Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. Suppose that the half-life of warfarin in the body is 35 hours.
Write a differential equation satisfied by Q:
How many days does it take for the drug level in the body to be reduced to 30 percent of its original level?
Well, well, well, looks like warfarin is taking a break from its role as an anticoagulant to participate in some mathematical modeling! Brace yourself for some differential equation fun, my friend.
Let's start by defining our variable Q as the quantity of warfarin remaining in the patient's body at any given time. According to the problem, the rate at which Q decreases is proportional to the quantity remaining.
Now, they also mentioned the half-life of warfarin is 35 hours. Thankfully, we can use this information to set up our differential equation. The half-life tells us that after 35 hours, Q will be reduced to half of its original level. Mathematically, we can express this as:
dQ/dt = -kQ
Where dQ/dt represents the rate of change of Q with respect to time and k is the constant of proportionality.
But wait, there's more! We need the drug level in the body to be reduced to 30 percent of its original level. That means we want Q to be 30 percent of its original value (let's say Q0). So, we can write:
Q/Q0 = 0.3
Now, we have a system to work with! Let's combine these equations and solve for the time it takes to reach 30 percent of the original level:
dQ/dt = -kQ
Q/Q0 = 0.3
First, let's solve the differential equation:
(dQ/Q) = -k * dt
Integrating both sides:
∫(dQ/Q) = ∫(-k * dt)
ln(Q) = -kt + C
Where C is the constant of integration. Now, we'll use the initial condition Q = Q0 at t = 0:
ln(Q0) = C
Substituting that back into our equation:
ln(Q) = -kt + ln(Q0)
To get rid of the natural logarithm, we'll take the exponent of both sides:
Q = e^(-kt + ln(Q0))
Simplifying:
Q = e^(-kt) * e^(ln(Q0))
Q = e^(-kt) * Q0
Finally, we can plug in our value for Q and solve for t:
0.3 = e^(-kt) * Q0
Dividing both sides by Q0:
0.3/Q0 = e^(-kt)
Taking the natural logarithm of both sides:
ln(0.3/Q0) = -kt
Solving for t:
t = -ln(0.3/Q0) / k
Voila! We have our solution for t. Plug in the values of Q0 and k, and you'll find the number of days it takes for the drug level to drop to 30 percent of its original level.
And remember, laughter is the best medicine, but warfarin's half-life equation is a close second!
To write the differential equation, let's assume that Q(t) represents the quantity of warfarin in the body at time t.
According to the given information, the quantity remaining in the patient's body decreases at a rate proportional to the quantity remaining. This can be expressed as:
dQ/dt = -k * Q
where dQ/dt is the rate of change of the quantity with respect to time, k is the proportionality constant, and Q is the quantity remaining.
Since we know that the half-life of warfarin in the body is 35 hours, we can use this information to find the value of k.
The half-life of a substance is determined by the formula:
T1/2 = (ln 2) / k
Substituting the given half-life (T1/2 = 35 hours) into the formula, we have:
35 = (ln 2) / k
Now, we can find the value of k:
k = (ln 2) / 35
To determine how many days it takes for the drug level in the body to be reduced to 30 percent of its original level, we can integrate the differential equation.
Integrating both sides of the equation, we have:
∫ dQ / Q = ∫ -k dt
ln |Q| = -kt + C
where C is the constant of integration.
Now, let's solve for Q at t = 0 (original level) and t = D (desired time when the drug level is reduced to 30 percent).
For Q at t = 0, we have:
ln |Q(0)| = -k * 0 + C
ln |Q(0)| = C
For Q at t = D, we have:
ln |Q(D)| = -kD + C
Since we are given that the drug level in the body is reduced to 30 percent of its original level, we have:
Q(D) = 0.3 * Q(0)
Substituting these values into the equation, we get:
ln |0.3 * Q(0)| = -k * D + ln |Q(0)|
Finally, we can solve for D, which will give us the number of days it takes for the drug level in the body to be reduced to 30 percent of its original level.
To write a differential equation satisfied by Q, we can use the fact that the quantity remaining decreases at a rate proportional to the quantity remaining.
Let Q(t) be the quantity of warfarin in the patient's body at time t. The rate of change of Q with respect to time is given by dQ/dt.
According to the question, the rate of decrease of the drug quantity is proportional to the quantity remaining. This can be represented by the equation:
dQ/dt = -kQ
where k is the proportionality constant. The negative sign indicates the decrease in the quantity remaining.
To find the half-life of warfarin, we can use the fact that the quantity remaining decreases to half its initial value after one half-life. Since the half-life is given as 35 hours, we can write the equation:
Q(35) = 0.5Q(0)
where Q(0) is the initial quantity of warfarin in the body.
To solve the differential equation, we can use the separation of variables method. Rearranging the equation, we have:
dQ/Q = -k dt
Integrating both sides, we get:
∫ dQ/Q = -k ∫ dt
ln|Q| = -kt + C
where C is the constant of integration.
To determine the constant C, we can substitute the initial condition Q(0) = Q0 and t = 0 into the equation:
ln|Q0| = C
Therefore, the equation becomes:
ln|Q| = -kt + ln|Q0|
Now, to find the time it takes for the drug level in the body to be reduced to 30% of its original level, we need to find the value of t when Q(t) = 0.3Q0.
Setting Q = 0.3Q0 and Q0 = 1 (assuming Q0 as the original level):
ln|0.3| = -kt + ln|1|
ln(0.3) = -kt
Solving for t, we have:
t = -ln(0.3) / k
Since the half-life is given as 35 hours, we can use this information to find the value of k by substituting t = 35 and Q(t) = 0.5Q0 into the equation:
ln(0.5Q0) = -k(35)
k = ln(0.5) / 35
Substituting this value of k into the equation for t, we can find the time it takes for the drug level in the body to be reduced to 30% of its original level.
dQ/dt = kQ
dQ/Q = k dt
lnQ = kt + c
Q = c*e^(kt)
since 2 = e^(ln2), that can of course be written by massaging the constants as
Q(t) = c*2^(-t/k)
where k=35
Q(t) = c*2^(-t/35)
Since we are given no actual amounts, we can set c=1, so that as a fraction of the initial amount,
Q(t) = 2^(-t/35)
So, just solve that when Q=.30