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2003 Paper 2 Q8
It is given that \(y\) satisfies $$ {{\d y} \over { \d t}} + k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;, $$ where \(k\) is a constant, and \(y=A \) when \(t=0\,\), where \(A\) is a positive constant. Find \(y\) in terms of \(t\,\), \(k\) and \(A\,\). Show that \(y\) has two stationary values whose ratio is \((3/2)^{6k}\e^{-5{k}/2}.\) Describe the behaviour of \(y\) as \(t \to +\infty\) for the case where \(k> 0\) and for the case where \(k<0\,.\) In separate diagrams, sketch the graph of \(y\) for \(t>0\) for each of these cases.
Solution: \begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.
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 2 Q8
Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]
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 \(\pm x = \sqrt{y^2-1}-\frac12\)
2001 Paper 1 Q7
In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]
- Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
- Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
- Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).
Solution:
- \(\,\)
Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
- Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
- Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.
2001 Paper 1 Q8
Given that \(y=x\) and \(y=1-x^2\) satisfy the differential equation $$ \frac{\d^2 {y}}{\d x^2} + \p(x) \frac{\d {y}}{\d x} + \q(x) {y}=0\;, \tag{*} $$ show that \(\p(x)= -2x(1+x^2)^{-1}\) and \(\q(x) = 2(1+x^2)^{-1}\). Show also that \(ax+b(1-x^2)\) satisfies the differential equation for any constants \(a\) and \(b\). Given instead that \(y=\cos^2(\frac{1}{2}x^2)\) and \(y=\sin^2(\frac{1}{2}x^2)\) satisfy the equation \((*)\), find \(\p(x)\) and \(\q(x)\).
Solution: \begin{align*} && y &= x \\ && y' &= 1 \\ && y'' &= 0 \\ \Rightarrow && 0 &= 0 + p(x) + xq(x) \tag{1} \\ \\ && y &= 1-x^2 \\ && y' &= -2x \\ && y'' &= -2 \\ \Rightarrow && 0 &= -2 -2x p(x)+(1-x^2)q(x) \tag{2}\\ \\ 2x*(1) +(2): && 2 &= (2x^2+1-x^2) q(x) \\ \Rightarrow && q(x) &= 2(1+x^2)^{-1} \\ \Rightarrow && p(x) &= -2x(1+x^2)^{-1} \tag{by (1)} \end{align*} \begin{align*} && \frac{\d^2}{\d x^2} \left (a x + b(1-x^2) \right) + p(x) \frac{\d}{\d x} \left (a x + b(1-x^2) \right)+q(x) \left (a x + b(1-x^2) \right) \\ &&= a \frac{\d^2 x}{\d x^2} + b \frac{\d^2}{\d x^2} \left ( 1- x^2 \right) + ap(x) \frac{\d x}{ \d x} + bp(x) \frac{\d }{\d x} \left ( 1- x^2 \right) + aq(x) x + bq(x)(1-x^2) \\ &&= a \left (\frac{\d^2 x}{\d x^2}+ p(x) \frac{\d x}{ \d x} +q(x)x\right)+b \left ( \frac{\d^2}{\d x^2} \left ( 1- x^2 \right)+ p(x) \frac{\d }{\d x} \left ( 1- x^2 \right)+q(x)(1-x^2)\right) &= 0 \end{align*} \begin{align*} && y &= \cos^2(\tfrac12 x^2) = \frac12 \left (1 + \cos(x^2) \right) \\ && y' &= -x \sin(x^2) \\ && y'' &= -2x^2 \cos(x^2)-\sin(x^2) \\ \Rightarrow && 0 &= -2x^2 \cos(x^2)-\sin(x^2)+p(x)(-x \sin(x^2)) +\frac12 \left (1 + \cos(x^2) \right)q(x) \\ \Rightarrow && 2x^2\cos(x^2)+\sin(x^2) &= -x \sin(x^2) p(x) + \frac12(1 + \cos(x^2)) q(x) \tag{3}\\ \\ && y &= \sin^2(\tfrac12 x^2) = \frac12 \left ( 1 - \cos (x^2) \right) \\ && y' &= x\sin(x^2) \\ && y'' &= 2x^2 \cos(x^2)+\sin(x^2) \\ \Rightarrow && 0 &= 2x^2 \cos(x^2)+\sin(x^2) +p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x)\\ \Rightarrow && -2x^2 \cos(x^2)-\sin(x^2) &= p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x) \tag{4}\\ (3)+(4): && 0 &= q(x) \\ \Rightarrow && p(x) &= -\frac{2x^2 \cos(x^2)+\sin(x^2)}{x \sin(x^2)} \end{align*}
2001 Paper 2 Q5
The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)
2001 Paper 2 Q8
The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).
- Give an example of such a function.
