Problems

Solution:

1. 

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   By symmetry, the centre of mass will lie on the \(y\) axis. Notice that a single slice (when revolved around the \(y\)-axis) has volume \(y \cdot \pi \cdot ((x+ \delta x)^2 - x^2) = 2 \pi x y \delta x\), and COM at height \(\frac12 y\) so we can conclude: \[ \overline{y} \sum_{\delta x} 2 \pi x y \delta x = \sum_{\delta x} \pi xy^2 \delta x\] \begin{align*} && \overline{y} \int_0^r 2xy \d x &= \int_0^r y^2 x \d x \\ \Rightarrow && \overline{y} 2\int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)x \d x &= \int_0^r \left (r - \frac{x^n}{r^{n-1}} \right)^2 x \d x \\ \Rightarrow && \overline{y} \left [r \frac{x^2}{2} - \frac{1}{r^{n-1}} \frac{x^{n+2}}{n+2} \right]_0^r &= \left [r^2 \frac{x^2}{2} - \frac{2}{r^{n-2}} \frac{x^{n+2}}{n+2} + \frac{1}{r^{2n-2}} \frac{x^{2n+2}}{2n+2} \right]_0^r \\ \Rightarrow && 2\overline{y} \left (\frac{r^3}{2} - \frac{r^3}{n+2} \right) &= \left (\frac12 r^4 - \frac{2}{n+2}r^4 + \frac{1}{2n+2}r^4 \right) \\ \Rightarrow && \overline{y}r^3 \frac{n}{(n+2)} &= r^4\frac{(n+1)(n+2)-2\cdot2\cdot(n+1)+(n+2)}{2(n+1)(n+2)} \\ \Rightarrow && \overline{y} \frac{n}{(n+2)} &= r \left ( \frac{n^2}{2(n+1)(n+2)} \right) \\ \Rightarrow && \overline{y} &= \frac{nr}{2(n+1)} \\ &&&= r \left (1 -p^n \right) \end{align*} as required.
2. \begin{align*} && r^{n-1}y &= r^n - x^n \\ \frac{\d}{\d x}: && r^{n-1} \frac{\d y}{\d x} &= -n x^{n-1} \\ && \frac{\d y}{\d x} &= -np^{n-1} \end{align*} Therefor the normal has the equation: \begin{align*} && \frac{y-r(1-p^n)}{x-rp} &= \frac{1}{np^{n-1}} \\ \Rightarrow && Y &= \frac{-rp}{np^{n-1}} + r(1-p^n) \\ &&&= r \left (1 - p^n - \frac{1}{np^{n-2}} \right) \end{align*} If \(n = 4\) then \begin{align*} && Y &= r\left (1 - p^4 - \frac{1}{4p^{2}} \right) \\ \Rightarrow && \frac{\d Y}{\d p} &= r \left (-4p^3 + \frac{1}{2p^3} \right) \end{align*} Therefore there is a stationary point if \(p^6 = \frac18 \Rightarrow p =2^{-1/2}\). Clearly this will be a maximum (sketch or second derivative) therefore, \(Y = r(1 - \frac14 - \frac{2}{4}) = \frac14 r\)
3. The centre of mass of this shape can be found using this table: \begin{array}{|c|c|c|} \hline \text{} & \overline{y} & \text{mass} \\ \hline r^3y = r^4 - x^4 & \frac{2r}{5} & \frac{4\pi r^3}{6} = \frac23 \pi r^3\\ ry = -(r^2 - x^2) & -\frac{r}{3}& \frac{2 \pi r^3}{4}=\frac12\pi r^3 \\ \text{combined} & \frac{(\frac25 \cdot \frac23-\frac13 \cdot \frac12)r^4}{\frac76 r^3} = \frac3{35}r & \frac76 \pi r^3\\ \hline \end{array} Normals to the surface through points on the upper surface will meet the \(y\)-axis between \((-\infty, \frac14 r)\), and since \(p = 0 \to -\infty\) and \(p = 1 \to -\frac14 r\), so normals will pass through \((0, \frac3{35}r)\) from two different points. Normals to the surface through points on the lower surface will go through \(-r(1 - p^2 - \frac12) =- r(\frac12 -p^2)\) which ranges monotonically from \(\frac12 r \to -\frac12 r\) so there will be one point where the normal goes through \(\frac3{35}r\). Therefore there are three angles where the vector \(\overrightarrow{AB}\) is not vertical but the normal to the surfaces runs through the centre of mass (ie the the solid can rest in equilibrium)

Solution:

1. Since we have \(h'' + \omega^2 h = 0\) we must have that \(h(t) = A \cos \omega t + B \sin \omega t\). The initial conditions tell us that \(A = 0\) and \(B = L\), so \(h(t) = L \sin \omega t\).
   

