Problem 40 - Applied Calculus

Question

The quantity of a drug in the bloodstream \(t\) hours after a tablet is swallowed is given, in \(\mathrm{mg}\), by $$q(t)=20\left(e^{-t}-e^{-2 t}\right)$$ (a) How much of the drug is in the bloodstream at time \(t=0 ?\) (b) When is the maximum quantity of drug in the bloodstream? What is that maximum? (c) In the long run, what happens to the quantity?

Step-by-Step Solution

Answer

(a) 0 mg at \( t=0 \); (b) Maximum of 5 mg at \( t=\ln(2) \); (c) Approaches 0 mg in the long run.

Step 1: Calculate Initial Quantity at t=0

To find the quantity of the drug at time \( t=0 \), substitute \( t=0 \) into the equation \( q(t) = 20(e^{-t} - e^{-2t}) \). This yields:\[ q(0) = 20(e^0 - e^0) = 20(1 - 1) = 0 \]Thus, there is no drug in the bloodstream immediately after the tablet is swallowed.

Step 2: Find the Time of Maximum Quantity

To determine when the maximum quantity is reached, we need to find the critical points by differentiating the function \( q(t) \) and setting the derivative equal to zero. The function is:\[ q(t) = 20(e^{-t} - e^{-2t}) \]Differentiating with respect to \( t \) gives:\[ q'(t) = 20(-e^{-t} + 2e^{-2t}) \]Set \( q'(t) = 0 \):\[ 20(-e^{-t} + 2e^{-2t}) = 0 \]\[ -e^{-t} + 2e^{-2t} = 0 \]\[ e^{-t} = 2e^{-2t} \]Divide both sides by \( e^{-2t} \):\[ e^{t} = 2 \]Take the natural logarithm of both sides:\[ t = \ln(2) \]This is the time when the maximum amount of drug is present.

Step 3: Calculate Maximum Quantity

Substitute \( t = \ln(2) \) back into the original function \( q(t) \) to find the maximum quantity:\[ q(\ln(2)) = 20(e^{-\ln(2)} - e^{-2\ln(2)}) \]Use the property \( e^{\ln(x)} = x \):\[ q(\ln(2)) = 20\left(\frac{1}{2} - \frac{1}{4}\right) = 20\left(\frac{1}{4}\right) = 5 \]Thus, the maximum quantity is 5 mg.

Step 4: Determine Long-term Behavior

The long-term behavior of the quantity can be analyzed as \( t \to \infty \). Observe the function:\[ q(t) = 20(e^{-t} - e^{-2t}) \]As \( t \to \infty \), \( e^{-t} \to 0 \) and \( e^{-2t} \to 0 \), hence:\[ q(t) \to 20(0 - 0) = 0 \]Therefore, in the long run, the quantity of the drug in the bloodstream approaches zero.

Key Concepts

Drug Concentration Model

The drug concentration model helps us understand how the amount of a drug changes in the bloodstream over time. This is important for determining how often a person might need to take their medication to maintain effective levels. The model uses a mathematical function to express the drug quantity as a function of time. In our exercise, the formula given is:

• \( q(t)=20(e^{-t}-e^{-2t}) \)

where \( q(t) \) is the quantity of the drug at time \( t \), and both exponential terms show how the concentration decreases over time.
Initially, when \( t = 0 \), the value of \( q(t) \) was found to be 0, suggesting that there's a delay before the drug begins to reach significant levels in the bloodstream. Understanding how the drug's amount changes helps in predicting the drug's therapeutic effectiveness and scheduling doses.

Derivative Calculation

Derivative calculation is key in analyzing how functions change, and is central to understanding the rate of change of the drug concentration over time. We use derivatives to identify critical points, which can help in determining maximum and minimum concentrations.
For the function \( q(t) \), the derivative is calculated as follows:

• \( q'(t) = 20(-e^{-t} + 2e^{-2t}) \)

This derivative tells us at what rate the concentration is increasing or decreasing. By setting \( q'(t) = 0 \), one can find the rate change is zero, indicating potential points where the drug reaches maximum or minimum concentrations. This is a common technique in differential calculus to find where a function hits its peak.

Exponential Decay

Exponential decay is a mathematical concept that describes how quantities decrease rapidly at first before leveling off. This is often seen in radioactive decay, population decline, and drug concentrations diminishing over time.
In our drug concentration model:

• \( e^{-t} \) and \( e^{-2t} \) represent exponential decay terms.

These terms imply that as \( t \) increases, both \( e^{-t} \) and \( e^{-2t} \) approach zero. As a result, the drug concentration in the bloodstream declines over time as well, moving from an initial high toward zero. Exponential decay is crucial for understanding how fast a drug will clear from the body, impacting how dosage intervals are set.

Critical Points

Critical points in a function are where the first derivative is zero or undefined, indicating potential maxima or minima. Finding these points helps reveal where key changes in behavior occur for the function.

• The derivative \( q'(t) = 20(-e^{-t} + 2e^{-2t}) \) was set to zero, leading to \( e^{-t} = 2e^{-2t} \).
• By simplifying, we found that \( t = \ln(2) \) is a critical point.

Plugging this back into the original function \( q(t) \) reveals the maximum drug level to be 5 mg. Understanding where these critical points are is vital in pharmacokinetics, ensuring the drug stays at effective levels while avoiding overdose.