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2012 Paper 1 Q10
D: 1500.0 B: 1500.0
constant acceleration free fall speed of sound vertical well kinematics surd manipulation approximation modelling

I stand at the top of a vertical well. The depth of the well, from the top to the surface of the water, is \(D\). I drop a stone from the top of the well and measure the time that elapses between the release of the stone and the moment when I hear the splash of the stone entering the water. In order to gauge the depth of the well, I climb a distance \(\delta\) down into the well and drop a stone from my new position. The time until I hear the splash is \(t\) less than the previous time. Show that \[ t = \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u\,, \] where \(u\) is the (constant) speed of sound. Hence show that \[ D = \tfrac12 gT^2\,, \] where \(T= \dfrac12 \beta + \dfrac \delta{\beta g}\) and \(\beta = t - \dfrac \delta u\,\). Taking \(u=300\,\)m\,s\(^{-1}\) and \(g=10\,\)m\,s\(^{-2}\), show that if \(t= \frac 15\,\)s and \(\delta=10\,\)m, the well is approximately \(185\,\)m deep.

Solution: \begin{align*} && s &= ut + \frac12at^2 \\ && D &= \frac12gt_D^2 \\ \Rightarrow && t_D &= \sqrt{\frac{2D}{g}} \\ \Rightarrow && t_{D-\delta} &= \sqrt{\frac{2(D-\delta}{g}} \end{align*} Therefore the difference in times of what I hear will be: \begin{align*} t &= \underbrace{\sqrt{\frac{2D}{g}}}_{\text{time for first stone to hit water}} + \underbrace{\frac{D}{u}}_{\text{time to hear about it}} - \left (\underbrace{\sqrt{\frac{2(D-\delta)}{g}}}_{\text{time for second stone to hit water}} + \underbrace{\frac{D-\delta}{u}}_{\text{time to hear about it}} \right) \\ &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \end{align*} \begin{align*} && t &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} + \frac \delta u \\ \Rightarrow && \beta &= \sqrt{\frac{2D}g} - \sqrt{\frac{2(D-\delta)}g} \\ && \beta^2 &= \frac{2D}{g} + \frac{2(D-\delta)}{g} - \frac{4}{g}\sqrt{D(D-g)} \\ &&&= \frac{4D}{g} - \frac{2\delta}{g} - \frac{4}{g} \sqrt{D(D-\delta)}\\ \Rightarrow && g\beta^2 &= 4D-2\delta -4\sqrt{D(D-\delta)}\\ \Rightarrow && (g \beta^2-4D+2\delta)^2 &= 16D(D-\delta) \\ \Rightarrow && g^2\beta^4 + 16D^2 + 4\delta^2 -8g\beta^2D+4g\beta^2 \delta -16D\delta &= 16D^2-16D\delta \\ \Rightarrow && 8g\beta^2D &= g\beta^4 +4\delta^2 +4g\beta^2 \delta \\ \Rightarrow && D &= \frac1{8g\beta^2} \left ( g^2\beta^4 +4\delta^2 +4g\beta^2 \delta\right) \\ &&&= \frac1{8g\beta^2} \left ( g\beta^2 +2\delta \right)^2 \\ &&&= \frac12g \left ( \frac{\beta}{2} + \frac{\delta}{g\beta} \right)^2 \end{align*} If \(u = 300, g = 10, t = \frac15, \delta = 10\), then \begin{align*} && \beta &= \frac15-\frac{10}{300}\\ &&&= \frac15 - \frac1{30} \\ &&&= \frac{1}{6}\\ && D &= \frac12 \cdot 10 \left ( \frac1{12} + 6 \right)^2 \\ &&&= 5\cdot (36 + 1 + \frac{1}{12^2}) \\ &&&\approx 37 \cdot 5 = 185 \end{align*}

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2011 Paper 1 Q7
D: 1500.0 B: 1500.0
differential equation first order separable partial fractions logarithmic approximation flow rate modelling integration

In this question, you may assume that \(\ln (1+x) \approx x -\frac12 x^2\) when \(\vert x \vert \) is small. The height of the water in a tank at time \(t\) is \(h\). The initial height of the water is \(H\) and water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.

  1. Suppose that water leaks out at a rate proportional to the height of the water in the tank, and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that \[ \frac {\d h}{\d t } = k( \alpha^2 H -h)\,, \] for some positive constant \(k\). Deduce that the time \(T\) taken for the water to reach height \(\alpha H\) is given by \[ kT = \ln \left(1+\frac1\alpha\right)\,, \] and that \(kT\approx \alpha^{-1}\) for large values of \(\alpha\).
  2. Suppose that the rate at which water leaks out of the tank is proportional to \(\sqrt h\) (instead of \(h\)), and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that the time \(T'\) taken for the water to reach height \(\alpha H\) is given by \[ cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln \left(1+\frac1 {\sqrt\alpha} \right)\right)\, \] for some positive constant \(c\), and that \(cT'\approx \sqrt H\) for large values of \(\alpha\).

Solution:

  1. \begin{align*} \frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}} \end{align*}. We also know that when \(h = \alpha^2 H\), \(\frac{\d h}{\d t} = 0\), ie \(c - k \alpha^2 H = 0\) therefore: \[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \] \begin{align*} && \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\ && \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\ && - \ln |\alpha^2H -h| &= kt + C \\ t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\ \Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\ && kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\ &&&= \ln \frac{1+\alpha}{\alpha} \\ &&&= \ln \left (1 + \frac1{\alpha} \right) \\ &&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\ &&&\approx \alpha^{-1} \end{align*}
  2. \begin{align*} && \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\ \Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\ u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\ &&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}} \right)\right)\\ &&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}} \right)\right)}\\ &&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha} \right)\right) \\ &&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\ &&&= \sqrt{H} \end{align*} as required.