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   Therefore we can see the angle at \(B\) is \(\omega t\) and so \(P\) has \(y\)-coordinate \((1-s)L \sin \omega t\) and \(x\)-coordinate \(sL \cos \omega t\)
2. If the position is \(\binom{sL \cos \omega t}{(1-s) L \sin \omega t}\) then the acceleration is \(-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}\)
3. 

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   \begin{align*} \text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\ \text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\ \Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix} &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\ &&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\ &&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\ &&&= mg (1 - k \sin \omega t) \cos \omega t \\ \Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\ &&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\ &&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\ &&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\ &&&= mgsk + N \tan \omega t \end{align*}
4. The particle will not slip if \(F < \tan \alpha N\). When \(t = 0\), \(N = mg, F = mgsk\), but clearly \(sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N\). The particle will slip when: \(F > \tan \alpha N\), but we have \(F = mgsk + N \tan \omega t\). Clearly when \(\omega t = \alpha\) we have reached a point where \(F > \tan \alpha N\). Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than \(\alpha\) to the horizontal. Notice also that \(N\) changes sign when \(1-k \sin \omega t = 0\), however, to do this \(N\) must become very small, smaller than \(mgsk\), therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when \(N\) is positive AND the plank still makes a shallow angle.

Solution:

1. Note that \(\displaystyle F(y) = \mathbb{P}(X_i < y) = \int_0^y \lambda e^{-\lambda t} \d t = 1-e^{-\lambda y}\). Notice also that \begin{align*} G(y) &= \mathbb{P}(Y < y) \\ &= \mathbb{P}(\max_i(X_i) < y) \\ &= \mathbb{P}(X_i < y \text{ for all }i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i < y) \\ &= \prod_{i=1}^n (1-e^{-\lambda y})\\ &= (1-e^{-\lambda y})^n \end{align*} as required.
2. \begin{align*} && \mathbb{P}(Y < L(\alpha)) &= \alpha \\ \Rightarrow && (1-e^{-\lambda L(\alpha)})^n &= \alpha \\ \Rightarrow && 1-e^{-\lambda L(\alpha)} &= \alpha^{\tfrac1n} \\ \Rightarrow && L(\alpha) &= -\frac{1}{\lambda}\ln \left (1-\alpha^{\tfrac1n} \right) \end{align*} Notice also: \begin{align*} && \mathbb{P}(Y > U(\alpha)) &= \alpha \\ \Rightarrow && 1 - (1-e^{-\lambda U(\alpha)})^n &= \alpha \\ \Rightarrow && U(\alpha) &= -\frac{1}{\lambda}\ln \left ( 1-(1-\alpha)^{\tfrac1n} \right) \end{align*}
3. \begin{align*} \lambda L(\alpha) &= -\ln \left (1-\alpha^{\tfrac1n} \right) \\ &= -\ln \left (1-e^{\tfrac1n \ln \alpha} \right) \\ &\approx - \ln \left ( 1 - 1 - \frac1n \ln \alpha\right) \tag{\(e^t \approx 1 + t\)} \\ &= -\ln \left ( \frac{1}{n} \ln \frac{1}\alpha \right) \\ &= - \ln \frac{1}{n} - \ln \left ( \ln \frac{1}{\alpha} \right )\\ &= \ln n - \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \end{align*} since if \(n\) is large, \(\frac{\ln \alpha}{n}\) is small.
4. The median is the value where \(\mathbb{P}(Y < M) = \frac12\), or in other words \(L(\frac12)\), but this is \(\approx \frac{\ln n - \ln (\ln 2)}{\lambda} \to \infty\). \begin{align*} && \lambda U(\alpha) &\approx \ln n - \ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right) \\ \Rightarrow && \lambda(U(\alpha) - L(\alpha)) &\approx -\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right)+ \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \\ \Rightarrow && U(\alpha) - L(\alpha) &\to \frac{1}{\lambda} \left ( \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right)-\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right ) \right) \end{align*} which doesn't depend on \(n\).
5. Suppose \(\alpha = \frac{1}{20}\) then \begin{align*} U(\alpha) - L(\alpha) &\approx \frac{1}{\lambda} \left (\ln \ln 20 - \ln \ln \frac{20}{19} \right) \\ &= \lambda^{-1} \left (\ln \ln 20 - \ln \ln (1 + \frac{1}{19}) \right) \\ &\approx \lambda^{-1} \left (\ln 3 - \ln \frac{1}{19} \right) \tag{\(\ln(1+t) \approx t\)} \\ &\approx \lambda^{-1} \ln 3 \cdot 19 \\ &\approx \lambda^{-1} (1 + 3) \\ &\approx 4\lambda^{-1} \end{align*} [Note that \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]

