2.7 Answers to the Problem Set

1. 1. Find the \(x\) which gives max\((dx/dt)\). Differentiate \(dx/dt\) with respect to \(x\), equate to zero and solve for \(x\): \[\frac{d}{dx}(\frac{dx}{dt})=A(N-2x)\] \[0=A(N-2x)\] \[x=N/2\] Since the second derivative is negative, i.e. \[\frac{d^2}{dx^2}(dx/dt)=-2A<0\] then at \(x=N/2\), \(dx/dt\) is maximal.   b. Find \(dR/dx\), set equal to zero, solve for \(x\): \[R=\frac{1}{x}[Ax(N-x)]=A(N-x)\]

\[\frac{dR}{dx}=-A<0\] So no relative max exists, and the maximum must be at the lower end of the domain of \(x\): since \(0\le x\le N\), then max \((R)\) occurs at \(x=0\). This model then implies that crowding effects are present whenever any animals exist.

2. \[35=2\pi\sqrt{(a^2+b^2)/2}\] \[(\frac{35}{2\pi})^2=\frac{a^2+b^2}{2}\] \[a=(\frac{35^2}{2\pi^2}-b^2)^{1/2}\] The area \(A\) in terms of \(b\) is then \[A=\pi(\frac{35^2}{2\pi^2}-b^2)^{1/2}b=\pi(\frac{35^2b^2}{2\pi^2}-b^4)^{1/2}\] Maximize \(A(b)\): \[\begin{align*} \frac{dA}{db}&=\frac{\pi}{2}(\frac{35^2b^2}{2\pi^2}-b^4)^{-1/2}(\frac{35^2b}{\pi^2}-4b^3) \\ 0&=\frac{35^2b}{\pi^2}-4b^3,\;\;\;\;\mbox{assume }b\not=0 \\ &=\frac{35^2}{\pi^2}-4b^2 \\ b&=\frac{35}{2\pi},\;\;\;\;\mbox{since }b>0 \\ a&=(\frac{35^2}{2\pi^2}-\frac{35^2}{4\pi^2})^{1/2}=\frac{35}{2\pi},\;\;\;\;\mbox{since }a>0 \end{align*}\] Thus the maximum area occurs when \(a=b\), i.e. a circle.

3. To convert (2.7) so \(Q\) is in ml/hr, we divide by 1000: \[Q=1.87\times10^{-3}(m+1)W^{0.7}\] \[\begin{align*} \frac{\partial Q}{\partial m}&=1.87\times10^{-3}W^{0.7} \\ &=0.234\;\;\;\;\mbox{for } W = 1000 \end{align*}\] The units are then ml O₂ per hour per unit of activity.

4.a. First solve for \(r(\theta)\):

\[\begin{align*} &r=\frac{D}{2\pi}\sin\theta \\ &v=(\frac{D}{2\pi}\sin\theta)^2(D\cos\theta) \\ &v(\theta)=\frac{D^3}{4\pi}\sin^2\theta \cos\theta \\ &v(0) = 0\;\;\mbox{since}\;\;\sin(0)=0 \\ &v(90^{\circ})= 0\;\;\mbox{since}\;\;\cos(90^{\circ})=0 \end{align*}\]

  b.

\[\begin{align*} &\frac{dv}{d\theta}=\frac{D^3}{4\pi}\{2\sin\theta \cos^2\theta-\sin^3\theta\} \\ &0=2\sin\theta \cos^2\theta-\sin^3\theta\;\;\;\;\mbox{Assume }\;\theta > 0 \\ &0=2\cos^2\theta - \sin^2\theta \\ &\sin^2\theta=2\cos^2\theta \\ &tan\theta=\sqrt{2} \\ &\theta = 54.74^{\circ} \end{align*}\]

