Problem 70
Question
A patient is injected with a drug every 12 hours. Immediately before each injection the concentration of the drug has been reduced by \( 90\% \) and the new dose in increase the concentration by 1.5 mg/L. (a) What is the concentration after three doses? (b) If \( C_n \) is the concentration after the \( n \)th dose, find a formula for \( C_n \) as a function of \( n. \) (c) What is the limiting value of the concentration?
Step-by-Step Solution
Verified Answer
(a) 1.665 mg/L. (b) \( C_n = 0.1 C_{n-1} + 1.5 \). (c) Limit is 1.67 mg/L.
1Step 1: Understanding Drug Concentration Change
When each new dose is given, the concentration increases by 1.5 mg/L, and the remaining drug concentration is 10% of the previous one before the new dose. We need to account for these changes step-by-step.
2Step 2: Calculate After First Dose
Initially, there is no drug concentration, therefore after the first dose, the concentration is simply the increase due to the dose: \[ C_1 = 1.5 \]
3Step 3: Calculate After Second Dose
By the time of the second dose, the concentration is 10% of \( C_1 \) plus 1.5 mg/L from the new dose. \[ C_2 = 0.1 \times 1.5 + 1.5 = 0.15 + 1.5 = 1.65 \]
4Step 4: Calculate After Third Dose
Before the third dose, the concentration is 10% of \( C_2 \). After the new dose, it becomes:\[ C_3 = 0.1 \times 1.65 + 1.5 = 0.165 + 1.5 = 1.665 \]
5Step 5: General Formula for Concentration After nth Dose
The pattern shows that each subsequent concentration is \[ C_n = 0.1 C_{n-1} + 1.5 \]. Thus, the formula is iterative, depending on the previous concentrations.
6Step 6: Express Convergent Series for the Limiting Value
As n approaches infinity, the concentration reaches a stable value, or limit. Set the limit as X:\[ X = 0.1X + 1.5 \]. Solving for X gives:\[ 0.9X = 1.5 \]\[ X = \frac{1.5}{0.9} = \frac{15}{9} = \frac{5}{3} \approx 1.67 \]
7Step 7: Conclusion
After three doses, the concentration is 1.665 mg/L. The formula for subsequent doses is \( C_n = 0.1 C_{n-1} + 1.5 \), and the concentration asymptotically approaches 1.67 mg/L.
Key Concepts
Drug Concentration DecayLimiting Value in SequencesIterative Formula
Drug Concentration Decay
When a drug is administered in cycles, such as every 12 hours, the concentration in the bloodstream doesn't remain constant. Instead, it experiences a phenomenon known as drug concentration decay. This refers to the reduction of the drug's concentration over time due to metabolism and excretion by the body. In the given problem, the concentration decreases by 90% between doses. This means that only 10% of the drug remains before the next dose is administered. Understanding this decay is vital for determining the eventual steady concentration levels of the drug, known as the limiting concentration.
- Every 12 hours, 90% of the drug concentration disappears.
- This leaves 10% of the drug before the next dose.
- A new dose increases the drug concentration by 1.5 mg/L.
Limiting Value in Sequences
The limiting value of a sequence is the number that a sequence approaches as the term number increases infinitely. In pharmacology, this is particularly important as it helps predict the stable concentration of a drug after several doses. For the problem at hand, the formula evolves with each dose according to the sequence: \[C_n = 0.1 C_{n-1} + 1.5\], which is a classical example of a recursively defined sequence.
To find the limiting value, we equate the stable concentration (\(X\)) to the next calculated concentration in the sequence:\[X = 0.1X + 1.5\]. Solving for \(X\), gives:\[0.9X = 1.5\]\[X = \frac{1.5}{0.9} = \frac{5}{3} \approx 1.67\].
This means that after many doses, the concentration will stabilize at approximately 1.67 mg/L. Understanding these limits ensures that drugs are used safely and effectively. It guarantees that over time, what's administered leads to the desired effect without inadvertently causing harm due to accumulation or lack thereof.
To find the limiting value, we equate the stable concentration (\(X\)) to the next calculated concentration in the sequence:\[X = 0.1X + 1.5\]. Solving for \(X\), gives:\[0.9X = 1.5\]\[X = \frac{1.5}{0.9} = \frac{5}{3} \approx 1.67\].
This means that after many doses, the concentration will stabilize at approximately 1.67 mg/L. Understanding these limits ensures that drugs are used safely and effectively. It guarantees that over time, what's administered leads to the desired effect without inadvertently causing harm due to accumulation or lack thereof.
Iterative Formula
An iterative formula allows us to compute each term of a sequence based on the previous one, making it easier to model processes like drug administration. In our scenario, we define the concentration after the nth dose with the iterative formula \[C_n = 0.1 C_{n-1} + 1.5\]. This formula accounts for both the decay of the drug concentration and the infusion of a new dose.
By using an iterative process, we can predict future concentrations after any number of doses:
By using an iterative process, we can predict future concentrations after any number of doses:
- Start with initial known values (e.g., \(C_1 = 1.5\) for the first dose).
- Compute each subsequent value using the iteration formula.
- Observe how the values approach the limiting value over time.
Other exercises in this chapter
Problem 69
For what values of \( r \) is the sequence \( \left\\{ nr^n \right\\} \) convergent?
View solution Problem 70
Use multiplication of division of power series to find the first three nonzero terms in the Maclaurin series for each function. \( y = e^x \ln (1 + x) \)
View solution Problem 70
(a) If \( \left \\{ a_n \right\\} \) is convergent, show that \( \displaystyle\lim_{n\to\infty} a_{n+1} = \displaystyle\lim_{n\to\infty} a_n \) (b) A sequence \
View solution Problem 71
Use multiplication of division of power series to find the first three nonzero terms in the Maclaurin series for each function. \( y = (\arctan x)^2 \)
View solution