Solution:

1. \begin{align*} \sum_{r=1}^{m+1} \left (f(r) \sum_{s=r-1}^m g(s) \right) &= \sum_{r=1}^{m+1} \sum_{s=r-1}^m f(r)g(s) \\ &= \sum_{(r,s) \in \{(r,s) : 1 \leq r \leq m+1, 0 \leq s \leq m, s \geq r-1\}} f(r)g(s) \\ &= \sum_{(r,s) \in \{(r,s) : 0 \leq s \leq m, 1 \leq r \leq m+1, r \leq s+1\}} f(r)g(s) \\ &= \sum_{s=0}^m \sum_{r=1}^{s+1} f(r)g(s) \\ &= \sum_{s=0}^m \left ( g(s) \sum_{r=1}^{s+1} f(r) \right) \end{align*}
2. \(X_1\) takes the values \(0, 1\) with equal probabilities (since \(X_0 = 0\)). Therefore \(\mathbb{E}(X_1) = \frac12\).
   1. \begin{align*} \mathbb{P}(X_2 = 0) &= \mathbb{P}(X_2 = 0 | X_1 = 0) \mathbb{P}(X_1 = 0) + \mathbb{P}(X_2 = 0 | X_1 = 1) \mathbb{P}(X_1 = 1) \\ &= \frac12 \cdot \frac12 + \frac13 \cdot \frac12 \\ &= \frac5{12} \\ \\ \mathbb{P}(X_2 = 1) &= \mathbb{P}(X_2 = 1 | X_1 = 0) \mathbb{P}(X_1 = 0) + \mathbb{P}(X_2 = 1 | X_1 = 1) \mathbb{P}(X_1 = 1) \\ &= \frac12 \cdot \frac12 + \frac13 \cdot \frac12 \\ &= \frac5{12} \\ \\ \mathbb{P}(X_2 = 3) &= 1 - \mathbb{P}(X_2 = 0) - \mathbb{P}(X_2 = 1) \\ &= 1 - \frac{10}{12} = \frac16 \\ \\ \mathbb{E}(X_2) &= \frac{5}{12} + 2\cdot \frac{1}{6} \\ &= \frac34 \end{align*}
   2. \begin{align*} \mathbb{P}(X_n = 0) &= \sum_{s=0}^{n-1} \mathbb{P}(X_n = 0 | X_{n-1} = s)\mathbb{P}(X_{n-1} = s) \\ &= \sum_{s=0}^{n-1} \frac{1}{s+2}\mathbb{P}(X_{n-1} = s) \\ \end{align*} as required. (Where \(\mathbb{P}(X_n = 0 | X_{n-1} = s) = \frac{1}{s+2}\) since if \(X_{n-1} = s\) there are \(0, 1, \ldots, s + 1\) values \(X_n\) can take with equal chance (ie \(s+2\) different values). \begin{align*} \mathbb{P}(X_n = r) &= \sum_{s=0}^{n-1} \mathbb{P}(X_n = r | X_{n-1} = s)\mathbb{P}(X_{n-1} = s) \\ &= \sum_{s=r-1}^{n-1} \frac{\mathbb{P}(X_{n-1}=s)}{s+2} \end{align*}
   3. \begin{align*} \mathbb{E}(X_n) &= \sum_{r=1}^{n} r \cdot \mathbb{P}(X_n = r) \\ &= \sum_{r=1}^{n} r \cdot \sum_{s=r-1}^{n-1} \frac{\mathbb{P}(X_{n-1}=s)}{s+2} \\ &= \sum_{s=0}^{n-1} \frac{\mathbb{P}(X_{n-1}=s)}{s+2} \sum_{r=1}^{s+1} r \\ &= \sum_{s=0}^{n-1} \frac{\mathbb{P}(X_{n-1}=s)}{s+2} \frac{(s+1)(s+2)}{2} \\ &= \frac12 \sum_{s=0}^{n-1} (s+1)\mathbb{P}(X_{n-1}=s) \\ &= \frac12 \sum_{s=0}^{n-1} s\mathbb{P}(X_{n-1}=s) + \frac12 \sum_{s=0}^{n-1} \mathbb{P}(X_{n-1}=s) \\\\ &= \frac12 \left ( \mathbb{E}(X_{n-1}) + 1 \right) \end{align*} Suppose \(\mathbb{E}(X_n) = 1-2^{-n}\), then notice that this expression matches for \(n = 0, 1, 2\) and also: \(\frac12(1 - 2^{-n} + 1) = 1-2^{-n-1}\) satisfies the recusive formula. Therefore by induction (or similar) we can show that \(\mathbb{E}(X_n) = 1- 2^{-n}\).

