- #1
vorcil
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Homework Statement
Newton's law of cooling says that the rate at which the emperature of any object changes is proportional to the temperature difference between the object and its surroundings. So if the temperature of the object is T, and its surroundings are at a constant temperature [tex] \hat{T} [/tex]
[tex] \frac{dT}{dt} = -k(T - \hat{T}) [/tex]
a) let [tex] \theta = T - \hat{T} [/tex], Show that [tex] \frac{d\theta}{dt} = -k\theta [/tex] and hence write down the general solution for [tex] \theta(t) [/tex]
b) given [tex] \theta(0) = 64 [/tex] and [tex] \theta(2) = 36 [/tex], find [tex] \theta(3) [/tex]
The Attempt at a Solution
a)
[tex] \frac{d\theta}{dt} = -k\theta [/tex]
[tex] \int \frac{d\theta}{dt} = \int -k\theta [/tex]
re arranging
[tex] \int \frac{d\theta}{\theta} = \int -k dt [/tex]
and integrating
[tex] ln(\theta) = -k t + C[/tex]
(I think I add a + C?)
exponentiating both sides
[tex] eln(\theta) = e^{-kt +c} = Ce^{-kt} [/tex]
though I'm not sure why the C goes out to the front, or why it is being multiplied by C
simplifying gives the solution
[tex] \theta(t) = Ce^{-kt} [/tex]
b)
[tex] \theta(0) = Ce^{-k * 0} = 64 [/tex]
since [tex] e^{-k * 0} = e^0 = 1 [/tex]
that makes
[tex] C*1 = 64 [/tex]
so C = 64
now
[tex] \theta(2) = 36 [/tex]
so
[tex] 64*e^{-k * 2} = 36 [/tex]
[tex] e^{-k * 2} = \frac{36}{64} [/tex]
taking the logarithim of each side
[tex] -k*2 = ln(\frac{36}{64} [/tex]
solving for k, I get, K = 0.28768then going to [tex] \theta(3) [/tex]
putting it into the equation, then a calculator
64*e^{-0.28768*3} = 27
I think this is right, since it is a real number / integer, and it sort of fits the pattern
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if someone could check, i'd be very happy, thank you :P