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.