Answer :
To find the approximate value of [tex]\( k \)[/tex], we need to use the provided information and apply it to the formula [tex]\( f(t) = C e^{-kt} + 60 \)[/tex].
1. Identify Given Values:
- Initial temperature of the object: [tex]\( 135^\circ \)[/tex]F
- Surrounding temperature: [tex]\( 60^\circ \)[/tex]F
- Temperature of the object after 6 minutes: [tex]\( 85^\circ \)[/tex]F
- Time, [tex]\( t \)[/tex]: 6 minutes
2. Determine the constant [tex]\( C \)[/tex]:
The constant [tex]\( C \)[/tex] represents the difference between the initial temperature of the object and the surrounding temperature:
[tex]\[ C = 135 - 60 = 75 \][/tex]
3. Set up the equation with known values:
At [tex]\( t = 6 \)[/tex] minutes, the temperature of the object is 85 degrees. Substitute these values into our formula:
[tex]\[ 85 = 75 e^{-6k} + 60 \][/tex]
4. Isolate the exponential term:
Subtract 60 from both sides to isolate the term involving [tex]\( e \)[/tex]:
[tex]\[ 85 - 60 = 75 e^{-6k} \][/tex]
Simplifying the left side, we get:
[tex]\[ 25 = 75 e^{-6k} \][/tex]
5. Solve for the exponential term:
Divide both sides by 75 to isolate [tex]\( e^{-6k} \)[/tex]:
[tex]\[ \frac{25}{75} = e^{-6k} \][/tex]
Simplifying the fraction:
[tex]\[ \frac{1}{3} = e^{-6k} \][/tex]
6. Solve for [tex]\( k \)[/tex] using natural logarithms:
To isolate [tex]\( k \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln\left(e^{-6k}\right) \][/tex]
Using the property [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln\left(\frac{1}{3}\right) = -6k \][/tex]
7. Isolate [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{1}{6} \ln\left(\frac{1}{3}\right) \][/tex]
8. Calculate the natural logarithm:
Using a calculator:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln(1) - \ln(3) = 0 - \ln(3) \approx -1.0986 \][/tex]
Therefore:
[tex]\[ k = -\frac{1}{6} \times -1.0986 = \frac{1.0986}{6} \approx 0.1831 \][/tex]
9. Round to three decimal places:
[tex]\[ k \approx 0.183 \][/tex]
Thus, the approximate value of [tex]\( k \)[/tex] is:
[tex]\[ k = \boxed{0.183} \][/tex]
1. Identify Given Values:
- Initial temperature of the object: [tex]\( 135^\circ \)[/tex]F
- Surrounding temperature: [tex]\( 60^\circ \)[/tex]F
- Temperature of the object after 6 minutes: [tex]\( 85^\circ \)[/tex]F
- Time, [tex]\( t \)[/tex]: 6 minutes
2. Determine the constant [tex]\( C \)[/tex]:
The constant [tex]\( C \)[/tex] represents the difference between the initial temperature of the object and the surrounding temperature:
[tex]\[ C = 135 - 60 = 75 \][/tex]
3. Set up the equation with known values:
At [tex]\( t = 6 \)[/tex] minutes, the temperature of the object is 85 degrees. Substitute these values into our formula:
[tex]\[ 85 = 75 e^{-6k} + 60 \][/tex]
4. Isolate the exponential term:
Subtract 60 from both sides to isolate the term involving [tex]\( e \)[/tex]:
[tex]\[ 85 - 60 = 75 e^{-6k} \][/tex]
Simplifying the left side, we get:
[tex]\[ 25 = 75 e^{-6k} \][/tex]
5. Solve for the exponential term:
Divide both sides by 75 to isolate [tex]\( e^{-6k} \)[/tex]:
[tex]\[ \frac{25}{75} = e^{-6k} \][/tex]
Simplifying the fraction:
[tex]\[ \frac{1}{3} = e^{-6k} \][/tex]
6. Solve for [tex]\( k \)[/tex] using natural logarithms:
To isolate [tex]\( k \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln\left(e^{-6k}\right) \][/tex]
Using the property [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln\left(\frac{1}{3}\right) = -6k \][/tex]
7. Isolate [tex]\( k \)[/tex]:
[tex]\[ k = -\frac{1}{6} \ln\left(\frac{1}{3}\right) \][/tex]
8. Calculate the natural logarithm:
Using a calculator:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln(1) - \ln(3) = 0 - \ln(3) \approx -1.0986 \][/tex]
Therefore:
[tex]\[ k = -\frac{1}{6} \times -1.0986 = \frac{1.0986}{6} \approx 0.1831 \][/tex]
9. Round to three decimal places:
[tex]\[ k \approx 0.183 \][/tex]
Thus, the approximate value of [tex]\( k \)[/tex] is:
[tex]\[ k = \boxed{0.183} \][/tex]