- The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
- Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)
Solution:
- \(f(x) = \lfloor x \rfloor+1\)
- Clearly \(\displaystyle F(x) = \int_0^x f(t) \d t\), in particular: \begin{align*} && F(n) &= \int_0^n f(t) \d t \\ &&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\ &&&= n \int_{0}^1 f(t) \d t\\ &&&= n F(1) \end{align*}
- \(\,\) \begin{align*} && 0 &= \frac{\d y}{\d x} +f(x) y \\ \Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\ \Rightarrow && -F(x) & = \ln y + C \\ x=0,y=1: && C &= -F(0) \\ \Rightarrow && y &= \exp(F(0)-F(x)) \end{align*} Well this \(F(0)-F(x)\) is equivalent to \(-F(x)\) where \(F(0) = 0\), in particular \(F(n) = nF(1)\), so \(y(n) = e^{-nF(1)}\) which tends to zero as long as \(F(1) > 0\), but since \(f(x) > 0\) for all \(x\) this must be true.
2001 Paper 2 Q9
A particle of unit mass is projected vertically upwards with speed \(u\). At height \(x\), while the particle is moving upwards, it is found to experience a total force \(F\), due to gravity and air resistance, given by \(F=\alpha \e^{-\beta x}\), where \(\alpha\) and \(\beta\) are positive constants. Calculate the energy expended in reaching this height. Show that \[ F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;, \] where \(v\) is the speed of the particle, and explain why \( \alpha = \frac12 \beta u^2 +g\), where \(g\) is the acceleration due to gravity. Determine an expression, in terms of \(y\), \(g\) and \(\beta\), for the air resistance experienced by the particle on its downward journey when it is at a distance \(y\) below its highest point.
Solution: Considering the energy of the particle, we have initial kinetic energy of \(\frac12 u^2\) and final energy is \(\frac12 v^2\), the change in energy is the work done by the force, \begin{align*} &&\text{Work done against resistance} &= \text{loss in kinetic energy} \\ &&\int F \, \d x &= \int \alpha e^{-\beta x} \, \d x \\ &&&= \frac{\alpha}{\beta} \l 1 - e^{-\beta x} \r \\ &&&= \frac{1}{\beta} \l \alpha - F\r \\ &&&= \frac12 u^2 - \frac12 v^2 \\ \Rightarrow && F &= \frac12 \beta v^2 + \alpha - \frac12 \beta u^2 \end{align*} When \(v = 0\) there is no air resistance, ie \(F = g\), but \(g = 0 + \alpha - \frac12 \beta u^2 \Rightarrow \alpha = g + \frac12 \beta u^2\) \(F = \frac12 \beta v^2 + g\), ie air resistance is \(\frac12 \beta v^2\) Looking at forces acting on the particle when it's descending, \begin{align*} && v \frac{dv}{dx} &= g - \frac12 \beta v^2 \\ \Rightarrow && \frac{v}{g - \frac12 \beta v^2} \frac{dv}{dx} &= 1 \\ \Rightarrow && \int \frac{v}{g - \frac12 \beta v^2} \, dv &= \int dx \\ \Rightarrow && \frac1{\beta}\l\ln(g - \frac12\beta v^2) - \ln(g)\r &= y\\ \Rightarrow && \ln \l 1 - \frac12 \frac{\beta}{g}v^2 \r &= \beta y \\ \Rightarrow && \frac{g}{\beta} \l 1-e^{-\beta y} \r = \frac12 v^2 \end{align*} Since force is the rate of change of work, we can say that the force is \(ge^{-\beta y}\) and the air resistance is \(g \l 1-e^{-\beta y} \r\)
2001 Paper 3 Q7
Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]
Solution:
2000 Paper 2 Q8
- Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
- Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.
2000 Paper 2 Q9
In an aerobatics display, Jane and Karen jump from a great height and go through a period of free fall before opening their parachutes. While in free fall at speed \(v\), Jane experiences air resistance \(kv\) per unit mass but Karen, who spread-eagles, experiences air resistance \mbox{\(kv + (2k^2/g)v^2\)} per unit mass. Show that Jane's speed can never reach \(g/k\). Obtain the corresponding result for Karen. Jane opens her parachute when her speed is \(g/(3{k})\). Show that she has then been in free fall for time \(k^{-1}\ln (3/2)\). Karen also opens her parachute when her speed is \(g/(3{k})\). Find the time she has then been in free fall.
