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2025 Paper 2 Q7
The differential equation \[\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}\] describes the motion of a particle with position \(x(t)\) at time \(t\). At \(t = 0\), \(x = a\), where \(a > 0\).
- Solve the differential equation in the case where \(\frac{dx}{dt} = a^2\) when \(t = 0\). What happens to the particle as \(t\) increases from 0?
- Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 + p\) when \(t = 0\), where \(p > 0\). What happens to the particle as \(t\) increases from 0?
- Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 - q^2\) when \(t = 0\), where \(q > 0\). What happens to the particle as \(t\) increases from 0? Give conditions on \(a\) and \(q\) for the different cases which arise.
Solution: Let \(v = \frac{\d x}{\d t}\) and notice that \(\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}\). Also notice that: \begin{align*} && v \frac{\d v}{\d x} &= 2x v \\ \Rightarrow && \frac{\d v}{\d x} &= 2x \\ \Rightarrow && v &= x^2 + C \\ \Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\ \end{align*}
- When \(t = 0, \frac{\d x}{\d t} = a^2\) so \(C = 0\), therefore \(\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k\) and so \(k = a^{-1}\) and \(x = \frac{a}{1-at}\). As \(t\) increases from \(0\) the particle heads to infinity at an increasing rate, `reaching' infinity around \(t=\frac{1}{a}\)
- When \(t = 0, \frac{\d x}{\d t} = a^2 + p\) so \(C = p\). Therefore \(\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c\). When \(c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)\), so \begin{align*} && t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\ &&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\ \Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\ \Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\ \Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\ \Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)} \end{align*} The particle heads to \(\frac{\sqrt{p}}{a}\).
- When \(t = 0, \frac{\d x}{\d t} = a^2-q^2\) so \(C = -q^2\). Therefore \begin{align*} && \frac{\d x}{\d t} &= x^2 -q^2 \\ \Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\ &&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\ &&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\ &&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\ \Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\ \underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\ \Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\ \Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\ \Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}} \end{align*}
2018 Paper 2 Q10
A uniform elastic string lies on a smooth horizontal table. One end of the string is attached to a fixed peg, and the other end is pulled at constant speed \(u\). At time \(t=0\), the string is taut and its length is \(a\). Obtain an expression for the speed, at time \(t\), of the point on the string which is a distance \(x\) from the peg at time \(t\). An ant walks along the string starting at \(t=0\) at the peg. The ant walks at constant speed \(v\) along the string (so that its speed relative to the peg is the sum of \(v\) and the speed of the point on the string beneath the ant). At time \(t\), the ant is a distance \(x\) from the peg. Write down a first order differential equation for \(x\), and verify that \[ \frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,. \] Show that the time \(T\) taken for the ant to reach the end of the string is given by \[uT = a(\e^k-1)\,,\] where \(k=u/v\). On reaching the end of the string, the ant turns round and walks back to the peg. Find in terms of \(T\) and \(k\) the time taken for the journey back.
