Exponential growth word problem, pie.

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SUMMARY

The discussion centers on solving an exponential decay problem related to the cooling of a pie using Newton's Law of Cooling. The standard formula for this scenario is T(t) = T_m + (T_0 - T_m)e^{-kt}, where T_m is the ambient temperature (70°F), and T_0 is the initial temperature of the pie. Participants clarify that the temperature after a certain time (e.g., f(2) = 150°F) is not the initial temperature but rather a value at a specific time. The conversation emphasizes the need to adjust the standard exponential growth formula to fit the context of exponential decay.

PREREQUISITES
  • Understanding of exponential functions and decay
  • Familiarity with Newton's Law of Cooling
  • Basic knowledge of logarithms for solving equations
  • Ability to interpret temperature data in a mathematical context
NEXT STEPS
  • Study Newton's Law of Cooling in detail
  • Practice solving exponential decay problems using T(t) = T_m + (T_0 - T_m)e^{-kt}
  • Learn how to derive initial conditions from temperature data
  • Explore applications of exponential decay in real-world scenarios
USEFUL FOR

Students studying calculus or physics, educators teaching exponential functions, and anyone needing to apply mathematical models to real-world cooling scenarios.

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​
 
Last edited:
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. ;)
 
Last edited:
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.
 
Last edited:

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