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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.Earth's atmospheric pressure $p$ is often modeled by assuming that the rate $d p / d h$ at which $p$ changes with the altitude $h$ above sea level is proportional to $p .$ Suppose that the pressure at sea level is 1013 millibars (about 14.7 pounds per square inch) and that the pressure at an altitude of $20 \mathrm{km}$ is 90 millibars. a. Solve the initial value problemDifferential equation: $\quad d p / d h=k p \quad$ (k a constant) Initial condition: $\quad p=p_{0} \quad$ when $\quad h=0$ to express $p$ in terms of $h .$ Determine the values of $p_{0}$ and $k$ from the given altitude-pressure data.b. What is the atmospheric pressure at $h=50 \mathrm{km} ?$c. At what altitude does the pressure equal 900 millibars?

a. $ p(h) = 1013 e^{ 20 ln( \frac{90}{1013} )h } $ b. $ p(50) \approx 1013e^{-2430} mbar \approx 0 $. c. $ h \approx 0.002 \ km $

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Campbell University

Harvey Mudd College

Idaho State University

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So they give us this differential equation model that is supposed to tell us about the change in pressure with respect to our altitude. And they first wants to solve this initial value problem that they give us. So let's go ahead and do that. And then they tell us to Saul or everything in terms of these two actual values that they give us. Let's first figure out what we have that we comply everything in from there. All right, so this is what we're going to solve. Well, that equation there is a slipper be differential equation. So let's go ahead and first divide over by P and multiply by D h. So we're gonna have one overpay dp is equal to, and so are constant k Th Now we can integrate each side and we're going to take the natural log. Are integrating the left hand side is gonna be the natural log of the absolute value of Pete is equal to and then integrating a constant here. Well, that should just be K h plus from constant C. All right, an actual hold a C one for right now. Now we want to go ahead and exponentially ate. Each side is it won't matter about the absolute values, coz in this sense it doesn't really make sense to talk about negative pressures. So we'll never have to plug that in or ah, pressure of zero doesn't really make much sense. Either of those council and there were still have p is equal to e to the K h must see one. And now we could rewrite this cause. Remember, when we have things being added as exponents, it would be e to the C one times e to the K h, but then eat. The C one is also just a constant so we can rewrite. This is just see two times e to the cage. All right, so this is what our solution is going to be. But we're also told this initial condition right here, let's go ahead and plug that in. So we know when p or when h is equal to zero Misha equal to Pena. So this is gonna be able to see to e to the k time zero so that she's gonna be eating zero, which ends up just being one Landau with C two is peanuts. So our equation that we're going to have is p of H is equal to it. Looks like p not times e to the K h. So this is just our general solution to it. So right here, general solution. But now they want us to use this information here. We'll see Level is when h is equal to zero. So this right here tells us what p not is. So we could just go ahead and plug that in. So we're gonna have 1013 that we have e to the K H. Now we want to solve for Kay, and then we can use this one here, so this is going to be a TSH. And this is going to be P S O p. Of h. Now plug and all that, and we should get 90 is equal to 1 13 mmm to the K times 20. And now we just want to first divide and then take the natural log on each side. So that's going to be 90/1 13. Naturally, zoom out a little bit, so 90 over 1000 and then we're gonna want to take the natural log. Like I was saying, and that should give us K times 20. Then we would divide each side by 20 to get K is equal to 20. Natural log. That should have been 10 13. So natural log of 90 over 1 30 and sold out. Lastly, we can plug everything in to this equation. Here, take it. That p of h is going to be equal to. And actually, let's just, uh, approximate this number actual. Plug it indirectly. So 10 13 e raised to the 20 natural log of 90/10 13 each. So this is what it is. Exactly. And if you wanted, you could just approximate our value for Kay here and let's see what that is. So natural log 0 90 divided by 10 13 times 20. So this is around negative, wordy. Thanks point. Or so you could just replace that with what we have in the longer if you don't have to write all that up. But I would say that is our solution for everything that they give us. Now we want to determine what the pressure is at a height of 50 kilometers, So we're gonna plug in 50. So that's gonna be P of 50 is equal to. So I'll say approximately equal to 10. 13 e to the A, And I'm gonna use that estimation this time. And then this is times 50. So we're gonna multiply that number by 50 and then exponentially ate it and actually give the super small value so it's actually rate that out. So this is going to be approximately in 13 times Hey, to the negative to or 30 And even when I plug this into my calculator, this just says it's approximately zero. So this is extremely small number. All right, so this may be so far out that the pressure is negligible ended. Lastly, we want to figure out when it would be 50 millibars and let's see. So that means we're going to set our function equal to 900. And in salt for h that would be 10 13 or actual approximately because I'm going to use the estimation for it again. So negative 48.4 times teeth. So first up is gonna be the same. We're going to go ahead and divided side by 10 13. Take the natural log that I should give us natural log of 900 over 10 13 is approximately equal to negative. 48 went or teeth. Then we would divide each side by negative 48.4. And that's gonna give us natural log of 900 over 10. 13 over negative, 48.4 at the yourself approximately here. And then let's see what we get for this number here. So natural log of 900 divided by 10. 13 and then divide that by negative 48.4. So this is about 0.0 two. Well, Mr Ned Beatty, that should be, uh, h let me replace on those cause this is supposed to be a height, so this would be in kilometers so you could multiply that by 1000 which would be about 2.4 meters. So this is about where that's pressure will occur.

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