Homework Help: Math

Posted by Lucy on Thursday, May 24, 2012 at 4:53am.

on a day when the air temperature is 34 degrees celsius, a can of soft drink is taken from the fridge at a temperature of 4 degrees celsius at 2pm. Thereafter the temperature of the can satisfies the differential equation; dT/dt=0.1(34-T), where t is the number of minutes after 2pm. What is the temperature of the can at 2:15pm? At what time does the temperature reach 30 degrees?

Math - MathMate, Thursday, May 24, 2012 at 8:04am
dT/dt=0.1(34-T)
dT/(34-T) = 0.1dt
Integrate both sides
-ln(34-T)=0.1t+c
34-T=Ce^(-0.1t)
T=34-Ce^(-0.1t)
Initial conditions:
t=0, T=4 => 4=34-C => C=30
so
T=34-30e^(0.1t)
A) Substitute t=15 to solve for T.
B) Substitute T=30 to solve for t.
Math - Lucy, Thursday, May 24, 2012 at 8:05am
wow thank you so much
Math :) - MathMate, Thursday, May 24, 2012 at 8:49am
You're welcome!

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