Answer :
- Calculate the time $t$ for each given temperature $T$ using the formula $t = \frac{\log(T/349.2)}{\log(0.98)}$.
- Interpolate the temperature from the table for each calculated time $t$.
- Calculate the absolute difference (error) between the given temperature and the interpolated temperature.
- Determine the temperature with the smallest error, which is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t)=349.2(0.98)^t$ which models the temperature of an oven after $t$ minutes of cooling. We want to determine which of the given temperatures (0, 100, 300, 400) the model most accurately predicts the time spent cooling. We have a table of oven temperatures at different times.
2. Finding Time for Each Temperature
First, we need to find the time $t$ for each given temperature using the model. We can rearrange the equation to solve for $t$: $$t = \frac{\log(T/349.2)}{\log(0.98)}$$ where $T$ is the given temperature.
3. Calculating Time Values
Now, we calculate the time $t$ for each temperature:
For $T = 0$: Since the logarithm of 0 or a negative number is undefined, we can consider the time to be infinite.
For $T = 100$: $t = \frac{\log(100/349.2)}{\log(0.98)} \approx 61.90$ minutes.
For $T = 300$: $t = \frac{\log(300/349.2)}{\log(0.98)} \approx 7.52$ minutes.
For $T = 400$: $t = \frac{\log(400/349.2)}{\log(0.98)} \approx -6.72$ minutes. Since time cannot be negative, this value is not physically meaningful.
4. Interpolating Temperatures from the Table
Next, we need to find the corresponding temperature from the table for each calculated time using linear interpolation. The table data is:
Time (minutes) | Oven temperature (°F)
---|---
5 | 315
10 | 285
15 | 260
20 | 235
25 | 210
For $t = 61.90$ minutes: Since this time is beyond the table's range, we will use the last temperature in the table, which is 210°F.
For $t = 7.52$ minutes: This time falls between 5 and 10 minutes. We can use linear interpolation: $$f(7.52) = 315 + \frac{7.52 - 5}{10 - 5} (285 - 315) = 315 + \frac{2.52}{5} (-30) = 315 - 15.12 = 299.88$$ So, the interpolated temperature is approximately 299.88°F.
For $t = -6.72$ minutes: Since time cannot be negative, we will use the first temperature in the table, which is 315°F. However, since the time is not physically meaningful, we will disregard this temperature.
5. Calculating the Errors
Now, we calculate the absolute difference (error) between the given temperature and the temperature obtained from the table using interpolation:
For $T = 0$: Since the time is infinite, the error is also considered infinite.
For $T = 100$: Error = $|100 - 210| = 110$.
For $T = 300$: Error = $|300 - 299.88| = 0.12$.
For $T = 400$: Since the time is not physically meaningful, we will disregard this temperature.
6. Determining the Most Accurate Temperature
Comparing the errors, we have:
Temperature | Error
---|---
0 | Infinite
100 | 110
300 | 0.12
400 | Disregarded
The smallest error is 0.12, which corresponds to the temperature 300°F.
7. Final Answer
Therefore, the model most accurately predicts the time spent cooling for the temperature of 300°F.
### Examples
Understanding how ovens cool is crucial in various culinary and industrial applications. For instance, in a bakery, knowing the cooling rate helps determine when to safely ice a cake or package bread. Similarly, in manufacturing, controlled cooling of materials like glass or metal prevents cracking or warping. This problem demonstrates how mathematical models can be used to predict and optimize cooling processes, ensuring product quality and safety.
- Interpolate the temperature from the table for each calculated time $t$.
- Calculate the absolute difference (error) between the given temperature and the interpolated temperature.
- Determine the temperature with the smallest error, which is $\boxed{300}$.
### Explanation
1. Understanding the Problem
We are given the function $f(t)=349.2(0.98)^t$ which models the temperature of an oven after $t$ minutes of cooling. We want to determine which of the given temperatures (0, 100, 300, 400) the model most accurately predicts the time spent cooling. We have a table of oven temperatures at different times.
2. Finding Time for Each Temperature
First, we need to find the time $t$ for each given temperature using the model. We can rearrange the equation to solve for $t$: $$t = \frac{\log(T/349.2)}{\log(0.98)}$$ where $T$ is the given temperature.
3. Calculating Time Values
Now, we calculate the time $t$ for each temperature:
For $T = 0$: Since the logarithm of 0 or a negative number is undefined, we can consider the time to be infinite.
For $T = 100$: $t = \frac{\log(100/349.2)}{\log(0.98)} \approx 61.90$ minutes.
For $T = 300$: $t = \frac{\log(300/349.2)}{\log(0.98)} \approx 7.52$ minutes.
For $T = 400$: $t = \frac{\log(400/349.2)}{\log(0.98)} \approx -6.72$ minutes. Since time cannot be negative, this value is not physically meaningful.
4. Interpolating Temperatures from the Table
Next, we need to find the corresponding temperature from the table for each calculated time using linear interpolation. The table data is:
Time (minutes) | Oven temperature (°F)
---|---
5 | 315
10 | 285
15 | 260
20 | 235
25 | 210
For $t = 61.90$ minutes: Since this time is beyond the table's range, we will use the last temperature in the table, which is 210°F.
For $t = 7.52$ minutes: This time falls between 5 and 10 minutes. We can use linear interpolation: $$f(7.52) = 315 + \frac{7.52 - 5}{10 - 5} (285 - 315) = 315 + \frac{2.52}{5} (-30) = 315 - 15.12 = 299.88$$ So, the interpolated temperature is approximately 299.88°F.
For $t = -6.72$ minutes: Since time cannot be negative, we will use the first temperature in the table, which is 315°F. However, since the time is not physically meaningful, we will disregard this temperature.
5. Calculating the Errors
Now, we calculate the absolute difference (error) between the given temperature and the temperature obtained from the table using interpolation:
For $T = 0$: Since the time is infinite, the error is also considered infinite.
For $T = 100$: Error = $|100 - 210| = 110$.
For $T = 300$: Error = $|300 - 299.88| = 0.12$.
For $T = 400$: Since the time is not physically meaningful, we will disregard this temperature.
6. Determining the Most Accurate Temperature
Comparing the errors, we have:
Temperature | Error
---|---
0 | Infinite
100 | 110
300 | 0.12
400 | Disregarded
The smallest error is 0.12, which corresponds to the temperature 300°F.
7. Final Answer
Therefore, the model most accurately predicts the time spent cooling for the temperature of 300°F.
### Examples
Understanding how ovens cool is crucial in various culinary and industrial applications. For instance, in a bakery, knowing the cooling rate helps determine when to safely ice a cake or package bread. Similarly, in manufacturing, controlled cooling of materials like glass or metal prevents cracking or warping. This problem demonstrates how mathematical models can be used to predict and optimize cooling processes, ensuring product quality and safety.
