Chapter 3: Q3.1-46E (page 95)

When forgetfulness is taken into account, the rate of memorization of a subject is given by
$\frac{d A}{d t} = k_{1} (M - A) - k_{2} A$
Where, $k_{1} > 0, k_{2} > 0, A (t)$ is the amount memorized in time t, M is the total amount to be memorized, and $M - A$ is the amount remaining to be memorized.
(a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of $A (t)$ as $t \to \infty$ . Interpret the result.
(b) Solve the DE subject to $A (0) = 0$ . Sketch the graph of $A (t)$ and verify your prediction in part (a).

Short Answer

Expert verified

a) We can see from the graph that as t goes to infinity A goes to $A = \frac{k_{1} M}{k_{1} + k_{2}}$ .

b) $A (t) = \frac{k_{1} M}{k_{1} + k_{2}} A (t) = \frac{k_{1} M}{k_{1} + k_{2}} - \frac{k_{1} M}{k_{1} + k_{2}} e^{- (k_{1} + k_{2}) t}$

Step by step solution

To Find the limiting value of A(t) as t⇒∞

a) This is the phase portrait and solution curve for the autonomous differential question

$\frac{d A}{d t} = k_{1} (M - A) - k_{2} A = K_{1} M - (k_{1} + k_{2}) A$

Where the limiting value of x equals $\frac{k_{1} M}{k_{1} + k_{2}}$ as $t \Rightarrow \infty$ , and we have the following cases

For $A = \frac{k_{1} M}{k_{1} + k_{2}}$ , we have $\frac{d A}{d t} = 0$

For $A > \frac{k_{1} M}{k_{1} + k_{2}}$ , we have $\frac{d A}{d t} < 0$

For $A < \frac{k_{1} M}{k_{1} + k_{2}}$ , we have $\frac{d A}{d t} > 0$

Phase Portrait and solution curve

Describe the rate of memorization

b) We have the rate of memorization A of a subject is described by the differential equation.

$\frac{d A}{d t} = k_{1} (M - A) - k_{2}$

Where $M - A$ is the amount remaining to be memorized, which is

$\frac{d A}{d t} = k_{1} M - (k_{1} + k_{2}) A$

With the condition

$A (0) = 0$

Solve the DE

And we have to solve it as the following,

This differential equation is a first order and separable D.E, then we can solve it as

$\begin{array}{rcl} \frac{d A}{k_{1} M - (k_{1} + k_{2}) A} & = & d t \\ \int \frac{1}{k_{1} M - (k_{1} + k_{2}) A} d A & = & \int d t \\ - \frac{1}{k_{1} + k_{2}} \int \frac{- (k_{1} + k_{2})}{k_{1} M - (k_{1} + k_{2}) A} d A & = & \int d t \\ - \frac{1}{k_{1} + k_{2}} I n (k_{1} M - (k_{1} + k_{2}) A) & = & - (k_{1} + k_{2}) t + c_{2} \\ e^{I n (k_{1} M - (k_{1} + k_{2}) A)} & = & e^{- (k_{1} + k_{2}) t + c_{2}} \\ k_{1} M - (k_{1} + k_{2}) A & = & e^{c_{2}} e^{- (k_{1} + k_{2}) t} \\ (k_{1} + k_{2}) A & = & k_{1} M - c e^{- (k_{1} + k_{2}) t} \end{array}$

Then we have

$\begin{array}{rcl} A (t) & = & \frac{k_{1}}{k_{1} + k_{2}} M - \frac{c}{k_{1} + k_{2}} e^{- (k_{1} + k_{2}) t} \\ = & \frac{k_{1} M}{k_{1} + k_{2}} - n e^{- (k_{1} + k_{2}) t} \end{array}$ ----------------- (1)

Find the DE

After that, to find the value of constant, we have to apply the point of condition $(A, t) = (0, 0)$ into equation (1), then we have

$0 = \frac{k_{1} M}{k_{1} + k_{2}} - n e^{0}$

Then we have,

$n = \frac{k_{1} M}{k_{1} + k_{2}}$

After that, substitute with the value of constant n into equation (1), then we have,

$A (t) = \frac{k_{1} M}{k_{1} + k_{2}} - \frac{k_{1} M}{k_{1} + k_{2}} e^{- (k_{1} + k_{2}) t}$ , is the amount of the drug in the bloodstream at time t.