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.
### 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.
