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
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?
1 answer