suppose m(0) milligrams of a drug are put in the blood of an injection. The amount of drug t hours after the injection is given by
m(t)=m(o)e^-kt, for t (=>) 0, where k is the rate constant, which is related to the half life. we also treat oral administration
of drugs as an injection, although the model is less accurate because of the drug must be absorbed into the blood through the stomach
1.
ibuprofen has a short half-life of 1.5 hours. find the rate constant k for ibuprofen and write the function that gives the drug level after t hours.
graph the drug function m with m0=400 for 0(<=)t(<=)10 hours. how much drug remains in the blood 4 hr and 8hr after a 400-mg dose is taken?
2.
show that if the half-life t1/2 of a drug is known, then its rate constant at k=ln 2/t 1/2
3.
how many hours after taking a dose of ibuprofen does the amount of drug in the
blood reach 1% of the amount of the initial dose?
4.
the sedative diazepam has a half-life of 7 hr. find the drug function m for diazepam.
graph the drug function with m(0)=5 for 0(<=) t (<=)48 hours.
how much drug remains in the blood 12 hr and 24 hr after a 5-mg dose is taken?
5.
the antibiotic tetracycline has a half-life of 9 hours. suppose a doctor wishes a patient to have a 100mg of tetracycline in the blood 18 hours after an injection. what initial does meets his requirement?
6.
twelvee hours after a 200 mg dose of a drug is injected. the drug level in thebloodd is 75mg. what is the approximate half-life drug?
1.
.5 = e^1.5k
ln .5 = 1.5 k
ln.5÷1.5 = k
m(4) = 400e^4(ln.5÷1.5)=62.996 mg
m(8) = 400e^8(ln.5÷1.5)=9.921 mg
2.
If half life =t1/2
.5 = e^kt1/2
ln.5 = kt1/2
k=ln.5/t1/2 hr
3.
.01 = e^t(ln.5÷1.5)
ln.01 =t( ln.5 ÷1.5)
ln .01/( ln.5 ÷1.5)=t
9.966 hr=t
To answer these questions, we need to use the given formulas for the drug level in the blood and the relationship between the rate constant and the half-life. Let's go through each question step by step:
1. Ibuprofen has a half-life of 1.5 hours.
To find the rate constant, we can use the formula: k = ln(2) / t1/2. Plugging in the values, we get k = ln(2) / 1.5, which is approximately 0.4621.
The function that gives the drug level after t hours is: m(t) = m(0) * e^(-kt). With m(0) = 400 and k = 0.4621, the function becomes: m(t) = 400 * e^(-0.4621t).
To graph the drug function, you can use a graphing calculator or software. Plot the function for t from 0 to 10 hours.
To find how much drug remains in the blood at 4 hours and 8 hours after a 400-mg dose is taken, substitute the respective values of t into the drug function. For t = 4, calculate m(4) = 400 * e^(-0.4621 * 4). For t = 8, calculate m(8) = 400 * e^(-0.4621 * 8).
2. To show that if the half-life of a drug is known, then its rate constant k = ln(2) / t1/2:
We start with the drug function: m(t) = m(0) * e^(-kt).
If we let t = t1/2, the drug level will be reduced to half of its initial amount, so m(t1/2) = m(0) / 2.
Substitute these values into the drug function and solve for the rate constant k:
m(t1/2) = m(0) * e^(-kt1/2)
m(0) / 2 = m(0) * e^(-k * t1/2)
1/2 = e^(-k * t1/2)
ln(1/2) = -k * t1/2
ln(2^-1) = -k * t1/2
-1 * ln(2) = -k * t1/2
k = ln(2) / t1/2.
Hence, if the half-life of a drug is known, we can find the rate constant k using k = ln(2) / t1/2.
3. To determine the number of hours after taking a dose of ibuprofen when the amount of drug in the blood reaches 1% of the initial dose:
We can set up the equation m(t) = 0.01 * m(0), where m(t) is the drug level at time t and m(0) is the initial dose.
Substitute m(t) = m(0) * e^(-kt) into the equation and solve for t:
m(0) * e^(-kt) = 0.01 * m(0)
e^(-kt) = 0.01
Using the property of logarithms, take the natural logarithm of both sides:
-kt = ln(0.01)
t = ln(0.01) / -k
Substitute the given rate constant k for ibuprofen into the equation to calculate the time t.
4. Diazepam has a half-life of 7 hours.
The drug function m(t) for diazepam is given by m(t) = m(0) * e^(-kt).
To find k, we can use k = ln(2) / t1/2, where t1/2 is the half-life. Plugging in t1/2 = 7, we get k = ln(2) / 7.
With m(0) = 5 and k = ln(2) / 7, the drug function becomes: m(t) = 5 * e^(-(ln(2) / 7) * t).
To find how much drug remains in the blood at 12 hours and 24 hours after a 5-mg dose is taken, substitute the respective values of t into the drug function. For t = 12, calculate m(12) = 5 * e^(-(ln(2) / 7) * 12). For t = 24, calculate m(24) = 5 * e^(-(ln(2) / 7) * 24).
5. To determine the initial dose of tetracycline to have 100 mg in the blood after 18 hours:
We need to solve the equation m(t) = 100, where m(t) is the drug level at time t.
Using the given drug function m(t) = m(0) * e^(-kt), we have:
m(0) * e^(-k * 18) = 100
Solve for the initial dose m(0) by dividing both sides by e^(-k * 18):
m(0) = 100 / e^(-k * 18)
Substitute the given half-life and calculate k using k = ln(2) / t1/2. Then substitute both k and t into the equation to determine the initial dose m(0).
6. To find the approximate half-life of the drug when the drug level is 75 mg after 12 hours:
We can rearrange the drug function m(t) = m(0) * e^(-kt) to solve for k:
k = ln(m(t) / m(0)) / -t
Given m(t) = 75 and m(0) = 200, substitute these values into the equation along with t = 12 to calculate the approximate rate constant k. To find the approximate half-life, use t1/2 = ln(2) / k.