Question

A potato is put into an oven that has been heated to 350 degrees Fahrenheit. Its temperature as a function of time is give by T(t)=a(1-e^(-kt))+b. The potato was 50 degrees F when it was first put into the oven. If the potato is 60 degrees F after 2 minutes, what is the value of k? Explain

Answer 1

\work with the difference in the temperatures
the initial difference is \(350-50=300\)
the difference after 2 minutes is \(350-60=290\)

Answer 2

this difference in temperatures will decay to zero, so the model for the differences will be something like \(300e^{kt}\)
you know if \(t=2\) you get \(290\) so you can find \(k\) by setting
\[300e^{2k}=290\] and solve for \(k\)

Answer 3

\[e^{2k}=\frac{29}{30}\]
\[2k=\ln(\frac{2}{30})\]
\[k=\frac{\ln(\frac{29}{30})}{2}\]