Differential equations in Calculus...plsssss help?

Suppose that represents the temperature of a cup of coffee set out in a room, where T is expressed in degrees Fahrenheit and t in minutes.
A physical principle known as Newton’s Law of Cooling tells us that
dT/dt = -1/15T+5
15T + 5.
a) Supposes that T(0) = 105. What does the differential equation give us for the
value of dT
dt |T=0? Explain in a complete sentence the meaning of these two
facts.
(b) Is T increasing or decreasing at t = 0?
(c) What is the approximate temperature at t = 1?
(d) On a graph, make a plot of dT/dt as a function of T.
(e)For which values of T does T increase?
(f) What do you think is the temperature of the room? Explain your thinking.
(g) Verify that T(t) = 75 + 30e^(-t/15) is the solution to the differential equation with initial value T(0) = 105. What happens to this solution after a long time?

asked by Bridget
  1. Hard to see just what you wrote, but I'm guessing

    dT/dt = -1/(15T+5)
    (15T+5) dT = -dt
    No, that can't be right, if you want an exponential function.
    How about

    dT/dt = 15T+5
    then we have
    dT/(3T+1) = 5dt
    ln(3T+1) = 5t+k
    3T+1 = ce^5t
    T = (ce^5t - 1)/3

    Hmm. Not that either. Let's do some answer analysis.

    T(t) = 75 + 30e^(-t/15)
    dT/dt = -2e^(-t/15) = -2(T-75)

    Well, if you fix up your T function, things should be pretty straightforward. Where do you get stuck?

    posted by Steve
  2. Sorry for the confusion, it's (-1/15)T+5.

    And the last part is 75+30e^(-t/15) the fraction is all part of the e and is not divided by each other.

    I am confused on a through e mainly and g at the end. Should I solve the equation and just plug in 105 and graph the function to solve the answers to the problems being asked? I just don't get what I'm supposed to do.

    posted by Bridget
  3. so. you have

    dT/dt = -T/15 + 5 = -1/15 (T-75)
    dT/(T-75) = -1/15 dt
    ln(T-75) = -t/15 + k
    T-75 = e^(-t/15 + k)
    T-75 = ce^(-t/15)
    T = 75 + ce^(-t/15)

    Now, you are told that T(0) = 105, so
    105 = 75 + ce^0
    c = 30

    T(t) = 75 + 30e^(-t/15)
    as desired in part (g)
    So, the coffee starts at 105 and decreases from there, ever more slowly, as the temperature difference becomes less and less.

    You can see that as t grows large, e^(-t/15) -> 0, so the coffee approaches 75° as a limit. That must be the temperature of the room.

    As for the increasing and decreasing stuff, you know that T(t) is increasing where dT/dt is positive.

    posted by Steve

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