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Differential Equation Proof

  1. Jan 13, 2006 #1
    I need some help.

    A pot of boiling water at 100C is removed from a stove at time t = o and left to cool in the kitchen. After 5 min, the water temperature has decreased to 80C, and another 5min later it has dropped to 65C. Assume Newtons law of cooling (dT/dt = k(M - T) ) applies, determain the constant temperature (M) of the kitchen.

    Okay, so the solution to the differential equation is - T= Ce^(kt) + M and we want to solve for M

    T(0) = C +M
    100 = C + M

    Equation 1.
    T(5) = Ce^(k5) + M
    80 = Ce^(k5) + M

    Equation 2.
    T(10) = Ce^(k10) + M
    65 = Ce^(k10) + M

    Therefore.. the 2 equations are
    80 = Ce^(k5) + M
    65 = Ce^(k10) + M

    and if I subtract them, I get

    15 = Ce^(k5) - Ce^(k10)

    And this is where im lost. I have no idea how to solve for M

    Can somebody help me please?

  2. jcsd
  3. Jan 13, 2006 #2


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    So you have:








    (minus c, plus c same dif it's arbitrary)

    so that:




    So you have 3 conditions, and 3 unknowns, kinda' messy but try and make those substitutions and then solve for something like M in terms of c, c in terms of k, then substitute that into one of them to solve for k, then the others.
    Last edited: Jan 13, 2006
  4. Jan 13, 2006 #3


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    You are given the temperature at 3 times because you have 3 unknown numbers (C, k, and M) and need to solve 3 equations. You wrote the third equation but didn't number it or use it: C+ M= 100.

    You can use that to write the two equations 80= Ce5k+ M and 65= Ce10k+ M as 80= (100-M)e5k+ M and 65= (100-M)e10k+ M. Now, you want to eliminate k, not M, from those equations. Since e10k= (e5k)2 one way to do that is to write (80- M)2= (100-M)2e10k and 65- M= (100-M)e10k and divide one equation by the other. You get
    [tex]\frac{(80-M)^2}{65-M}= 100- M[/tex].
    That's a quadratic equation for M.
  5. Jan 14, 2006 #4
    Salty dog, when i originally did solved the differential equation, i got the same answer as you, T= M - Ce^(-kt)

    However, my text book solution gives T= Ce^(kt) + M.

    Now im really confused. which one is right?

    Well... I did land at the same answer for M for both answers, so i assume they are both correct
    Last edited: Jan 14, 2006
  6. Jan 14, 2006 #5


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    Since C, k, and M are arbitrary constants their algebraic sign is arbitrary until you have determined their value.
  7. Jan 14, 2006 #6


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    Ok guys. Didn't notice that.

    They are the same equation when solved for the constants:


  8. Jan 14, 2006 #7
    okay i have one more differential equation....

    m(dv/dt) = mg - bv

    where m is the mass, g is the acceleration of gravity, and b>0 is a constant.
    if m = 100kg, g = 9.8m/s^2 and b=5 kg/sec, and v(0) = 10m/s, solve for v(t). what is the terminal velocity of the object.

    So firstly, i want to solve for this differential equation.
    But i cannot find a way to seperate it. (i mean, its not seperable?)

    m(dv/dt) = mg - bv
    (dv/dt) = g - bv/m

    then i pluged in the numbers and got

    (dv/dt) = 9.8 - 0.05v

    and thats about as far as i could get.

    could somebody please give me a little push

  9. Jan 14, 2006 #8
    It is seperable, I'm not sure that's the
    way i would solve it but:

    (dv/dt) = g - (b/m)v => dv/(g - vb/m)= dt

    then a change of variables v -> vb/m turns this
    into a simple integral.
  10. Jan 14, 2006 #9
    oh actually you are right...

    dv/(g - vb/m)= dt and you sub in the constants to get

    dv/(9.8-0.05v) = dt and solve

    (-1/0.05)ln|9.8 - 0.05v| = t + c

    and isolate for v to get

    v(t) = 196 - 20e^(-0.05t)(c)

    and given that v(0) = 10, you can solve for C to get

    v(t) = 196 - 186e^(-0.05t)

    Now.. the problem asks "What is the limiting (i.e, terminal) velocity of the object?"

    Would that just be the limit t-> infinity ? that would get 10m/s?
  11. Jan 14, 2006 #10
    the idea is right:

    v_term = limit (t->infinity) 196 - 186 exp(- t/20) = 196.

    given v(0) = v0 and m dv/dt = mg - bv we get:
    v(t) = gm/b + (v0 - gm/b)exp(-bt/m)

    and v_terminal = limit (t-> infinity) v(t).
    thus v_terminal = gm/b

    [ as lim t->infinity exp(-k*t) = 0 if k>0 ]

    we can check the answer, since at v_terminal
    there is no acceleration. that is dv/dt = 0 at v_terminal
    and the equation of motion is:

    m dv/dt = 0 = m g - b v_terminal
    => v_terminal = mg/b.

    i always like problems that have built in checks
    to see whether or not i get the right answer.
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