Problems

A plane is inclined at an angle \(\arctan \frac34\) to the horizontal and a small, smooth, light pulley~\(P\) is fixed to the top of the plane. A string, \(APB\), passes over the pulley. A particle of mass~\(m_1\) is attached to the string at \(A\) and rests on the inclined plane with \(AP\) parallel to a line of greatest slope in the plane. A particle of mass \(m_2\), where \(m_2>m_1\), is attached to the string at \(B\) and hangs freely with \(BP\) vertical. The coefficient of friction between the particle at \(A\) and the plane is \( \frac{1}{2}\). The system is released from rest with the string taut. Show that the acceleration of the particles is \(\ds \frac{m_2-m_1}{m_2+m_1}g\). At a time \(T\) after release, the string breaks. Given that the particle at \(A\) does not reach the pulley at any point in its motion, find an expression in terms of \(T\) for the time after release at which the particle at \(A\) reaches its maximum height. It is found that, regardless of when the string broke, this time is equal to the time taken by the particle at \(A\) to descend from its point of maximum height to the point at which it was released. Find the ratio \(m_1 : m_2\). \noindent [Note that \(\arctan \frac34\) is another notation for \(\tan^{-1} \frac34\,\).]

Find the general solution of the differential equation \(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{xy}{x^2+a^2}\;\), where \(a\ne0\,\), and show that it can be written in the form \(\displaystyle y^2(x^2+a^2)= c^2\,\), where \(c\) is an arbitrary constant. Sketch this curve. Find an expression for \(\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} (x^2+y^2)\) and show that \[ \frac{\mathrm{d^2}}{\mathrm{d}x^2} (x^2+y^2) = 2\left(1 -\frac {c^2}{(x^2+a^2)^2} \right) + \frac{8c^2x^2}{(x^2+a^2)^3}\;. \]

1. Show that, if \(0 < c < a^2\), the points on the curve whose distance from the origin is least are \(\displaystyle \l 0\,,\;\pm \frac{c}{a}\r\).
2. If \(c > a^2\), determine the points on the curve whose distance from the origin is least.

Show Solution

---

Solution: \begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).

+ - ↺

\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}

1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)

+ - ↺

Five independent timers time a runner as she runs four laps of a track. Four of the timers measure the individual lap times, the results of the measurements being the random variables \(T_1\) to \(T_4\), each of which has variance \(\sigma^2\) and expectation equal to the true time for the lap. The fifth timer measures the total time for the race, the result of the measurement being the random variable \(T\) which has variance \(\sigma^2\) and expectation equal to the true race time (which is equal to the sum of the four true lap times). Find a random variable \(X\) of the form \(aT+b(T_1+T_2+T_3+T_4)\), where \(a\) and \(b\) are constants independent of the true lap times, with the two properties:

1. whatever the true lap times, the expectation of \(X\) is equal to the true race time;
2. the variance of \(X\) is as small as possible.

Prove that, if \(\vert \alpha\vert < 2\sqrt{2},\) then there is no value of \(x\) for which \begin{equation} x^2 -{\alpha}\vert x \vert + 2 < 0\;. \tag{\(*\)} \end{equation} Find the solution set of \((*)\) for \({\alpha}=3\,\). For \({\alpha} > 2\sqrt{2}\,\), the sum of the lengths of the intervals in which \(x\) satisfies \((*)\) is denoted by \(S\,\). Find \(S\) in terms of \({\alpha}\) and deduce that \(S < 2{\alpha}\,\). Sketch the graph of \(S\,\) against \(\alpha \,\).

Show Solution

---

Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)

+ - ↺

\(\,\) \setlength{\unitlength}{1cm}

1. An attempt is made to move a rod of length \(L\) from a corridor of width \(a\) into a corridor of width~\(b\), where \(a \ne b.\) The corridors meet at right angles, as shown in Figure 1 and the rod remains horizontal. Show that if the attempt is to be successful then $$ L \le a \cosec {\alpha} + b \sec {\alpha} \;, $$ where \({\alpha}\) satisfies $$ \tan^3\alpha =\frac a b \;. $$
2. An attempt is made to move a rectangular table-top, of width \(w\) and length \(l\), from one corridor to the other, as shown in the Figure 2. The table-top remains horizontal. Show that if the attempt is to be successful then $$ l\le a \cosec {\beta} + b \sec {\beta} -2w \cosec 2{\beta}, $$ where \({\beta}\) satisfies $$ w= \left(\frac {a -b \tan^3 \beta} {1 - \tan^2 \beta} \right) \cos \beta \;. $$

A bag contains \(b\) balls, \(r\) of them red and the rest white. In a game the player must remove balls one at a time from the bag (without replacement). She may remove as many balls as she wishes, but if she removes any red ball, she loses and gets no reward at all. If she does not remove a red ball, she is rewarded with \pounds 1 for each white ball she has removed. If she removes \(n\) white balls on her first \(n\) draws, calculate her expected gain on the next draw and show that %her expected total reward would be the same as before it is zero if \(\ds n = {b-r \over r+1}\,\). Hence, or otherwise, show that she will maximise her expected total reward if she aims to remove \(n\) balls, where \[ n = \mbox{ the integer part of } \ds {b + 1 \over r + 1}\;. \] With this value of \(n\), show that in the case \(r=1\) and \(b\) even, her expected total reward is \(\pounds {1 \over 4}b\,\), and find her expected total reward in the case \(r=1\) and \(b\) odd.

