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Chapter 5: Problem 58

Medication can be administered to a patient in different ways. For a given method, let \(c(t)\) denote the concentration of medication in the patient's bloodstream (measured in \(\mathrm{mg} / \mathrm{L}\) ) thours after the dose is given. Over the time interval \(0 \leq t \leq b\) the area between the graph of \(c=c(t)\) and the interval \([0, b]\) indicates the "availability" of the medication for the patient's body over that time period. Determine which method provides the greater availability over the given interval. $$\begin{aligned}&\text { Method } 1: c(t)=5\left(e^{-0.2 t}-e^{-t}\right)\\\&\text { Method } 2: c(t)=4\left(e^{-0.2 t}-e^{-3 t}\right) ; \quad[0,24]\end{aligned}$$

Short Answer

Expert verified
Method 2 provides greater availability over the interval \([0, 24]\).

Step by step solution

01

Understand the Availability Calculation

The availability of medication for a method over a time interval is calculated using the definite integral. For each method, we integrate the concentration function, \(c(t)\), over the interval \([0, 24]\).We are given \(c_1(t) = 5(e^{-0.2t} - e^{-t})\) for Method 1 and \(c_2(t) = 4(e^{-0.2t} - e^{-3t})\) for Method 2.
02

Calculate the Integral for Method 1

For Method 1, we need to compute the integral: \[ \int_{0}^{24} 5(e^{-0.2t} - e^{-t}) \, dt \]First, compute the antiderivative:- The antiderivative of \(e^{-0.2t}\) is \(-\frac{1}{0.2} e^{-0.2t} = -5e^{-0.2t}\).- The antiderivative of \(e^{-t}\) is \(-e^{-t}\).Thus, the antiderivative of \(5(e^{-0.2t} - e^{-t})\) is:\[ -5(-5e^{-0.2t} - e^{-t}) = 25e^{-0.2t} - 5e^{-t} \].
03

Evaluate the Integral for Method 1

Evaluate the integral \(\left[25e^{-0.2t} - 5e^{-t}\right]_{0}^{24}\):- Substitute \(t = 24\): \[ 25e^{-0.2 \times 24} - 5e^{-24} \approx 25 \times 0.000335 - 5 \times 0.000000006\]- Substitute \(t = 0\):\[ 25e^{0} - 5e^{0} = 25 - 5 = 20\]Calculate the difference:\( 25 \times 0.000335 - 5 \times 0.000000006 - 20 \approx 20.008375 - 0 \equiv \approx 5.008375\).
04

Calculate the Integral for Method 2

For Method 2, compute the integral: \[ \int_{0}^{24} 4(e^{-0.2t} - e^{-3t}) \, dt \]First, compute the antiderivative:- The antiderivative of \(e^{-0.2t}\) is \(-5e^{-0.2t}\).- The antiderivative of \(e^{-3t}\) is \(-\frac{1}{3}e^{-3t}\).Thus, the antiderivative of \(4(e^{-0.2t} - e^{-3t})\) is:\[ -20e^{-0.2t} + \frac{4}{3}e^{-3t} \].
05

Evaluate the Integral for Method 2

Evaluate the integral \(\left[-20e^{-0.2t} + \frac{4}{3}e^{-3t}\right]_{0}^{24}\):- Substitute \(t = 24\):\[ -20e^{-0.2 \times 24} + \frac{4}{3}e^{-3 \times 24} \approx -20 \times 0.000335 + 0 \] (since \(e^{-72}\) effectively approaches 0)- Substitute \(t = 0\):\[ -20e^{0} + \frac{4}{3}e^{0} = -20 + \frac{4}{3} \approx -18.6667 \]Calculate the difference:\( -20 \times 0.000335 + 0 + 18.6667 \approx 18.6667\).
06

Compare the Availability of Both Methods

Now that we have computed the areas (availability) for both methods:- Method 1: Approximately 5.008375- Method 2: Approximately 18.6667Since 18.6667 is greater than 5.008375, Method 2 provides greater availability over the time interval \([0, 24]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Concentration Functions
Concentration functions represent the amount of a drug in a patient's bloodstream over time. When calculating the concentration of medication, the function is denoted by \(c(t)\), where \(t\) represents time in hours. For this exercise, we examine how two different methods of drug administration affect the concentration functions over the time interval from 0 to 24 hours.
  • Method 1 uses the function \(c_1(t) = 5(e^{-0.2t} - e^{-t})\).
  • Method 2 uses the function \(c_2(t) = 4(e^{-0.2t} - e^{-3t})\).
Each concentration function provides a unique curve depicting how the drug's availability changes over time. The goal is to find out which method maintains a higher availability, represented by the area under the curve. Finding the area under these curves involves calculating definite integrals, which sum up the continuous change in drug concentration across the time interval.
The Role of the Antiderivative in Integration
In calculus, an antiderivative is a function that reverses differentiation. We find the antiderivative of a function to determine the area beneath its curve. For calculating the availability of drugs, finding the antiderivative of the concentration function is key.For Method 1, the concentration function is \(5(e^{-0.2t} - e^{-t})\). We find:
  • The antiderivative of \(e^{-0.2t}\) is \(-5e^{-0.2t}\).
  • The antiderivative of \(e^{-t}\) is \(-e^{-t}\).
Hence, the overall antiderivative for Method 1 is:\[25e^{-0.2t} - 5e^{-t} \].For Method 2, the function is \(4(e^{-0.2t} - e^{-3t})\). Performing similar operations leads us to:
  • The antiderivative of \(e^{-0.2t}\) is \(-5e^{-0.2t}\).
  • The antiderivative of \(e^{-3t}\) is \(-\frac{1}{3}e^{-3t}\).
Thus, the antiderivative for Method 2 is:\[-20e^{-0.2t} + \frac{4}{3}e^{-3t} \].Calculating these antiderivatives helps in evaluating the definite integral over the desired time interval to find the total availability of medication.
Method Comparison: Which Provides Greater Availability?
After calculating the definite integrals for both methods, we can compare the results to determine which method provides greater drug availability. For Method 1, evaluating the definite integral gives us an availability of approximately 5.008375 units. This value is derived by substituting the upper and lower bounds of the time interval, \([0, 24]\), into the antiderivative function and finding the difference.For Method 2, the process results in an availability of approximately 18.6667 units. The same method of substitution into the respective antiderivative is employed.By comparing these two values:
  • Method 1: 5.008375 units
  • Method 2: 18.6667 units
We conclude that Method 2 provides a significantly higher drug availability over the 24-hour period. This evaluation highlights not just the importance of knowing how to compute definite integrals but also illustrates how different dosage methods can result in varying effectiveness of drug administration.

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Most popular questions from this chapter

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