A drug is administered intravenously to a patient at the rate of 11 mg/h. The patient's body eliminates the drug over time so that after t hours the proportion that remains is
e−0.26t.
If the patient currently has 50 mg of the drug in her bloodstream, how much of the drug is present eight hours from now? (Round your answer to one decimal place.)

To determine the amount of the drug present in the patient's bloodstream after 8 hours, we need to account for both the administration of the drug and the elimination of the drug over time.

**Step-by-Step Solution:**

1. **Drug Elimination:**

   The proportion of the drug that remains after \( t \) hours is given by \( e^{-0.26t} \).
   
   After 8 hours, the amount of the initial 50 mg of drug remaining is:
   \[
   50 \times e^{-0.26 \times 8}
   \]
   Calculating the exponent:
   \[
   -0.26 \times 8 = -2.08
   \]
   So, the remaining drug from the initial dose after 8 hours is:
   \[
   50 \times e^{-2.08}
   \]

2. **Continuous Drug Administration:**

   The drug is administered at a constant rate of 11 mg/h. To find the total amount of drug administered over 8 hours, we integrate the rate of administration while accounting for the decay factor:
   \[
   \int_{0}^{8} 11 \cdot e^{-0.26(8-t)} \, dt
   \]
   
   We can rewrite the integral as:
   \[
   11 \int_{0}^{8} e^{-0.26(8-t)} \, dt = 11 \int_{0}^{8} e^{-2.08} e^{0.26t} \, dt
   \]
   Simplifying the exponent:
   \[
   11 \cdot e^{-2.08} \int_{0}^{8} e^{0.26t} \, dt
   \]
   
3. **Evaluate the Integral:**

   The integral of \( e^{0.26t} \) is:
   \[
   \int e^{0.26t} \, dt = \frac{1}{0.26} e^{0.26t} = \frac{1}{0.26} e^{0.26t}
   \]
   
   Evaluating this from 0 to 8:
   \[
   \frac{1}{0.26} \left[ e^{0.26 \times 8} - e^{0.26 \times 0} \right]
   \]
   Simplifying further:
   \[
   \frac{1}{0.26} \left[ e^{2.08} - 1 \right]
   \]
   
4. **Combining Terms:**

   The total amount of drug after 8 hours is the sum of the remaining initial dose and the accumulated dose from continuous administration:
   \[
   50 \times e^{-2.08} + 11 \times e^{-2.08} \times \frac{1}{0.26} \left[ e^{2.08} - 1 \right]
   \]
   
5. **Calculating Numerical Values:**

   Using the approximate value of \( e^{-2.08} \approx 0.124 \) and \( e^{2.08} \approx 8.01 \):
   \[
   50 \times 0.124 \approx 6.2
   \]
   \[
   11 \times 0.124 \times \frac{1}{0.26} \left[ 8.01 - 1 \right] = 11 \times 0.124 \times 3.846 \times 7.01 \approx 37.2
   \]
   
   Therefore, the total amount of drug after 8 hours is:
   \[
   6.2 + 37.2 = 43.4 \text{ mg}
   \]

So, after 8 hours, the amount of the drug present in the patient's bloodstream is approximately **43.4 mg**.

Would you like more details or have any questions?

---

### Additional Questions:
1. How would the drug concentration change if the administration rate was increased to 15 mg/h?
2. What would be the remaining drug amount after 12 hours instead of 8 hours?
3. How would the solution change if the elimination rate was different, say \( e^{-0.30t} \)?
4. What is the integral of the drug concentration rate over an infinite time period?
5. How does changing the initial amount of drug in the bloodstream affect the result?
6. Can we generalize this approach for any given rate of administration and elimination factor?
7. How would you approach this problem if the drug administration was not constant but variable over time?
8. What assumptions are made in this model, and how might real-life factors alter the calculations?

### Tip:
When dealing with problems involving exponential decay or growth, it's useful to remember the integral of \( e^{kt} \) is \( \frac{1}{k} e^{kt} \), which can simplify solving related integrals.

Learn how to calculate drug concentration over time with intravenous administration and exponential decay. Solve problems involving drug elimination rates and constant administration rates. Suitable for advanced high school students.

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