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. 

   + - ↺
   \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 \(\mathbb{E}(X_n) = 1- 2^{-n}\).