QuestionA drug starts at $70 \mathrm{mg}$ and drops to $49 \mathrm{mg}$ in 1 hour. Find the hourly percentage removed.

Studdy Solution

STEP 1

Assumptions1. The initial amount of drug in the blood is $a_{0}=70 \mathrm{mg}$ . The amount of drug in the blood after one hour is $a_{1}=49 \mathrm{mg}$
3. No drug is added to the blood between $t=0$ and $t=1$
4. The drug has first order elimination kinetics, which means the rate of elimination is proportional to the amount of drug in the body

STEP 2

First, we need to find the amount of drug that is removed in one hour. We can do this by subtracting the amount of drug at $t=1$ from the initial amount of drug.
$Drug\, removed = a_{0} - a_{1}$

STEP 3

Now, plug in the given values for $a_{0}$ and $a_{1}$ to calculate the amount of drug removed.
$Drug\, removed =70 \mathrm{mg} -49 \mathrm{mg}$

STEP 4

Calculate the amount of drug removed.
$Drug\, removed =70 \mathrm{mg} -49 \mathrm{mg} =21 \mathrm{mg}$

STEP 5

Now that we have the amount of drug removed, we can find the percentage of drug that is removed each hour. This can be calculated by dividing the amount of drug removed by the initial amount of drug and then multiplying by100 to convert to percentage.
$Percentage\, removed = \left(\frac{Drug\, removed}{a_{0}}\right) \times100$

STEP 6

Plug in the values for the drug removed and $a_{0}$ to calculate the percentage removed.
$Percentage\, removed = \left(\frac{21 \mathrm{mg}}{70 \mathrm{mg}}\right) \times100$

STEP 7

Calculate the percentage of drug removed.
$Percentage\, removed = \left(\frac{21 \mathrm{mg}}{70 \mathrm{mg}}\right) \times100 =30\%$ The percentage of drug removed each hour is30%.

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