Chapter 7.3, Problem 53E

a.

To determine

To formulate: The differential equation and show that the tissue grows fastest when $A\left(t\right)=\frac{1}{3}M$ .

Answer to Problem 53E

  $\frac{dA}{dt}=K\left[\sqrt{A}\cdot \left(M-A\right)\right]\text{ [K is a constant]}$

Explanation of Solution

Given information :

Rate of growth of the area is jointly proportional to $\sqrt{A}$ and $\left(M-A\left(t\right)\right)$

Concept used :

If $f^{\prime }\left(x\right)=0$ and $f^{″}\left(x\right)<0$ then it can be said that $f\left(x\right)$ is maximum at $x=0$

Calculation:

By the problem,

  $\begin{array}{l}\frac{dA}{dt}\propto \sqrt{A}\cdot \left(M-A\right)\\ \Rightarrow \frac{dA}{dt}=K\left[\sqrt{A}\cdot \left(M-A\right)\right]\text{ [K is a constant]}\end{array}$

  $\frac{dA}{dt}$ represents the rate of growth.

  $\begin{array}{c}\frac{d}{dt}\left(\frac{dA}{dt}\right)=\frac{d}{dt}\left[K\left[\sqrt{A}\cdot \left(M-A\right)\right]\right]\\ =K\left[\sqrt{A}\frac{d\left(M-A\right)}{dt}+\left(M-A\right)\frac{d\sqrt{A}}{dt}\right]\\ =K\left[-\sqrt{A}\frac{dA}{dt}+\left(M-A\right)\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{A}}\cdot \frac{d\sqrt{A}}{dt}\right]\\ =K\left[-\sqrt{A}\left[K\sqrt{A}\cdot \left(M-A\right)\right]+\frac{M-A}{2\sqrt{A}}\left[K\sqrt{A}\cdot \left(M-A\right)\right]\right]\\ =K\left[-KA\left(M-A\right)+\frac{K}{2}{\left(M-A\right)}^2\right]\end{array}$

  $\frac{d^{2}A}{d{t}^2}=K\left[-KA\left(M-A\right)+\frac{K}{2}{\left(M-A\right)}^2\right]$

If $\frac{d^{2}A}{d{t}^2}=0$ then we can write

  $K\left[-KA\left(M-A\right)+\frac{K}{2}{\left(M-A\right)}^2\right]=0$

  $\begin{array}{l}\Rightarrow \frac{K}{2}{\left(M-A\right)}^2=KA\left(M-A\right)\\ \Rightarrow M-A=2A\\ \Rightarrow M=3A\\ \Rightarrow A=\frac{1}{3}M\end{array}$

  $\frac{d^{2}A}{d{t}^2}=0$ for $A=\frac{1}{3}M$ . So $\frac{dA}{dt}$ has the maxima or minima at $A=\frac{1}{3}M$ .

  $\begin{array}{c}\frac{dA}{dt}=K\left[\sqrt{A}\cdot \left(M-A\right)\right]\text{ [K is a constant]}\\ \frac{d^{2}A}{d{t}^2}=K\left[-KA\left(M-A\right)+\frac{K}{2}{\left(M-A\right)}^2\right]\\ \frac{d}{dt}\left(\frac{d^{2}A}{d{t}^2}\right)=\frac{d}{dt}\left[K\left[-KA\left(M-A\right)+\frac{K}{2}{\left(M-A\right)}^2\right]\right]\\ =K\left[-K\left[A\frac{d\left(M-A\right)}{dt}+\left(M-A\right)\frac{dA}{dt}\right]+\frac{K}{2}\left[-2\left(M-A\right)\frac{dA}{dt}\right]\right]\\ =K\left[KA\frac{dA}{dt}-K\left(M-A\right)\frac{dA}{dt}-K\left(M-A\right)\frac{dA}{dt}\right]\\ =K^2\left[A\frac{dA}{dt}-2\left(M-A\right)\frac{dA}{dt}\right]\\ =K^2\left[AK\sqrt{A}\cdot \left(M-A\right)-2\left(M-A\right)K\cdot \sqrt{A}\cdot \left(M-A\right)\right]\\ =K^3\left[A^{\frac{3}{2}}\left(M-A\right)-2\sqrt{A}{\left(M-A\right)}^2\right]\\ =K^3\left(M-A\right)A^{\frac{1}{2}}\left[A-2\left(M-A\right)\right]\end{array}$

