Problem 35 - Calculus an Applied Approach

Question

Health The temperature \(T\) (in degrees Fahrenheit) of a person during an illness can be modeled by the equation \(T=-0.0375 t^{2}+0.3 t+100.4,\) where \(t\) is time in hours since the person started to show signs of a fever. (a) Use a graphing utility to graph the function. Be sure to choose an appropriate window. (b) Do the slopes of the tangent lines appear to be positive or negative? What does this tell you? (c) Evaluate the function for \(t=0,4,8,\) and \(12 .\) (d) Find \(d T / d t\) and explain its meaning in this situation. (e) Evaluate \(d T / d t\) for \(t=0,4,8,\) and \(12 .\)

Step-by-Step Solution

Answer

The graph of the given function opens downward, with the slope initially positive and later becoming negative. This indicates an initial increase and subsequent decrease in temperature matching a typical fever cycle. The function evaluated at \(t=0\), \(4\), \(8\) and \(12\) gives temperature at these instances. The derivative, which represents the rate of change of temperature, decreases over time, showing a decrease in the speed at which the temperature changes.

Step 1: Graphing the function

Use graphing utility software to plot the given function \(T= -0.0375t^{2} + 0.3t + 100.4\). A suitable window would be from \(t=0\) to \(t=12\) on x-axis as these are the specified time periods.

Step 2: Analyzing the graph

While the graph is visual, it can be deduced that initially, the slope of the tangent lines would be positive (as the temperature goes up when fever starts) and then becomes negative (as the temperature starts going down over time). This indicates the function is a downward-opening parabola, confirming the negative coefficient of the quadratic term.

Step 3: Evaluating the Function

Evaluate the function at \(t=0\), \(4\), \(8\), and \(12\). Calculation shows that \(T(0) = 100.4F\), \(T(4) = 101F\), \(T(8) = 100.4F\), \(T(12) = 98.6F\). These temperatures correspond to measurements at 0, 4, 8 and 12 hours.

Step 4: Finding the Derivative and Interpreting

Calculate the derivative of the function to find rate of change of temperature with respect to time. The derivative is \(d T / d t = -0.075t + 0.3\). This shows how fast the temperature is rising or falling.

Step 5: Evaluating the Derivative

Evaluate the derivative at \(t=0\), \(4\), \(8\), and \(12\). It comes out that \(dT/dt (0) = 0.3\), \(dT/dt (4) = 0.0\), \(dT/dt (8) = -0.3\), \(dT/dt (12) = -0.6\). The derivative at these instances, indicate the change in rate of body's temperature over time.

Key Concepts

Graphing utilities

Graphing utilities are powerful tools that help visualize functions, making it easier to analyze their behavior. In this exercise, we use graphing utilities to plot the function that represents a person's temperature over time during an illness, modeled by the equation: \( T = -0.0375t^2 + 0.3t + 100.4 \). To effectively use a graphing utility:

• Input the equation correctly.
• Choose an appropriate window that covers the relevant range of \(t\), such as from \(t = 0\) to \(t = 12\).
• Examine the plot to understand the shape of the curve and identify points of interest.

By using the graphing utility, it's easier to see the overall trend of the temperature changes over the 12-hour period. This visualization supports further analysis and interpretation.

Tangent lines

Tangent lines provide valuable information about the behavior of functions. They touch a curve at just one point and indicate the immediate rate of change at that point. In this context, analyzing the slope of tangent lines helps understand how the temperature changes at different times. Initially, for \(t\) values close to zero, the tangent line slopes should be positive. This implies that the body's temperature is increasing, indicating the beginning stages of a fever.

• Positive slope: Rising temperature.
• Negative slope: Falling temperature as the person begins to recover.

Later on, the tangent lines become negative, confirming the downward trend as the temperature begins to fall. Recognizing these slope changes allows us to interpret physical changes in the patient's condition over time.

Derivative

The derivative of a function provides crucial insights into how a quantity changes over time. By finding the derivative of the temperature function, represented as \(\frac{dT}{dt} = -0.075t + 0.3\), we determine the rate at which the temperature changes as time progresses.The steps for finding the derivative in this exercise were straightforward since the function is a simple quadratic:

• The derivative of \(-0.0375t^2\) is \(-0.075t\).
• The derivative of the linear term \(0.3t\) is \(0.3\).

Adding these components gives our expression. This derivative tells us how fast the temperature is rising or falling at any point \(t\). Its interpretation is vital for understanding the dynamics of the temperature change experienced during the fever.

Rate of change

The rate of change is an essential concept when analyzing any time-dependent process. Here, it refers to how the temperature changes as hours progress. Evaluating the derivative, \(\frac{dT}{dt}\), at specific time points (\(t = 0, 4, 8, 12\)) reveals:

• At \(t = 0\), \(\frac{dT}{dt} = 0.3\), indicating a steady increase in temperature at the beginning.
• At \(t = 4\), \(\frac{dT}{dt} = 0.0\), suggesting a turning point where the temperature stabilizes briefly.
• At \(t = 8\), \(\frac{dT}{dt} = -0.3\), a falling temperature suggests the fever is subsiding.
• At \(t = 12\), \(\frac{dT}{dt} = -0.6\), indicating a rapid decrease, suggesting recovery.

These evaluations highlight the body's response to an illness over time, providing a clear view of how the fever develops and eventually subsides.