Problem 17 - Calculus for Biology and Medicine

Question

17\. Air Density The density of air changes with height. Under some conditions density \(\rho\), depends on height \(z\), and temperature \(T\) according to the equation: $$ \rho(z, T)=\rho_{0} e^{-\lambda z / T} $$ where \(\rho_{0}\) and \(\lambda\) are both constants. A meteorological balloon ascends (i.e., starts at \(z=1\) and gains height) over the course of several hours. (a) Assuming that the balloon ascends at a speed \(v\) (i.e., \(d z / d t=\) \(v\) ) and that the temperature changes over time (i.e., that \(T\) is given by a function \(T(t))\), derive, using the chain rule, an expression for the rate of change of air density, as measured by the weather balloon. (b) Assume that \(v=1, \rho_{0}=1\), and \(\lambda=1\) and that when \(t=0, T=1\). Are there any conditions under which the density, as measured by the balloon will not change in time? That is, find a differential equation that \(T\) must satisfy, if \(d \rho / d t=0\), and solve this differential equation.

Step-by-Step Solution

Answer

T(t) = e^t for density to remain constant at the balloon's measured height.

Step 1: Differentiate Air Density with Respect to Time

We have the equation for air density \( \rho(z, T) = \rho_0 e^{-\lambda z / T} \). We need to find \( \frac{d \rho}{d t} \), the rate of change of air density over time. Using the chain rule, we can express \( \frac{d \rho}{d t} \) in terms of \( \frac{\partial \rho}{\partial z} \) and \( \frac{\partial \rho}{\partial T} \). Specifically, \( \frac{d \rho}{d t} = \frac{\partial \rho}{\partial z} \cdot \frac{d z}{d t} + \frac{\partial \rho}{\partial T} \cdot \frac{d T}{d t} \).

Step 2: Find Partial Derivatives of Air Density

We first find \( \frac{\partial \rho}{\partial z} \), which is \(-\frac{\lambda \rho_0}{T} e^{-\lambda z / T} \). Next, we find \( \frac{\partial \rho}{\partial T} \), which is \( \frac{\lambda \rho_0 z}{T^2} e^{-\lambda z / T} \).

Step 3: Express the Rate of Change of Air Density

Substitute the partial derivatives into the chain rule expression: \[ \frac{d \rho}{d t} = \left( -\frac{\lambda \rho_0}{T} e^{-\lambda z / T} \right) \cdot v + \left( \frac{\lambda \rho_0 z}{T^2} e^{-\lambda z / T} \right) \cdot \frac{d T}{d t} \]. Simplifying, it becomes: \[ \frac{d \rho}{d t} = -\frac{\lambda \rho_0 v}{T} e^{-\lambda z / T} + \frac{\lambda \rho_0 z}{T^2} e^{-\lambda z / T} \cdot \frac{d T}{d t} \].

Step 4: Determine Condition for Unchanging Air Density

For the density to remain constant, \( \frac{d \rho}{d t} = 0 \). Therefore, \[ -\frac{\lambda \rho_0 v}{T} + \frac{\lambda \rho_0 z}{T^2} \cdot \frac{d T}{d t} = 0 \].

Step 5: Find the Differential Equation for \( T \)

Simplify the equation to find the condition on \( T \): \[ \frac{d T}{d t} = \frac{v T}{z} \]. Substitute \( v = 1, z = 1, \lambda = 1, \rho_0 = 1 \); thus, the differential equation becomes: \( \frac{d T}{d t} = T \).

Step 6: Solve the Differential Equation

The equation \( \frac{d T}{d t} = T \) is a first-order linear differential equation. Separate variables and integrate: \( \int \frac{1}{T} d T = \int d t \). This results in \( \ln |T| = t + C \). Exponentiating both sides, we have \( T = e^{t + C} = Ce^t \). For \( t = 0, T = 1 \), \( C = 1 \). Thus, \( T(t) = e^t \).

Key Concepts

Differential Equations

Differential equations are a key concept in calculus that deal with functions and their rates of change. They are like the mathematicians' way of using equations to describe how things evolve over time. In our exercise, we want to understand how the air density, which depends on height and temperature, changes as a meteorological balloon ascends. The heart of solving this problem is converting the relationship between the different variables into an equation that expresses how temperature and height impact air density.

Differential equations can often involve finding unknown functions that satisfy certain conditions. In simpler terms, this means figuring out a rule that tells us how quantities are related. When we derive a differential equation from our air density problem, we're aiming to express this rate of change with respect to time. This effort helps us to solve for the temperature condition under which the density doesn't change over time. Essentially, it boils down to turning a problem into a format we can manipulate mathematically, giving us insight into how the various parts—height, temperature, and density—interact.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. Put simply, it helps us take the derivative of a function that is nested inside another function. It's like peeling layers of an onion to get to the rate at which something changes.

In our particular problem about air density, the chain rule is crucial for finding the rate of change of density over time. This rate is influenced by changes in height and temperature as the balloon ascends. By using the chain rule, we can express this as:

• First, differentiate with respect to each variable (height and temperature).
• Then, multiply these derivatives by the rates at which the respective variables change over time.

By combining these rates, the chain rule gives us a complete picture of how air density is changing as the balloon rises through different altitudes and experiences varying temperatures.

This approach is powerful because it allows us to connect how independent changes (like height and temperature) collectively affect dependent outcomes (air density), enabling us to derive meaningful relationships like the differential equation solved in our exercise.

Air Density

Air density, denoted by the symbol \(\rho\), is a measure of air's mass per unit volume. It's crucial in understanding various atmospheric phenomena and calculations, such as how a balloon ascends into the sky. In the formula given in the problem, \(\rho(z, T) = \rho_{0} e^{-\lambda z / T}\), air density changes with height \(z\) and temperature \(T\).

As a meteorological balloon moves upward, the density of air changes due to a decrease in atmospheric pressure and changes in temperature. It’s like climbing a mountain: the higher you go, the "thinner" or less dense the air becomes.

• Density decreases with height (an inverse relationship).
• Temperature also affects density; warmer air tends to be less dense.

Understanding these relationships is important for predicting how things like balloons or planes interact with the atmosphere.

For students, grasping why and how air density changes with altitude and temperature is key, as it interlinks with principles in meteorology and flight dynamics. Knowing the mathematical representation allows us to make these theoretical foundations applicable to real-world scenarios.