2}\right)\right|^2 = \frac{r^2}{(|a|^2-r^2)^2}\) ie \(\omega\) lies on a circle centre \(\frac{a^*}{|a|^2-r^2}\), radius \(\frac{r}{||a|^2-r^2|}\). If these are the same circle then \(r = \frac{r}{||a|^2-r^2|} \Rightarrow (|a|^2-r^2)^2 = 1\) and \(a = \frac{a^*}{|a|^2-r^2} \Rightarrow a = \pm a^*\), ie \(a\) is purely real or imaginary.
2. This is the same story, except we end up with centre \(\frac{a}{|a|^2-r^2}\), so we do not end up with the same conditions

Solution:

1. Suppose \(a=b\), ie \begin{align*} && y^2(y^2-a^2) &= x^2(x^2-a^2) \\ \Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\ &&&= (x^2-y^2)(x^2+y^2-a^2) \end{align*} Therefore we have the lines \(y = \pm x\) and a circle radius \(a\).
   

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2. 1. Since \(x^4 - 4x^2 - y^2(y^2-5)= 0\), we must have \(0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0\), therefore \(0 \leq y \leq 1\) or \(y \geq 2\) (since we are only considering positive values of \(y\)).
   2. When \((x, y) \approx 0\) the equation is more like \(4x^2 \approx 5y^2\) or \(y \approx \frac{2}{\sqrt{5}}x\) If \(|x|, |y|\) are very large, it is more like \(x^4 \approx y^4\), ie \(y \approx x\)
   3. \(\,\) \begin{align*} && (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\ \Rightarrow && (4y^3-10y)y' &= 4x^3-8x \end{align*} Therefore the gradient is parallel to the \(x\)-axis when \(x = 0, x = \sqrt{2}\). We need \(x = 0, y \neq 0\), ie \(y = \sqrt{5}\), so \((0, \sqrt{5})\) and \((\sqrt{2}, 0)\) It is parallel to the \(y\)-axis when \(y = 0\) or \(y = \sqrt{\frac52}\), ie \((2, 0)\)
   

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3. \(\,\)
   

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Solution:

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1. Let \(A = (0,0,0)\) and then \(B = 22b \mathbf{i}, D = 2d\mathbf{j}, C = 2b\mathbf{i}+2d\mathbf{j}\) and \(V = b \mathbf{i} + d\mathbf{j} + h\mathbf{k}\) We also have \begin{align*} && \tan \alpha &= \frac{h}{d}\\ && \tan \beta &= \frac{d}{b} \\ && \vec{AV} \times \vec{VB} &= \begin{pmatrix} b \\ d \\ h \end{pmatrix} \times \begin{pmatrix} -b \\ d \\ h \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ -2bh \\ 2db \end{pmatrix} \\ &&&= 2b \begin{pmatrix} 0 \\ -d \tan \alpha \\ d \end{pmatrix} \\ &&&= k \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \end{align*} similarly for the vector perpendicular to the other face it must be \(\begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix}\) Looking at the angle between these perpendicular (to find the angles between the faces we see: \begin{align*} \begin{pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix}-\sin \beta \\ 0 \\ \cos \beta \end{pmatrix} &= \cos \alpha \cos \beta \end{align*} But this is also \(\pi -\) the angle between the planes, ie \(\cos \theta = \cos \alpha \cos \beta\)
2. \(\,\) \begin{align*} && \cot^2 \phi &= \frac{b^2+d^2}{h^2} \\ && \cot^2 \alpha &= \frac{d^2}{h^2} \\ && \cot^2 \beta &= \frac{b^2}{h^2} \\ \Rightarrow && cot^2 \phi &= \cot^2 \beta+\cot^2 \alpha \end{align*} \begin{align*} && \cos^2 \phi &= \frac{b^2+d^2}{b^2+d^2+h^2} \\ && \cos^2 \alpha &= \frac{d^2}{d^2+h^2} \\ && \cos^2 \beta &= \frac{b^2}{b^2+h^2} \\ && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &= \frac{\frac{d^2}{d^2+h^2}+\frac{b^2}{b^2+h^2}-2\cdot \frac{d^2}{d^2+h^2} \cdot \frac{b^2}{b^2+h^2}}{1 - \frac{d^2}{d^2+h^2} \cdot\frac{b^2}{b^2+h^2}} \\ &&&= \frac{d^2(b^2+h^2)+b^2(d^2+h^2)-2d^2b^2}{(d^2+h^2)(b^2+h^2)-d^2b^2} \\ &&&= \frac{h^2(b^2+d^2)}{h^2(b^2+d^2+h^2)} \\ &&&= \frac{b^2+d^2}{b^2+d^2+h^2} \\ &&&= \cos^2\phi \end{align*} Also notice that \begin{align*} && \cos^2 \alpha + \cos^2 \beta &\underbrace{\geq}_{AM-GM} 2 \cos \alpha \cos \beta \\ &&&= 2 \cos \theta \\ \Rightarrow && \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1-\cos^2 \theta} &\geq \frac{2 \cos \theta - 2\cos^2 \theta}{1-\cos^2 \theta} \\ &&&= \frac{2\cos \theta}{1+\cos \theta} = \cos \theta \frac{2}{1+\cos \theta} \\ &&&> \cos^2 \theta \\ \Rightarrow && \phi &< \theta \end{align*}

