Chapter 2.9, Problem 48E

To determine

To calculate: To find the value of $\frac{dp}{dt}$ .

Answer to Problem 48E

The value of $p$ increases then $\frac{dp}{dt}$ decreases.

Explanation of Solution

Given information:

Consider the equation $E=c_1{(pM)}^{0.75}+c_2{((1-p)M)}^{0.75}$

$c_1=0.022,M=30.0,\frac{dM}{dt}=2.0;\frac{dE}{dt}=0.01$

Calculation:

Given that the equation is $E=c_1{(pM)}^{0.75}+c_2{((1-p)M)}^{0.75}$ ,

Where $c_1=0.022,M=30.0,\frac{dM}{dt}=2.0;\frac{dE}{dt}=0.01$

Assume,

$E=c_1{(pM)}^{0.75}+c_2{((1-p)M)}^{0.75}$ -----(1)

Since, $p$ is the fraction of tissue of the first type, $c_1=0.022,M=30.0,\frac{dM}{dt}=2.0;\frac{dE}{dt}=0.01$

Differentiating equation (1) with respect to t,

$\begin{array}{l}\frac{dE}{dt}=\frac{d}{dt}\left\{c_1{(pM)}^{0.75}\right\}+\frac{d}{dt}\left\{c_2{((1-p)M)}^{0.75}\right\}\\ =c_1\left\{p^{0.75}0.75M^{-0.25}\frac{dM}{dt}+M^{0.75}0.75p^{-0.25}\frac{dp}{dt}\right\}\\ +c_2\left\{{(1-p)}^{0.75}0.75M^{-0.25}\frac{dM}{dt}+M^{0.75}0.75{(1-p)}^{-0.25}\frac{d(1-p)}{dt}\right\}\end{array}$

Substitute the values, $c_1=0.022,M=30.0,\frac{dM}{dt}=2.0;\frac{dE}{dt}=0.01$

$\begin{array}{l}0.01=(0.01)\left\{p^{0.75}0.75{(30)}^{-0.25}(2)+{(30)}^{0.75}0.75p^{-0.25}\frac{dp}{dt}\right\}\\ +(0.022)\left\{{(1-p)}^{0.75}0.75{(30)}^{-0.25}(2)-{(30)}^{0.75}0.75{(1-p)}^{-0.25}\frac{dp}{dt}\right\}\end{array}$

The simplified equation is,

$\begin{array}{l}\frac{dp}{dt}=\frac{(0.01)-(2\times 0.75\times {(30)}^{-0.25}(0.01\times {(p)}^{0.75}+0.022{(1-p)}^{0.75}))}{{(30)}^{0.75}\times 0.75(0.01\times {(p)}^{-0.25}-0.022{(1-p)}^{0.25})}---(2)\\ \end{array}$

Substitute $p$ =0.1 in equation (2),

$\begin{array}{l}\frac{dp}{dt}=\frac{(0.01)-(2\times 0.75\times {(30)}^{-0.25}(0.01\times {(0.1)}^{0.75}+0.022{(1-0.1)}^{0.75}))}{{(30)}^{0.75}\times 0.75(0.01\times {(0.1)}^{-0.25}-0.022{(1-0.1)}^{0.25})}\\ \\ =0.09\end{array}$

Thus, the fraction of the first type of metabolic tissue is increasing.

Substitute $p$ =0.5 in equation (2),

$\begin{array}{l}\frac{dp}{dt}=\frac{(0.01)-(2\times 0.75\times {(30)}^{-0.25}(0.01\times {(0.5)}^{0.75}+0.022{(1-0.5)}^{0.75}))}{{(30)}^{0.75}\times 0.75(0.01\times {(0.5)}^{-0.25}-0.022{(1-0.5)}^{0.25})}\\ \\ =0.016\end{array}$

Thus, the fraction of the first type of metabolic tissue is increasing.

Substitute $p$ =0.9 in equation (2),

$\begin{array}{l}\frac{dp}{dt}=\frac{(0.01)-(2\times 0.75\times {(30)}^{-0.25}(0.01\times {(0.9)}^{0.75}+0.022{(1-0.9)}^{0.75}))}{{(30)}^{0.75}\times 0.75(0.01\times {(0.9)}^{-0.25}-0.022{(1-0.9)}^{0.25})}\\ \\ =-0.00566\end{array}$

Thus, the fraction of the first type of metabolic tissue is decreasing.

Hence, as $p$ value increases, the rate of increase of the $p$ decreases.

$\therefore$ The value of $p$ increases then $\frac{dp}{dt}$ decreases.