Problem 41 - 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

Initially, 0 mg. Maximum of 5 mg at \( t = \ln(2) \). Long-term, it approaches 0 mg.

Step 1: Evaluate Initial Quantity

To find the quantity of the drug at time \( t=0 \), substitute \( t=0 \) into the function \( q(t) \):\[q(0) = 20(e^{0} - e^{0}) = 20(1 - 1) = 20 \times 0 = 0 \text{ mg}\]This means initially, no drug is in the bloodstream.

Step 2: Determine Maximum Quantity

To find when the drug is at its maximum, we take the derivative \( q'(t) \) and set it to zero,\[q'(t) = 20(-e^{-t} + 2e^{-2t})\]Setting the derivative equal to zero,\[-20e^{-t} + 40e^{-2t} = 0\]This can be simplified to:\[e^{-t}(20 - 40e^{-t}) = 0\]Solving for \( e^{-t} = \frac{1}{2} \), we find:\[t = -\ln(\frac{1}{2}) = \ln(2)\]Thus, the maximum occurs at \( t = \ln(2) \) hours.

Step 3: Evaluate Maximum Quantity

Substitute \( t = \ln(2) \) back into the original function to find the maximum quantity:\[q(\ln(2)) = 20(e^{-\ln(2)} - e^{-2\ln(2)})\]Using the properties of logarithms and exponents, we get:\[q(\ln(2)) = 20(\frac{1}{2} - \frac{1}{4}) = 20(\frac{1}{4}) = 5 \text{ mg}\]Thus, the maximum quantity of drug is 5 mg.

Step 4: Long-term Behavior

Examine the limiting behavior as \( t \to \infty \). Since both \( e^{-t} \) and \( e^{-2t} \) approach zero,\[q(t) = 20(e^{-t} - e^{-2t}) \to 0 - 0 = 0\]In the long run, the drug quantity in the bloodstream approaches zero.

Key Concepts

Exponential Decay

Exponential decay is a pattern of reduction that we often see in contexts like drug concentration in the bloodstream. In the function given in the exercise, \( q(t) = 20(e^{-t} - e^{-2t}) \), the term \( e^{-t} \) signifies that the quantity decreases over time.

Exponential functions like \( e^{-t} \) decrease quickly at first and then level off slowly. This mirrors how substances like drugs dissipate in the body: they are rapidly processed initially, and then less so over time as their concentration dwindles.

• The term \( e^{-t} \) represents a decline at a rate proportional to its current value.
• The double exponent \( e^{-2t} \) shows an even steeper decay, suggesting parts of the drug dissolve more quickly.

Understanding how exponential decay fits into real-world scenarios helps in predicting the effects and planning the timing of drug doses.

Calculus Derivatives

Derivatives are a fundamental tool in calculus used to find how a function changes. In this exercise, finding the derivative of the function \( q(t) \) reveals the rate of change of the drug quantity at any given time. The derivative \( q'(t) = 20(-e^{-t} + 2e^{-2t}) \) shows how rapidly the drug concentration in the bloodstream changes.

Setting \( q'(t) = 0 \) helps us find critical points, which might indicate maximum or minimum values of the function. This process is central in finding **maximum values** of drug concentration:

• Finding where the derivative equals zero tells us where the concentration doesn't change further, marking a peak.
• From the solution \( -20e^{-t} + 40e^{-2t} = 0 \), solving it finds the time at which this maximum occurs, which is \( t = \ln(2) \).

Understanding derivatives allows us to predict critical points where changes in drug concentration reach their peak or trough.

Maximum and Minimum Values

Spotting the **maximum quantity** of drug in the bloodstream involves using derivatives to identify changes. When we find \( q'(t) = 0 \), it indicates possible maximum or minimum values of the function. For this particular case:

• The function peaks at \( t = \ln(2) \) hours, derived through critical point analysis of \( q'(t) \).
• Plugging this time back into the original function, \( q(\ln(2)) = 5 \text{ mg} \), determines the maximum concentration of the drug in the bloodstream.

Understanding how maximum values reflect the peak concentration reached is crucial for effective drug dosage planning. This point of inflection is where the body's absorption and drug processing are most effective.

Limits and Asymptotic Behavior

In mathematical analysis, investigating the behavior of a function as it approaches infinity is essential to understanding its **long-term behavior**. For this exercise:

• The function \( q(t) = 20(e^{-t} - e^{-2t}) \) approaches zero as time \( t \to \infty \).
• Both exponential terms \( e^{-t} \) and \( e^{-2t} \) tend towards zero, leading the overall drug concentration to diminish to zero eventually.

This asymptotic behavior tells us that no matter how slowly the drug is processed after a point, it will eventually be entirely out of the bloodstream.

Understanding these limits aids in long-term treatment strategies, ensuring medicine does not linger ineffectively in the body.