The triangle \(OAB\) is isosceles, with \(OA = OB\) and angle \(AOB = 2 \alpha\) where \(0< \alpha < {\pi \over 2}\,\). The semi-circle \(\mathrm{C}_0\) has its centre at the midpoint of the base \(AB\) of the triangle, and the sides \(OA\) and \(OB\) of the triangle are both tangent to the semi-circle. \(\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \ldots\) are circles such that \(\mathrm{C}_n\) is tangent to \(\mathrm{C}_{n-1}\) and to sides \(OA\) and \(OB\) of the triangle. Let \(r_n\) be the radius of \(\mathrm{C}_n\,\). Show that \[ \frac{r_{n+1}}{r_n} = \frac{1-\sin\alpha}{1+\sin\alpha}\;. \] Let \(S\) be the total area of the semi-circle \(\mathrm{C}_0\) and the circles \(\mathrm{C}_1\), \(\mathrm{C}_2\), \(\mathrm{C}_3\), \(\ldots\;\). Show that \[ S = {1 + \sin^2 \alpha \over 4 \sin \alpha} \, \pi r_0^2 \;. \] Show that there are values of \(\alpha\) for which \(S\) is more than four fifths of the area of triangle~\(OAB\).

A particle is projected with speed \(V\) at an angle \(\theta\) above the horizontal. The particle passes through the point \(P\) which is a horizontal distance \(d\) and a vertical distance \(h\) from the point of projection. Show that \[ T^2 -2kT + \frac{2kh}{d}+1=0\;, \] where \(T=\tan\theta\) and \(\ds k= \frac{V^2}{gd}\,\). %Derive an equation relating \(\tan \theta\), \(V\), \(g\), \(d\) and \(h\). Show that, if \(\displaystyle {kd > h + \sqrt {h^2 + d^2}}\;\), there are two distinct possible angles of projection. Let these two angles be \(\alpha\) and \(\beta\). Show that \(\displaystyle \alpha + \beta = \pi - \arctan ( {d/ h}) \,\).

Consider the equations \begin{alignat*}{2} ax-&y- \ z && =3 \;,\\ 2ax -&y -3z && = 7 \;,\\ 3ax-&y-5z && =b \;, \end{alignat*} where \(a\) and \(b\) are given constants.

1. In the case \(a=0\,\), show that the equations have a solution if and only if \(b=11\,\).
2. In the case \(a\ne0\) and \(b=11\,\) show that the equations have a solution with \(z=\lambda\) for any given number \(\lambda\,\).
3. In the case \(a=2\) and \(b=11\,\) find the solution for which \(x^2+y^2+z^2\) is least.
4. Find a value for \(a\) for which there is a solution such that \(x>10^6\) and \(y^2+z^2<1\,\).

Point \(B\) is a distance \(d\) due south of point \(A\) on a horizontal plane. Particle \(P\) is at rest at \(B\) at \(t=0\), when it begins to move with constant acceleration \(a\) in a straight line with fixed bearing~\(\beta\,\). Particle \(Q\) is projected from point \(A\) at \(t=0\) and moves in a straight line with constant speed \(v\,\). Show that if the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\), then \[ v^2 \ge ad \l 1 - \cos \beta \r\;. \] Show further that if \(v^2 >ad(1-\cos\beta)\) then the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\) before \(P\) has moved a distance \(d\,\).

Show that \((a+b)^2\le 2a^2+2b^2\,\). Find the stationary points on the curve $y=\big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\,$, where \(a\) and \(b\) are constants. State, with brief reasons, which points are maxima and which are minima. Hence prove that \[ \vert a\vert +\vert b \vert \le \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12} \le \big(2a^2+2b^2\big)^{\frac12} \;. \]

Show Solution

---

Solution: \begin{align*} && 2a^2+2b^2 &= a^2 + b^2 + (a^2+b^2) \\ &&&\underbrace{\geq}_{AM-GM} a^2+b^2+2\sqrt{a^2b^2} \\ &&&= a^2+b^2 + 2|a||b| \\ &&&\geq a^2+b^2 + 2ab \\ &&&= (a+b)^2 \end{align*} Assume \(a^2 \neq b^2\), otherwise the curve is a constant. \begin{align*} && y & = \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\\ && \frac{\d y}{\d \theta} &= \tfrac12 \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} \cdot (2 \sin \theta \cos \theta (b^2 - a^2)) + \tfrac12 (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12} \cdot (2 \sin \theta \cos \theta (a^2 - b^2) \\ &&&= \tfrac12\sin2 \theta (b^2 - a^2) \left ( \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} - (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12}\right) \\ \therefore \frac{\d y}{\d x} = 0 \Rightarrow && \sin 2\theta = 0 & \text{ or } a^2\cos^2\theta +b^2\sin^2\theta = a^2\sin^2\theta +b^2\cos^2\theta \\ \Rightarrow && \theta &= 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \\ && (a^2-b^2) \cos ^2\theta &= (a^2-b^2) \sin^2 \theta \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \end{align*} WLOG \(b^2 - a^2 > 0\), then the two parts of the derivative look like:

