Problem 88 - Calculus and its applications

Question

he temperature \(T\) of a person during an illness is given by $$ T(t)=-0.1 t^{2}+1.2 t+98.6 $$ where \(T\) is the temperature, in degrees Fahrenheit, at time \(t,\) in days. a) Find the rate of change of the temperature with respect to time. b) Find the temperature at \(t=1.5\) days. c) Find the rate of change at \(t=1.5\) days.

Step-by-Step Solution

Answer

a) Rate of change is \(-0.2t + 1.2\). b) Temperature at 1.5 days is 100.175°F. c) Rate of change at 1.5 days is 0.9°F/day.

Step 1: Find the Derivative

To determine the rate of change of the temperature with respect to time, we need to differentiate the function with respect to \( t \). The given function is \( T(t) = -0.1t^2 + 1.2t + 98.6 \). The derivative, \( \frac{dT}{dt} \), will be calculated as follows: \( \frac{d}{dt}[-0.1t^2 + 1.2t + 98.6] = -0.2t + 1.2 \). Thus, the rate of change with respect to time is \( -0.2t + 1.2 \).

Step 2: Evaluate the Temperature at Specific Time

To find the temperature at \( t = 1.5 \) days, substitute \( t = 1.5 \) into the original temperature equation \( T(t) = -0.1t^2 + 1.2t + 98.6 \). Calculate: \( T(1.5) = -0.1(1.5)^2 + 1.2(1.5) + 98.6 = -0.1(2.25) + 1.8 + 98.6 = -0.225 + 1.8 + 98.6 = 100.175 \). Therefore, the temperature at \( t = 1.5 \) days is \( 100.175 \) degrees Fahrenheit.

Step 3: Evaluate the Rate of Change at Specific Time

Now, substitute \( t = 1.5 \) into the derivative we found in Step 1: \( -0.2t + 1.2 \). Calculate: \( -0.2(1.5) + 1.2 = -0.3 + 1.2 = 0.9 \). Thus, the rate of change of temperature at \( t = 1.5 \) days is \( 0.9 \) degrees Fahrenheit per day.

Key Concepts

Rate of Change

Understanding the rate of change is fundamental in calculus, especially when examining how one quantity changes in relation to another. In this scenario, we are interested in how a person's temperature changes over time during an illness.
To find the rate of change, we take the derivative of the temperature function with respect to time. The derivative provides us with an equation that describes how the temperature varies as time progresses.
The given temperature function is \( T(t) = -0.1t^2 + 1.2t + 98.6 \). By differentiating this function with respect to \( t \), we get:

• \( \frac{dT}{dt} = -0.2t + 1.2 \)

This result tells us that the rate at which the temperature changes depends linearly on time. The rate is expressed in degrees Fahrenheit per day. Knowing how to compute this allows us to predict how fast (or slow) the temperature will change as time moves forward.

Temperature Function

The temperature function is a mathematical representation that helps model real-world situations, like an individual's body temperature over time. This function helps us to understand not just temperatures at specific points in time, but also how temperatures change over a period.
Consider the given function: \( T(t) = -0.1t^2 + 1.2t + 98.6 \).
* The term \(-0.1t^2\)* suggests the curve of the graph, creating a parabola that opens downward.

• It implies that the rate of temperature increase slows down, and may eventually decrease if observed continuously over a longer time span.

* The term \(1.2t\)* indicates a linear increase in temperature with time, especially at the start.

• The constant 98.6 represents the base temperature, aligning with typical human body temperatures in degrees Fahrenheit.

By substituting different values of \( t \), we can find temperatures at particular times, which can be useful in predicting trends and aiding healthcare decisions.

Evaluating Functions

Evaluating functions involves determining the exact value of the function for a particular input. This allows us to find precise points on the graph of the function.
For instance, if we want to find the temperature at precisely \( t = 1.5 \) days, we substitute \( t = 1.5 \) into our function \( T(t) = -0.1t^2 + 1.2t + 98.6 \) and compute:

• \( T(1.5) = -0.1(1.5)^2 + 1.2(1.5) + 98.6 \)

Breaking this down:

• \(-0.1(2.25) = -0.225\)
• \(1.2(1.5) = 1.8\)
• Therefore, \(-0.225 + 1.8 + 98.6 = 100.175\)

The result \( 100.175 \) gives us the temperature in degrees Fahrenheit at that specific time point. Evaluating functions like this is essential in calculus and data analysis as it allows for precise and actionable insights.