Solution: Points always maintain a constant fraction of the distance from the start, so the point distance \(x\) from the start at time \(t\) is moving with speed \(\frac{x}{a+ut} u\) The point is moving with speed \(v+\frac{x}{a+ut} u\) or in other words \begin{align*} && \frac{\d x}{\d t} &= v + \frac{x}{a+ut}u \\ \Rightarrow && \frac{\d x }{\d t} - \frac{u}{a+ut} x &= v \\ \Rightarrow && \frac{1}{a+ut} \frac{\d x}{\d t} - \frac{u}{(a+ut)^2} x &= \frac{1}{a+ut} v\\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{x}{a+ut} \right) &= \frac{v}{a+ut} \\ \Rightarrow && \frac{x}{a+ut} &=\frac{v}{u} \ln (a + ut) + C \\ t = 0, x = 0: && 0 &= \frac{v}{u} \ln a + C \\ \Rightarrow && x &= (a+ut) \frac{v}{u} \ln \left ( \frac{a+ut}{a} \right) \\ \\ \Rightarrow && 1 &= \frac{v}{u} \ln \left ( \frac{a+uT}{a} \right) \\ \Rightarrow && e^k &= 1 + \frac{uT}{a} \\ \Rightarrow && uT &= a(e^k-1) \end{align*} On the return journey, we have \begin{align*} && \frac{\d x}{\d t} &= \frac{x}{a+ut}u - v \\ \Rightarrow && \frac{\d x}{\d t} - \frac{u}{a+ut} x &= - v \\ \Rightarrow && \frac{\d }{\d x} \left ( \frac{x}{a+ut} \right) &= -\frac{v}{a+ut} \\ \Rightarrow &&f &= -\frac{v}{u} \ln(a+ut) + K \\ t = T, f = 1: && 1 &= -\frac{v}{u}\ln(a+uT) + K \\ \Rightarrow && f &= 1+\frac{v}{u}\ln \left ( \frac{a+uT}{a+ut} \right) \\ \Rightarrow && 0 &= 1+\frac{v}{u} \ln \left ( \frac{a+uT}{a+uT_2} \right) \\ \Rightarrow && e^k &= \frac{a+uT_2}{a+uT}\\ \Rightarrow && uT_2 &= (a+uT)e^k - a \\ \Rightarrow && T_2 - T &= \frac{1}{u} \left ( (a+uT)e^k - a - uT\right) \\ &&&= \frac{1}{u} \left ((a+a(e^k-1))e^k-a-a(e^k-1) \right) \\ &&&= \frac{1}{u} \left (ae^{2k} -ae^k \right) \\ &&&= \frac{ae^k}{u} \left ( e^k-1 \right) \\ &&&= Te^k \end{align*}
2008 Paper 2 Q7
- By writing \(y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}\), where \(u\) is a function of \(x\), find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = xy + \frac x {1+x^2} \] for which \(y=1\) when \(x=0\).
- Find the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^2y + \frac {x^2 } {1+x^3} \] for which \(y=1\) when \(x=0\).
- Give, without proof, a conjecture for the solution of the equation \[ \frac 1 y \frac{\d y} {\d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n} \] for which \(y=1\) when \(x=0\), where \(n\) is an integer greater than 1.
2004 Paper 2 Q8
Let \(x\) satisfy the differential equation $$ \frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n} $$ and the condition \(x=0\) when \(t=0 \,\).
- Solve the equation in the case \(n=1\) and sketch the graph of the solution for \(t > 0 \,\).
- Prove that \(1-x < (1-x^2)^{1/2} \) for \(0 < x < 1 \,\). Use this result to sketch the graph of the solution in the case \(n=2\) for \(0 < t < \frac12 \pi \,\), using the same axes as your previous sketch. By setting \(x=\sin y\,\), solve the equation in this case.
- Use the result (which you need not prove) \[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \] to sketch, without solving the equation, the graph of the solution of the equation in the case \(n=3\) using the same axes as your previous sketches. Use your sketch to show that \(x=1\) at a value of \(t\) less than \(\frac12 \pi \,\).
Solution:
- \(\,\) \begin{align*}
&& \dot{x} &= (1-x) \\
\Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\
\Rightarrow && -\ln |1-x| &= t + C \\
t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\
\Rightarrow && -\ln|1-x| &= t \\
\Rightarrow && 1-x&= e^{-t} \\
\Rightarrow && x &= 1-e^{-t}
\end{align*}
- Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
\begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
- \(\,\)
We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)
2004 Paper 3 Q11
Particles \(P\), of mass \(2\), and \(Q\), of mass \(1\), move along a line. Their distances from a fixed point are \(x_1\) and \(x_2\), respectively where \(x_2>x_1\,\). Each particle is subject to a repulsive force from the other of magnitude \(\displaystyle {2 \over z^3}\), where \(z = x_2-x_1 \,\). Initially, \(x_1=0\), \(x_2 = 1\), \(Q\) is at rest and \(P\) moves towards \(Q\) with speed 1. Show that \(z\) obeys the equation \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = {3 \over z^3}\). By first writing \(\displaystyle {\mathrm{d}^2 z \over \mathrm{d}t^2} = v {\mathrm{d}v \over \mathrm{d}z} \,\), where \(\displaystyle v={\mathrm{d}z \over \mathrm{d}t}\,\), show that \(z=\sqrt{4t^2-2t+1}\,\). By considering the equation satisfied by \(2x_1+x_2\,\), find \(x_1\) and \(x_2\) in terms of \(t \,\).
