Answer :
To solve this problem, we use the function [tex]\( f(t) = C e^{-kt} + 33 \)[/tex], which models the temperature of the object over time. We'll follow these steps:
1. Identify the Initial Conditions:
- The initial temperature of the object is 75 degrees Fahrenheit.
- The temperature of the lake is 33 degrees Fahrenheit.
2. Determine the Constant [tex]\( C \)[/tex]:
- At [tex]\( t = 0 \)[/tex], the temperature of the object is 75 degrees. Plug this into the function:
[tex]\[
f(0) = C \cdot e^{-k \cdot 0} + 33 = 75
\][/tex]
Since [tex]\( e^{0} = 1 \)[/tex], this simplifies to:
[tex]\[
C + 33 = 75
\][/tex]
Thus, [tex]\( C = 75 - 33 = 42 \)[/tex].
3. Find the Constant [tex]\( k \)[/tex]:
- After 2 minutes, the temperature of the object is 55 degrees. Use this information to find [tex]\( k \)[/tex]:
[tex]\[
f(2) = 42 \cdot e^{-2k} + 33 = 55
\][/tex]
Rearrange this to solve for [tex]\( e^{-2k} \)[/tex]:
[tex]\[
42 \cdot e^{-2k} = 55 - 33
\][/tex]
[tex]\[
42 \cdot e^{-2k} = 22
\][/tex]
[tex]\[
e^{-2k} = \frac{22}{42}
\][/tex]
Take the natural logarithm on both sides:
[tex]\[
-2k = \ln\left(\frac{22}{42}\right)
\][/tex]
[tex]\[
k = -\frac{1}{2} \ln\left(\frac{22}{42}\right)
\][/tex]
4. Find the Temperature After 4 Minutes:
- Now, using the calculated value of [tex]\( k \)[/tex] and the function, we determine the temperature after 4 minutes:
[tex]\[
f(4) = 42 \cdot e^{-4k} + 33
\][/tex]
Substituting back [tex]\( k \)[/tex]:
[tex]\[
f(4) = 42 \cdot e^{-4 \left(-\frac{1}{2} \ln\left(\frac{22}{42}\right)\right)} + 33
\][/tex]
5. Calculate and Round:
- Compute this expression to find the temperature after 4 minutes. The temperature of the object is approximately 44.5 degrees Fahrenheit after rounding to the nearest tenth.
Therefore, after 4 minutes, the temperature of the object will be approximately 44.5 degrees Fahrenheit.
1. Identify the Initial Conditions:
- The initial temperature of the object is 75 degrees Fahrenheit.
- The temperature of the lake is 33 degrees Fahrenheit.
2. Determine the Constant [tex]\( C \)[/tex]:
- At [tex]\( t = 0 \)[/tex], the temperature of the object is 75 degrees. Plug this into the function:
[tex]\[
f(0) = C \cdot e^{-k \cdot 0} + 33 = 75
\][/tex]
Since [tex]\( e^{0} = 1 \)[/tex], this simplifies to:
[tex]\[
C + 33 = 75
\][/tex]
Thus, [tex]\( C = 75 - 33 = 42 \)[/tex].
3. Find the Constant [tex]\( k \)[/tex]:
- After 2 minutes, the temperature of the object is 55 degrees. Use this information to find [tex]\( k \)[/tex]:
[tex]\[
f(2) = 42 \cdot e^{-2k} + 33 = 55
\][/tex]
Rearrange this to solve for [tex]\( e^{-2k} \)[/tex]:
[tex]\[
42 \cdot e^{-2k} = 55 - 33
\][/tex]
[tex]\[
42 \cdot e^{-2k} = 22
\][/tex]
[tex]\[
e^{-2k} = \frac{22}{42}
\][/tex]
Take the natural logarithm on both sides:
[tex]\[
-2k = \ln\left(\frac{22}{42}\right)
\][/tex]
[tex]\[
k = -\frac{1}{2} \ln\left(\frac{22}{42}\right)
\][/tex]
4. Find the Temperature After 4 Minutes:
- Now, using the calculated value of [tex]\( k \)[/tex] and the function, we determine the temperature after 4 minutes:
[tex]\[
f(4) = 42 \cdot e^{-4k} + 33
\][/tex]
Substituting back [tex]\( k \)[/tex]:
[tex]\[
f(4) = 42 \cdot e^{-4 \left(-\frac{1}{2} \ln\left(\frac{22}{42}\right)\right)} + 33
\][/tex]
5. Calculate and Round:
- Compute this expression to find the temperature after 4 minutes. The temperature of the object is approximately 44.5 degrees Fahrenheit after rounding to the nearest tenth.
Therefore, after 4 minutes, the temperature of the object will be approximately 44.5 degrees Fahrenheit.
