Question

The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of Q0 is injected directly into the blood, 18% is left in the blood after 3 hours. (a) Write and solve a differential equation for the quantity, Q, of the drug in the blood after t hours. Use k as the positive proportionality constant for this question. Round your answer for k to four decimal places.

Answer 1

Answer:

(a) $\dfrac{dQ}{dt} =-k Q$

k=0.5716

$Q(t)=Q_0e^{-0.5716t}$

(b)18 mg

Step-by-step explanation:

The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time.

$\dfrac{dQ}{dt} \propto Q$

Since the drug is leaving the bloodstream, it is reducing and hence we introduce a negative sign and our proportionality constant k.

$\dfrac{dQ}{dt} =-k Q$

Solving this equation:

$\dfrac{dQ}{Q} =-k dt\\ Taking the integral of both sides with respect to t \\\int \dfrac{dQ}{Q} =-\int k dt\\ln Q=-kt+ln C\\ Taking exponential of both sides \\Q(t)=Ce^{-kt}$

At time t=0, the quantity of the drug in the blood, $Q(t)=Q_0$

$Q_0=Ce^{-k*0}\\Q_0=C$

Therefore:

$Q(t)=Q_0e^{-kt}$

If after 3 hours, 18% of the initial amount is left in the blood stream, then:

$0.18Q_0=Q_0e^{-3k}\\0.18=e^{-3k}\\ Take the natural logarithm of both sides \\ln(0.18)=ln [e^{-3k}]\\ln(0.18)=-3k\\k=\frac{ln(0.18)}{-3}\\k=0.5716$

Thus, the quantity, Q(t) of drug left in the bloodstream after t hours is:

$Q(t)=Q_0e^{-0.5716t}$

(b)Since no application is given, a typical example is presented below.

How much of this drug is in a patient's body after 6 hours if the patient is given 100 mg initially?

In this case, $Q_0=100mg, t=6 hours$

Therefore, the Quantity of the drug left:

$Q(t)=Q_0e^{-0.5716t}\\=100\cdot e^{-0.5716*3}\\\approx 18 mg$