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Calculus I help?

S

SLCentral

Diamond Member
Not a extremely difficult problem, I just can't remember for the life of me how to do it. Here goes:

A pizza cools at a rate proportional to its difference from room temperature.

a). When the pizza temperature is 200C, is the rate of change of temperature positive or negative? The room temperature is 20C.

b). Suppose the temperature of the pizza at time t is known to be given by f(t) = 250e^(-kt) +c, for some numbers c and k. Also suppose that the room temperature is 20C. Determine the number c.

Any ideas? I have a final this week and this is on the review sheet. Obviously I'm not a math major.
 
If you can wait till tomorrow, maybe I can dig some of my old calc stuff out. I know this problem has to do with Newton's Law of Cooling though, look into that.
 
Is t = 0 and pizza temperature an initial condition?

Is that the entire problem statement? You have two unknowns (c and k) and only one equation/initial condition, so I'm not sure how you can solve this like that.

edit: oh yeah that's right, i forgot the e in there, you can take the limit of that.
 
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You know at time t=infinity, that the pizza will be at room temperature, 20C. With that knowledge you can find C.

With these types of problems, think about what you know at time t=0 or t=infinity.
 
You know at time t=infinity, that the pizza will be at room temperature, 20C. With that knowledge you can find C.

With these types of problems, think about what you know at time t=0 or t=infinity.
 
gwai lo said:
Is t = 0 and pizza temperature an initial condition?

Is that the entire problem statement? You have two unknowns (c and k) and only one equation/initial condition, so I'm not sure how you can solve this like that.
Click to expand...

Thats the entire thing...thats why I'm stuck. Maybe I'm supposed to imply that when t=0, the pizza is 200C? I'm not sure. This is a Calc I class for non-science/engineer majors, so it shouldn't be too hard.
 
That's what I thought too, you have c and k both as unknowns. Unless there's another formula/law given you're supposed to use.. I dunno how to work this out with what you gave.
 
gwai lo said:
Remember your limits for e, you've got enough here to solve the problem.
Click to expand...

So since e^0 = 1, so I need to make kt = 0, right? But its two different variables so I'm still stuck. Maybe I've just been at the library for too long...going on 13 hours.
 
When t=infinity, f(t)=20 - since f(t) is the temperature of the pizza, and since this cannot be cooled below ambient using ambient (any watercooling enthusiast should be able to tell you), the lowest it can be cooled to is 20C. Therefore, given this:

f(infinity) = 20 = 250e^(-k*infinity) + c. Since e^(-infinity) = 0, f(infinity) = c = 20.

This is not calculus. This is basic algebra, wrapped up in a lot of numbers and symbols to confuse you.
 
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The pizza is already burned. Why would it's cooling rate matter? Even a stoner wouldn't eat a pizza cooked to 200C
 
Mr. Pedantic said:
When t=infinity, f(t)=20 - since f(t) is the temperature of the pizza, and since this cannot be cooled below ambient using ambient (any watercooling enthusiast should be able to tell you), the lowest it can be cooled to is 20C. Therefore, given this:

f(infinity) = 20 = 250e^(-k*infinity) + c. Since e^(infinity) = 0, f(infinity) = c = 20.

This is not calculus. This is basic algebra, wrapped up in a lot of numbers and symbols to confuse you.
Click to expand...

AH, that makes everything a lot clearer. For some reason I took the first derivative and I just got stuck.

One more problem if anyone else wants to help...:

13. In this problem, f(x) = e^(-k/x), where k is a positive constant, and x >0. The quotient -k/x is all in the exponent). Find the first and second derivative using the chain rule.

I know this isn't hard, but we haven't done derivatives for a few months now and I just can't remember how to do this guy.
 
fwiw, the first problem was the solution of the heat equation, taking the derivative would've given you the heat equation back. the only time i can think where you would do this is if one of your initial values is a derivative (in case you see a similar problem on your final, one easy way to give a question would be to switch up the initial values/boundary conditions).

for the second problem, i would google up what the chain rule is...i've seen some useful youtube videos for these simpler things. as a hint, u = k/x
 
That is the solution to dT/dt=k(T-Tm)
where T(t) is the objects temp and Tm is the ambient temp.

T(0)=250e^(-kt) +c=200
For k you would need more info. 😛
 
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Oh, by the way, what you quoted was wrong. I edited my post slightly, you might want to read it before you start copying what you quoted.

Now for the problem:

I'm going to say f(x) = y here.

dy/dx = dt/dx * dy/dt

t = -k/x
y = e^t

dt/dx = d(-kx^(-1))/dx = kx^(-2)

dy/dt = e^t

dy/dx = (ke^t)/x^2 = (ke^(-k/x))/(x^2).

That's the first derivative. I'll leave it to you to figure out the second. You can choose to make t any letter and stand for any part of the equation, but I chose -k/x for simplicity.

you are missing info for solving how they want you to, I think...
That is the solution to dT/dt=k(T-Tm)
where T(t) is the objects temp and Tm is the ambient temp.
Click to expand...
Really overthinking the problem, there.
 
I'm a med student. I just like maths too. Though, to be honest, the problems aren't hard. The first ones were just algebra. Year-9 stuff, just with AS-level content. The second one was probably AS maths stuff. Or at least I did chain rule in AS maths...
 
That's neat. Yeah, you made it seem pretty easy but to be honest I don't remember much of the process at all. Chain rule was over 6-7 years ago to me.

Wouldn't you be able to apply calculus into the medical field? Like for the basis of MRI's, red blood cell count, hemolysis coagulaton and specific test based results like that.
 
Yes, I suppose you can. But I'm not really that interested in that, to be honest. In my opinion, if I'm studying medicine, I would rather interact with patients and deal with them, rather than with their protons.
 
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