Solution: \begin{align*} \text{N2}: && 2\ddot{x}_1 &= -\frac{2}{(x_2-x_1)^3}\\ \text{N2}: && \ddot{x}_2 &= \frac{2}{(x_2-x_1)^3}\\ \Rightarrow && \ddot{x}_2 - \ddot{x}_1 &= \frac{3}{(x_1-x_2)^3} \\ \Rightarrow && \frac{\d^2 z}{\d t^2} &= \frac{3}{z^3} \\ \Rightarrow && v \frac{\d v}{\d z} &= \frac{3}{z^3} \\ \Rightarrow && \int v \d v &= \int \frac{3}{z^3} \d z \\ \Rightarrow && \frac{v^2}{2} &= -\frac{3}{2}z^{-2} + C \\ \Rightarrow && v^2 &= -3 z^{-2} + C' \\ t=0,z=1,v=-1: && 1 &= -3+C \Rightarrow C = 4 \\ \Rightarrow && \frac{\d z}{\d t} &= -\sqrt{4-3z^{-2}} \\ \Rightarrow && \int \d t &= -\int \frac{1}{\sqrt{4-3z^{-2}}} \d z \\ \Rightarrow && t &= \int \frac{z}{\sqrt{4z^2-3}} \d z \\ \Rightarrow && t &= -\frac14\sqrt{4z^2-3} + C \\ t=0, z = 1: && 0 &= -\frac14+C \\ \Rightarrow && C &= \frac14\\ \Rightarrow && 4t -1 &= -\sqrt{4z^2-3} \\ \Rightarrow && 16t^2+1-8t &= 4z^2-3 \\ \Rightarrow && z &= \sqrt{4t^2-2t+1} \end{align*} \begin{align*} && 2\ddot{x}_1 + \ddot{x}_2 &= 0 \\ \Rightarrow && 2x_1+x_2 &= At + B \\ t = 0, v = -1: && 2x_1+x_2 &= -t+1 \\ \\ \Rightarrow && x_2-x_1 &= \sqrt{4t^2-2t+1}\\ && 2x_1+x_2 &= 1-t \\ \Rightarrow && x_1 &= \frac13 \left (1-t-\sqrt{4t^2-2t+1} \right) \\ && x_2 &= \frac13(1-t + \sqrt{4t^2-2t+1}) \end{align*} This method of considering the relative position and considering the motion of the centre of mass is extremely common for solving systems of particles problems.
2003 Paper 3 Q10
A particle moves along the \(x\)-axis in such a way that its acceleration is \(kx \dot{x}\,\) where \(k\) is a positive constant. When \(t = 0\), \(x = d\) (where \(d>0\)) and \(\dot{x} =U\,\).
- Find \(x\) as a function of \(t\) in the case \(U = kd^2\) and show that \(x\) tends to infinity as \(t\) tends to \(\displaystyle \frac{\pi }{2 dk}\,\).
- If \(U < 0\), find \(x\) as a function of \(t\) and show that it tends to a limit, which you should state in terms of \(d\) and \(U\,\), as \(t\) tends to infinity.
