Chapter 3.EA, Problem 1EA

A $500\text{-mg}$ dose of a drug is administered by rapid injection to a patient. The half-life of the drug is $9$ hours.

a. Find a model for the amount of drug in the bloodstream $t$ hours after the drug is administered.

b. Find the average rate of change of drug in the bloodstream between $t=0$ and $t=2$. Repeat for $t=9$ and $t=11$.

To determine

a.

A model for the amount of drug in the bloodstream $t$ hours after the drug is administered if a $500mg$ dose is administered by rapid injection to a patient and the half-life of the drug is $9$ hours.

Answer to Problem 1EA

Solution:

A model for the amount of drug in the bloodstream $t$ hours after the drug is administered is $\text{A}\left(t\right)=500e^{\left(-0.07701\right)t}$.

Explanation of Solution

Given information:

The dose is $500mg$ and half-life of the drug is $9$ hours.

Formula used:

The growth-decay is $\text{A}\left(t\right)=D{e}^{kt}$,

$D$ is the size of the dose administered, and

$k$ is the exponential decay constant for the drug.

Calculation:

The amount of drug in the bloodstream $t$ hours after a single rapid injection can be modeled using an exponential decay function $\text{A}\left(t\right)=D{e}^{kt}$.

Here, $D=500mg$

The function has the form $\text{A}\left(t\right)=500e^{kt}$.

The general equation of the half-life $T$ in terms of the decay constant $k$ is $T=-\frac{\ln 2}{k}$.

Solving this equation for $k$,

$k=-\frac{\ln 2}{T}$

Since the half-life of this drug is $9$ hours,

$k=-\frac{\ln 2}{T}=-\frac{\ln 2}{9}\approx -0.07701$

Therefore, the model for the amount of drug in the bloodstream $t$ hours after the drug is administered is $\text{A}\left(t\right)=500e^{\left(-0.07701\right)t}$.

To determine

b.

The average rate of change of the drug in the bloodstream between $t=0$ and $t=2$

And the average rate of change of drug in the blood stream between $t=9$ and $t=11$.

Answer to Problem 1EA

Solution:

The average rate of change from $t=0$ and $t=2$ is $35.5$ mg per hour.

The average rate of change from $t=9$ and $t=11$ is $18$ mg per hour.

Explanation of Solution

Given information:

The dose is $500mg$ and half-life of the drug is $9$ hours.

Formula used:

The average rate of change of function at $x=a$ and $x=b$ is $\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Calculation:

A model for the amount of drug in the bloodstream $t$ hours after the drug is administered is $\text{A}\left(t\right)=500e^{\left(-0.07701\right)t}$.

The average rate of change of drug in the bloodstream between $t=0$ and $t=2$ is $\begin{array}{l}\frac{\text{A}\left(2\right)-\text{A}\left(0\right)}{2-0}=\frac{428.6274-500}{2-0}\\ =-\frac{71.3726}{2}=-35.6863\end{array}$

Therefore, the average rate of change of drug in the bloodstream between $t=0$ and $t=2$ is $-35.6863$ mg per hour.

And the average rate of change of drug in the bloodstream between from $t=9$ and $t=11$ is

$\begin{array}{l}\frac{\text{A}\left(11\right)-\text{A}\left(9\right)}{11-9}=\frac{214.3259-250.01429}{2}\\ =-\frac{35.68839}{2}=-17.844195\end{array}$

Therefore, the average rate of change of amount of the drug in the bloodstream from $t=9$ to $t=11$ is $-17.844195$ mg per hour