## 2.7 Answers to the Problem Set

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.

1. $35=2\pi\sqrt{(a^2+b^2)/2}$ $(\frac{35}{2\pi})^2=\frac{a^2+b^2}{2}$ $a=(\frac{35^2}{2\pi^2}-b^2)^{1/2}$ 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.

2. 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*}

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

2. 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.

1. 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.