Atmospheric pressure The earth's atmospheric pressure p is...

Atmospheric pressure The 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. \begin{equation} \begin{array}{l}{\text { a. Solve the initial value problem }} \\ {\text { Differential equation: } d p / d h=k p \quad(k \text { a constant) }} \\ {\text { Initial condition: } \quad p=p_{0} \quad \text { when } h=0} \\ {\text { to express } p \text { in terms of } h \text { . Determine the values of } p_{0} \text { and } k} \\ {\text { from the given altitude-pressure data. }}\\{\text { b. What is the atmospheric pressure at } h=50 \mathrm{km} ?} \\ {\text { c. At what altitude does the pressure equal } 900 \text { millibars? }}\end{array} \end{equation}

Atmospheric pressure The 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.
\begin{equation}
\begin{array}{l}{\text { a. Solve the initial value problem }} \\ {\text { Differential equation: } d p / d h=k p \quad(k \text { a constant) }} \\ {\text { Initial condition: } \quad p=p_{0} \quad \text { when } h=0} \\ {\text { to express } p \text { in terms of } h \text { . Determine the values of } p_{0} \text { and } k} \\ {\text { from the given altitude-pressure data. }}\\{\text { b. What is the atmospheric pressure at } h=50 \mathrm{km} ?} \\ {\text { c. At what altitude does the pressure equal } 900 \text { millibars? }}\end{array}
\end{equation}

Key Concepts

Parameter Estimation

Parameter estimation is the process of using data to determine the constants or parameters within a mathematical model. In the context of differential equations, this involves using given conditions or data points to solve for unknown constants, ensuring that the model accurately reflects the observed system behavior.

Mathematical Modeling

Mathematical modeling is the process of representing real-world situations with mathematical expressions and equations. It involves making assumptions, formulating equations such as differential equations based on underlying principles, and then using these models to analyze, predict, and understand complex systems in fields like physics, engineering, and biology.

Exponential Models

Exponential models describe processes where the rate of change of a quantity is proportional to the quantity itself. This proportionality leads to growth or decay that can be represented mathematically by exponential functions. Such models are common in various natural and applied scenarios, including population growth, radioactive decay, and atmospheric pressure variations.

Separable Differential Equations

Separable differential equations are a class of first-order differential equations that can be expressed as a product of a function of the dependent variable and a function of the independent variable. This separability allows the equation to be rearranged so that all terms involving the dependent variable are on one side of the equation and all terms involving the independent variable on the other, facilitating straightforward integration to obtain the solution.

Initial Value Problem

An initial value problem (IVP) involves a differential equation together with a specified value of the unknown function at a given point. The IVP framework ensures that, under appropriate conditions, there exists a unique solution that satisfies both the differential equation and the initial condition, setting the stage for modeling dynamic systems where the initial state is known.

Transcript

00:01 So for this problem, we're looking at atmospheric pressure.

00:04 And so we're given that the earth's atmospheric pressure p is often modeled by assuming that the rate dpdh, which p changes without the h of c level, is proportional to p, and we're given some initial values.

00:17 So for the part a, we're supposed to solve the initial value problem.

00:20 So we're given that dp -d -h is equal to kp.

00:31 So we already know from the book that, so this is an exponential decay.

00:36 Problem.

00:37 So we've right now that the solution to these types of problems is just going to be p is equal to p .0 .e to the k h.

00:53 Again, this is seen in your section in the book.

00:58 So now looking at some of the initial values they've given us, i first see that it says suppose of the pressure at sea level is 1 ,013 ,000.

01:09 So pressure at sea level.

01:13 So that's going to be my initial value.

01:16 So given we already know that our initial value at sea level is 1013.

01:27 So that's an answer to one of my problems that we were supposed to find.

01:33 Next it says that the pressure at an altitude of 20 kilometers is 90 millibars.

01:41 So that is telling us that.

01:44 That when we solve p for an altitude at 20 kilometers, we get that the pressure is 90 millibars.

02:00 So we're gonna use this given information to find everything else that we need.

02:08 So the first thing i'm gonna do is express the p in terms of h.

02:14 So we're almost there, and we just now plug in some things that we know.

02:20 So i know what my p.

02:22 Not is, so i have p.

02:23 It's equal to 1013e to the kh.

02:33 And so now the last thing i need to find is what this value of k is.

02:38 So to find that, i'm going to use this other given information that we were given in the problem.

02:46 So i'm going to plug in 20 for my h and 90 for my p.

02:53 So i have 90 is equal to 1013 e to the 20.

03:00 Okay.

03:02 So as we learned in the last section, solve this, i need to first take the natural log of both sides.

03:09 So i have natural log of 90.

03:10 It's equal to natural log of this, which by product rule is just 1013 plus the natural log of e to 20k...

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