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Homework Help: Exponential rate? when will the pressure be 2500 millibar?

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    [10] The air pressure in an automobile's spare tire was initially 3000 millibar.
    Unfortunately, the tire had a slow leak. After 10 days the pressure in the tire had declined to 2800 millibar. If P(t) is the air pressure in the tire at time t,then P(t) satis es the di erential equation
    dP/dt = −k(P(t)−A);
    where k is a constant and A is the atmospheric pressure. For simplicity, take
    atmospheric pressure to be 1000 millibar. When will the pressure in the tire be
    2500 millibar?


    2. Relevant equations

    3. The attempt at a solution

    P(t) = Ce^-kt
    0 = 3000e^-k(0)
    2800 = 3000 e^-k(10days)

    k = -[ln 28/30]/10

    2500 = 3000 e^[ln 28/30]t/10

    t = 10 [ln 25/30]/[ln 28/30] wrong answer

    dP/dt = 0.1 ln(28/30) (P(t) - 1000)

    I can't figure out what is the function P(t) = ? or where to put in variable t, time. And where does the dP/dt equation come in? this is a complicated exam question worth 10 marks with 20 minutes allocated to it.

    thank you
  2. jcsd
  3. May 21, 2013 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    The function P(t) must satisfy the differential equation. Have you checked to see if this is the case?
  4. May 22, 2013 #3


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    Science Advisor

    Your problem appears to be that you have neglected the atmospheric pressure, A.
    From "dP/dt = −k(P(t)−A)" with A= 1000, dP/(P- 1000)= -kdt so that ln(P- 1000)= -kt+ c and then
    [itex]P- 1000= Ce^{-kt}[/itex]

    Alternatively, define a new dependent variable, Q= P- 1000, the difference between the pressure in the tire and atmospheric pressure, so that your differential equation becomes
    "dQ/dt= -kQ" and you CAN use "Q= Ce^{-kt}", but, of course remember to subtract 1000 from each of the given tire pressures: that is, initially Q= 3000- 1000= 2000, after 10 days, Q= 2800- 1000= 1800, and the question is asking for t when Q= 2500- 1000= 1500.
  5. May 22, 2013 #4
    I'm closer to the answer now, but still something wrong

    [itex]P- 1000= Ce^{-kt}[/itex]

    I tried using this equation, but my answer was something like t = 10*ln 0.5 / ln(3/5)

    Tried it again using another equation you gave me:

    dP/dt = −k(P(t)−A)
    dP/[P(t)-A] = -kdt

    took the integral of both sides

    ln [P(t)-1000] = -kt + C

    P(t)-1000 = e^-kt+c

    3000-1000 = e^-k(0)+c
    c = ln2000

    2800 - 1000 = e^-k(10)+ln2000
    k = -(ln9/10)/10

    2500 - 1000 = e^(ln9/10)/10 (t)+ln2000
    t = 10ln0.75/ln0.9

    My fractions seem to be the inverses of the answer's fractions, what did I do wrong?

  6. May 22, 2013 #5
    \frac{\ln \left( \frac{3}{4} \right)}{\ln \left( \frac{9}{10} \right)} = \frac{\ln(3) - \ln(4)}{\ln(9) - \ln(10)} = \frac{-(\ln(4) - \ln(3))}{-(\ln(10) - \ln(9))} = \frac{\ln \left( \frac{4}{3} \right)}{\ln \left( \frac{10}{9} \right)}
    Your result is correct but your derivation is hard to follow, e.g.:
    P(t)-1000 = e^-kt+c

    should be:
    P(t)-1000 = e^(-kt+c)

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