Problem 71 - Calculus: Early Transcendentals

Question

A patient takes 150 mg of a drug at the same time every day. Just before each tablet is taken, \( 5% \) of the drug remains in the body. (a) What quantity of the drug is in the body is in the body after the third tablet? After the \( n \)th tablet? (b) What quantity of the drug remains in the body in the long run?

Step-by-Step Solution

Answer

After the third tablet, 7.89375 mg remains; after the nth tablet, use the formula. Long term remains at 157.8947 mg.

Step 1: Determine Drug Retention After One Day

Since only 5% of the drug remains in the body after one day, we calculate the amount remaining of the initial dosage of 150 mg. Therefore, the remaining drug after one day is \( 0.05 \times 150 = 7.5 \) mg.

Step 2: Calculate Drug Sum After Second Tablet

On the second day, before taking another 150 mg, there is 7.5 mg from the first dose. After taking the second dose, the total in the body becomes \( 7.5 + 150 = 157.5 \) mg. But at the end of this day, 5% of it will remain, so \( 0.05 \times 157.5 = 7.875 \) mg remains after the second day.

Step 3: Calculate Drug Sum After Third Tablet

On the third day, before taking another tablet, 7.875 mg remains in the body. After taking the third tablet, the total becomes \( 7.875 + 150 = 157.875 \) mg. Thus, at the end of the third day, \( 0.05 \times 157.875 = 7.89375 \) mg remains after the third day.

Step 4: Develop General Formula for Drug After \( n \) Tablets

The remaining drug after the \( n \)th tablet can be modeled by the recursive formula: \( A_n = 150 + 0.05 \times A_{n-1} \). This is a geometric progression where each new amount considers the residue from the previous day.

Step 5: Determine Drug Amount in the Long Term

In the long term, the daily intake equals the daily loss, leading to a stable state. The stable amount is the solution where \( A_{n+1} = A_n = L \). Solving \( L = 150 + 0.05L \), we find \( L = \frac{150}{0.95} = 157.8947 \) mg.

Key Concepts

Geometric Progression

When discussing drug retention in the body over time, understanding a geometric progression is crucial. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number. In our case, the drug remaining in the body follows this pattern. The drug amount before taking a new dose reduces to 5% of the prior day's total due to the body's natural processes. For example:

• Day 1: Initially, 150 mg is taken. After one day, only 5% remains, so 7.5 mg of the drug stays in the system.
• Day 2: At the beginning of this day, 7.5 mg is retained from the previous day, and another 150 mg is added when the next dose is taken, following the geometric sequence of previous drug residues.

Geometric progressions are a clear mirror of how residue accumulation during consecutive drug doses works, progressively building up yet defined by the constant shrinking rate each day.

Recursive Formula

A recursive formula is a powerful tool in mathematics that helps us calculate something based on earlier terms. In the context of drug retention, this formula explains how the quantity of the drug in the body evolves day by day. Here, the recursive formula for drug retention is given by:\[A_n = 150 + 0.05 imes A_{n-1}\]- \(A_n\) represents the drug amount in the body after the \(n\)th tablet.- \(A_{n-1}\) is the quantity from the previous analysis (\(n-1\)th day).- The term \(150\) is added to account for the new dose taken every day.- The factor \(0.05\) reflects the retention percentage.This recursive nature offers a systematic approach to predict the drug's quantity for any day \(n\). It's an iterative process, building upon each day's remaining amount.

Long-term Stability

In the long run, many systems naturally settle into a pattern or achieve stability after a period of fluctuation. The concept of long-term stability in drug retention refers to the point where the daily intake of the drug equals the drug lost. This balance results in the remaining amount in the body stabilizing over time. To find this equilibrium, we use the equation from the recursive formula:\[L = 150 + 0.05L\]Here, \(L\) symbolizes the stable amount. We solve for \(L\) to find:

• Subtract \(0.05L\) from both sides: \(0.95L = 150\)
• Divide by \(0.95\): \(L = \frac{150}{0.95} = 157.8947\)

This solution, 157.8947 mg, represents the stabilization point, where intake and residue loss total to the same amount daily. Understanding long-term stability is crucial in pharmacokinetics, ensuring that therapeutic levels of a drug are maintained without risking over-accumulation.