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. a. Solve the initial value problem Differential equation: $d p / d h=k p \quad(k$ 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?

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.
a. Solve the initial value problem Differential equation: $d p / d h=k p \quad(k$ 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?

Key Concepts

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.

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.

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.

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.

Transcript

00:01 Given this differential equation, then we could derive that the original function is going to be an exponential.

00:08 And so this will be a function of height, and this will give us the pressure.

00:15 And so it's going to be equal to the initial pressure, multiplied by an exponential times the decay factor, multiplied by not time, but height.

00:27 And then we're given some initial conditions here when the height is zero or at sea level.

00:34 We have a pressure of 1013 millibars.

00:39 And we could go ahead and substitute these in.

00:43 And so we'll end up with 1013 on the left side is equal to p.

00:49 Not multiplied by the exponential to the power zero since h is zero.

00:59 And then that knocks out the exponential, and therefore p -not is 1013.

01:07 And next we're going to solve 4k, so we're going to use that pressure function, and we're going to use the information that the pressure is 90 milligrams, and of course we have the initial pressure here, but the pressure is 90 milligrams at 20 kilometers.

01:31 So we'll go ahead and put that as our h and then multiply that by rk.

01:40 And we want to, again, solve for our k value here.

01:44 And we can do this in two steps here.

01:46 We're going to first divide the 1013 over.

01:50 So we'll have 90 over 1013.

01:58 Then we have e on the right side.

02:00 So at this point, we're going to take the natural log of both sides.

02:04 And that'll knock out the exponential on the right side.

02:08 And so then that just leaves us with 20.

02:11 Okay.

02:15 And then, of course, we want to solve for k, so we're going to divide by 20.

02:23 And just for reference, this is equal to about negative 0 .121...