Solution: \(y = (-\tan \theta)(x-1)+k\) so when \(x = 0\), \(y = k + \tan \theta\), so \(Y = (0, k+\tan \theta)\). When \(y = 0\), \(x = 1 + \frac{k}{\tan \theta}\)

1. \(A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)\) Notice that \(x + \frac{k^2}{x} \geq 2 k\) by AM-GM, so the minimum is \(k + \frac12 \cdot 2k = 2k\)
2. \(\,\) \begin{align*} L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\ &= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta) \end{align*} \begin{align*} && \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\ \Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\ \Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\ &&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\ &&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\ &&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\ \Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\ &&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\ &&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2}{1-\sin \alpha} \end{align*}

Solution: \begin{align*} && \frac{\d y}{\d x} &= \frac{\frac{\d y}{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{6t^2}{6t} = t \\ \Rightarrow && \frac{y-2p^3}{x - 3p^2} &= p \\ \Rightarrow && y &= px-3p^3+2p^3 \\ && y &= px - p^3 \end{align*} The two lines will be \begin{align*} && y &= px - p^3 \\ && y &= qx - q^3 \\ \Rightarrow && p^3-q^3 &= (p-q)x \\ \Rightarrow && x &= p^2+pq+q^2 \\ && y &= p(p^2+pq+q^2)-p^3 \\ &&&= pq(p+q) \\ && (x,y) &= (p^2+pq+q^2,pq(p+q)) \\ \end{align*} If the tangents are \(\perp\) then \(pq=-1\), so we have \begin{align*} && (x,y) &= (p^2+2pq+q^2-pq, pq(p+q)) \\ &&&= ((p+q)^2-1, -(p+q)) \\ &&&= (u^2-1, -u) \end{align*} We have \(x = y^2+1\) and \(\left ( \frac{x}{3} \right)^3 = \left ( \frac{y}{2}\right)^2 \Rightarrow y^2 = \frac{4}{27}x^3\) so \begin{align*} && 0 &= \frac{4}{27}x^3-x+1 \\ &&0&=4x^3-27x+27 \\ &&&= (x+3)(2x-3)^2 \end{align*} So we have the points \((x,y) = \left (\frac32, \pm\frac{1}{\sqrt{2}}\right)\)

