Problem 536 - Precalculus

Question

For the following exercises, use the given rational function to answer the question. The concentration \(C\) of a drug in a patient's bloodstream \(t\) hours after injection in 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

Answer

As \(t\) increases, the concentration \(C(t)\) approaches 0.

Step 1: Understanding the Function

The given rational function is \(C(t)=\frac{2t}{3+t^{2}}\). This represents the concentration of a drug over time \(t\). As \(t\) increases, we want to analyze what happens to \(C(t)\).

Step 2: Analyze the Denominator

As \(t\) increases, the denominator \(3 + t^2\) becomes very large because \(t^2\) grows quickly. This increase in the denominator causes the overall value of the fraction \(C(t)\) to decrease.

Step 3: Analyze the Numerator

In the numerator, \(2t\) increases linearly as \(t\) increases. However, because \(t^2\) in the denominator increases faster than \(t\) in the numerator, the increase in \(2t\) is outpaced by the increase in \(3 + t^2\).

Step 4: Determine the Limit

To find the behavior of \(C(t)\) for very large \(t\), analyze the function as \(t\to\infty\). The highest power of \(t\) in both numerator and denominator is what matters: \(\frac{2t}{t^2} = \frac{2}{t}\). As \(t\to\infty\), \(\frac{2}{t}\to 0\).

Step 5: Conclusion

As \(t\) increases, \(C(t)\) approaches 0. This means the concentration of the drug decreases over time and becomes negligible as time goes on.

Key Concepts

End Behavior

When we talk about the end behavior of a rational function, we are considering what happens to the function's value as its input becomes very large or very small. For the drug concentration function given as \( C(t) = \frac{2t}{3 + t^2} \), we're interested in what happens when \( t \) becomes very large—in this case, as time continues to pass after an injection.

As \( t \) increases, the function's end behavior is driven by the growth rates of the numerator and denominator. Since the denominator \( 3 + t^2 \) grows more rapidly due to the \( t^2 \) term than the linear \( 2t \) in the numerator, the overall fraction becomes very small. Therefore, the end behavior shows the concentration approaching zero, meaning that the drug's presence in the bloodstream diminishes over time.

Numerator Analysis

Analyzing the numerator of a rational function helps us understand part of its changing behavior. In our given function, \( C(t) = \frac{2t}{3 + t^2} \), the numerator is \( 2t \). This is a simple linear expression, which implies that it grows in direct proportion to \( t \).

As \( t \) increases, \( 2t \) also increases, doubling each time \( t \) increases by one unit. However, this linear growth is relatively slow compared to quadratic growth such as \( t^2 \). Therefore, while \( 2t \) does indeed increase, it is not enough to maintain the function's value as the denominator also increases, and it does so much faster.

Denominator Analysis

The denominator of our function \( C(t) = \frac{2t}{3 + t^2} \), \( 3 + t^2 \), plays a crucial role in defining the function's overall behavior. The inclusion of \( t^2 \) is significant because it grows quadratically—meaning that as \( t \) increases, \( t^2 \) increases rapidly, much faster than any linear term can increase.

Even though there is a constant term 3, its effect becomes negligible as \( t^2 \) becomes larger and larger. This rapid increase in the denominator makes the fraction \( C(t) = \frac{2t}{3 + t^2} \) get smaller over time, causing the concentration of the drug to decrease.

Limits

Limits help us formally define the behavior of rational functions as the inputs become very large or very small. For our concentration function \( C(t) = \frac{2t}{3 + t^2} \), we use limits to describe what happens as \( t \to \infty \).

To compute this limit, we focus on the highest power of \( t \) in both the numerator and the denominator. In this case, the dominant term in the denominator is \( t^2 \) and the dominant term in the numerator is \( t \). The simplified form of the limit expression becomes \( \frac{2t}{t^2} = \frac{2}{t} \).

As \( t \to \infty \), \( \frac{2}{t} \to 0 \). Hence, the concentration of the drug approaches zero as time progresses. This limit tells us that eventually, the drug's concentration will become negligible.