Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Atmospheric density at altitude \(h\) meters is approximately \(1.225 e^{-0.000101 h}\) \(\mathrm{kg} / \mathrm{m}^{3}\) for \(0 \leq a \leq 5000\) meters. Compute the mass of air in a vertical one-square meter column between 0 and 5000 meters.

Step-by-Step Solution

Answer

The mass of air is approximately 4782.5 kg.

Step 1: Understanding the Problem

We need to compute the mass of air in a vertical column with a cross-sectional area of 1 square meter and a height from 0 to 5000 meters. To do so, we must integrate the atmospheric density function over this height.

Step 2: Define the Density Function

The given atmospheric density function is \( \rho(h) = 1.225 e^{-0.000101 h} \) kg/m³. This function describes how the density varies with altitude \( h \).

Step 3: Set up the Integral for Mass

The mass of air over a height can be found by integrating the density function. The differential mass \( dm \) can be defined as \( \rho(h) \cdot A \cdot dh \), where \( A \) is the area (1 m²). Thus, the integral for mass \( M \) is: \[ M = \int_{0}^{5000} 1.225 e^{-0.000101 h} \, dh \]

Step 4: Evaluate the Integral

To find the solution to the integral, we solve \[ M = 1.225 \int_{0}^{5000} e^{-0.000101 h} \, dh \].The integral of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \), so the evaluated integral becomes: \[ M = 1.225 \left[ -\frac{1}{0.000101} e^{-0.000101 h} \right]_{0}^{5000} \].

Step 5: Compute the Definite Integral

Evaluate the definite integral from 0 to 5000:\[ M = 1.225 \left( -\frac{1}{0.000101} \right) \left( e^{-0.000101 \times 5000} - e^0 \right) \]Simplify using \( e^0 = 1 \) and compute numerically:\( M = 1.225 \times -9900 \times (e^{-0.505} - 1) \).

Step 6: Calculate and Simplify

Calculate the exponential term \( e^{-0.505} \approx 0.603028 \),then compute \[ M = 1.225 \times -9900 \times (0.603028 - 1) \approx 1.225 \times 9900 \times 0.396972 \].Finally, \[ M \approx 4782.5 \].

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in calculus, focusing on the accumulation of quantities and understanding the changes across intervals. It is often visualized as the area under a curve on a graph. In this particular problem, integral calculus helps us compute the total mass of air within a specified atmospheric column.By integrating a function, you essentially add up an infinite number of infinitesimally small quantities. Here, the relevant function is the atmospheric density, represented by \( \rho(h) = 1.225 e^{-0.000101 h} \), which varies with altitude \( h \). The integral \[ \int_{0}^{5000} 1.225 e^{-0.000101 h} \, dh \] essentially sums up the weighted height of the air column.One key concept to grasp with integrals is that they provide a means to handle continuous data, as opposed to discrete sums. Such analysis is useful in physics, engineering, and many other fields where understanding cumulative quantities is crucial.

Atmospheric Density

Atmospheric density is a measure of how much mass of air is present in a given volume. It varies with altitude due to changes in pressure and temperature. In this exercise, atmospheric density is modeled as a function of altitude \( h \), given by \( \rho(h) = 1.225 e^{-0.000101 h} \) kg/m³.Density decreases exponentially with altitude, which is why the density function includes an exponential component. This behavior aligns with physical principles, as higher altitudes have lower air pressure and, consequently, a lower density. Unlike a constant density scenario, this functional relationship allows us to better understand how air density impacts the overall mass in different layers of the atmosphere.By integrating this function over the interval from 0 to 5000 meters, we calculate the mass of air in a column, showing how this seemingly simple concept of density can be used to resolve complex atmospheric behavior in real-world applications.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a e^{bx} \), where \( e \) is the base of the natural logarithm. These functions grow or decay at rates proportional to their current value, hence they are ideal for modeling processes like radioactive decay, population growth, and atmospheric pressure and density.In this problem, the atmospheric density is modeled as an exponential function \( \rho(h) = 1.225 e^{-0.000101 h} \). Here, \( e^{-0.000101 h} \) indicates that the density decreases exponentially as altitude increases. The negative exponent signifies decay, illustrating how atmospheric characteristics change with elevation.Exponential functions are vital tools in mathematics and science due to their property of providing scalable and adaptable models. They capture both the fast-decreasing and flattening-off behaviors observed in natural processes. Understanding these functions helps illustrate how crucial decay and growth rates are, providing insight into the dynamics affecting the mass of atmospheric layers.