Problem 24 - Thomas Calculus

Question

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}

Step-by-Step Solution

Answer

Pressure at 50 km is approximately 0.987 millibars. The altitude for 900 millibars is approximately 1.12 km.

Step 1: Set up the Differential Equation

The differential equation given in the problem is \( \frac{dp}{dh} = kp \), where \( k \) is a proportionality constant. This equation signifies that the rate of change of pressure with respect to altitude is proportional to the current pressure.

Step 2: Solve the Differential Equation

To solve \( \frac{dp}{dh} = kp \), we use separation of variables. We rearrange the equation to \( \frac{dp}{p} = k \, dh \). Integrating both sides gives \( \ln |p| = kh + C \), where \( C \) is the integration constant. Exponentiating both sides results in \( p = e^{kh+C} = e^C e^{kh} = C'e^{kh} \), where \( C' = e^C \).

Step 3: Determine the Initial Condition

The initial condition is \( p = p_0 \) when \( h = 0 \). Given \( p_0 = 1013 \) millibars at \( h = 0 \), we substitute these into \( p = C'e^{kh} \) to find \( C' = 1013 \). Therefore, \( p = 1013e^{kh} \).

Step 4: Determine the Constant k

Given that \( p = 90 \) millibars at \( h = 20 \, \mathrm{km} \), substitute into the equation: \( 90 = 1013e^{20k} \). Solve for \( k \): \( e^{20k} = \frac{90}{1013} \), then \( 20k = \ln \left( \frac{90}{1013} \right) \). So, \( k = \frac{1}{20} \ln \left( \frac{90}{1013} \right) \).

Step 5: Pressure at 50 km

To find the pressure at \( h = 50 \, \mathrm{km} \), use the expression \( p = 1013e^{kh} \) with \( h = 50 \). Substitute \( k \) and calculate: \( p = 1013e^{50\left(\frac{1}{20} \ln \left( \frac{90}{1013} \right)\right)} \). Simplifying this, it can be written \( p = 1013 \left( \frac{90}{1013} \right)^{\frac{50}{20}} \).

Step 6: Altitude for 900 millibars

To find the altitude where \( p = 900 \) millibars, substitute into \( 900 = 1013e^{kh} \). Solve for \( h \): \( e^{kh} = \frac{900}{1013} \), \( kh = \ln \left( \frac{900}{1013} \right) \). Substituting for \( k \), \( h = \frac{1}{k} \ln \left( \frac{900}{1013} \right) \). Simplify it using the expression for \( k \): \( h = 20 \times \frac{\ln \left( \frac{900}{1013} \right)}{\ln \left( \frac{90}{1013} \right)} \).

Key Concepts

Exponential Growth and Decay

Exponential growth and decay are fundamental concepts in differential equations, depicting how quantities change at rates proportional to their current values. In this context, the focus is on decay since atmospheric pressure decreases with increasing altitude.
In the problem, the differential equation \(\frac{dp}{dh} = kp\) describes how pressure \(p\) changes with altitude \(h\). The constant \(k\) indicates whether pressure increases or decreases as altitude changes. Here, \(k\) will be negative because pressure decays with height.
The solution to this differential equation, when solved, takes the form \(p = C'e^{kh}\). This equation is an example of exponential decay, where the pressure decreases exponentially as the altitude \(h\) increases. By understanding this concept, one can predict how pressure changes in atmospheric conditions at different altitudes.

Separation of Variables

Separation of variables is a method used to solve differential equations like \(\frac{dp}{dh} = kp\). This technique involves rearranging the equation so that each variable and its differential is on opposite sides of the equation.
For our atmospheric pressure problem, we start by rewriting the differential equation as \(\frac{dp}{p} = k \, dh\). This step separates variables \(p\) and \(h\) into distinct sides of the equation, making it easier to solve.

• The next step is integrating both sides: \( \int \frac{dp}{p} = \int k \, dh \).
• Integration yields \( \ln |p| = kh + C \), where \(C\) is the integration constant.
• By exponentiating both sides, we arrive at the general form of the exponential function describing the pressure: \( p = e^{kh+C} = C'e^{kh} \).

Separation of variables makes it straightforward to derive solutions to exponential growth and decay problems by isolating and integrating the relevant terms.

Initial Value Problem

An initial value problem (IVP) in differential equations has additional conditions that the solution must satisfy at a particular point. This is key to solving real-world scenarios where initial conditions are known.
In our exercise, the initial condition is that the pressure \(p\) equals 1013 millibars at sea level (\(h = 0\)). This condition helps determine the constant \(C'\) in the exponential solution \(p = C'e^{kh}\).
To find \(C'\), substitute the initial condition into the equation: \(1013 = C'e^{0}\). Since \(e^0 = 1\), we find \(C' = 1013\). With \(C'\) known, the pressure equation becomes \(p = 1013e^{kh}\).
The initial value problem persuasively combines integration and initial conditions, providing a unique solution that fits the specific constraints of the problem, like altitude and pressure in this case.