Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Drug dosage A drug is eliminated from the body through urine. Suppose that for an initial dose of 10 milligrams, the amount \(A(t)\) in the body \(t\) hours later is given by \(A(t)=10(0.8)^{t}\) (a) Estimate the amount of the drug in the body 8 hours after the initial dose. (b) What percentage of the drug still in the body is eliminated each hour?

Step-by-Step Solution

Answer

(a) 1.68 mg; (b) 20% per hour.

Step 1: Understand the Function

The function given is \( A(t) = 10(0.8)^t \) where \( A(t) \) represents the amount of drug in milligrams present in the body after \( t \) hours. We can see that it's an exponential decay model because the base of the exponential term, \( 0.8 \), is less than 1.

Step 2: Calculate Amount After 8 Hours

We need to find \( A(8) \). Substitute \( t = 8 \) into the function: \( A(8) = 10(0.8)^8 \). Now, calculate \( (0.8)^8 \):\[ (0.8)^8 \approx 0.16777216 \].Then calculate \( A(8) = 10 \times 0.16777216 \approx 1.6777216 \) mg.

Step 3: Interpret Initial Conditions

Initially, when \( t = 0 \), the drug amount is \( A(0) = 10(0.8)^0 = 10 \) mg. Every hour, the amount of drug is multiplied by \( 0.8 \).

Step 4: Determine Percentage Elimination Each Hour

Since each hour the amount is multiplied by \( 0.8 \), this means \( 20\% \) of the drug is eliminated each hour. This is because \( 1 - 0.8 = 0.2 \) which represents \( 20\% \).

Key Concepts

Drug Elimination

When a drug is ingested into the body, it doesn't stay there indefinitely; it goes through a process of elimination. This often occurs through the urine, where the body filters out and removes the drug over time.
In our situation, the elimination process can be described mathematically using an exponential decay function. This function makes it possible to predict how much of a drug remains in the body at any time. Understanding this process is important, especially for medications that need to be maintained at certain levels in the bloodstream.
Drug elimination rates can vary depending on factors like kidney function, but the exponential function gives a general idea of how the concentration decreases over time.

Exponential Function

An exponential function is used to describe processes that decrease or increase at a consistent rate, like our drug elimination case. The function given is: \[ A(t) = 10(0.8)^t \]Here,

• \( A(t) \) represents the remaining drug amount,
• \( \text{10} \) is the initial amount of the drug, and
• \( 0.8 \) is the decay factor.

This "decay factor" of \( 0.8 \) is less than 1, showing that it represents a decrease in the drug amount over time.
Each hour, the current drug amount is multiplied by \( 0.8 \), reducing its value. So, the original amount decreases steadily as time progresses. Such functions are very helpful in modeling natural or human activities where something diminishes or accumulates over time.

Percentage Elimination

To understand how much of the drug is eliminated hourly, we look at the decay factor in the exponential function.
The current factor of \( 0.8 \) in \( A(t) = 10(0.8)^t \) means that 80% remains each hour. Hence, the percentage eliminated is the difference from 100%:
\(1.0 - 0.8 = 0.2\), or \(20\%\).

• This tells us that 20% of the drug is removed from the body every hour.
• The body's efficient process steadily minimizes drug concentration.

This information is crucial for calculating proper dosages and timings for medication, ensuring it remains effective without reaching toxic levels.