Solution:
- \(\,\) \begin{align*} && \ddot{x} &= kx \dot{x} \\ \Rightarrow && \frac{\d v}{\d x} \dot{x} &= k x \dot{x} \\ \Rightarrow && \int \d v &= \int k x \d x \\ \Rightarrow && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = kd^2: && kd^2 &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2+d^2) \\ \Rightarrow && \frac{\d x}{\d t} &= \frac12k(x^2+d^2) \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2+d^2)} \d x \\ &&&= \frac{2}{kd}\tan^{-1} \frac{x}{d} \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d} + C' \\ t = 0, x = d: && 0 &= \frac{\pi}{2kd} + C' \\ \Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d}-\frac{\pi}{2kd} \end{align*} As \(x \to \infty\), \(t \to \frac{2}{kd} \frac{\pi}{2} - \frac{\pi}{2kd} = \frac{\pi}{2kd} \)
- \(\,\) \begin{align*} && v &= \frac12kx^2 + C \\ t=0, x = d, \dot{x} = U && U &= \frac12kd^2 + C \\ \Rightarrow && \dot{x} &= \frac12k(x^2-d^2)+U \\ \Rightarrow && \frac{\d x}{\d t} &=\frac12k(x^2-d^2)+U \\ \Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2-d^2)+U} \d x \\ && &=\frac{2}{k} \int \frac{1}{x^2-d^2+\frac{2U}k} \d x \\ &&&= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} + C'' \\ t = 0, \dot{x} = d: && 0 &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-d}{d+\sqrt{d^2-\frac{2U}k}} + C'' \\ \Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \left ( \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \frac{d+\sqrt{d^2-\frac{2U}k}}{ \sqrt{d^2-\frac{2U}k}-d} \right ) \end{align*} as \(t \to \infty\) the denominator needs to head to \(0\), ie \(x \to -\sqrt{d^2-\frac{2U}k}\)
2002 Paper 3 Q6
Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points
- \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
- \(\l 0, \, \frac{1}{2}\sqrt5 \r\).
Solution: \begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}
- Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
- Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)
1994 Paper 3 Q4
Find the two solutions of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}=4y \] which pass through the point \((a,b^{2}),\) where \(b\neq0.\) Find two distinct points \((a_{1},1)\) and \((a_{2},1)\) such that one of the solutions through each of them also passes through the origin. Show that the graphs of these two solutions coincide and sketch their common graph, together with the other solutions through \((a_{1},1)\) and \((a_{2},1)\). Now sketch sufficient members of the family of solutions (for varying \(a\) and \(b\)) to indicate the general behaviour. Use your sketch to identify a common tangent, and comment briefly on its relevance to the differential equation.
1992 Paper 3 Q4
A set of curves \(S_{1}\) is defined by the equation \[ y=\frac{x}{x-a}, \] where \(a\) is a constant which is different for different members of \(S_{1}.\) Sketch on the same axes the curves for which \(a=-2,-1,1\) and \(2\). A second of curves \(S_{2}\) is such that at each intersection between a member of \(S_{2}\) and a member of \(S_{1}\) the tangents of the intersecting curves are perpendicular. On the same axes as the already sketched members of \(S_{1},\) sketch the member of \(S_{2}\) that passes through the point \((1,-1)\). Obtain the first order differential equation for \(y\) satisfied at all points on all members of \(S_{1}\) (i.e. an equation connecting \(x,y\) and \(\mathrm{d}y/\mathrm{d}x\) which does not involve \(a\)). State the relationship between the values of \(\mathrm{d}y/\mathrm{d}x\) on two intersecting curves, one from \(S_{1}\) and one from \(S_{2},\) at their intersection. Hence show that the differential equation for the curves of \(S_{2}\) is \[ x=y(y-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}. \] Find an equation for the member of \(S_{2}\) that you have sketched.
