Exponential growth word problem, pie.

AI Thread Summary
The discussion centers on understanding a problem related to the exponential cooling of a pie, specifically how to apply the correct formula. The standard formula for exponential decay, T(t) = T_m + (T_0 - T_m)e^{-kt}, is emphasized as more appropriate than the initially mentioned growth formula. The participant struggles with the concept, particularly how the pie's temperature relates to the surrounding temperature of 70°F. They are advised to adjust their approach to reflect the decay of temperature over time, using logarithms to solve for unknown variables. Ultimately, the correct application of Newton's Law of Cooling is crucial for solving the problem accurately.
Matriculator
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This whole chapter has been tripping me. My professor made-up another way of doing this, which I'm having a hard time understanding. He's a really intelligent guy(even other professors say it) so he can do this, but I'm having a hard time understanding his way. I went to tutoring today, they couldn't help me with his way. So I'll just try the standard way and see if he'll let it pass.

How would I normally do this? I know that the standard formula is f(t)=Aert?

Would it start like f(2)=150er2? That 70 degrees is my biggest problem. I know that it can't go under 70 degrees, I'm having a hard time knowing how that'll fit into this equation. Thank you in advance.
 

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Matriculator said:
This whole chapter has been tripping me. My professor made-up another way of doing this, which I'm having a hard time understanding. He's a really intelligent guy(even other professors say it) so he can do this, but I'm having a hard time understanding his way. I went to tutoring today, they couldn't help me with his way. So I'll just try the standard way and see if he'll let it pass.

How would I normally do this? I know that the standard formula is f(t)=Aert?

Would it start like f(2)=150er2? That 70 degrees is my biggest problem. I know that it can't go under 70 degrees, I'm having a hard time knowing how that'll fit into this equation. Thank you in advance.
attachment.php?attachmentid=56108&d=1361843020.png


The temperature of the pie [STRIKE]cools[/STRIKE] decreases exponentially from whatever temperature it had coming out of the oven to room temperature (70° F). You will need a somewhat different fuction than what you showed.

If f(t) represents the temperature of the pie and t is number of hours after leaving the oven, then:
f(2) = 150°F

f(5) = 130°F​
 
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Matriculator said:
This whole chapter has been tripping me. My professor made-up another way of doing this, which I'm having a hard time understanding. He's a really intelligent guy(even other professors say it) so he can do this, but I'm having a hard time understanding his way. I went to tutoring today, they couldn't help me with his way. So I'll just try the standard way and see if he'll let it pass.

How would I normally do this? I know that the standard formula is f(t)=Aert?

Would it start like f(2)=150er2?
No, it wouldn't. 150°F is not the initial temperature of the pie. It's the temperature after 2 hours. What you would have is 150 = Ae^{2r}.

Are you sure you're supposed to use the formula f(t)=Aert? The way I remember learning it, this is Newton's Law of Cooling, and here was the formula:
T(t) = T_m + (T_0 - T_m)e^{-kt},
where
Tm is the temperature of the surrounding medium, and
T0 is the initial temperature of the object.EDIT: Beaten to it. ;)
 
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SammyS said:
attachment.php?attachmentid=56108&d=1361843020.png


The temperature of the pie cools exponentially from whatever temperature it had coming out of the oven to room temperature (70° F). You will need a somewhat different fuction than what you showed.

If f(t) represents the temperature of the pie and t is number of hours after leaving the oven, then:
f(2) = 150°F

f(5) = 130°F​

A decaying one? Such as f(t)=Ae-rt? I got this from online. I got that original formula from online. I wasn't too sure of how it worked. Since it allowed me to take the natural log when solving for time on simpler problems, I used it.
 
eumyang said:
No, it wouldn't. 150°F is not the initial temperature of the pie. It's the temperature after 2 hours. What you would have is 150 = Ae^2r.

Are you sure you're supposed to use the formula f(t)=Aert? The way I remember learning it, this is Newton's Law of Cooling, and here was the formula:
T(t) = T_m + (T_0 - T_m)e^{-kt},
where
Tm is the temperature of the surrounding medium, and
T0 is the initial temperature of the object.EDIT: Beaten to it. ;)

I'm not too sure. Since the chapter is on exponential growth I think that he gave one general formula of some sort to cover all types of exponential growth(interest, bacteria etc..)
 
This is not a problem in exponential growth, it is a problem in exponential decay. The thing that is decaying is the difference between the temperature of the pie, and the temperature of the air: (T - 70)
So,

time T T-70
2 150 80
5 130 60

For exponential decay, the constant r in your equation is negative.
 
I would suggest modifying your temperature function. Try \displaystyle \ \ f(t) - 70 = Ae^{-rt}\,, \ where f(t) is the temperature of the pie, and t is time in hours, after the pie has been removed from the oven .

You will need to use logarithms to solve this.
 
Using
eumyang said:
The way I remember learning it, this is Newton's Law of Cooling, and here was the formula:
T(t) = T_m + (T_0 - T_m)e^{-kt},
where
Tm is the temperature of the surrounding medium, and
T0 is the initial temperature of the object.

the only things you do not know are ##k## and ##T_0##. You can then insert the data you have about temperatures and times and you have a system with two equations and two variables (i.e. ##k## and ##T_0##). Solve the system and you will have your solution with all parameters.

Finally you just need to insert 7 (hours) and compute the result.
 
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