Problem 68 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The concentration (in millimeters/cubic centimeter) of a certain drug in a patient's bloodstream \(t\) hr after injection is given by $$ C(t)=\frac{0.2 t}{t^{2}+1} $$ Sketch the graph of the function \(C\) and interpret your results.

Step-by-Step Solution

Answer

The graph of \(C(t) = \frac{0.2t}{t^2 + 1}\) has a critical point at approximately \(t \approx 0.56\) hours, where the concentration of the drug in the patient's bloodstream is at its highest. The function is increasing up until that point, then decreasing thereafter, with a horizontal asymptote of \(y = 0\). This indicates that the drug's concentration initially increases rapidly, peaks at about 0.56 hours post-injection, and then gradually decreases, approaching but never actually reaching zero.

Step 1: Identify the function and its domain

Given function: \[ C(t) = \frac{0.2t}{t^2 + 1} \] The function is a rational function, and its domain is all real numbers, since the denominator never equals zero.

Step 2: Calculate the critical points

The critical points are the points where the derivative of the function is either equal to zero or not defined. Let's find the derivative: \[ C'(t) = \frac{d}{dt} \left( \frac{0.2t}{t^2 + 1} \right) \] We can use the quotient rule: \[ C'(t) = \frac{(t^2+1)(0.2)-(0.2t)(2t)}{(t^2+1)^2} \] Simplify the derivative: \[ C'(t) = \frac{0.2t^2 + 0.2 - 0.4t^3}{(t^2 + 1)^2} \] Find the points where the derivative equals zero or is not defined: The derivative is always defined since the denominator has no zeros. To find where the derivative is equal to zero, set the numerator equal to zero: \[ 0.2t^2 + 0.2 - 0.4t^3 = 0 \] This equation is difficult to solve analytically, but we can graph it and find the approximate critical points.

Step 3: Analyze the behavior of the function

Using graphing software or a graphing calculator, we find that the critical point occurs at approximately \(t \approx 0.56\) hours. We can now analyze the increasing and decreasing intervals of the function. Since there is only one critical point, we can test values on either side of that point: - Left of the critical point: Choose \(t = 0\), and \(C'(0) = 0.2 > 0\), hence the function is increasing in this interval. - Right of the critical point: Choose \(t = 1\), and \(C'(1) = -0.1 < 0\), hence the function is decreasing in this interval. Finally, we'll analyze the horizontal asymptotes of the function. Since the degree of the denominator is greater than the degree of the numerator, the function will have a horizontal asymptote at \(y = 0\).

Step 4: Sketch and interpret the graph

From the above analysis, we can sketch the graph of the function with the following characteristics: - Domain: All real numbers - Critical point: Approximately at \(t \approx 0.56\) hours - Increasing interval: \((-\infty, 0.56)\) - Decreasing interval: \((0.56, \infty)\) - Horizontal asymptote: \(y = 0\) Interpretation: The concentration of the drug in the patient's bloodstream increases rapidly at first, reaching its highest concentration 0.56 hours after the injection. From that point onwards, the concentration decreases over time, never reaching zero but getting infinitely close to zero as time passes. Note: It's highly recommended to use graphing software or a graphing calculator to verify these results and get an accurate graph.

Key Concepts

Critical Points

Critical points in a rational function like this one help us understand where the behavior of the function changes from increasing to decreasing, or vice versa. To find these critical points, we first take the derivative of the given function. In this exercise, the derivative was calculated using the quotient rule, a handy tool when you have fractions involved.

The critical points are found where the derivative equals zero or where it is undefined. In our exercise, the derivate was noted to be always defined because the denominator of the derivative has no zeros. So, we focus on the numerator to find where it equals zero. Solving this equation led us to the approximate critical point around 0.56 hours.

• This means at time = 0.56 hours, the behavior of the function changes.
• Before this point, the function is increasing.
• After this point, it starts to decrease.

By testing points from both intervals (before and after 0.56), we confirmed this change. This information is crucial for sketching the graph and understanding how the drug concentration changes over time.

Horizontal Asymptotes

Horizontal asymptotes are lines that the graph of a function approaches as the input either increases or decreases indefinitely. For rational functions like the one in this exercise, the degree of the numerator and the degree of the denominator play a crucial role in determining if there are horizontal asymptotes and where they are.

In this case, we consider the degrees of the polynomial expressions. The function given was:\[ C(t) = \frac{0.2t}{t^2 + 1} \]Here, the numerator has degree 1 and the denominator has degree 2. Since the degree of the denominator is greater than that of the numerator, the horizontal asymptote of the function is at \( y = 0 \).

• This indicates that as \( t \) approaches both positive and negative infinity, the value of \( C(t) \) gets closer and closer to 0.
• The drug concentration, therefore, never fully reaches zero but approaches it over a long period.

Understanding horizontal asymptotes is vital to visualizing and interpreting rational functions because they tell us about the behavior of the function far from the starting point.

Derivative Calculation

Calculating the derivative of a rational function is key to understanding its rate of change over time. In this exercise, we used the quotient rule, which is suitable when differentiating functions in the form of a fraction, consisting of one function divided by another.

The quotient rule states that for two functions \( u(t) \) and \( v(t) \), their derivative is:\[ \frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{v(t) \cdot u'(t) - u(t) \cdot v'(t)}{v(t)^2} \]Applying this to our function:\[ C(t) = \frac{0.2t}{t^2 + 1} \]We designated \( u(t) = 0.2t \) and \( v(t) = t^2 + 1 \). Calculating the derivatives of these parts and substituting them into the quotient rule gave us the derivative function \( C'(t) \).

• This resulted in a complex expression, but it allows us to identify critical points by setting the derivative equal to zero.
• Knowing the derivative can help predict the function's behavior at various stages.

Being adept at calculating derivatives is essential for thoroughly analyzing the properties of rational functions and other calculus-related problems.