1990 Paper 2 Q11
A disc is free to rotate in a horizontal plane about a vertical axis through its centre. The moment of inertia of the disc about this axis is \(mk^{2}.\) Along one diameter is a narrow groove in which a particle of mass \(m\) slides freely. At time \(t=0,\) the disc is rotating with angular speed \(\Omega,\) and the particle is at a distance \(a\) from the axis and is moving towards the axis with speed \(V\), where \(k^{2}V^{2}=\Omega^{2}a^{2}(k^{2}+a^{2}).\) Show that, at a later time \(t,\) while the particle is still moving towards the axis, the angular speed \(\omega\) of the disc and the distance \(r\) of the particle from the axis are related by \[ \omega=\frac{\Omega(k^{2}+a^{2})}{k^{2}+r^{2}}\qquad\mbox{ and }\qquad\frac{\mathrm{d}r}{\mathrm{d}t}=-\frac{\Omega r(k^{2}+a^{2})}{k(k^{2}+r^{2})^{\frac{1}{2}}}. \] Deduce that \[ k\frac{\mathrm{d}r}{\mathrm{d}\theta}=-r(k^{2}+r^{2})^{\frac{1}{2}}, \] where \(\theta\) is the angle through which the disc has turned at time \(t\). By making the substitution \(u=1/r\), or otherwise, show that \(r\sinh(\theta+\alpha)=k,\) where \(\sinh\alpha=k/a.\) Hence, or otherwise, show that the particle never reaches the axis.
1989 Paper 1 Q6
The normal to the curve \(y=\mathrm{f}(x)\) at the point \(P\) with coordinates \((x,\mathrm{f}(x))\) cuts the \(y\)-axis at the point \(Q\). Derive an expression in terms of \(x\), \(\mathrm{f}(x)\) and \(\mathrm{f}'(x)\) for the \(y\)-coordinate of \(Q\). If, for all \(x\), \(PQ=\sqrt{\mathrm{e}^{x^{2}}+x^{2}}\), find a differential equation satisfied by \(\mathrm{f}(x)\). If the curve also has a minimum point \((0,-2)\), find its equation.
Solution: The normal to the curve \(y = f(x)\) has gradient \(-\frac{1}{f'(x)}\) and so has equation: \begin{align*} && \frac{Y - f(x)}{X - x} &= -\frac{1}{f'(x)} \\ \Rightarrow && Y &= -\frac{1}{f'(x)}X + \frac{x}{f'(x)}+f(x) \end{align*} Hence the \(Q\) is \(\displaystyle \left (0, f(x) + \frac{x}{f'(x)} \right)\). \begin{align*} && |PQ|^2 &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && x^2 + e^{x^2} &= x^2 + \frac{x^2}{(f'(x))^2} \\ \Rightarrow && (f'(x))^2 &=x^2 e^{-x^2} \end{align*} Therefore \(f'(x) = \pm x e^{-x^2/2}\). If \(f(x)\) has a minimum at \((0,-2)\) then \(f''(0) > 0\), and \(f''(x) = \pm (e^{-x^2/2} - x^2e^{-x^2/2}) = \pm e^{-x^2/2}(1-x^2)\) so we should take the positive branch of the solution, ie \(f'(x) = xe^{-x^2/2}\). Therefore \(f(x) = - e^{-x^2/2}+C\). Since \(f(0) = -2\) we must have \(-2 = -1 + C\), ie \(C = -1\). Therefore \(f(x) = -1 - e^{-x^2/2}\)
1989 Paper 1 Q10
A spaceship of mass \(M\) is travelling at constant speed \(V\) in a straight line when it enters a force field which applies a resistive force acting directly backwards and of magnitude \(M\omega(v^{2}+V^{2})/v\), where \(v\) is the instantaneous speed of the spaceship, and \(\omega\) is a positive constant. No other forces act on the spaceship. Find the distance travelled from the edge of the force field until the speed is reduced to \(\frac{1}{2}V\). As soon as the spaceship has travelled this distance within the force field, the field is altered to a constant resistive force, acting directly backwards, whose magnitude is within 10% of that of the force acting on the spaceship immediately before the change. If \(z\) is the extra distance travelled by the spaceship before coming instantaneously to rest, determine limits between which \(z\) must lie.
Solution: Using Newton's second law, we have: \begin{align*} && -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\ \Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\ \Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\ &&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\ &&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\ &&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\ \Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r \end{align*}. The resistive force just before the field changes is \(M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega\). Therefor the constant resistive force is between \(\frac{11}4MV\omega\) and \(\frac{9}{4}MV \omega\) and acceleration is \(\frac{11}{4}V\omega, \frac{9}{4}V\omega\). Since \(v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}\) therefore \(z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]\)