Problem 42 - Applied Calculus

Question

The quantity of a drug in the bloodstream \(t\) hours after a tablet is swallowed is given, in 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; (b) Maximum is 10 mg at \( t = \,\ln(2) \); (c) Quantity approaches 0 mg as \( t \to \infty \).

Step 1: Substitute to Find Initial Quantity

To find the quantity of the drug at time \( t = 0 \), we substitute \( t = 0 \) into the function \( q(t) = 20(e^{-t} - e^{-2t}) \). This gives: \[ q(0) = 20(e^{0} - e^{0}) = 20(1 - 1) = 0 \] mg.

Step 2: Find the Derivative for Maximum Quantity

To find the maximum quantity of the drug, we first differentiate \( q(t) = 20(e^{-t} - e^{-2t}) \) with respect to \( t \). The derivative is: \[ q'(t) = 20(-e^{-t} + 2e^{-2t}) = 20(e^{-2t} - e^{-t}) \].

Step 3: Set the Derivative to Zero

We find the critical points by setting \( q'(t) = 0 \): \[ 20(e^{-2t} - e^{-t}) = 0 \] which simplifies to \( e^{-2t} = e^{-t} \). Solving gives \( t = \, \ln(2) \).

Step 4: Verify Maximum Using Second Derivative Test

The second derivative \( q''(t) \) is found by differentiating \( q'(t) \): \[ q''(t) = 20(-2e^{-2t} + e^{-t}) \]. At \( t = \, \ln(2) \), \( q''(t) \) is negative, confirming a local maximum.

Step 5: Find Maximum Quantity

Substitute \( t = \, \ln(2) \) back into \( q(t) \) to find the maximum quantity: \[ q(\ln(2)) = 20(e^{-ln(2)} - e^{-2\ln(2)}) = 20\left(\frac{1}{2} - \frac{1}{4}\right) = 10 \text{ mg} \].

Step 6: Evaluate Long-term Behavior

As \( t \to \infty \), both \( e^{-t} \) and \( e^{-2t} \) approach zero, thus \( q(t) = 20(e^{-t} - e^{-2t}) \to 0 \). Therefore, the quantity of the drug approaches zero as time increases.

Key Concepts

Drug Absorption Modeling

In the world of pharmacology, understanding how drugs enter and behave in the body is crucial. This process is often modeled as a function of time, which describes how much of a drug remains in the bloodstream over a period. This model helps in determining effective dosages and timing for medication.
The formula used in this scenario is:\[ q(t) = 20(e^{-t} - e^{-2t}) \]where \( q(t) \) represents the quantity of the drug in the bloodstream, measured in milligrams, and \( t \) represents time in hours. The equation relies on exponential terms, which we'll delve into next.   In this model:

• The term \( e^{-t} \) represents the natural exponential decay of the drug, indicating how the concentration decreases over time.
• The term \( e^{-2t} \) shows a faster rate of decay, often associated with how quickly the drug is being metabolized and eliminated from the body.

Through this type of modeling, it is possible to assess when the drug will achieve its peak concentration in the bloodstream, as well as its depletion over time. Understanding these dynamics is vital for ensuring that the drug remains effective without causing toxicity.

Exponential Functions

Exponential functions like \( e^{-t} \) and \( e^{-2t} \) are fundamental to modeling decay processes such as drug absorption. These are functions where the variable \( t \) is an exponent, making them unique in how they change over time. In this context, they are used to effectively simulate the rates at which the drug concentration decreases in the bloodstream.
The natural exponential function \( e^x \) has special properties:

• As \( x \) increases, \( e^x \) grows rapidly, but for negative \( x \), \( e^{-x} \) diminishes towards zero.
• This property is particularly useful for modeling processes like radioactive decay or, in our case, drug elimination.

 With the given drug model equation \( q(t) = 20(e^{-t} - e^{-2t}) \), it's the difference between two decay rates that gives the characteristic shape of the concentration over time. Understanding exponential functions is critical to comprehending how mathematical modeling can predict real-world phenomena effectively.

Optimization Problems

Optimization problems involve finding the most efficient or optimal solution to a given problem. In applied calculus, these problems often require finding maximum or minimum values of a function. The given exercise is an example of such a problem where we aim to find the maximum concentration of drug in the bloodstream.
To solve optimization problems:

• First, we differentiate the function to determine the rate of change. For our drug model, we found \( q'(t) = 20(e^{-2t} - e^{-t}) \).
• Next, set the derivative equal to zero to find critical points. This step is crucial because it helps identify potential maxima or minima.
• Finally, use the second derivative test or other methods to confirm that the identified critical points are indeed a maximum or minimum.

In this exercise, the maximum drug concentration occurs at \( t = \ln(2) \), and the value is found to be 10 mg.This approach allows us to determine when intervention or adjustment might be necessary in drug administration to keep the treatment effective.