foram said:
Newtons Law of Cooling. It's in the Applications of Math to the physical world bit. The place with projectile motion and SHM I think. I dont have the fitzpatric book with me now, but i'm guessing its about there.
Also i don't believe this question can be asked in a test. It's beyond the syllabus, i'm sure, because you need to introduce an extra value for the heating at 5/min, and they MUST give you the general solution for a question of that sort.
It is possible that you may need to use dT/dt = 5 + k(T-P).
I would start from dT/dt = 5 + k(T-P), and maybe integrate it?? Either way, as t-> infinity, T=120, and at t=0, T=20... Ahhg this is an evil question. But i'm sure about using dT/dt = 5 + k(T-P).
I THINK.
Oh sweet! Of course.
Does P = 120? Ambient temp?
dT/dt = 5 + k(T-120)
let y(t)=T(t)-120
dy/dt = dT/dt = 5+ky
So dy/dt - ky = 5 (a non-homogenous first order linear differential equation).
Use integrating factor u=e^(Int -k dt) = e^(-kt)
udy/dt -uky = 5u
udy/dt - ydu/dt = 5u
d(uy)/dt = 5u
d(uy) = 5u dt
uy = Int 5u dt
e^(-kt)*(T-120) = Int -5e^(-kt)/k dt = 5e^(-kt)/k^2 + C
T(t) = 5/k^2 + Ce^(kt) + 120
T(infinity) = fuck. It's exactly the same damn function as I got using logs instead, below:
dT/dt = 5 + k(T-120) = 5 - 120k + kT
dT/(5-120k+kT) = dt
kln(5-120k+kT) = t + C (C = some constant)
5-120k+kT(t) = Je^t (J = some constant)
T(t) = (J/k)e^t +120k - 5
Aw man I can't apply a limit to infinity cause I need a negative constant in the power of the exponential. Bah. Hmm actually I could do that but it would mean j is zero hence this isn't an exponential question.