  $\frac{d^{3}A}{d{t}^3}=K^3\left(M-A\right)A^{\frac{1}{2}}\left[A-2\left(M-A\right)\right]$

When, $A=\frac{1}{3}M$ the value of $\frac{d^{3}A}{d{t}^3}$ will be

  $\begin{array}{c}{{\displaystyle \frac{d^{3}A}{d{t}^3}|}}_{A=\frac{1}{3}M}=K^3\left(M-\frac{1}{3}M\right){\left(\frac{1}{3}M\right)}^{\frac{1}{2}}\left[\frac{1}{3}M-2M+\frac{2}{3}M\right]\\ =K^3\left(\frac{2}{3}M\right){\left(\frac{1}{3}M\right)}^{\frac{1}{2}}\left(-M\right)\end{array}$

It is now evident that the value of $\frac{d^{3}A}{d{t}^3}$ will be negative.

So, it can be told that $\frac{dA}{dt}$ shows the maximum value when $A=\frac{1}{3}M$ .

b.

To determine

To find: The solution of differential equation.

Answer to Problem 53E

  $A={\left[\frac{\sqrt{M}\left(1+E{e}^{\sqrt{M}kt}\right)}{\left(1-E{e}^{\sqrt{M}kt}\right)}\right]}^2$

Explanation of Solution

Given information:

  $\frac{dA}{dt}=K\left[\sqrt{A}\cdot \left(M-A\right)\right]\text{ [K is a constant]}$

Concept Used:

To solve the equation first we separate the variables and then integrate.

Calculation:

  $\frac{dA}{dt}=K\left[\sqrt{A}\cdot \left(M-A\right)\right]\text{ [K is a constant]}$

  $\Rightarrow \frac{dA}{\sqrt{A}\cdot \left(M-A\right)}=Kdt$

Integrating both sides we get,

  ${\displaystyle \int \frac{dA}{\sqrt{A}\cdot \left(M-A\right)}}={\displaystyle \int Kdt}$

  $\Rightarrow \frac{\ln \left(\frac{\left|\sqrt{A}-\sqrt{M}\right|}{\sqrt{A}+\sqrt{M}}\right)}{\sqrt{M}}=kt+C$

  $\begin{array}{l}\Rightarrow \ln \left(\frac{\left|\sqrt{A}-\sqrt{M}\right|}{\sqrt{A}+\sqrt{M}}\right)=\sqrt{M}kt+\sqrt{M}C\\ \Rightarrow \frac{\left|\sqrt{A}-\sqrt{M}\right|}{\sqrt{A}+\sqrt{M}}=e^{\sqrt{M}kt+\sqrt{M}C}\\ \Rightarrow \frac{\sqrt{A}-\sqrt{M}}{\sqrt{A}+\sqrt{M}}=\pm e^{\sqrt{M}C}\cdot e^{\sqrt{M}kt}\\ \Rightarrow \frac{\sqrt{A}-\sqrt{M}}{\sqrt{A}+\sqrt{M}}=E{e}^{\sqrt{M}kt}\left[E=\pm e^{\sqrt{M}C}\right]\\ \Rightarrow \sqrt{A}-\sqrt{M}=\sqrt{A}E{e}^{\sqrt{M}kt}+\sqrt{M}E{e}^{\sqrt{M}kt}\\ \Rightarrow \sqrt{A}\left[1-E{e}^{\sqrt{M}kt}\right]=\sqrt{M}\left[1+E{e}^{\sqrt{M}kt}\right]\\ \Rightarrow \sqrt{A}=\left[\frac{\sqrt{M}\left(1+E{e}^{\sqrt{M}kt}\right)}{\left(1-E{e}^{\sqrt{M}kt}\right)}\right]\\ \Rightarrow A={\left[\frac{\sqrt{M}\left(1+E{e}^{\sqrt{M}kt}\right)}{\left(1-E{e}^{\sqrt{M}kt}\right)}\right]}^2\end{array}$