Sure! Let's find the temperature of the tea after 5 minutes step-by-step.
1. Understand the given function:
The function [tex]\( f(t) = C \cdot e^{-kt} + 70 \)[/tex] describes how the temperature of the tea changes over time, where:
- [tex]\( t \)[/tex] is time in minutes,
- [tex]\( C \)[/tex] and [tex]\( k \)[/tex] are constants,
- The room temperature is 70 degrees Fahrenheit.
2. Initial conditions:
- At [tex]\( t = 0 \)[/tex] minutes, the temperature of the tea is 120 degrees Fahrenheit:
[tex]\( f(0) = 120 = C \cdot e^{-k \cdot 0} + 70 \)[/tex].
Since [tex]\( e^0 = 1 \)[/tex], this simplifies to [tex]\( 120 = C + 70 \)[/tex].
Solving for [tex]\( C \)[/tex]:
[tex]\[
C = 120 - 70 = 50.
\][/tex]
3. Using the temperature after 3 minutes to find [tex]\( k \)[/tex]:
- At [tex]\( t = 3 \)[/tex] minutes, the temperature of the tea is 100 degrees Fahrenheit:
[tex]\( f(3) = 100 = 50 \cdot e^{-3k} + 70 \)[/tex].
Isolate the exponential term:
[tex]\[
100 - 70 = 50 \cdot e^{-3k} \implies 30 = 50 \cdot e^{-3k}
\][/tex]
Divide both sides by 50:
[tex]\[
e^{-3k} = \frac{30}{50} = 0.6
\][/tex]
Take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[
-3k = \ln(0.6)
\][/tex]
Therefore:
[tex]\[
k = -\frac{\ln(0.6)}{3}
\][/tex]
4. Use [tex]\( k \)[/tex] to determine the temperature after 5 minutes:
- Now we want to find the temperature after 5 minutes, [tex]\( f(5) \)[/tex]:
[tex]\[
f(5) = 50 \cdot e^{-k \cdot 5} + 70
\][/tex]
Substituting [tex]\( k = -\frac{\ln(0.6)}{3} \)[/tex]:
[tex]\[
f(5) = 50 \cdot e^{\ln(0.6)/3 \cdot (-5)} + 70
\][/tex]
Simplify the exponent:
[tex]\[
f(5) = 50 \cdot e^{\ln(0.6) \cdot (-5/3)} + 70
\][/tex]
Simplifying further inside the exponent:
[tex]\[
e^{\ln(0.6) \cdot (-5/3)} = e^{\ln((0.6)^{-5/3})} = (0.6)^{-5/3}
\][/tex]
Compute the value inside:
[tex]\[
f(5) = 50 \cdot (0.6)^{-5/3} + 70 \approx 91.3
\][/tex]
Thus, the temperature of the tea after 5 minutes will be approximately 91.3 (rounded to the nearest tenth).
