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March 26, 2017

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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 - ,

    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 - ,

    wow thank you so much

  • Math :) - ,

    You're welcome!

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