Exponential growth word problem, pie.

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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 [itex]150 = Ae^{2r}[/itex].

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:
[itex]T(t) = T_m + (T_0 - T_m)e^{-kt}[/itex],
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 [itex]150 = Ae^2r[/itex].

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:
[itex]T(t) = T_m + (T_0 - T_m)e^{-kt}[/itex],
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..)
 
I would suggest modifying your temperature function. Try [itex]\displaystyle \ \ f(t) - 70 = Ae^{-rt}\,, \[/itex] 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:
[itex]T(t) = T_m + (T_0 - T_m)e^{-kt}[/itex],
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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