/*! 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 49 The concentration \(C(t)\) of a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The concentration \(C(t)\) of a drug in a patient's bloodstream \(t\) hours after administration is given by $$C(t)=\frac{4 t}{3+t^{2}}$$ where \(C(t)\) is in milligrams per liter. During what time interval will the concentration be greater than 1 milligram per liter?

Short Answer

Expert verified
The concentration of the drug will be greater than 1 milligram per liter between 1 and 3 hours after administration.

Step by step solution

01

Identify when concentration is equal to 1 mg/L

For \(C(t) = 1\), solve for \(t\). So, we have \(1 = \frac{4t}{3+t^2}\). Cross multiplication yields \(4t = 1 \times (3 + t^2)\), or \(4t = 3 + t^2\). Rearranging, we have \(t^2 - 4t + 3 = 0\). This is a simple quadratic equation which can be solved.
02

Solve the equation

Solve \(t^2 - 4t + 3 = 0\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -4, c = 3\). Therefore, \(t = 1,3\).
03

Test the intervals

We now have three intervals: \(t < 1\), \(1 < t < 3\), and \(t > 3\). Choose test values in these intervals, for instance, 0, 2, and 4. If \(C(t) > 1\) for a test value, then \(C(t) > 1\) for all \(t\) in that interval.
04

Conclusion

Testing gives \(C(0) = 0\), \(C(2) = 1.33\), and \(C(4) = 0.571\). Hence, the concentration is above 1 milligram per liter only in the interval \(1 < t < 3\) hours.

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Key Concepts

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

Quadratic Equations
A quadratic equation is a second-order polynomial equation in a single variable x, with a coefficient that is not zero. It is of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). The quadratic equation represents a parabola on a graph, and it has distinctive properties such as a vertex, axis of symmetry, and can have either two real roots, one real root (if the parabola touches the x-axis), or no real roots (if the parabola doesn't intersect the x-axis).

In the context of the exercise, we are dealing with the quadratic equation \(t^2 - 4t + 3 = 0\) to find the points in time where the concentration equals 1 milligram per liter. Going through this process helps us understand how the concentration changes over time and at what specific points it hits a certain value. Recognizing and solving these kinds of equations is fundamental to many areas of science and engineering, especially in pharmacokinetics where we analyze drug concentration in the bloodstream.
Solving Quadratic Formula
To solve the quadratic equation \(ax^2 + bx + c = 0\), you can use the quadratic formula, which is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). This formula provides an exact solution for the roots of any quadratic equation.

For the exercise in question, by plugging in \(a = 1\), \(b = -4\), and \(c = 3\) into the formula, we find the two time points (roots) when the concentration is precisely 1 milligram per liter. The linear term \(b\) determines the symmetry of the parabola, while the constant term \(c\) indicates where the parabola crosses the y-axis. The discriminant \(b^2 - 4ac\) is crucial as it informs us about the nature of the roots. In this case, the discriminant is positive, indicating two distinct real roots, which correctly correspond to the times when the concentration is 1 milligram per liter. This process is imperative for determining the precise time intervals for specific drug concentrations.
Function Analysis
Function analysis involves studying the properties of a function to understand its behavior. This includes analyzing intervals for which the function is increasing or decreasing, its maximum and minimum values, and points where the function equals certain values.

In our exercise, analyzing the function \(C(t)=\frac{{4t}}{{3+t^2}}\) involves finding when the drug concentration is above a particular threshold—in this case, greater than 1 milligram per liter. After using the quadratic formula to find potential crossing points, we test intervals to determine where the concentration exceeds that threshold. By selecting test points within each interval, we can effectively gauge the behavior of the function and identify that the concentration \(C(t)\) only surpasses the desired 1 milligram per liter within the time interval between 1 and 3 hours. Such analysis is vital for practical applications, such as determining the effective duration of a medication in a patient's body or scheduling subsequent doses.

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