Problem 26 - Applied Calculus

Question

The concentration, \(C,\) in \(\mathrm{ng} / \mathrm{ml},\) of a drug in the blood as a function of the time, \(t,\) in hours since the drug was administered is given by \(C=15 t e^{-0.2 t} .\) The area under the concentration curve is a measure of the overall effect of the drug on the body, called the bioavailability. Find the bioavailability of the drug between \(t=0\) and \(t=3\).

Step-by-Step Solution

Answer

The bioavailability of the drug between \(t=0\) and \(t=3\) is approximately 84.24 ng/ml-hr.

Step 1: Understanding the Problem

We are provided with the concentration function of a drug over time, \( C(t) = 15t e^{-0.2t} \). The goal is to calculate the bioavailability, which is the area under the curve of this function from \( t=0 \) to \( t=3 \). This requires integration of the function over the specified interval.

Step 2: Set Up the Integral

Bioavailability is found by integrating the concentration function over the given interval \([0, 3]\). The integral to solve is: \[ \int_{0}^{3} 15t e^{-0.2t} \, dt \] This integral will give us the total area under the concentration curve between the specified times.

Step 3: Use Integration by Parts

Integration by parts is required here, as the integrand is a product of functions. Let \( u = 15t \) and \( dv = e^{-0.2t} \, dt \). Then \( du = 15 \, dt \) and \( v = -5e^{-0.2t} \) (since integration of \( e^{-0.2t} \) gives \( -5e^{-0.2t} \)). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 4: Solve Using Integration by Parts

Apply the formula: \[ \int 15t e^{-0.2t} \, dt = 15t(-5e^{-0.2t}) \Big|_0^3 + 5 \int e^{-0.2t} \, dt \Bigg|_0^3 \] Calculate the definite integral for \( \int e^{-0.2t} \, dt \), which results in \( -5e^{-0.2t} \).

Step 5: Evaluate the Expressions

Calculate \[ 15t(-5e^{-0.2t}) \Big|_0^3 = [-75t e^{-0.2t}]_0^3 \] and \[ 5 \int e^{-0.2t} \, dt \Bigg|_0^3 = [5(-5e^{-0.2t})]_0^3 \]. These are computed: 1. At \( t = 3 \): First term: \[ -75(3)e^{-0.6} = -225e^{-0.6} \] Second term: \[ 25 (-e^{-0.6} + 1) \] 2. At \( t = 0 \): First term is 0, Second term is also 0. Compute each expression, factor in \( e^{-0.6} \) using a calculator.

Step 6: Final Calculation and Result

Combine the results from Step 5: \[ -225e^{-0.6} + 25(-e^{-0.6} + 1) = -200e^{-0.6} + 25 \]Calculate using a numerical value for \( e^{-0.6} \approx 0.5488 \) to find \[ -200 \times 0.5488 + 25 \approx -109.76 + 25 \approx 84.24 \]. Thus, the bioavailability of the drug is \( 84.24 \) ng/ml-hr.

Key Concepts

Integration by Parts

The method of Integration by Parts arises from the product rule of differentiation. It's particularly useful when you have an integral of a product of two functions. The basic idea here is to transform the integral of a product into simpler terms. Given two functions, say \( u(x) \) and \( v(x) \), the method is based on the formula: \[\int u \, dv = uv - \int v \, du\]When applying integration by parts, the goal is to carefully choose \( u \) and \( dv \) such that differentiating \( u \) and integrating \( dv \) simplifies the problem. In this exercise, for the function \( C(t) = 15t e^{-0.2t} \), we let \( u = 15t \) and \( dv = e^{-0.2t} \, dt \). Subsequently, differentiating and integrating respectively, yields \( du = 15 \, dt \) and \( v = -5e^{-0.2t} \). Applying the formula, the integral becomes easier to evaluate step by step. Always remember:

• Choose \( u \) as a function that simplifies upon differentiation.
• \( dv \) should be easy to integrate.
• Calculate \( uv \) and \( \int v \, du \), then substitute back to find the solution.

Definite Integral

A definite integral is used to compute the total accumulation of quantities, and in many physical contexts, it represents the total area under a curve between two points on the x-axis. Unlike indefinite integrals which result in a family of functions, definite integrals provide a numerical value as a result. It is expressed with limits of integration, for example, \( \int_{a}^{b} f(x) \, dx \). In our context, we aim to find the bioavailability of a drug by calculating the area under the concentration curve from \( t = 0 \) to \( t = 3 \) using the concentration function \( C(t) = 15t e^{-0.2t} \). Steps to evaluating a definite integral include:

• Find the antiderivative of the function.
• Evaluate the antiderivative at the upper limit of integration \( b \).
• Subtract the evaluation of the antiderivative at the lower limit \( a \).

When computed correctly, this gives the total area under the curve for the interval \([0, 3]\), which represents the drug's bioavailability.

Concentration Function

The concentration function describes how the concentration of a drug changes over time in the bloodstream after being administered. This function can often be depicted as an exponential decay due to the way substances are metabolized and eliminated by the body. In our exercise, the concentration of the drug is provided by the function: \( C(t) = 15t e^{-0.2t} \).Here's why understanding this function is crucial:

• Concentration functions tend to show an initial increase when the drug enters the bloodstream.
• The exponential term \( e^{-0.2t} \) signifies a decrease or decay rate over time.
• Calculating the area under this curve helps determine the bioavailability, reflecting the effectiveness and duration the drug stays active in the body.

In practical terms, studying the concentration function gives insights into dosage and frequency for achieving desired drug effects. Careful analysis using tools like integration ensures optimal therapeutic outcomes with minimal side effects.