+ - ↺

And so \(\frac{\pi}{4}, \frac{3\pi}{4}, \cdots\) are maxima, and the others minima. The maxima are where \(\sin^2 \theta = \cos^2 \theta = \frac12\), so \(y(\frac{\pi}{4}) = 2\left ( \frac{a^2+b^2}{2} \right)^{\frac12} = (2a^2+2b^2)^{\frac12}\) and the maxima are \(\cos^2 \theta = 1, \sin^2 \theta = 0\) and vice versa, ie \(y = |a| + |b|\), therefore we obtain our desired result.

The lines \(l_1\), \(l_2\) and \(l_3\) lie in an inclined plane \(P\) and pass through a common point \(A\). The line \(l_2\) is a line of greatest slope in \(P\). The line \(l_1\) is perpendicular to \(l_3\) and makes an acute angle \(\alpha\) with \(l_2\). The angles between the horizontal and \(l_1\), \(l_2\) and \(l_3\) are \(\pi/6\), \(\beta\) and \(\pi/4\), respectively. Show that \(\cos\alpha\sin\beta = \frac12\,\) and find the value of \(\sin\alpha \sin\beta\,\). Deduce that \(\beta = \pi/3\,\). The lines \(l_1\) and \(l_3\) are rotated in \(P\) about \(A\) so that \(l_1\) and \(l_3\) remain perpendicular to each other. The new acute angle between \(l_1\) and \(l_2\) is \(\theta\). The new angles which \(l_1\) and \(l_3\) make with the horizontal are \(\phi\) and \(2\phi\), respectively. Show that \[ \tan^2\theta = \frac{3+\sqrt{13}}2\;. \]

A particle is projected from a point \(O\) on a horizontal plane with speed \(V\) and at an angle of elevation \(\alpha\). The vertical plane in which the motion takes place is perpendicular to two vertical walls, both of height \(h\), at distances \(a\) and \(b\) from \(O\). Given that the particle just passes over the walls, find \(\tan\alpha\) in terms of \(a\), \(b\) and \(h\) and show that \[ \frac{2V^2} g = \frac {ab} h +\frac{ (a+b)^2 h}{ab} \;. \] The heights of the walls are now increased by the same small positive amount \(\delta h\,\). A second particle is projected so that it just passes over both walls, and the new angle and speed of projection are \(\alpha +\delta \alpha \) and \(V+\delta V\), respectively. Show that \[ \sec^2 \alpha \, \delta \alpha \approx \frac {a+b}{ab}\,\delta h \;, \] and deduce that \(\delta \alpha >0\,\). Show also that \(\delta V\) is positive if \(h> ab/(a+b)\) and negative if \(h

A competitor in a Marathon of \(42 \frac38\) km runs the first \(t\) hours of the race at a constant speed of 13 km h\(^{-1}\) and the remainder at a constant speed of \(14 + 2t/T\) km h\(^{-1}\), where \(T\) hours is her time for the race. Show that the minimum possible value of \(T\) over all possible values of \(t\) is 3. The speed of another competitor decreases linearly with respect to time from 16~km~h\(^{-1}\) at the start of the race. If both of these competitors have a run time of 3 hours, find the maximum distance between them at any stage of the race.

A rigid straight beam \(AB\) has length \(l\) and weight \(W\). Its weight per unit length at a distance \(x\) from \(B\) is \(\alpha Wl^{-1} (x/l)^{\alpha-1}\,\), where \(\alpha\) is a positive constant. Show that the centre of mass of the beam is at a distance \(\alpha l/(\alpha+1)\) from \(B\). The beam is placed with the end \(A\) on a rough horizontal floor and the end \(B\) resting against a rough vertical wall. The beam is in a vertical plane at right angles to the plane of the wall and makes an angle of \(\theta\) with the floor. The coefficient of friction between the floor and the beam is \(\mu\) and the coefficient of friction between the wall and the beam is also \(\mu\,\). Show that, if the equilibrium is limiting at both \(A\) and \(B\), then \[ \tan\theta = \frac{1-\alpha \mu^2}{(1+\alpha)\mu}\;. \] Given that \(\alpha =3/2\,\) and given also that the beam slides for any \(\theta<\pi/4\,\) find the greatest possible value of \(\mu\,\).