Problem 65 - Single Variable Calculus

Question

According to the American Heart Association, the treatment benefit for heart attacks depends on the time until treatment and is described by the function $$f(t)=\frac{-16.94 t+203.28}{t+2.0328} \quad 0 \leq t \leq 12$$ where \(t\) is measured in hours and \(f(t)\) is expressed as a percent. a. Use a graphing utility to graph the function \(f\) using the viewing window \([0,13] \times[0,120]\). b. Use the numerical derivative capability of a graphing utility to find the derivative of \(f\) when \(t=0\) and \(t=2\). c. Interpret the results obtained in part (b).

Step-by-Step Solution

Answer

In summary, we found the derivative of the given function \(f(t)\) which represents the treatment benefit for heart attacks depending on the time until treatment, as \(f'(t) = \frac{-237.745472}{(t + 2.0328)^2}\). Evaluating it at \(t=0\) and \(t=2\), we found the slopes to be approximately -57.51 and -14.61 respectively. This indicates that the treatment benefit decreases over time and it is more effective when administered sooner.

Step 1: Graph the function using a graphing utility

Use a graphing utility to input the function \(f(t)=\frac{-16.94 t+203.28}{t+2.0328}\) with the viewing window \([0,13]\times[0,120]\). After ploting the graph, you will see the behavior of the function within the given window.

Step 2: Find the derivative of the function

To find the derivative of \(f(t)\), use the quotient rule, which states that if \(g(x) = \frac{u(x)}{v(x)}\), then \(g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}\). Let \(u(t) = -16.94t + 203.28\) and \(v(t) = t + 2.0328\). We have: \(u'(t) = -16.94\), \(v'(t) = 1\). Now we apply the quotient rule: \(f'(t) = \frac{(-16.94)(t + 2.0328) - (-16.94t + 203.28)(1)}{(t + 2.0328)^2}\) Simplify the expression: \(f'(t) = \frac{-16.94t - 34.465472 + 16.94t - 203.28}{(t + 2.0328)^2}\) \(f'(t) = \frac{-237.745472}{(t + 2.0328)^2}\)

Step 3: Evaluate the derivative at t=0 and t=2

Now, let's find the values of the derivative when \(t = 0\) and \(t = 2\). For \(t = 0\): \(f'(0) = \frac{-237.745472}{(0 + 2.0328)^2} = \frac{-237.745472}{4.1321984} \approx -57.51\) For \(t = 2\): \(f'(2) = \frac{-237.745472}{(2 + 2.0328)^2} = \frac{-237.745472}{16.2627984} \approx -14.61\)

Step 4: Interpret the results

The results of part (b) represent the rate of change (slope) of the treatment benefit at \(t = 0\) and \(t = 2\). When \(t = 0\), the derivative is approximately -57.51, which means that the treatment benefit is decreasing rapidly at the start. The decreasing benefit might be related to the immediate effects of the treatment on the patient. When \(t = 2\), the derivative is approximately -14.61, which means that the treatment benefit is still decreasing, but at a slower rate. This suggests that the treatment's effectiveness begins to decrease over time, but it is still beneficial for the patient. Overall, the results indicate that the sooner the treatment is administered, the greater the benefits will be in terms of the percent efficacy of the treatment.

Key Concepts

Quotient Rule

The quotient rule is a technique in calculus for finding the derivative of a quotient of two functions. If you have a function expressed as a ratio such as \(g(x) = \frac{u(x)}{v(x)}\), you can find its derivative using the formula:

• \(g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)}\)

This means you first take the derivative of the numerator \(u\) and multiply it by the denominator \(v\), then subtract the product of the numerator \(u\) and the derivative of the denominator \(v\). Finally, you divide the whole thing by the square of the denominator.
In the given function \(f(t) = \frac{-16.94 t + 203.28}{t+2.0328}\), the top part of the fraction (numerator) is \(-16.94t + 203.28\), and the bottom part (denominator) is \(t + 2.0328\). The derivatives would be:

• \(u'(t) = -16.94\)
• \(v'(t) = 1\)

By substituting these derivatives into the quotient rule, you simplify down to find the derivative of the original function, \(f'(t)\).
The quotient rule helps break down complexities of finding a derivative into manageable parts, particularly useful for functions like \(f(t)\), where simply multiplying the derivative components yields an accurate result.

Graphing Utility

A graphing utility is an electronic tool, such as a graphing calculator or computer software, used to visualize mathematical functions and their characteristics. It allows you to input a function and quickly plot its graph within a specified viewing window, helping you understand the function's behavior over a certain range.
The benefit of using a graphing utility becomes apparent when dealing with functions such as \(f(t) = \frac{-16.94 t + 203.28}{t+2.0328}\). By specifying the viewing window as \([0,13] \, \text{by} \, [0,120]\), you can observe how the function behaves within that time range \(0 \leq t \leq 12\). With a clear visual, patterns and trends, like where the function increases or decreases, become easier to spot.
Moreover, most graphing utilities offer features like zooming, tracing, and locating specific points on the graph, making it a versatile tool in exploring functions. Feedback from graphing utilities often includes numerical derivative capabilities, enabling automatic computation of derivative values at particular points, expediting calculations when working with complex derivatives.

Derivative

The derivative of a function represents the rate at which the function's value changes at any given point. It tells you how steep the curve of the function is—how quickly or slowly \(f(t)\) is changing with respect to \(t\).
To find the derivative of the function \(f(t) = \frac{-16.94t + 203.28}{t+2.0328}\), we apply the quotient rule, as this function is presented as a fraction of two functions. The resulting derivative, \(f'(t) = \frac{-237.745472}{(t + 2.0328)^2}\), provides a formula for calculating the rate of change of \(f(t)\) for any specific time \(t\).
Evaluating this derivative at \(t = 0\) and \(t = 2\) gave us the rates of change \(-57.51\) and \(-14.61\) respectively. These values suggest how effective changes over small intervals in time directly after a heart attack can dramatically affect treatment outcomes. The steep slope at \(t = 0\) indicates a large immediate impact, while the reduced slope at \(t = 2\) signifies a gradual decline in effectiveness over time.

Rate of Change

The rate of change is an important concept in calculus that measures how one variable changes with respect to another. In the context of the given function, it helps us understand how the treatment benefit changes as time progresses.
In calculus, the derivative \(f'(t)\) is used to determine the rate of change at any point \(t\). Here, for \(f(t)\), the rate of change gives the effectiveness of heart attack treatment over time.
When we found \(f'(0) \approx -57.51\), it meant the treatment benefit was decreasing rapidly, suggesting that immediate treatment could have profound positive effects initially. On the other hand, \(f'(2) \approx -14.61\) indicates a still decreasing but slower rate, meaning if treatment is delayed, the benefits reduce more slowly than initially.
Overall, analyzing the rate of change through derivatives enhances our understanding of dynamic processes such as this, showing the importance of time in optimizing medical treatment outcomes.