Solution:

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1. \(\mathbf{r} = (a \sin \theta - s) \mathbf{i}+a\cos \theta\mathbf{j}\), so \begin{align*} && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j}\\ \\ \text{COM}(\rightarrow): && 0 &= M(-\dot{s}) + m(a \dot{\theta} \cos \theta - \dot{s}) \\ \Rightarrow && \dot{s} &= \frac{ma \dot{\theta} \cos \theta}{m+M} \\ &&&= \left ( 1- \frac{M}{m+M} \right) a\dot{\theta} \cos \theta \\ &&&= (1 - k) a\dot{\theta} \cos \theta \\ \\ \Rightarrow && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= (a \dot{\theta} \cos \theta - (1 - k) a\dot{\theta} \cos \theta) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= a\dot{\theta} \left ( k \cos \theta \mathbf{i} - \sin \theta \mathbf{j} \right) \end{align*}
2. \(\,\) \begin{align*} COE: &&\underbrace{0}_{\text{k.e.}}+ \underbrace{mga}_{\text{GPE}} &= \underbrace{\frac12 m \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}}_{\text{k.e. }P} + \underbrace{mg a\cos \theta}_{\text{GPE}} + \underbrace{\frac12 M \dot{s}^2}_{\text{k.e. hemisphere}} \\ \Rightarrow && 2amg(1-\cos \theta) &= a^2m \dot{\theta}^2(k^2 \cos^2 \theta + \sin^2 \theta)+ M(1 - k)^2 a^2\dot{\theta}^2 \cos^2 \theta \\ \Rightarrow && 2mg(1-\cos \theta) &= a \dot{\theta}^2 \left (m\sin^2 \theta + (mk^2 + M(1-k)^2)\cos^2 \theta \right) \\ &&&= a \dot{\theta}^2 \left (m\sin^2 \theta + mk\cos^2 \theta \right) \\ \Rightarrow && 2g(1-\cos \theta) &= a \dot{\theta}^2 \left (\sin^2 \theta + k\cos^2 \theta \right) \\ \end{align*}
3. The equation of motion is \(m \ddot{\mathbf{r}} = \mathbf{R} - mg\mathbf{j}\) and the particle will leave the surface when \(\mathbf{R} = 0\). If we take the component in the directions suggested: \begin{align*} && \ddot{\mathbf{r}} &= a\ddot{\theta}(k \cos \theta \mathbf{i}- \sin \theta \mathbf{j}) + a \dot{\theta}(-k\dot{\theta} \sin \theta \mathbf{i}- \dot{\theta} \cos \theta \mathbf{j}) \\ &&&= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \mathbf{i} -a(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta) \mathbf{j} \\ \Rightarrow && \mathbf{\ddot{r}} \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \sin \theta -ak(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta)\cos \theta \\ &&&= - ak \dot{\theta}^2 \\ && (-g\mathbf{j}) \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= -gk \cos \theta \\ \mathbf{R} = 0: && gk \cos \theta &= ak \dot{\theta}^2 \\ \Rightarrow && g \cos \theta &= a \dot{\theta}^2 \end{align*}
4. \(\,\) \begin{align*} && 2g(1-\cos \theta) &= a \dot{\theta}^2(k \cos^2 \theta + \sin^2 \theta) \\ && a \dot{\theta}^2 &= g \cos \alpha \\ \Rightarrow && 2g(1-\cos \alpha) &= g \cos \alpha(k \cos^2 \alpha + (1-\cos^2 \alpha)) \\ \Rightarrow && 0 &= g(k-1)c^3+3gc-2g \\ \Rightarrow && 0 &= (k-1)c^3+3c - 2 \end{align*} When \(c =1, f(c) = k > 0\) when \(c = \frac23, f(c) = k-1 < 0\). Therefore there is a root with \(\cos \alpha > \frac23\)

