Problems

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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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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 \((\pm \sqrt{c-a^2}, c)\)

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Solution: We have \(\dot{A} = -kVxy\) or \(\dot{x}V = -kVxy\), ie \(\dot{x} = -kxy\) and similarly \(\dot{y} = kxy = k(1-y)y\). \begin{align*} && \frac{\d y}{\d t} &= ky(1-y) \\ \Rightarrow && \int k \d t &= \int \frac{1}{y(1-y)} \d y \\ \Rightarrow && kt &= \int \left ( \frac{1}{y} + \frac{1}{1-y} \right) \d y \\ &&&= \ln y - \ln (1-y) + C\\ \Rightarrow && kt &= \ln \frac{y}{D(1-y)} \\ \Rightarrow && De^{kt} &= \frac{y}{1-y} \\ \Rightarrow && y(1+De^{kt}) &= De^{kt} \\ \Rightarrow && y &= \frac{De^{kt}}{1+De^{kt}} \end{align*} As \(t \to \infty\) \(y \to \frac{D}{D} = 1\) so depending on how fine grained we want to go we might say that 'yes it completely converts' when there is an immeasurably small amount of \(A\) left, or we might say it doesn't since it only tends to \(1\) and never actually reaches it.

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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.