Newton's law of cooling:
Frank's automobile engine runs at 105C. On a day when the outside temperature is 24C, he turns off the ignition and notes that five minutes later, the engine has cooled to 77C.

Question

anonymous · Answer

I think its $$T' = -k(T(t) - Ta)$$
where Ta is ambient temp
correct?

anonymous · Answer

yes thats correct.

anonymous · Answer

then $$T(t) = Ta + (T0 - Ta)*e^{kt}$$

anonymous · Answer

so $$77 = 24 + (105 − 24)e^{k(5)}$$

anonymous · Answer

perfect :)

anonymous · Answer

$$k = (ln((77 - 24)/(105 - 24))/5$$

anonymous · Answer

that was correct except it's negative

anonymous · Answer

i guess it should be negative because we expect the temp of the engine to cool?  (im not 100% familiar with this kinda of stuff)

anonymous · Answer

I havent had a course in differential equations yet, but ive watched all the lectures from MIT's Open Courseware.
This lecture talkes about it (skip to about 14 mins in)

anonymous · Answer

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

anonymous · Answer

its quite confusing.  the output of the ln is negative, but in the derivative eq, the negative is outside the k, which means we need to negate the negaive ln to make it positive so it can be negative again in the dif eq.  0.0

anonymous · Answer

K. got correct answers. TY