Exponential rate? when will the pressure be 2500 millibar?

[i_m_mimi]

Homework Statement

[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 dierential 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?

answer:
10*[ln(4/3)]/[ln(10/9)]

Homework Equations

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

 

[SteamKing]

The function P(t) must satisfy the differential equation. Have you checked to see if this is the case?

 

[HallsofIvy]

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
P- 1000= Ce^{-kt}

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.

 

[i_m_mimi]

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
P- 1000= Ce^{-kt}

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.

I'm closer to the answer now, but still something wrong

P- 1000= Ce^{-kt}

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?

thanks

 

[milesyoung]

<br /> \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)}<br />
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)

etc.