Problem 82
Question
For the following exercises, express a rational function that describes the situation. The concentration \(C\) of a drug in a patient's bloodstream \(t\) hours after injection is given by \(C(t)=\frac{2 t}{3+t^{2}}\). What happens to the concentration of the drug as \(t\) increases?
Step-by-Step Solution
Verified Answer
As \(t\) increases, the concentration \(C(t)\) approaches zero.
1Step 1: Analyze the Rational Function
The function given is \(C(t) = \frac{2t}{3 + t^2}\). We know that this function describes how the concentration of a drug changes over time \(t\). Because it is a rational function, we should consider the behavior as \(t\) approaches large values (i.e., as \(t\) becomes very large).
2Step 2: Determine the Behavior of the Denominator
The denominator, \(3 + t^2\), increases as \(t\) increases because \(t^2\) becomes very large. This suggests that as \(t\) increases, the value of the denominator \(3 + t^2\) will be dominated by \(t^2\).
3Step 3: Analyze the Numerator
The numerator is \(2t\), and as \(t\) increases, \(2t\) also increases but at a linear rate, unlike the quadratic rate of growth of the denominator.
4Step 4: Evaluate the Limit as \(t\) Approaches Infinity
We want to see what happens to \(C(t)\) as \(t\) goes to infinity. The expression simplifies to finding the limit: \[\lim_{t \to \infty} \frac{2t}{3 + t^2}.\] To do this, we simplify by dividing the numerator and the denominator by \(t^2\): \[\lim_{t \to \infty} \frac{\frac{2}{t}}{\frac{3}{t^2} + 1}.\] As \(t\) approaches infinity, \(\frac{2}{t} \to 0\) and \(\frac{3}{t^2} \to 0\), leaving \(\frac{0}{1} = 0\).
5Step 5: Interpret the Result
The result \(\lim_{t \to \infty} C(t) = 0\) means that over time, the concentration of the drug in the bloodstream decreases and approaches zero as time \(t\) increases.
Key Concepts
Drug ConcentrationLimits and InfinityBehavior Analysis
Drug Concentration
Drug concentration refers to the amount of a drug present in the bloodstream at any given time. In this problem, the function \(C(t) = \frac{2t}{3+t^2}\) models how this concentration changes after a drug has been injected. Here ‘\(t\)’ represents time in hours post-injection. This rational function captures how both the absorption and elimination processes affect drug levels over time.
- The numerator \(2t\) implies that initially, as time increases, the drug concentration in the blood starts to increase.
- The denominator \(3 + t^2\) represents influences that slow down or mitigate the increase of concentration over time, such as metabolic processes or excretion.
Limits and Infinity
Limits explain the behavior of a function as the input grows exceptionally large or small. In this rational function \(C(t) = \frac{2t}{3 + t^2}\), we use limits to understand the behavior of drug concentration over a long period.As \(t\) approaches infinity, the emphasis is on how both the numerator and the denominator behave:
- \(2t\) increases linearly, but is outpaced by \(t^2\) in the denominator, a quadratic term.
- To compute \(\lim_{t \to \infty} \frac{2t}{3 + t^2}\), we simplify by dividing both parts by \(t^2\).
- This simplification yields \(\frac{\frac{2}{t}}{1 + \frac{3}{t^2}}\), reducing further to \(\lim_{t \to \infty} \frac{0}{1} = 0\).
Behavior Analysis
Behavior analysis of the function is about understanding how the concentration changes over time. The rational function \(C(t) = \frac{2t}{3 + t^2}\) shows several key behaviors:
- For small \(t\), the concentration increases because \(2t\) grows faster than \(3 + t^2\), reflecting a rising effect of the drug post-administration.
- Over time, the quadratic term \(t^2\) in the denominator dominates, reducing the overall concentration despite the linear growth in the numerator.
- The ultimate behavior, as indicated by the limit at infinity, is a drop towards zero concentration. This suggests that the drug is gradually cleared from the bloodstream.
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