5 problems found
A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).
A pain-killing drug is injected into the bloodstream. It then diffuses into the brain, where it is absorbed. The quantities at time \(t\) of the drug in the blood and the brain respectively are \(y(t)\) and \(z(t)\). These satisfy \[ \dot y = - 2(y-z)\,, \ \ \ \ \ \ \ \dot z = - \dot y -3z\, , \] where the dot denotes differentiation with respect to \(t\). Obtain a second order differential equation for \(y\) and hence derive the solution \[ y= A\e^{-t} + B\e ^{-6t}\,, \ \ \ \ \ \ \ z= \tfrac12 A \e^{-t} - 2 B \e^{-6t}\,, \] where \(A\) and \(B\) are arbitrary constants. \begin{questionparts} \item Obtain the solution that satisfies \(z(0)=0\) and \(y(0)= 5\). The quantity of the drug in the brain for this solution is denoted by \(z_1(t)\). \item Obtain the solution that satisfies $ z(0)=z(1)= c$, where \(c\) is a given constant. %\[ %C=2(1-\e^{-1})^{-1} - 2(1-\e^{-6})^{-1}\,. %\] The quantity of the drug in the brain for this solution is denoted by \(z_2(t)\). \item Show that for \(0\le t \le 1\), \[ z_2(t) = \sum _{n=-\infty}^{0} z_1(t-n)\,, \] provided \(c\) takes a particular value that you should find. \end {questionparts}
Solution:
What is the general solution of the differential equation \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+2k\frac{\mathrm{d}x}{\mathrm{d}t}+x=0 \] for each of the cases: (i) \(k>1;\) (ii) \(k=1\); (iii) \(0 < x < 1\)? In case (iii) the equation represents damped simple harmonic motion with damping factor \(k\). Let \(x(0)=0\) and let \(x_{1},x_{2},\ldots,x_{n},\ldots\) be the sequence of successive maxima and minima, so that if \(x_{n}\) is a maximum then \(x_{n+1}\) is the next minimum. Show that \(\left|x_{n+1}/x_{n}\right|\) takes a value \(\alpha\) which is independent of \(n\), and that \[ k^{2}=\frac{(\ln\alpha)^{2}}{\pi^{2}+(\ln\alpha)^{2}}. \]
Solution: The auxiliary equation is \(\lambda^2 + 2k\lambda + 1 = (\lambda + k)^2+1-k^2 = 0\) (i) If \(k > 1\) then the solution is \(A\exp \left ({(-k + \sqrt{k^2-1})t} \right)+B\exp\left((-k-\sqrt{k^2-1})t \right)\). (ii) If \(k = 1\) then the solution is \(x = (A+Bt)e^{-kt}\) (iii) If \(k < 1\) then the solution is \(x = Ae^{-kt} \sin \left ( \sqrt{1-k^2} t \right)+Be^{-kt} \cos \left ( \sqrt{1-k^2} t \right)\) If \(x(0) = 0\) then \begin{align*} && x &= Ae^{-kt} \sin(\sqrt{1-k^2}t)\\ && \dot{x} &= Ae^{-kt} \left (-k \sin(\sqrt{1-k^2}t)+\sqrt{1-k^2} \cos(\sqrt{1-k^2}t) \right) \\ (\dot{x} =0): && \tan (\sqrt{1-k^2}t) &= \frac{\sqrt{1-k^2}}{k}\\ \end{align*} Therefore maxima and minima occur every \(\frac{\pi}{\sqrt{1-k^2}}\), so \begin{align*} && \frac{x_{n+1}}{x_n} &= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \frac{\sin\left (\sqrt{1-k^2}\left(t+\frac{\pi}{\sqrt{1-k^2}}\right)\right)}{\sin(\sqrt{1-k^2}t)} \\ &&&= \exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \left (-1+0 \right)\\ &&&= -\exp\left ( -\frac{k\pi}{\sqrt{1-k^2}} \right) \\ \Rightarrow && \ln \alpha &= - \frac{k\pi}{\sqrt{1-k^2}} \\ \Rightarrow && (\ln \alpha)^2 &= \frac{k^2\pi^2}{1-k^2} \\ \Rightarrow && (1-k^2)(\ln \alpha)^2 &= k^2 \pi^2 \\ \Rightarrow && k^2(\pi^2+(\ln \alpha)^2) &= (\ln \alpha)^2 \\ \Rightarrow && k^2 &= \frac{(\ln \alpha)^2}{\pi^2 + (\ln \alpha)^2} \end{align*}
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*}