Q. 66 - Calculus

Question

Suppose $X\left(t\right)$ is the number of milligrams of the drug Xenaphoril that is present in the body $t$ hours after it is ingested. As the drug is absorbed, the quantity of the drug decreases at a rate proportional to the amount of the drug in the body.

(a) Set up a differential equation describing $\frac{dX}{dt}$and solve it to get a formula for $X\left(t\right)$. Your answer will involve two constants.

(b) The half-life of a drug is the number of hours that it takes for the quantity of the drug to decrease by half. In an exponential decay model, the half-life will be the same no matter when we start measuring the amount of the drug. If Xenaphoril has a half-life of $4$ hours, what is the constant of proportionality for this model?

(c) Given the constant of proportionality you found in part (b), how much of a $20$-mg dose of Xenaphoril remains after $10$ hours?

Step-by-Step Solution

Answer

Part$\left(a\right)$: A formula for $X\left(t\right)$ is, $X\left(t\right)=A{e}^{-kt}$.

Part$\left(b\right)$: The constant of proportionality for this model is, $0.1732868$.

Part $\left(c\right)$: Approximately $3.5$mg of drug will remain in the body after $10$ hours.

Part a Step 1 . Given information

$X\left(t\right)$is the number of milligrams of the drug Xenaphoril that is present in the body $t$ hours after it is ingested.

The quantity of the drug decreases at a rate proportional to the amount of the drug in the body.

Part a Step 2 . Note that the amount of drug present in the body is continuously decreasing with time.

This means that the rate of change of drug in the body is negative. Hence, the mathematical equation representing this model will follow the model of negative growth rate.

From the given information,

If $X\left(t\right)$ is the amount of drug remaining at any time $t$, then the decay rate $\frac{dX}{dt}$ is proportional to $X\left(t\right)$. Hence, the model representing the decay rate of the drug is,

$$\begin{gathered} \frac{dX}{dt}\alpha X \\ =-kX \end{gathered}$$

Part a Step src="data:image/svg+xml;base64,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" role="math" localid="1649326833364" 3 . In the above relation k is the decay constant.

Now solve the differential equation by antidifferentiation method.

$$\begin{gathered} \int \frac{dX}{X}=-k\int dt \\ \ln \left|X\right|=-kt+C \\ X=e^{-kt+C} \\ =A{e}^{-kt} [\because e^C=A] \end{gathered}$$

So, the solution for the model for the decay rate of the drug is, 

$X\left(t\right)=A{e}^{-kt}$

Note that the solution contains the two constants, namely the decay constant or constant of proportionality $k$ and the constant $A$.

Part b Step 1 . Suppose the quality of the drug administered is X , then as per the given information,

The quantity of drug remaining in the body after $4$ hours will be $\frac{X}{2}$. So, take $X\left(0\right)=X$ and $X\left(4\right)=\frac{X}{2}$ in the solution to get,

$$\begin{gathered} X=A \\ \frac{X}{2}=A{e}^{-4k} \end{gathered}$$

Solve the above equations for $k$.

$$\begin{gathered} \frac{X}{2}=X{e}^{-4k} \\ {e}^{4k}=2 \\ k=\frac{1}{4}\ln 2 \\ =0.1732868 \end{gathered}$$

Part c Step 1 . Substitute the value of k obtained in part b to get the solution of the decay model as,

$X\left(t\right)=A{e}^{-0.1732868t}$

For denoting the initial quantity substitute $A=X_0$ in the above equation.

$X\left(t\right)=X_0{e}^{-0.1732868t}$

As per the given information, $20$ mg of dose has been administered and find the amount of drug remaining in the body after $10$ hours.

So, substitute $X_0=20,t=10$ in the obtained equation.

$$\begin{gathered} X\left(10\right)=20e^{-0.1732868\left(10\right)} \\ =20e^{-1.732868} \\ \approx 3.5 \end{gathered}$$