/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 Many people take aspirin on a re... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 59

Many people take aspirin on a regular basis as a preventive measure for heart disease. Suppose a person takes \(80 \mathrm{mg}\) of aspirin every 24 hours. Assume also that aspirin has a half-life of 24 hours; that is, every 24 hours, half of the drug in the blood is eliminated. a. Find a recurrence relation for the sequence \(\left\\{d_{n}\right\\}\) that gives the amount of drug in the blood after the \(n\) th dose, where \(d_{1}=80\). b. Using a calculator, determine the limit of the sequence. In the long run, how much drug is in the person's blood? c. Confirm the result of part (b) by finding the limit of \(\left\\{d_{n}\right\\}\) directly.

Short Answer

Expert verified
Answer: The long-term amount of aspirin in the person's blood is 160mg.

Step by step solution

01

Determine the initial conditions and dosage information.

Given the initial dosage \(d_1 = 80mg\) and the half-life of aspirin as 24 hours, we can define the recurrence relation as \(d_n = d_{n-1} + 80\), where \(d_{n-1}\) is the amount of aspirin left in the person's blood from the previous day.
02

Calculate the amount of aspirin remaining after each day.

Since the half-life of aspirin is 24 hours, every 24 hours half of the drug in the blood is eliminated. Therefore, we can write the amount of aspirin remaining after n days as: \(d_n = \frac{1}{2}d_{n-1} + 80\). This is the recurrence relation for the amount of drug in the blood after the \(n\)-th dose. #b. Limit of the sequence#
03

Determine the limit of the sequence.

To find the limit of the sequence as \(n\) approaches infinity, we can first recognize that it is a geometric sequence with a common ratio of \(\frac{1}{2}\) and a constant term of 80. We can find the limit by using the formula for the sum of an infinite geometric series: L = \(\frac{initial\_term}{1 - common\_ratio} = \frac{80}{1 - \frac{1}{2}} = \frac{80}{\frac{1}{2}} = 160\). In the long run, there will be 160mg of aspirin in the person's blood. #c. Confirm the result#
04

Confirm the result of part (b) by finding the limit of \(\left\\{d_{n}\right\\}\) directly.

To confirm the result in part (b), we need to find the limit of the sequence \(\left\\{d_{n}\right\\}\) as \(n\) approaches infinity. We can rewrite the recurrence relation as \(d_n - \frac{1}{2}d_{n-1} = 80\), and as \(n\) approaches infinity, both \(d_n\) and \(d_{n-1}\) will approach the same limit, which we can denote as L: \(L - \frac{1}{2}L = 80\). Solving for L, we get: \(\frac{1}{2}L = 80\). \(L = 80 \times 2 = 160\). This confirms the result of part (b), that in the long run, there will indeed be 160mg of aspirin in the person's blood.

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