anonymous
  • anonymous
we take milk carton out of the refrigerator and put it on the table in room temp. we assume that the temperature of the carton, T, as a function of time t. following Newtons cooling law dT/dt = k(T-a). where a is the environmental temperature, and k is Konstan. we assume that the temperature of the environment is a = 20 and T (0) = 4. show that T (t) = 20-16eKt
Differential Equations
katieb
  • katieb
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Slickster256
  • Slickster256
oh but uh uh
anonymous
  • anonymous
Separate the variables, \[\frac{dT}{dt}=k(T-a)\]\[\frac{dT}{T-a}=kdt\] Now you are given the limits \[t=0 \space \& \space T=4 \space \space \space ; \space \space \space t=t \space \& \space T=T\] So integrate within the limits \[\int\limits_{4}^{T}\frac{dT}{T-a}=k \int\limits_{0}^{t}dt\] \[[\log (T-a)]_{4}^{T}=k[t]_{0}^{t}\] \[\log(T-a)-\log(4-a)=k(t-0)\]\[\log(\frac{T-a}{4-a})=kt\] \[\frac{T-a}{4-a}=e^{kt}\] \[\frac{T-20}{4-20}=e^{kt}\]\[\frac{T-20}{-16}=e^{kt}\]\[T-20=-16e^{kt}\] \[T=20-16e^{kt}\]
anonymous
  • anonymous
thanx :)

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