Chapter 7, Problem 58RE

a.

To determine

To Solve:

To solve the equation for $y\left(t\right)$ , using $y_0=y\left(0\right)$ .

Answer to Problem 58RE

The value of $y\left(t\right)$ is $y=c+\left(y_0-c\right)e^{-\frac{kA}{V}t}$ .

Explanation of Solution

Given information:

The result of the movement of a dissolved substances across a cell membrane is described by the equation,

  $\frac{dy}{dt}=k\frac{A}{V}\left(c-y\right)$ where $y$ -concentration of the substance inside the cell;

  $\frac{dy}{dt}$ -rate with which $y$ changes over time, $A$ -constants, $k$ - permeability co-efficient, $V$ - cell’s volume; $c$ - concentration of the substance outside the cell.

The result of the movement of a dissolved substances across a cell membrane is described by the equation,

  $\frac{dy}{dt}=k\frac{A}{V}\left(c-y\right)$ where $y$ -concentration of the substance inside the cell;

  $\frac{dy}{dt}$ -rate with which $y$ changes over time, $A$ -constants, $k$ - permeability co-efficient, $V$ - cell’s volume; $c$ - concentration of the substance outside the cell.

  $\begin{array}{l}\frac{dy}{dt}=k\frac{A}{V}\left(c-y\right)\\ \frac{dy}{c-y}=\frac{kA}{V}dt\end{array}$

Differentiating y with respect to t,

  $\begin{array}{l}-\ln \left|y-c\right|=\frac{kA}{V}t+C\\ y-c=C{e}^{-\frac{kA}{V}t}\\ y=c+C{e}^{-\frac{kA}{V}t}\\ C=y_0-c\\ y=c+\left(y_0-c\right)e^{-\frac{kA}{V}t}\end{array}$

Therefore, the value of $y\left(t\right)$ is $y=c+\left(y_0-c\right)e^{-\frac{kA}{V}t}$ .

b.

To determine

To Find:

To find the steady state concentration ${\lim }_{t\to \infty }y\left(t\right)$ .

Answer to Problem 58RE

The value of ${\lim }_{t\to \infty }y\left(t\right)$ is ${\lim }_{t\to \infty }y\left(t\right)=c$ .

Explanation of Solution

Given information:

The result of the movement of a dissolved substances across a cell membrane is described by the equation,

  $\frac{dy}{dt}=k\frac{A}{V}\left(c-y\right)$ where $y$ -concentration of the substance inside the cell;

  $\frac{dy}{dt}$ -rate with which $y$ changes over time, $A$ -constants, $k$ - permeability co-efficient, $V$ - cell’s volume; $c$ - concentration of the substance outside the cell.

From part (a) we know that the value of $y\left(t\right)$ is $y=c+\left(y_0-c\right)e^{-\frac{kA}{V}t}$ .

Now taking the limit $t\to \infty$ ,

  $\begin{array}{l}{\lim }_{t\to \infty }y\left(t\right)={\lim }_{t\to \infty }\left[c+\left(y_0-c\right)e^{-\frac{kA}{V}t}\right]\\ {\lim }_{t\to \infty }y\left(t\right)=c+\left(y_0-c\right)e^{-\infty }\\ {\lim }_{t\to \infty }y\left(t\right)=c+0\\ {\lim }_{t\to \infty }y\left(t\right)=c\end{array}$

Therefore, the value of ${\lim }_{t\to \infty }y\left(t\right)$ is ${\lim }_{t\to \infty }y\left(t\right)=c$ .