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2012 Paper 3 Q11
D: 1700.0 B: 1500.0

One end of a thin heavy uniform inextensible perfectly flexible rope of length \(2L\) and mass \(2M\) is attached to a fixed point \(P\). A particle of mass \(m\) is attached to the other end. Initially, the particle is held at \(P\) and the rope hangs vertically in a loop below \(P\). The particle is then released so that it and a section of the rope (of decreasing length) fall vertically as shown in the diagram.

\psset{xunit=1.0cm,yunit=0.9cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(0.13,-0.26)(3.26,5.51) \psline(1,5)(3,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(1.52,0)(1.52,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(2.53,3.2)(2.53,5) \psline(2.1,3.18)(2.06,0.25) \psline(2,5)(2.02,0.26) \psline(2.02,0.26)(2.03,0) \psline(2.03,0)(2.06,0.25) \rput[tl](1.94,5.45){\(P\)} \rput[tl](2.6,4.25){\(x\)} \rput[tl](0.2,2.85){\(L+\frac{1}{2}x\)} \begin{scriptsize} \psdots[dotsize=4pt 0,dotstyle=*](2.1,3.18) \end{scriptsize} \end{pspicture*}
You may assume that each point on the moving section of the rope falls at the same speed as the particle. Given that energy is conserved, show that, when the particle has fallen a distance \(x\) (where \(x< 2L\)), its speed \(v\) is given by \[ v^2 = \frac { 2g x \big( mL +ML - \frac14 Mx)}{mL +ML - \frac12 Mx}\,. \] Hence show that the acceleration of the particle is \[ g + \frac{ Mgx\big(mL+ML- \frac14 Mx\big)}{2\big(mL +ML -\frac12 Mx\big)^2}\, \,.\] Deduce that the acceleration of the particle after it is released is greater than \(g\).

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2012 Paper 3 Q12
D: 1700.0 B: 1469.4

  1. A point \(P\) lies in an equilateral triangle \(ABC\) of height 1. The perpendicular distances from \(P\) to the sides \(AB\), \(BC\) and \(CA\) are \(x_1\), \(x_2\) and \(x_3\), respectively. By considering the areas of triangles with one vertex at \(P\), show that \(x_1+x_2+x_3=1\). Suppose now that \(P\) is placed at random in the equilateral triangle (so that the probability of it lying in any given region of the triangle is proportional to the area of that region). The perpendicular distances from \(P\) to the sides \(AB\), \(BC\) and \(CA\) are random variables \(X_1\), \(X_2\) and \(X_3\), respectively. In the case \(X_1= \min(X_1,X_2,X_3)\), give a sketch showing the region of the triangle in which \(P\) lies. Let \(X= \min(X_1,X_2,X_3)\). Show that the probability density function for \(X\) is given by \[ \f(x) = \begin{cases} 6(1-3x) & 0 \le x \le \frac13\,, \\ 0 & \text{otherwise}\,. \end{cases} \] Find the expected value of \(X\).
  2. A point is chosen at random in a regular tetrahedron of height 1. Find the expected value of the distance from the point to the closest face. \newline [The volume of a tetrahedron is \(\frac13 \times \text{area of base}\times\text{height}\) and its centroid is a distance \(\frac14\times \text{height}\) from the base.]

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2012 Paper 3 Q13
D: 1700.0 B: 1484.0

  1. The random variable \(Z\) has a Normal distribution with mean \(0\) and variance \(1\). Show that the expectation of \(Z\) given that \(a < Z < b\) is \[ \frac{\exp(- \frac12 a^2) - \exp(- \frac12 b^2) } {\sqrt{2\pi\,} \,\big(\Phi(b) - \Phi(a)\big)}, \] where \(\Phi\) denotes the cumulative distribution function for \(Z\).
  2. The random variable \(X\) has a Normal distribution with mean \(\mu\) and variance \(\sigma^2\). Show that \[ \E(X \,\vert\, X>0) = \mu + \sigma \E(Z \,\vert\,Z > -\mu/\sigma). \] Hence, or otherwise, show that the expectation, \(m\), of \(\vert X\vert \) is given by \[ m= \mu \big(1 - 2 \Phi(- \mu / \sigma)\big) + \sigma \sqrt{2 / \pi}\; \exp(- \tfrac12 \mu^2 / \sigma^2) \,. \] Obtain an expression for the variance of \(\vert X \vert\) in terms of \(\mu \), \(\sigma \) and \(m\).

Solution:

  1. \(\,\) \begin{align*} && \mathbb{E}(Z| a < Z < b) &= \mathbb{E}(Z\mathbb{1}_{(a,b)}) /\mathbb{E}(\mathbb{1}_{(a,b)}) \\ &&&= \int_a^b z \phi(z) \d z \Big / (\Phi(b) - \Phi(a)) \\ &&&= \frac{\int_a^b \frac{1}{\sqrt{2 \pi}}z e^{-\frac12 z^2} \d z}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left [-e^{-\frac12 z^2} \right]_a^b}{\Phi(b) - \Phi(a)} \\ &&&= \frac{\frac1{\sqrt{2\pi}} \left (e^{-\frac12 a^2}-e^{-\frac12 b^2} \right)}{\Phi(b) - \Phi(a)} \\ \end{align*}
  2. \(\,\) \begin{align*} && \mathbb{E}(X |X > 0) &= \mathbb{E}(\mu + \sigma Z | \mu + \sigma Z > 0) \\ &&&= \mathbb{E}(\mu + \sigma Z | Z > -\tfrac{\mu}{\sigma}) \\ &&&= \mathbb{E}(\mu| Z > -\tfrac{\mu}{\sigma})+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ &&&= \mu+ \sigma \mathbb{E}(Z | Z > -\tfrac{\mu}{\sigma})\\ \end{align*} Hence \begin{align*} &&\mathbb{E}(|X|) &= \mathbb{E}(X | X > 0)\mathbb{P}(X > 0) - \mathbb{E}(X | X < 0)\mathbb{P}(X < 0) \\ &&&=\left ( \mu+ \sigma \mathbb{E}(Z | Z > -\mu /\sigma)\right)(1-\Phi(-\mu/\sigma)) - \left ( \mu+ \sigma \mathbb{E}(Z | Z < -\mu /\sigma)\right)\Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2\pi}(1-\Phi(-\mu/\sigma))}(1-\Phi(-\mu/\sigma)) + \sigma \frac{e^{-\frac12\mu^2/\sigma^2}}{\sqrt{2 \pi} \Phi(-\mu/\sigma)} \Phi(-\mu/\sigma) \\ &&&= \mu(1 - 2\Phi(-\mu/\sigma)) + \sigma \sqrt{\frac{2}{\pi}} \exp(-\tfrac12 \mu^2/\sigma^2) \end{align*} Finally, \begin{align*} && \textrm{Var}(|X|) &= \mathbb{E}(|X|^2) - [\mathbb{E}(|X|)]^2 \\ &&&= \mu^2 + \sigma^2 - m^2 \end{align*}

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