Problem 27 - 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 194.22 ng/ml.

Step 1: Understand the Problem

We need to determine the bioavailability of a drug, which is the area under the concentration curve of the drug from \(t=0\) to \(t=3\). The concentration function is given as \(C = 15t e^{-0.2t}\). This requires calculating the definite integral of the function from 0 to 3.

Step 2: Set Up the Integral

The definite integral that represents the area under the concentration curve from \(t=0\) to \(t=3\) is given by \[ \int_{0}^{3} 15t e^{-0.2t} \, dt\]. This integral will give us the bioavailability.

Step 3: Apply Integration by Parts

To solve the integral \(\int 15t e^{-0.2t} \, dt\), we use integration by parts. Let- \(u = t\) and \(dv = 15e^{-0.2t} \, dt\), then- \(du = dt\) and \(v = -75 e^{-0.2t}\).Apply integration by parts formula: \(\int u \, dv = uv - \int v \, du\).

Step 4: Calculate the Parts

Substitute the parts into the integration by parts formula:\[ = -75t e^{-0.2t} - \int -75 e^{-0.2t} \, dt\].The new integral \(\int 75 e^{-0.2t} \, dt = -375 e^{-0.2t} + C\) after integrating.

Step 5: Simplify the Solution

Combine the expressions:\[ -75t e^{-0.2t} + (-375 e^{-0.2t})\] Evaluate the definite integral at the limits 0 and 3.

Step 6: Evaluate the Definite Integral

Plug the limits \(t=3\) and \(t=0\) into the simplified expression:\[ \left[ -75(3)e^{-0.6} - 375e^{-0.6} \right] - \left[ -75(0)e^{0} - 375e^{0} \right]\].Calculate this to find the bioavailability.

Step 7: Compute the Numerical Result

Calculate the numeric results,- For \(t=3\), compute \(-75 \times 3 \times e^{-0.6} - 375 \times e^{-0.6}\)- For \(t=0\), compute \(-375 \times e^{0} = -375\). Solving this gives:\(-225e^{-0.6} - 375e^{-0.6} + 375\).

Step 8: Final Calculation

Simplify the expression further:\[= (375 - 600e^{-0.6}) \].Calculate this using a calculator to find the precise numerical value of the bioavailability.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus. It helps us find the area under a curve, which in the context of this problem, represents the bioavailability of a drug over a time period. When we talk about a definite integral, we are essentially calculating the accumulation of quantities over a given interval. In this exercise, we are calculating the integral of the concentration function from time \(t = 0\) to \(t = 3\).

The notation for this operation often looks like this: \int_{a}^{b} f(x) \, dx\. Here, \(a\) and \(b\) are the limits of integration, representing the interval over which we are measuring. In our case, \(a = 0\) and \(b = 3\).

• The definite integral provides a numerical value for the area under the graph of the function.
• In physical applications, this area can correspond to total accumulated quantities like distance, volume, or, as here, drug concentration.

Understanding the definite integral is crucial for solving real-world problems where accumulation matters. In this scenario, it allows us to determine the total effect a drug has over time.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is used to simplify the integration of products of functions, which are harder to integrate directly. The formula is given by \int u \, dv = uv - \int v \, du\. It's particularly useful when dealing with products like \(te^{-0.2t}\), seen in the concentration function of the drug.

In this exercise, we set \(u = t\) and \(dv = 15e^{-0.2t} \, dt\). By differentiating \(u\) and integrating \(dv\), we find \(du = dt\) and \(v = -75e^{-0.2t}\).

• Using the formula, we substitute these into: \int u \, dv = uv - \int v \, du\.
• The function transforms into simpler integrable expressions.
• This method requires experience to choose \(u\) and \(dv\) correctly, which can often save computation time.

Integration by parts offers a strategic approach to tackle challenging integrals that arise when functions are multiplied together, streamlining the process of integration.

Concentration Function

In this exercise, the concentration function \(C = 15t e^{-0.2t}\) represents how the concentration of the drug changes over time. It's crucial to understand this function's behavior to accurately evaluate the bioavailability of the drug.

The concentration function is important because:

• It shows how quickly the drug's presence in the bloodstream changes.
• By plotting this function, one can visualize peaks and decay in concentration.
• The pattern informs medical professionals about the drug's dynamics and duration in the body.

Analyzing the concentration function helps in predicting how much of the drug remains in the system over time, which can be vital for dosing schedules and treatment plans. In mathematical terms, the concentration function is given by an exponential decay influenced by time after an initial increase, due to the \(te^{-0.2t}\) form of the function. Understanding this type of exponential function is key in many scientific and engineering applications, particularly in pharmacokinetics.