5. Maximize \(A\) by minimizing the denominator: \[0=\frac{d}{dn}[(s-4\pi^2n^2m)^2+(2\pi nK)^2]=2(s-4\pi^2n^2m)(-8\pi^2mn)+8\pi^2K^2n\] \[0=-2(ms-4\pi^2m^2n^2)+K^2\] \[2ms-K^2=8\pi^2m^2n^2\] \[n=\sqrt{\frac{2ms-K^2}{8\pi^2m^2}}\]

6. First calculate the required partial derivatives: \[\frac{\partial z}{\partial x}=3x^2-3y,\;\;\;\;\frac{\partial z}{\partial y}=3y^2-3x\] \[\frac{\partial^2 z}{\partial x^2}=6x,\;\;\;\;\frac{\partial^2 z}{\partial x\partial y}=-3,\;\;\;\;\frac{\partial^2z}{\partial y^2}=6y\] Now find the critical point(s) by simultaneously solving \[\frac{\partial z}{\partial x}=0\;\;\;\mbox{and}\;\;\;\frac{\partial z}{\partial y}=0\]

Thus \[3x^2-3y=0\;\;\;\;3y^2-3x=0\] From the first equation we obtain: \[y=x^2\] Substitute into the second equation and solve for \(x\): \[3(x^2)^2-3x=0\] \[3x(x^3-1)=0\] Thus \(x=0,\;1\). The minimum is said to be at \((1,1)\).
With \(x=1\) we evaluate \(y\): \[y=(1)^2=1\] To classify this critical point, we evaluate \[(\frac{\partial^2 z}{\partial x^2})(\frac{\partial^2z}{\partial y^2})-(\frac{\partial^2z}{\partial x\partial y})^2\;\;\;\mbox{at}\;\;\;(x,y)=(1,1)\] \[(6\cdot1)(6\cdot1)-(-3)^2=36-9>0\] Since \[\frac{\partial^2 z}{\partial x^2}=6>0\;\;\;\mbox{at}\;\;\;(x,y)=(1,1)\] then \((1,1)\) is indeed a relative minimum.

7. First show the boundary conditions to be satisfied: \[\begin{align*} T(t,0)&=T_a+\theta_0e^{-0/d}\cos(\omega t-0/d) \\ &=T_a+\theta_0\cos\omega t \\ T(t,\infty)&=T_a+\theta_0e^{-\infty/d}\cos(\omega t-\infty/d) \\ &=T_a+0 \\ &=T_a \end{align*}\] Now show that the diffusion equation is satisfied:

\[\begin{align*} \frac{\partial T}{\partial t}&=-\theta_0e^{-z/d}\omega \sin(\omega t-z/d) \\ \frac{\partial T}{\partial z}&=-\frac{1}{d}\theta_0e^{-z/d}\cos(\omega t-z/d)-\theta_0e^{-z/d}(-\frac{1}{d})\sin(\omega t-z/d) \\ &=\frac{\theta_0e^{-z/d}}{d}[\sin(\omega t-z/d)-\cos(\omega t-z/d)] \\ \frac{\partial^2 T}{\partial z^2}&=(-\frac{1}{d})\frac{\theta_0}{d}e^{-z/d}[\sin(\omega t-z/d)-\cos(\omega t-z/d)] \\ &-(\frac{1}{d})\frac{\theta_0}{d}e^{-z/d}[\cos(\omega t-z/d)+\sin(\omega t-z/d)] \\ &=-\frac{2\theta_0}{d^2}e^{-z/d}\sin(\omega t-z/d) \end{align*}\] Substituting into the diffusion equation: \[-\theta_0e^{-z/d}\omega \sin(\omega t-z/d)=-a\frac{2\theta_0}{d^2}e^{-z/d}\sin({\omega t-z/d})\] The equation is certainly true when \(\sin(\omega t-z/d)=0\). Now assume that \(\sin(\omega t-z/d)\not=0\) and divide both sides by \(-\theta_0e^{-z/d}\sin(\omega t-z/d\): \[\omega = 2a/d^2\] Substituting for \(d\), we obtain \[\omega = 2a/[(2a/\omega)^{1/2}]^2=\omega\] Thus the equation is indeed satisfied.