Question

Differential Equation help.


Newtons’ law of cooling tells us that the rate at which a body cools is proportional to the difference between the cooling body and the ambient temperature, To .
This gives the equation dT/dt=k(T-To)

As T is decreasing k<0. If T=60 at t=0 T=50 at t=20 and T=45 when t=40 find the temperature when t=30.

Answer 1

dont you need some kind of t on left side? I think?

Answer 2

You need to separate the variables. Int(dT/(T-T0))=Int(k)dt and solve for the constant to find the equation

Answer 3

t-Intat(1/k(_a-To), _a = T(t))+_C1 = 0

Answer 4

hmm i think this question is worded weirdly, do you think they meant that To is t?

Answer 5

T0 probably means the first number in the sequence.

Answer 6

so after doing the integrals i got ln(T-To)=kt+c

I can sub in T and t but not sure what to put for To?

Answer 7

and further simplification i get T=Ae^(kt)+To

Answer 8

oh i think i can get it from now given that i have three temps to play with i can find A and To

Answer 9

Yes, I would substitute in. My guess is that T0 is the initial position or whatever T is defined as.

Answer 10

equation for T is apparently, T=40+20e^(tln2/20)....so hopefully i can get there. Does look like To=t here....just not sure how that makes sense...

Answer 11

no no To is not t.... dont know what i was thinking!

Answer 12

I end up with this system of equations.

60=A+To
50=Ae^(20k)+To
45=Ae^40k)+To

but im stuck as to how to then get k, To and A

Answer 13

hang on might be on to it...sorry....

Answer 14

ok got it. Thanks heaps for getting me started!! So for anyone interested in the conclusion, equations are rearranged so that To=60-A, and A=(-10/(e^(20k)-1) and k is solved by solving for 45 when the rhs has only the k variable. This gives A=20, k=-ln(2)/20 and To=40 which in T, is T=40+20e^(-tln(2)/20). And finally, when t=30, T=40+5sqrt2

Thanks again!