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1996 Paper 1 Q7
D: 1484.0 B: 1469.7
differential equation first order integrating factor exponential decay modelling separable equation constant coefficients

  1. At time \(t=0\) a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time \(t\) is \(y\). Show that there is a constant \(b < 1\) such that \(y=b^{t}.\)
  2. Suppose instead that the tank contains one unit of water at time \(t=0,\) but that in addition to water flowing out as described, water is added at a steady rate \(a>0.\) Show that \[ \frac{\mathrm{d}y}{\mathrm{d}t}-y\ln b=a, \] and hence find \(y\) in terms of \(a,b\) and \(t\).

Solution:

  1. Since water flows out a rate proportional to the water in the tank we must have \(\dot{y} = -ky\), ie \(y = Ae^{-k t}\). Since \(t = 0, y = 1\) we have \(y = e^{-kt} = (e^{-k})^t\), so call \(b = e^{-k}\) and we have the result. (Since \(k > 0 \Rightarrow b < 1\)
  2. Notice that \begin{align*} && \dot{y} &= -\underbrace{ky}_{\text{flow out}} + \underbrace{a}_{\text{flow in}} \\ &&&= y\ln b + a \\ \Rightarrow && \dot{y} - y \ln b &= a \\ \\ \text{CF}: && y &= Ae^{\ln b t} = Ab^t\\ \text{PI}: && y &= -\frac{a}{\ln b} \\ t = 0, y = 1: && 1 &= A-\frac{a}{\ln b} \\ \Rightarrow && y &= \frac{a}{\ln b} \left ( b^t - 1 \right)+b^t \end{align*}

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1993 Paper 1 Q9
D: 1500.0 B: 1484.0
differential equation first order linear integrating factor mixing problem modelling exponential decay particular integral

In the manufacture of Grandma's Home Made Ice-cream, chemicals \(A\) and \(B\) pour at constant rates \(a\) and \(b-a\) litres per second (\(0 < a < b\)) into a mixing vat which mixes the chemicals rapidly and empties at a rate \(b\) litres per second into a second mixing vat. At time \(t=0\) the first vat contains \(K\) litres of chemical \(B\) only. Show that the volume \(V(t)\) (in litres) of the chemical \(A\) in the first vat is governed by the differential equation \[ \dot{V}(t)=-\frac{bV(t)}{K}+a, \] and that \[ V(t)=\frac{aK}{b}(1-\mathrm{e}^{-bt/K}) \] for \(t\geqslant0.\) The second vat also mixes chemicals rapidly and empties at the rate of \(b\) litres per second. If at time \(t=0\) it contains \(L\) litres of chemical \(C\) only (where \(L\neq K\)), how many litres of chemical \(A\) will it contain at a later time \(t\)?

Solution: The total volume in the first vat at time \(t\) is always \(K\), since \(b\) litres per second are coming in and \(b\) litres per second are going out. \begin{align*} &&\frac{\d V}{\d t} &= \underbrace{a}_{\text{incoming chemical }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{\frac{V(t)}{K}}_{\text{fraction of outgoing which is }A} \\ &&&= a - b \frac{V}{K} \\ \Rightarrow && \int \frac{1}{a-b\frac{V}{K}}\d V &= \int \d t \\ && - \frac{K}{b} \ln |a - b \frac{V}{K}| &= t +C\\ (t,V) = (0,0): && -\frac{K}{b} \ln a &= C \\ \Rightarrow && 1-\frac{b}{a} \frac{V}{K} &= e^{-bt/K} \\ \Rightarrow && V &= \frac{aK}{b} (1 - e^{-bt/K}) \end{align*} \begin{align*} &&\frac{\d W}{\d t} &= \underbrace{b}_{\text{incoming volume}} \cdot \underbrace{\frac{a}{b} (1 - e^{-bt/K})}_{\text{incoming fraction }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{ \frac{W(t)}{L}}_{\text{fraction of outgoing which is }A} \\ &&&= a (1 - e^{-bt/K}) - b \frac{W}{L} \\ \Rightarrow && \frac{\d W}{\d t} + \frac{b}{L} W &= a (1-e^{-bt/K}) \\ && \frac{\d}{\d t} \left ( e^{b/L t} W\right) &= ae^{b/L t}(1-e^{-bt/K}) \\ \Rightarrow && W &= e^{-bt/L} \left ( \frac{aL}{b}e^{b/Lt} - \frac{a}{\frac{b}{L} - \frac{b}{K}}e^{b/L t - b/K} \right) + Ce^{-bt/L} \\ &&&= \frac{aL}{b} \left (1 - \frac{K}{K-L}e^{-b/Kt} \right)+ Ce^{-bt/L} \\ (t,W) = (0,0): && 0 &= \frac{aL}{b} \frac{-L}{K-L} + C \\ \Rightarrow && C &= \frac{aL^2}{b(K-L)} \\ \Rightarrow && W &= \frac{aL}{b} \left (1 - \frac{K}{K-L} e^{-bt/K} + \frac{L}{K-L} e^{-bt/L} \right) \end{align*}

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