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Solution: \begin{align*} \int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{1 - \sin^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{\cos^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \sec^2 x - \sec x \tan x dx \\ &= \left [\tan x-\sec x \right]_0^{\frac{1}{4}\pi} \\ &= 2 - \frac{1}{\sqrt{2}} \end{align*} \begin{align*} \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} \d x &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1-\sec x}{1 - \sec^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{\sec x-1}{\tan^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \cot x \cosec x-\cot^2 x\d x \\ &= \left [ -\cosec x +x+\cot x\right]_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \\ &= \l -\frac{2}{\sqrt3}+\frac{\pi}{3}+\frac{1}{\sqrt{3}}\r - \l-\sqrt{2}+\frac{\pi}{4}+1 \r \\ &= \frac{\pi}{12}-\frac{1}{\sqrt{3}}+\sqrt{2}-1 \end{align*} \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{(1-\sin^2 x)^2} \d x \\ &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{\cos^4 x} \d x \\ \end{align*} Splitting this up into: \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{-2\sin x}{\cos^4 x} \d x &= -\frac23 \left [ \frac{1}{\cos^3 x}\right]_0^{\frac{1}{3}\pi} \\ &= -\frac{16}3+\frac23 \\ &= -\frac{14}3 \end{align*} and \begin{align*} && \int_0^{\frac{1}{3}\pi} \frac{1+\sin^2x}{\cos^4 x} \d x &= \int_0^{\frac{1}{3}\pi} (\sec^2 x + \tan^2 x) \sec^2 x \d x \\ &&&= \int_0^{\frac{1}{3}\pi} (1+ 2\tan^2 x) \sec^2 x \d x \\ u = \tan x, \d u = \sec^2 x \d x&&&= \int_0^{\sqrt{3}}(1+2u^2) \d u \\ &&&= \left [u + \frac23 u^3 \right]_0^{\sqrt{3}} \\ &&&= \sqrt{3} + 2\sqrt{3} \\ &&&= 3\sqrt{3} \end{align*} And so our complete integral is: \[ \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x = 3\sqrt{3} - \frac{14}3\]

Solution:

1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

Solution:

1. If \(\vec{PQ} = \vec{SR}\) we have a parallelogram. \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
2. If the four points lie in a plane then \((\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0\), so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
   1. \(\,\) \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is \(\frac14 (p+1+r, q+s+1, 2)\) which is clearly \(\frac12(1,1,1)\) iff those equations are true.
   2. \(\,\) \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have \(r = 1-p\), and \(-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p\) and \(q = \frac12 + p\) \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For \(PQRS\) to be a square \(\cos PQR = 0\), ie \(p = \frac14\) and so \((p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)\) and \(|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4\), notice that \(\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}\) (\(441 < 450\)) therefore the sides are at least as long as \(\frac{21}{20}\)

Solution:

1. \(\,\) \begin{align*} && y &= 9x^2 - 12x \cos \theta + 4 \\ &&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\ &&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta \end{align*} Therefore the minimum is \(4\sin^2 \theta\) when \(x = \frac23 \cos \theta\). \begin{align*} && y &= 12x^2 \sin \theta - 9x^4 \\ &&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2 \end{align*} Therefore the maximum is \(4\sin^2 \theta\) when \(x^2 = \frac23\sin \theta\) Therefore \begin{align*} && 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\ && \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta} \end{align*} Therefore the equality cases must be achieved in both cases, ie \(x = \frac23 \cos \theta\) and \(x^2 = \frac23 \sin \theta\) \begin{align*} && x^2 &= \frac49\cos^2 \theta \\ &&&= \frac49(1-\sin^2 \theta) \\ &&&= \frac49(1-\frac94 x^2) \\ \Rightarrow && 2x^2 &= \frac49 \\ \Rightarrow && x &= \pm \frac{\sqrt{2}}3\\ \Rightarrow && \cos \theta &=\pm \frac32 \frac{\sqrt{2}}3 \\ &&&= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\ \Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right), \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right) \end{align*}
2. Sketching we obtain, noticing we can find the turning point by: \begin{align*} && \frac{x^2}{x-\theta} &= \lambda \\ \Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\ \Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta \\ \Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0 \end{align*}
   

   + - ↺
   Notice that \(\sin^2 \theta \cos^2 x \leq 1\) and \(1 + cos^2 \theta \sin^2 x \geq 1\) and therefore we must have the inequality desired. \begin{align*} && \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\ \text{both}=0: && x = 0 &, \sin \theta = 0 \\ \text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\ && 1 &= \cos^2 2 \theta \\ &&&= (1-2 \sin^2 \theta)^2 \\ &&&= 1 \\ \Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right) \end{align*}