High School

[tex] F(t) = 74.4 - 38.8 e^{-0.05 t} [/tex]

a. What is the long-term behavior of the function [tex] g(t) = e^{-0.05 t} [/tex]? Why?

b. What is the long-term behavior of the function [tex] F(t) = 74.4 - 38.8 e^{-0.05 t} [/tex]? What is the meaning of this value in the physical context of the problem?

c. What is the temperature of the refrigerator? Why?

d. Compute the average rate of change of [tex] F [/tex] on the intervals [tex] [10, 20] [/tex], [tex] [20, 30] [/tex], and [tex] [30, 40] [/tex]. Write a careful sentence, with units, to explain the meaning of each, and write an additional sentence to describe any overall trends in how the average rate of change of [tex] F [/tex] is changing.

Answer :

Let's explore the solutions to the given questions in a detailed manner.

### a. Long-term behavior of the function [tex]\( g(t) = e^{-0.05 t} \)[/tex]

The function [tex]\( g(t) = e^{-0.05 t} \)[/tex] represents an exponential decay with a decay constant of [tex]\( 0.05 \)[/tex]. To determine the long-term behavior as [tex]\( t \)[/tex] approaches infinity, we analyze the limit:
[tex]\[
\lim_{t \to \infty} e^{-0.05 t} = 0
\][/tex]
This means that as [tex]\( t \)[/tex] becomes very large, the value of [tex]\( g(t) \)[/tex] approaches 0. Therefore, the long-term behavior of [tex]\( g(t) \)[/tex] is that it tends to 0 as [tex]\( t \)[/tex] approaches infinity.

### b. Long-term behavior of the function [tex]\( F(t) = 74.4 - 38.8 e^{-0.05 t} \)[/tex]

To determine the long-term behavior of [tex]\( F(t) \)[/tex], we look at the limit as [tex]\( t \)[/tex] approaches infinity:
[tex]\[
\lim_{t \to \infty} F(t) = \lim_{t \to \infty} \left( 74.4 - 38.8 e^{-0.05 t} \right) = 74.4 - 38.8 \cdot 0 = 74.4
\][/tex]
As [tex]\( t \)[/tex] becomes very large, the term [tex]\( e^{-0.05 t} \)[/tex] approaches 0, thus [tex]\( F(t) \)[/tex] approaches 74.4. In the physical context of the problem, this value 74.4°F represents the long-term equilibrium temperature that the system will reach.

### c. Temperature of the refrigerator

The temperature of the refrigerator in the long term is given by the long-term behavior of [tex]\( F(t) \)[/tex]. As determined from part (b), the long-term temperature [tex]\( F(t) \)[/tex] approaches 74.4°F. Therefore, the temperature of the refrigerator is 74.4°F. This is because, over time, the temperature stabilizes and approaches this equilibrium value.

### d. Average rate of change of [tex]\( F \)[/tex] on the intervals [10,20], [20,30], and [30,40]

To calculate the average rate of change of [tex]\( F \)[/tex] over an interval [tex]\([t_1, t_2]\)[/tex], we use the formula:
[tex]\[
\text{Average Rate of Change} = \frac{F(t_2) - F(t_1)}{t_2 - t_1}
\][/tex]

#### Interval [tex]\([10, 20]\)[/tex]
[tex]\[
\text{Average Rate of Change}_{[10, 20]} = \frac{F(20) - F(10)}{20 - 10} = 0.925966727939821 \, \text{°F per hour}
\][/tex]
This means that between 10 and 20 hours, the temperature [tex]\( F(t) \)[/tex] increases, on average, by approximately 0.926°F per hour.

#### Interval [tex]\([20, 30]\)[/tex]
[tex]\[
\text{Average Rate of Change}_{[20, 30]} = \frac{F(30) - F(20)}{30 - 20} = 0.561627210369289 \, \text{°F per hour}
\][/tex]
Between 20 and 30 hours, the temperature [tex]\( F(t) \)[/tex] increases, on average, by approximately 0.562°F per hour.

#### Interval [tex]\([30, 40]\)[/tex]
[tex]\[
\text{Average Rate of Change}_{[30, 40]} = \frac{F(40) - F(30)}{40 - 30} = 0.340644122417851 \, \text{°F per hour}
\][/tex]
Between 30 and 40 hours, the temperature [tex]\( F(t) \)[/tex] increases, on average, by approximately 0.341°F per hour.

### Overall Trend

Notice that the average rate of change of [tex]\( F(t) \)[/tex] is decreasing over these intervals; specifically, it reduces from about 0.926°F per hour (between 10 and 20 hours) to about 0.341°F per hour (between 30 and 40 hours). This indicates that the temperature is increasing at a slower rate as time progresses, which suggests that the system is approaching the equilibrium temperature more gradually over time.