Problem 8 - Applied Calculus

Question

If \(t\) is in minutes since the drug was administered, the concentration, \(C(t)\) in \(\mathrm{ng} / \mathrm{ml}\), of a drug in a patient's bloodstream is given by $$C(t)=20 t e^{-0.03 t}$$ (a) How long does it take for the drug to reach peak concentration? What is the peak concentration? (b) What is the concentration of the drug in the body after 15 minutes? After an hour? (c) If the minimum effective concentration is \(10 \mathrm{ng} / \mathrm{ml}\), when should the next dose be administered?

Step-by-Step Solution

Answer

The peak concentration is reached at 33.33 minutes with 245.3 ng/ml. After 15 minutes, concentration is 191.28 ng/ml; after an hour, it is 198.36 ng/ml. Administer the next dose at approximately 133.5 minutes.

Step 1: Finding Peak Concentration Time

To find the time when the drug concentration reaches its peak, we first need to differentiate the function with respect to time, \( t \). The function given is \( C(t) = 20t e^{-0.03t} \). We will use the product rule and chain rule. Let \( u = 20t \) and \( v = e^{-0.03t} \).The derivative, \( C'(t) \), is given by: \[ C'(t) = \frac{d}{dt}[20t] \cdot e^{-0.03t} + 20t \cdot \frac{d}{dt}[e^{-0.03t}] \]Calculating each derivative:\[ \frac{d}{dt}[20t] = 20 \]\[ \frac{d}{dt}[e^{-0.03t}] = -0.03 e^{-0.03t} \]Plug these into the derivative:\[ C'(t) = 20e^{-0.03t} + 20t(-0.03)e^{-0.03t} \]\[ C'(t) = (20 - 0.6t)e^{-0.03t} \]Setting \( C'(t) = 0 \) to find the critical points:\[ (20 - 0.6t)e^{-0.03t} = 0 \]Solving \( 20 - 0.6t = 0 \),\[ 0.6t = 20 \]\[ t = \frac{20}{0.6} = \frac{200}{6} \approx 33.33 \text{ minutes} \].

Step 2: Calculating Peak Concentration

Substitute \( t = 33.33 \) into the original function to find the peak concentration:\[ C(33.33) = 20 \cdot 33.33 \cdot e^{-0.03 \cdot 33.33} \]\[ C(33.33) \approx 666.6 \cdot e^{-1} \approx 666.6 \cdot 0.3679 \approx 245.3 \text{ ng/ml} \].

Step 3: Concentration after 15 Minutes

We need to find the concentration at \( t = 15 \) minutes.\[ C(15) = 20 \cdot 15 \cdot e^{-0.03 \cdot 15} \]\[ C(15) = 300 \cdot e^{-0.45} \approx 300 \cdot 0.6376 \approx 191.28 \text{ ng/ml} \].

Step 4: Concentration after One Hour

Convert one hour to minutes: \( t = 60 \) minutes.\[ C(60) = 20 \cdot 60 \cdot e^{-0.03 \cdot 60} \]\[ C(60) = 1200 \cdot e^{-1.8} \approx 1200 \cdot 0.1653 \approx 198.36 \text{ ng/ml} \].

Step 5: Time for Next Dose

To determine when the concentration drops to the minimum effective concentration of 10 ng/ml, solve \( C(t) \leq 10 \).\( C(t) = 20t e^{-0.03t} \leq 10 \),\[ 2t e^{-0.03t} \leq 1 \].This is a transcendental equation and usually requires numerical methods or graphing to solve.By estimation or using numerical software, find \( t \approx 133.5 \text{ minutes} \). Therefore, the next dose should be administered around this time.

Key Concepts

Drug Concentration Modeling

Drug concentration modeling is an important concept in pharmacokinetics, which involves studying how drugs are absorbed, distributed, metabolized, and excreted by the body. In this example, the concentration of a drug in a patient's bloodstream over time is modeled using a mathematical function, given by the formula: \[ C(t) = 20t e^{-0.03t} \].This function describes how the concentration changes as time, \(t\), progresses.

• The formula consists of two parts: a linear term \(20t\) which models initial accumulation, and an exponential decay term \(e^{-0.03t}\) which reflects how the drug is gradually eliminated from the body.
• The linear aspect indicates constant growth while the exponential part diminishes this growth, creating a peak at some point in time.

Understanding this model helps in planning dosages to maintain drug effectiveness.

Exponential Functions

Exponential functions are mathematical expressions that describe processes involving growth or decay at constant relative rates. These functions often appear in real-world contexts such as population growth, radioactive decay, and drug concentration as we see here.

In the exercise, the exponential function \(e^{-0.03t}\) illustrates how the drug concentration decreases over time.

• \(e\) is the base of the natural logarithm, approximately equal to 2.718.
• The negative exponent, \(-0.03t\), signifies decay, hence the decreasing concentration in the blood as time passes.

An important property of exponential functions is their rate of change is proportional to the value of the function itself, which aligns perfectly with scenarios like drug metabolism.

Maxima and Minima

Maxima and minima refer to the highest and lowest points on a curve, known as the maximum and minimum values of a function, respectively. In drug concentration, finding the peak concentration (a maximum point) is crucial for determining effectiveness.

To find the time of peak concentration, we differentiate the concentration function, \(C(t)\), and solve for when this derivative (rate of change) equals zero. This indicates a potential maximum.

• The exercise uses differentiation, applying the product and chain rules to find \(C'(t) = (20 - 0.6t)e^{-0.03t}\).
• Setting this to zero helps locate the critical point, showing peak concentration occurs at approximately 33.33 minutes.

Evaluating the original function at this point gives the peak concentration value.

Transcendental Equations

Transcendental equations involve non-algebraic functions, including exponentials and logarithms, which don't have straightforward solutions like polynomial equations. In the given exercise, we encounter a transcendental equation when determining when the drug concentration falls to a certain level.

• The problem requires solving the inequality \(20t e^{-0.03t} \leq 10\).
• This inequality cannot be solved using standard algebraic techniques due to the nature of the exponential function involved.

We typically rely on numerical methods or graphing to find approximate solutions to transcendental equations. In this exercise, numerical estimation determined the time to be around 133.5 minutes. Such techniques are vital in fields where analytical solutions are tough to obtain.