Answer :
Given the problem statement:
- [tex]\( x \)[/tex] represents the time that has elapsed in minutes.
- [tex]\( y \)[/tex] represents the temperature of the kettle in degrees Celsius.
Let's walk through the observations and reasoning that could be related to the situation where a boiling kettle cools down in a warm kitchen.
### Step 1: Initial Conditions
When the kettle just starts to cool:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 100 \)[/tex] degrees Celsius (assuming it's boiling water).
### Step 2: Newton's Law of Cooling
The temperature of the kettle will decrease over time. Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its current temperature and the ambient temperature of the environment.
- The ambient temperature of the warm kitchen (let's suppose it's [tex]\( T \)[/tex] degrees Celsius) remains constant.
The general form of Newton's Law of Cooling is:
[tex]\[ \frac{dy}{dx} = - k (y - T) \][/tex]
where:
- [tex]\( k \)[/tex] is a positive constant specific to the situation.
### Step 3: Solution of the Differential Equation
Integrating the differential equation gives:
[tex]\[ y - T = (100 - T) e^{-kx} \][/tex]
When simplified, the temperature as a function of time can be written as:
[tex]\[ y = T + (100 - T) e^{-kx} \][/tex]
This represents an exponential decay function where the kettle's temperature approaches the kitchen's ambient temperature over time.
### Step 4: Interpreting the Function
- [tex]\( e^{-kx} \)[/tex] decreases as [tex]\( x \)[/tex] increases.
- Initially, when [tex]\( x = 0 \)[/tex], [tex]\( y = 100 \)[/tex].
- As [tex]\( x \)[/tex] grows larger, [tex]\( e^{-kx} \)[/tex] approaches 0, so [tex]\( y \)[/tex] approaches [tex]\( T \)[/tex].
### Step 5: Applying the Information
Given that:
[tex]\[ x \) represents the elapsed time in minutes. \[ y \) represents the temperature of the kettle in degrees Celsius. You can see that the temperature of the kettle cools down exponentially and asymptotically approaches the ambient temperature \( T \) of the kitchen. ### Result Now, let's directly address the question using our understanding of the cooling process: - \( A, B, C, D \) appear to represent some distinct points of interest. Without specific points or values given by the problem statement, the particular details and the numeric final answer obtained might simply signify a possible outcome of this cooling process. Thus, considering the numerical result previously obtained reflects a situation covered by the mathematical model, we conclude: The result is as follows: \[ y = T \][/tex]
This implies that the cooling process eventually stabilizes at the ambient temperature of the kitchen.
Conclusively, the temperatures in the kettle are best modeled using the exponential cooling approach, representing the kettle gradually cooling to the ambient room temperature evident by real-world observations.
- [tex]\( x \)[/tex] represents the time that has elapsed in minutes.
- [tex]\( y \)[/tex] represents the temperature of the kettle in degrees Celsius.
Let's walk through the observations and reasoning that could be related to the situation where a boiling kettle cools down in a warm kitchen.
### Step 1: Initial Conditions
When the kettle just starts to cool:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 100 \)[/tex] degrees Celsius (assuming it's boiling water).
### Step 2: Newton's Law of Cooling
The temperature of the kettle will decrease over time. Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its current temperature and the ambient temperature of the environment.
- The ambient temperature of the warm kitchen (let's suppose it's [tex]\( T \)[/tex] degrees Celsius) remains constant.
The general form of Newton's Law of Cooling is:
[tex]\[ \frac{dy}{dx} = - k (y - T) \][/tex]
where:
- [tex]\( k \)[/tex] is a positive constant specific to the situation.
### Step 3: Solution of the Differential Equation
Integrating the differential equation gives:
[tex]\[ y - T = (100 - T) e^{-kx} \][/tex]
When simplified, the temperature as a function of time can be written as:
[tex]\[ y = T + (100 - T) e^{-kx} \][/tex]
This represents an exponential decay function where the kettle's temperature approaches the kitchen's ambient temperature over time.
### Step 4: Interpreting the Function
- [tex]\( e^{-kx} \)[/tex] decreases as [tex]\( x \)[/tex] increases.
- Initially, when [tex]\( x = 0 \)[/tex], [tex]\( y = 100 \)[/tex].
- As [tex]\( x \)[/tex] grows larger, [tex]\( e^{-kx} \)[/tex] approaches 0, so [tex]\( y \)[/tex] approaches [tex]\( T \)[/tex].
### Step 5: Applying the Information
Given that:
[tex]\[ x \) represents the elapsed time in minutes. \[ y \) represents the temperature of the kettle in degrees Celsius. You can see that the temperature of the kettle cools down exponentially and asymptotically approaches the ambient temperature \( T \) of the kitchen. ### Result Now, let's directly address the question using our understanding of the cooling process: - \( A, B, C, D \) appear to represent some distinct points of interest. Without specific points or values given by the problem statement, the particular details and the numeric final answer obtained might simply signify a possible outcome of this cooling process. Thus, considering the numerical result previously obtained reflects a situation covered by the mathematical model, we conclude: The result is as follows: \[ y = T \][/tex]
This implies that the cooling process eventually stabilizes at the ambient temperature of the kitchen.
Conclusively, the temperatures in the kettle are best modeled using the exponential cooling approach, representing the kettle gradually cooling to the ambient room temperature evident by real-world observations.
