Assume that the pressure in the tire, P , decreases at a rate proportional to the difference between the tire pressure and the atmospheric pressure, which is a constant 15 lbs. On Monday at 6:00 pm, she pumped up the tire pressure to 85 lbs. Now, on Tuesday at 6:00 pm, the pressure is down to 75 lbs. Let t be the number of days since Monday at 6:00 pm, and let k be the constant of proportionality.

a. Translate the problem into a differential equation and state the given initial values in terms of the variables.

b. Find the general solution to the differential equation symbolically (algebraically).

c. Use the initial values to find the solution to the initial value problem.

d. If she doesn’t pump up her tire, when will the tire pressure reach 40 lbs? Give the
answer in the number of days, t , since Monday at 6:00 pm.

Let P(t) be the pressure in the tire at time t, and let Pa be the atmospheric pressure. We are told that the rate at which the P(t) decreases is proportional to the difference between P(t) and Pa, so:

dP(t)/dt = -k*(P(t) - Pa), P(0) = Po = 85 lbs/in^2

where k is a positive constant of proportionality.

This is the initial value problem that you need to solve (the answer to part a).

[I'm going to drop the explicit reference to P as a function of t in what follows to save typing, i.e., I'll write P(t) simply as P.]

This is a separable differential equation:

dP/(P - Pa) = -k dt

Integrate

ln(P - Pa) - ln(Po - Pa) = -k*t

where Po = P(0) is the pressure at time t = 0.

ln((P - Pa)/(Po - Pa)) = -k*t

(P - Pa)/(Po - Pa) = exp(-k*t)

P(t) = Pa + (Po - Pa)*exp(-k*t)

This is the general solution to the problem (answer b).

Plugging in the values for Pa and Po, we have:

P(t) = [15 + 70*exp(-k*t)]*lbs/in^2

Now use the fact that at t = 1 day, P(1 day) = 75lbs/in^2 to solve for the constant, k:

75 = 15 + 70*exp(-k)

6/7 = exp(-k*day)

k = ln(7/6)/day = 0.154/day

So:

P(t) = [15 + 70*exp(-0.154*t)]*lbs/in^2

For part (d), you want to know when P(t) = 40 lbs/in^2:

40 = 15 + 70*exp(-0.154*t/day)

25/70 = exp(-ln(7/6)*t/day)

ln(25/70)*day = -ln(7/6)*t

t = ln(70/25)day/ln(7/6) = 6.68 days