An intravenous line provides a continuous flow of drug directly into the blood. Assuming no initial drug in the blood, the amount of drug in the blood t hours after the dosing begins
in m(t) - (a/k)(1-e^-kt), for t (=>)0, where k is the rate constant (again related to half life) and A is the rate at which drug flows into the blood (in units of mg/hr)
7.
Suppose an antibiotic with a half life of 12 hours is given to a patient intravenously at a rate of A=50mg/hr.
find the rate constant k
8.
what is the steady-state level of the antibiotic in step 7? that is, evaluate lim(t-> infinity)m(t).
9.
In general, what is the steady state level of a drug delivered by infusion in terms of A and K?
in general at what time does the drug level reach 90% of the steady state level, in terms of A and K?
10.
Based on patients weight, a doctor targets a steady state level of tetracycline of 100mg through infusion.
what infusion rate A should be used? The half life of tetracycline is 9 hr.
11.
in step 10, at what time does the drug level reach 90% of the steady state level? At that time,
how much drug has actually been delivered?
12.
Suppose a patient has been on infusion of tetracycline for 72 hours with infusion rate as found in step 10,
when the delivery is terminated. How long does it take for the drug level in the blood to reach 2mg?
7. To find the rate constant k, we can use the half-life formula:
t(1/2) = (ln(2))/k
Given that the half-life is 12 hours, we can substitute the values:
12 = (ln(2))/k
Rearranging the equation, we get:
k = (ln(2))/12
Therefore, the rate constant k is (ln(2))/12.
8. The steady-state level of the antibiotic can be found by evaluating the limit of m(t) as t approaches infinity:
lim(t-> infinity)m(t) = lim(t-> infinity)((A/k)(1 - e^(-kt)))
As t approaches infinity, the exponential term e^(-kt) approaches zero. Therefore, the equation simplifies to:
lim(t-> infinity)m(t) = (A/k)
So, the steady-state level of the antibiotic is A/k.
9. In general, the steady-state level of a drug delivered by infusion is equal to the rate of drug flow into the blood (A) divided by the rate constant (k). So, the steady-state level is A/k.
The drug level reaches 90% of the steady-state level when the equation m(t) = 0.9(A/k) is true. Solving for t:
0.9(A/k) = (A/k)(1 - e^(-kt))
Canceling out (A/k) on both sides:
0.9 = 1 - e^(-kt)
Rearranging the equation:
e^(-kt) = 0.1
Taking the natural logarithm (ln) of both sides:
-ln(10) = -kt
Simplifying:
t = (ln(10))/k
So, in general, the drug level reaches 90% of the steady-state level at t = (ln(10))/k.
10. To find the infusion rate A that should be used to target a steady-state level of 100mg for tetracycline with a half-life of 9 hours, we can use the steady-state level formula:
A/k = 100
We have already calculated the rate constant k in step 7:
k = (ln(2))/12
Substituting the values, we get:
A/((ln(2))/12) = 100
Rearranging the equation:
A = (100*(ln(2)))/12
Therefore, the infusion rate A that should be used is (100*(ln(2)))/12.
11. In step 10, we found that the infusion rate A that achieves a steady-state level of 100mg for tetracycline is (100*(ln(2)))/12.
To find the time at which the drug level reaches 90% of the steady-state level, we can use the formula derived in step 9:
t = (ln(10))/k
Substituting the value of k from step 7:
t = (ln(10))/((ln(2))/12)
Simplifying:
t = 12*ln(10)/ln(2)
To calculate how much drug has actually been delivered at that time, substitute t into the formula for m(t):
m(t) = (A/k)(1 - e^(-kt))
m(t) = ((100*(ln(2)))/12)/((ln(2))/12)(1 - e^(-((ln(2))/12)*t))
Simplifying:
m(t) = 100(1 - e^(-((ln(2))/12)*t))
So, m(t) gives the amount of drug delivered at time t.
12. If a patient has been on infusion of tetracycline for 72 hours, with the infusion rate calculated in step 10, we can find how long it takes for the drug level in the blood to reach 2mg by solving for t in the equation:
2 = (A/k)(1 - e^(-kt))
Substituting the values:
2 = ((100*(ln(2)))/12)/((ln(2))/12)(1 - e^(-((ln(2))/12)*72))
Simplifying:
2 = 100(1 - e^(-((ln(2))/12)*72))
Now, we can solve this equation to find the value of t.
7. To find the rate constant (k), we need to use the information given in the question. We are told that the half-life (T) of the drug is 12 hours and the rate of infusion (A) is 50 mg/hr.
The half-life of a drug is defined as the time it takes for the concentration of the drug to decrease by half. In this case, the half-life (T) is given as 12 hours.
We can use the formula for half-life to find the rate constant:
T = (ln(2)) / k
Plugging in the values:
12 = (ln(2)) / k
To find k, we can rearrange the equation:
k = (ln(2)) / 12
Therefore, the rate constant (k) is (ln(2)) / 12.
8. The steady-state level of the antibiotic (m(t)) can be found by evaluating the limit as t approaches infinity of m(t).
lim(t-> infinity) m(t) = lim(t-> infinity) (A / k)
Substituting the values from step 7:
lim(t-> infinity) m(t) = (50 mg/hr) / [(ln(2)) / 12]
Evaluating this expression will give you the steady-state level of the antibiotic.
9. In general, the steady-state level of a drug delivered by infusion (m(t)) is given by the expression A / k.
10. To find the infusion rate A that should be used to target a steady-state level of 100 mg of tetracycline, we need to use the given half-life (T) of 9 hours.
Using the formula for half-life:
T = (ln(2)) / k
Plugging in the given values:
9 = (ln(2)) / k
Solving for k:
k = (ln(2)) / 9
We can now use the formula for steady-state level (m(t) = A / k) and substitute the desired steady-state level (m(t) = 100 mg) to solve for A:
100 = A / [(ln(2)) / 9]
Solving for A will give you the required infusion rate.
11. To find the time it takes for the drug level to reach 90% of the steady-state level, we need to find the time (t) when m(t) = 0.9 * steady-state level.
Using the equation for m(t):
m(t) = (A / k) * (1 - e^(-kt))
Setting m(t) = 0.9 * steady-state level and solving for t will give you the time it take for the drug level to reach 90% of the steady-state level.
To find the amount of drug actually delivered at that time, you can substitute the value of t into the equation m(t) = (A / k) * (1 - e^(-kt)) and solve for m(t).
12. To find the time it takes for the drug level in the blood to reach 2 mg after the infusion is terminated, we need to solve for t when m(t) = 2 mg.
Using the equation for m(t):
m(t) = (A / k) * (1 - e^(-kt))
Setting m(t) = 2 and solving for t will give you the time it takes for the drug level in the blood to reach 2 mg after the infusion is terminated.