Solution: Looking at the forces on Jane, \(kv < g \Rightarrow v < \frac{g}{k}\). For Karen we have \begin{align*} kv + (2k^2/g)v^2 &< g\\ -g^2 + gkv + (2k^2)v^2 &< 0 \\ (2kv-g)(kv+g) &< 0\\ \Rightarrow v &< \frac{g}{2k} \end{align*} \begin{align*} && \dot{v} &= g - kv \\ \Rightarrow && \frac{\dot{v}}{g - kv} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv\\ && &= \int_0^{g/(3k)} \frac{1}{g - kv} dv \\ && &= \left [-\frac{1}{k} \ln \l g - kv \r \right ]_0^{g/(3k)} \\ && &= \frac{1}{k} \ln \l g \r - \frac{1}{k} \ln \l \frac{2}{3}g \r\\ &&&= \frac{1}{k} \ln \l \frac{3}{2} \r \end{align*} \begin{align*} && \dot{v} &= g - kv - (2k^2/g)v^2 \\ \Rightarrow && \frac{\dot{v}}{g - kv - (2k^2/g)v^2} &= 1 \\ \Rightarrow && T &= \int_0^{g/(3k)} \frac{1}{g - kv - (2k^2/g)v^2} dv \\ && &= \int_0^{g/(3k)} \frac{g}{(g-2kv)(kv+g)} dv\\ && &= \int_0^{g/(3k)} \l \frac{2}{3(g-2kv)} + \frac{1}{3(kv+g)} \r dv\\ && &= \left [ \l -\frac{1}{3k} \ln (g-2kv) + \frac{1}{3k}\ln(kv+g) \r \right ]_0^{g/(3k)} \\ && &= \left [ \l -\frac{1}{3k}\ln \l \frac{g}{3} \r + \frac{1}{3k}\ln \l \frac{4g}{3} \r \r \right ] - \left [- \frac1{3k} \ln(g) + \frac{1}{3k} \ln (g) \right ] \\ && &= \frac{1}{3k} \ln \l 4 \r \end{align*} NB: \(\sqrt[3]{4} \approx 1.58 > \frac{3}{2}\) so Karen has been in free-fall for longer, but not \emph{much} longer than Jane.
1999 Paper 2 Q9
In the \(Z\)--universe, a star of mass \(M\) suddenly blows up, and the fragments, with various initial speeds, start to move away from the centre of mass \(G\) which may be regarded as a fixed point. In the subsequent motion the acceleration of each fragment is directed towards \(G\). Moreover, in accordance with the laws of physics of the \(Z\)--universe, there are positive constants \(k_1\), \(k_2\) and \(R\) such that when a fragment is at a distance \(x\) from \(G\), the magnitude of its acceleration is \(k_1x^3\) if \(x < R\) and is \(k_2x^{-4}\) if \(x \ge R\). The initial speed of a fragment is denoted by \(u\).
- For \(x < R\), write down a differential equation for the speed \(v\), and hence determine \(v\) in terms of \(u\), \(k_1\) and \(x\) for \( x < R\).
- Show that if \(u < a\), where \(2a^2=k_1 R^4\), then the fragment does not reach a distance \(R\) from \(G\).
- Show that if \(u \ge b\), where $ 6b^2= 3k_1R^4 + 4k_2 /R^3, $ then from the moment of the explosion the fragment is always moving away from \(G\).
- If \(a < u < b\), determine in terms of \(k_2\), \(b\) and \(u\) the maximum distance from \(G\) attained by the fragment.
1998 Paper 1 Q8
Fluid flows steadily under a constant pressure gradient along a straight tube of circular cross-section of radius \(a\). The velocity \(v\) of a particle of the fluid is parallel to the axis of the tube and depends only on the distance \(r\) from the axis. The equation satisfied by \(v\) is \[\frac{1}{r}\frac{{\mathrm d}\ }{{\mathrm d}r} \left(r\frac{{\mathrm d}v}{{\mathrm d}r}\right) =-k,\] where \(k\) is constant. Find the general solution for \(v\). Show that \(|v|\rightarrow\infty\) as \(r\rightarrow 0\) unless one of the constants in your solution is chosen to be~\(0\). Suppose that this constant is, in fact, \(0\) and that \(v=0\) when \(r=a\). Find \(v\) in terms of \(k\), \(a\) and \(r\). The volume \(F\) flowing through the tube per unit time is given by \[F=2\pi\int_{0}^{a}rv\,{\mathrm d}r. \] Find \(F\).
1998 Paper 2 Q10
A wedge of mass \(M\) rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle \(\alpha\) to the horizontal. A particle of mass \(m\) slides down the face of the wedge, starting from rest. At a later time \(t\), the speed \(V\) of the wedge, the speed \(v\) of the particle and the angle \(\beta\) of the velocity of the particle below the horizontal are as shown in the diagram.
- \(V\sin\alpha=v\sin(\beta-\alpha)\);
- \(\tan\beta=(1+m/M)\tan\alpha\);
- \(2gy=v^2(M+m\cos^2\beta)/M\).