Solution:

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1. Since the impulse is along the line of centres, the velocities are as show in the diagram. Additionally, vertical velocity is unchanged, so: \(v \sin (\theta + \alpha) = u \sin \alpha\) \begin{align*} \text{COM}(\rightarrow): && u \cos\alpha &= v \cos(\alpha + \theta) + w \\ \Rightarrow && w &= u \cos \alpha - v \cos (\alpha + \theta) \end{align*}
2. Since the approach speed (horizontally) is \(u \cos \alpha\) the speed of separation is \(e u \cos \alpha\), in particular \(w - v \cos(\theta + \alpha) = e u \cos \alpha\) or \(w = v \cos (\theta + \alpha) + e u \cos \alpha\). \begin{align*} && w &= w \\ && v \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - v \cos (\alpha + \theta) \\ \Rightarrow && \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\alpha + \theta) \\ \Rightarrow && \sin \alpha \cos(\theta + \alpha) + e \sin (\alpha+\theta)\cos \alpha &= \sin(\alpha+\theta) \cos \alpha - \cos(\alpha+\theta)\sin \alpha \\ &&&= \sin ((\alpha+\theta)-\alpha) \\ &&&= \sin \theta \end{align*} as required. \begin{align*} && \sin \theta &= \cos(\theta+ \alpha)\sin \alpha + e \sin (\theta + \alpha) \cos \alpha \\ &&&= \cos \theta \cos \alpha \sin \alpha - \sin \theta \sin^2 \alpha + e \sin \theta \cos ^2 \alpha + e \cos \theta \sin \alpha \cos \alpha \\ \Rightarrow && \tan \theta \sec^2 \alpha &= \tan \alpha - \tan \theta \tan^2 \alpha + e \tan \theta + e \tan \alpha \\ \Rightarrow && \tan \theta (1 + \tan^2 \alpha+\tan^2 \alpha-e) &= \tan \alpha + e \tan \alpha \\ \Rightarrow && \tan \theta &= \frac{(1+e)\tan \alpha}{1-e + 2\tan^2 \alpha} \end{align*} We seek to maximise \(y = \frac{x}{c+2x^2}\), \begin{align*} && \frac{\d y}{\d x} &= \frac{c+2x^2-4x^2}{(c+2x^2)^2} \\ &&&= \frac{c-2x^2}{(c+2x^2)^2} \end{align*} Therefore the maximum will occur at \(x = \sqrt{c/2}\), ie \(\tan \alpha = \sqrt{(1-e)/2}\) and theta will be \(\displaystyle \frac{(1+e)\sqrt{(1-e)/2}}{2(1-e)} =\frac{1}{2\sqrt{2}} \frac{1+e}{\sqrt{1-e}}\)