Posted by **Anonymous** on Thursday, May 22, 2008 at 1:10am.

Time (min)0___10___20____30__40____50____60

Temp (F)68_72.64_76.97__80_82.9_85.38__87.5

Use the Trapezoidal Rule to find the best approximation you can for the average temperature in the trailer from 10:00 until 11:00.

(10/60)/2 (68 + 2(72.64) + 2(76.97) + 2(80) + 2(82.9) + 2(85.38) + 87.5)= 79.273

The temperature inside the trailer has been increasing since 10:00 AM at a rate proportional to the difference between it and the temperature outside the trailer (a cool 100 deg F). Write a differential equation and solve it. Use any of the items in the table to help you find the constants C and K.

dy/dt=k(100-y)

dy/(100-y)=kdt

-ln(100-y)=kt

100-y=e^-(kt)

100-y=Ce^-k (what do I do now and afterwards for the following questions?) Thanks for the help!

Use your diff-eq solution to find the average temp in the trailer between 10:00 and 11:00.

At 11:00 AM the temp outside was 100*F. Then it began to rise at a rate of 0.10 degrees per minute. I read somewhere that the cost of cooling a trailer accumulates at a rate of $0.005 per minute for each degree the outside temp exceeds 78*F. How much would you say it cost the school to keep you cool in that trailer from 11:15 until 11:45 AM?

- Calculus -
**Damon**, Thursday, May 22, 2008 at 4:48pm
100-y=Ce^-k

yes agree so y = 100 - C e^-kt

let's call t = 0 at 10 am

so at 10 am

68 = 100 - Ce^0 = 100 -C

so

C = 32

now

y = 100 -32 e^-kt

well, at t = 60 min which is 11 am, y = 87.5

so

87.5 = 100 - 32 e^-60 k

32 e^-60 k = 12.5

e^-60 k = .3906

-60 k = ln .3906 = - .94

k = .0157

so in the end

y = 100 - 32 e^-.0157 t

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