Question

I already had this question answered and webwork says the answers for the 4th and 5th question are incorrect. This is their work and I need help with this one

1 point) Susan tinds an allen antfact in the desert, where there are temperature variations from a lowin the 30s at night to a high in the 100s n the day. She is interested in how the aritact will (hotly pursued by the matary police who patrol Area 51), and slicks it in an "ovent-that its, a closed bex whose lemperature she can contror precisely Let T t)be the lemperature ot the anifact Newton's law of cooling says that Tt) changes at a rate proporional to the difesence between the temperature of the environment and the temperature of the artifact This says that there is a constant k, not dependent on time, such that T'= k(E-T).where Esthetemperature ofthe enmonmentme Before collecting the artact from the desert, Susan measured its temperature at a couple of tmes, and she has determined tuat tor the alen ant,ct, k = 0.55 Susan preheats her oven to 70 degrees Fahrenhet (she has stubtomy vetused to pn the metoc wons, A1 metOe oven is at exacty 70 degrees and is heating up, and the oven nuns through a tempesature cycle every 2t minutes, n wch ts temperature vaes by 20 derees above and 20degees below 70 degrees Let E(t) be the temperature of the oven ater t minutes Ad me t 0, when the antact is at a temperature of 40 degnees, she puts t n the oven Let Tt) be the temperature of the arttact at ume t Then T()-1 40 degrees Wite a dinerential equation which models the temperatune of the artfact 0 55(70- Note: Use T'rather than Tt) snce the er contuses the computer Dont enter unts or ths equaton Type here to search の cs

Answer 1

Solution:

We have the differential equation,

dT 0.55 (70+20sint- T) dt

dT 0.55T-0.55 (70-20 sin t) dt

The above is a first order linear differential equation we get whose integrating factor is,

0.55dt055t

multiplying the integrating factor on both sides we get,

>e0 55t(at +0.557-0.55 (70-20 sin t)

dT e0.5St0 dt + 0.55e0.55t T 0.55 (70 +20sint) e0.56

d (Te0.3 t ) dt 20 sin t) e,0.55t 0.55 (70 =

separating variables we get,

> d (Te0.st) = 0.55 (70+ 20 sint) e.Dtdt

0.55t dt d (Te0.55t) 0.55 (70+20 sin t)є

0.55t 10e zf (242 sin (t) 440 cos (t)3647) >Te 521

10 (242 sin (t)-140 cos (t)+3647) 521 ,-0.55t >T=

when t=0,T=40

> 40 61.5547 C

$=>C=-21.5547$

therefore,

when t tends to infinity,

> Tx = 70 9.6381 sin (t-61.19の

Therefore, T infinity lies in the range,