Problem 57 - Applied Calculus

Question

a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. Drug Dosage A drug taken orally is absorbed into the bloodstream at the rate of te \(^{-0.5 \mathrm{t}}\) milligrams per hour, where \(t\) is the number of hours since the drug was taken. Find the total amount of the drug absorbed during the first 5 hours.

Step-by-Step Solution

Answer

3.507 mg of the drug is absorbed in the first 5 hours.

Step 1: Define the Rate Function

The rate at which the drug is absorbed is given by the function \( R(t) = t e^{-0.5t} \), where \( t \) is the time in hours. We need to determine the total drug absorbed over a set period by integrating this function.

Step 2: Set Up the Integral

The total amount of drug absorbed from time \( t = 0 \) to \( t = 5 \) is given by \( \int_{0}^{5} t e^{-0.5t} \, dt \). This integral will calculate the accumulated quantity of the substance absorbed over the specified duration.

Step 3: Apply Integration by Parts

To solve the integral \( \int t e^{-0.5t} \, dt \), we need to use integration by parts. Let \( u = t \) and \( dv = e^{-0.5t} \, dt \). Then we have \( du = dt \) and \( v = -2e^{-0.5t} \).

Step 4: Integration by Parts Formula

Using the formula for integration by parts \( \int u \, dv = uv - \int v \, du \), we calculate:\[\int t e^{-0.5t} \, dt = -2te^{-0.5t} - \int (-2)e^{-0.5t} \, dt\]Thus, simplify the second integral.

Step 5: Integrate \( v \, du \)

Integrate \( 2e^{-0.5t} \, dt \) to find:\[\int 2e^{-0.5t} \, dt = -4e^{-0.5t} + C\]

Step 6: Solve the Integral from 0 to 5

Substitute back into the integration by parts result:\[\int_{0}^{5} t e^{-0.5t} \, dt = [-2te^{-0.5t} + 4e^{-0.5t}]_{0}^{5}\]Calculate the definite integral by evaluating at \( t = 5 \) and \( t = 0 \).

Step 7: Evaluate the Bounds

Substitute \( t = 5 \): \(-2(5)e^{-2.5} + 4e^{-2.5}\)Substitute \( t = 0 \): \(-2(0)e^{0} + 4e^{0} = 4\)Simplify:\[(-10 + 4)e^{-2.5} + 4 - 0 = -6e^{-2.5} + 4 \]

Step 8: Calculate the Final Value

Calculate the numerical value:\(-6e^{-2.5} + 4 \approx -6(0.0821) + 4 = -0.493 + 4 = 3.507\)Thus, the total drug absorbed is approximately 3.507 mg.

Key Concepts

Definite Integral

Definite integrals are a powerful mathematical tool that help us calculate the total accumulation of a quantity over a specified interval. In the context of drug absorption, we often want to know how much of a drug has been absorbed by the body over a certain period of time. This is exactly what a definite integral helps us find: the total amount absorbed between two points in time.

For example, given the rate of absorption function as a continuous function, integrating this function from the initial time to some later time gives us the total amount of the substance that has been absorbed. In our exercise, we integrate the rate function from time \(t = 0\) to \(t = 5\) hours to find the total absorption over this period.

The mathematical representation is:

• The rate of absorption function \(R(t) = t e^{-0.5t}\)
• The definite integral setup: \( \int_{0}^{5} t e^{-0.5t} \, dt \)

Through this integral, we calculate the amount of drug absorbed from hour 0 to hour 5. Thus, definite integrals enable us to quantify real-world phenomena, like drug absorption, with precision.

Exponential Function

Exponential functions play a vital role in modeling various natural processes, including drug absorption in the body. An exponential function has the form \(a e^{bt}\), where \(e\) is the base of the natural logarithms, \(a\) is a constant multiplier, and \(b\) indicates the rate of growth or decay.

In the exercise, the rate of drug absorption is represented by the function \(t e^{-0.5t}\). Here, the exponential part \(e^{-0.5t}\) models the decay of the drug absorption rate over time.

Key characteristics of exponential functions:

• Decay: When \(b\) is negative, the function represents an exponential decay, meaning the process, such as drug absorption, decreases over time.
• Natural Base \(e\): The number \(e\) (~2.718) is crucial in continuous growth and decay models, making it ideal for representing how quickly a drug is absorbed and released by the body.

By understanding exponential functions, we gain insights into how the absorption rate of a drug can decrease over time, reflecting real biological processes.

Drug Absorption

Drug absorption is a crucial concept in pharmacokinetics, describing how a drug enters the bloodstream after being administered. This process is influenced by factors like the rate at which the drug dissolves and passes through biological membranes.

In our scenario, the drug is taken orally, and the absorption rate is modeled as \(R(t) = t e^{-0.5t}\), where \(t\) is the time in hours. This function captures how the drug absorption rate initially increases, peaks, and then decreases as time progresses.

Important points about drug absorption:

• Dynamic Process: Initially, the absorption rate increases, reaching a peak before it starts to taper off, reflecting the body's metabolic processing.
• Impact of Time: The exponential decay component of the function indicates how absorption slows over time, a common pattern in biological systems.

By understanding these dynamics, we can better grasp how drug dosage and timing affect effectiveness, ensuring that medications are administered in a manner that maximizes their therapeutic benefits.