Math whizzes--please help me with this calculus problem! It's bugging me like crazy! o_O?

Question

Hello. This problem is for homework and it's not graded for correctness, but I don't know how to solve it and it's quite bugging me. I've even asked my math tutor and she wasn't sure either... T__T
This is a problem from Calculus AB, in the chapter titled "Logarithmic, Exponential, and Other Transcendental Functions"

Here is the problem:
Air pressure~
Under ideal conditions, air pressure decreases continuously wiht height above sea level at a rate proportional to the pressure at that height. If the barometer reads 30 inches at sea level and 15 inches at 18,000 feet, find the barometric pressure at 35, 000 feet.

*I tried using proportions...but that wasn't the right answer. According to the book, the correct answer is "about 7.79 inches"
Thank you very very much in advance!

Answer 1

Hello, according to your problem, your starting point is as follows:

1. dP/dH = -kP (meaning pressure decreases with proportion to current height, k is a rate constant)

2. dP/P = -kdH (seperation of variables)

3. ln(P) = -kH + C (Integration of both sides)

4. P = e^(-kH + C) (Putting everything to the e)

5. P = Ce^(-kH) (Exponent property, with C as an arbitrary constant)
**** Note P = P(H), just writing P for convienence***


6. Given 30 inches at sea level (where H = 0):
30 = Ce^(0) = C

7. Given 15 inches at 18000 ft:
15 = 30e^(-k*18000)
Solving for k gives 3.85*10^-5

8. Given H = 35000:
P(at 35000) = 30e^(-k*35000) = 7.79 inches


Hope that helps!

Answer 2

Let p = the pressure and h = the height above sea level.

The first part of the question tells you the following:

dp/dh = -kp(h)

Even if you don't know how to solve differential equations such as this, you can use the Leibniz notation above to solve the problem. First, "multiply both sides by 'dh' and divide by p(h) noting that p(h) can never be zero:

dp/p = -kdh

Now, take the antiderivative of both sides:

ln p = -kh + c

where c is the constant of integration. Taking the exponential of each side gives:

exp(ln p) = exp(-kh + c)
p = exp(-kh + c) = exp(-kh)*exp(c) = C*exp(-kh)

where C = exp(c)

Now, you have the following equation for the pressure as a function of height:

p(h) = C*exp(-kh)

Now, we have to solve for k and h. We have two unknowns and two equations, given by the data points:

p(18,000) = 15 = C*exp(-k*18000)
p(sea level) = p(0) = 30 = C*exp(-k*0)

The second equation tells you that C = 30 = the pressure at height zero. Plugging this into the first equation gives:

15 = 30*exp(-k*18,000)
exp(-k*18,000) = 0.5

Take the logs of both sides and you get:

-k*18000 = ln(0.5) = -0.693 (approximate)

k = 0.693/18000 = 0.0000385

Now, we have the full equation for the pressure as a function of height. To fine the pressure at 35,000 ft we simply have to plug in 35,000 ft as the height:

p(35,000) = C*exp(-k*35,000) = 30*exp(-0.0000385*35,000)
p (35,000) = 30*exp(-1.3475) = 7.79

So, the answer is that the pressure at 35,000 ft is 7.79 inches (actually closer to 7.8)

Answer 3

dp/dz = -kp

ln(p) = -kz + lnC
p = Cexp(-kz)

at z = 0, p = 30 so C = 30

p = 30 exp(-kz)

at z = 18000, p = 15

15 = 30 exp(-k*18000)
ln(.5) = -k*18000
k = 3.851 e-5

p = 30 exp (-3.851 e-5 z)

So at z = 35000 feet p = 7.79