Solve the problem.
-A single injection of a drug is administered to a patient. The amount Q in the body then decreases at a rate proportional to the amount present, and for this particular drug the rate is 3% per hour. Thus, with where t is time in hours. If the initial injection is 4 milliliters about how many hours
After the drug is given will there be 2 milliliters of the drug remaining in the body? (Round answer to the
Nearest tenth of an hour.)
A) 69.3 hours
B) 21.3 hours
C) 11.3 hours
D) 23.1 hours

Question

Solve the problem.
-A single injection of a drug is administered to a patient. The amount Q in the body then decreases at a rate proportional to the amount present, and for this particular drug the rate is 3% per hour. Thus,  with  where t is time in hours. If the initial injection is 4 milliliters  about how many hours
After the drug is given will there be 2 milliliters of the drug remaining in the body? (Round answer to the
Nearest tenth of an hour.)
A) 69.3 hours 
B) 21.3 hours 
C) 11.3 hours 
D) 23.1 hours

Quizplus · Accepted Answer

Using the given information, we can set up a differential equation to model the situation:
$\frac{dQ}{dt} = kQ$
where $k$ is the proportionality constant, which is given as 0.03 (since the rate of decrease is 3% per hour).

Solving the differential equation, we get:
$Q(t) = Q_0 e^{-kt}$

At time $t$, we want to find when $Q(t) = 2$, and we know $Q_0 = 4$. Substituting these values, we get:
$2 = 4e^{-0.03t}$
$e^{0.03t} = 2$
$t \approx 23.1$ (rounded to the nearest tenth)

Therefore, the best choice is D, which corresponds to 23.1 hours. Using the given information, we can write the differential equation:
$\frac{dQ}{dt}=kQ$ where $k=-0.03$ (since the rate of decrease is proportional to the amount present)
This is a separable differential equation:
\begin{align*}
\frac{dQ}{Q}&=k\,dt\
\int \frac{dQ}{Q}&=\int k\,dt\
\ln|Q|&=kt+C\
Q&=ce^{kt}
\end{align*}
where $C$ is the constant of integration, and $c$ is the initial amount of the drug in the patient's body.
To find $c$, we know that the initial injection was 4 milliliters, so $c=4$.
To find the time when there are 2 milliliters remaining, we need to solve $Q=2$ for $t$. So we have:
\begin{align*}
2&=4e^{-0.03t}\
\frac{1}{2}&=e^{-0.03t}\
-\ln(2)&=-0.03t\
t&=\frac{\ln(2)}{0.03}\approx 23.1 	ext{ hours}
\end{align*}
So the answer is choice D, and the time taken for 2 milliliters of the drug to be remaining in the body is about 23.1 hours.

Solve the Problem. -A Single Injection of a Drug Is Administered to a Administered