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A solid object with an initial temperature of 75 degrees Fahrenheit is dropped into a lake with a temperature of 33 degrees Fahrenheit. The function [tex]f(t) = C e^{-kt} + 33[/tex] represents the situation, where [tex]t[/tex] is time in minutes, [tex]C[/tex] is a constant, and [tex]k[/tex] is a constant.

After 2 minutes, the object has a temperature of 55 degrees. What will be the temperature of the object, in degrees Fahrenheit, after 4 minutes? Round your answer to the nearest tenth. Do not include units.



Answer :

GinnyAnswer
To determine the temperature of the object after 4 minutes, we start with the given information:

- Initial temperature of the object: [tex]\( 75 \)[/tex] degrees Fahrenheit
- Temperature of the lake: [tex]\( 33 \)[/tex] degrees Fahrenheit
- Temperature of the object after 2 minutes: [tex]\( 55 \)[/tex] degrees Fahrenheit

The function given to represent the situation is:

[tex]\[ f(t) = C e^{-kt} + 33 \][/tex]

Here, [tex]\( t \)[/tex] is time in minutes, [tex]\( C \)[/tex] and [tex]\( k \)[/tex] are constants.

### Step 1: Determine Constant [tex]\( C \)[/tex]
Using the initial condition ([tex]\( t = 0 \)[/tex]):

[tex]\[ f(0) = 75 \][/tex]

Substitute [tex]\( t = 0 \)[/tex] into the function:

[tex]\[ 75 = C e^{0} + 33 \][/tex]

Since [tex]\( e^0 = 1 \)[/tex]:

[tex]\[ 75 = C + 33 \][/tex]

Solve for [tex]\( C \)[/tex]:

[tex]\[ C = 75 - 33 \][/tex]
[tex]\[ C = 42 \][/tex]

### Step 2: Determine Constant [tex]\( k \)[/tex]
Using the temperature after 2 minutes ([tex]\( t = 2 \)[/tex]):

[tex]\[ f(2) = 55 \][/tex]

Substitute [tex]\( t = 2 \)[/tex] and [tex]\( C = 42 \)[/tex]:

[tex]\[ 55 = 42 e^{-2k} + 33 \][/tex]

Subtract 33 from both sides:

[tex]\[ 22 = 42 e^{-2k} \][/tex]

Divide both sides by 42:

[tex]\[ e^{-2k} = \frac{22}{42} \][/tex]
[tex]\[ e^{-2k} = \frac{11}{21} \][/tex]

Take the natural logarithm of both sides:

[tex]\[ -2k = \ln\left(\frac{11}{21}\right) \][/tex]

Solve for [tex]\( k \)[/tex]:

[tex]\[ k = -\frac{1}{2} \ln\left(\frac{11}{21}\right) \][/tex]

Using the numerical value obtained:

[tex]\[ k \approx 0.32331358246252623 \][/tex]

### Step 3: Calculate the Temperature after 4 Minutes
Using [tex]\( t = 4 \)[/tex]:

[tex]\[ f(4) = 42 e^{-4k} + 33 \][/tex]

Substitute [tex]\( k \approx 0.32331358246252623 \)[/tex]:

[tex]\[ f(4) = 42 e^{-4 \cdot 0.32331358246252623} + 33 \][/tex]

Simplify the exponent:

[tex]\[ f(4) \approx 42 e^{-1.29325432985} + 33 \][/tex]

Calculate the value of the exponential term:

[tex]\[ f(4) \approx 42 \cdot e^{-1.29325432985} + 33 \][/tex]
[tex]\[ f(4) \approx 42 \cdot 0.272471615 + 33 \][/tex]
[tex]\[ f(4) \approx 11.523809523 + 33 \][/tex]

Add the constants:

[tex]\[ f(4) \approx 44.523809523809526 \][/tex]

Round the result to the nearest tenth:

[tex]\[ f(4) \approx 44.5 \][/tex]

Thus, the temperature of the object after 4 minutes is approximately:

[tex]\[ 44.5 \][/tex]

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