Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

The nitrogen partial pressure in a muscle of a scuba diver is initially 0.8 atm. She descends to 30 meters and immediately the \(\mathrm{N}_{2}\) partial pressure in her blood is \(2.4 \mathrm{~atm},\) and remains at 2.4 atm while she remains at 30 meters. Each minute the \(\mathrm{N}_{2}\) partial pressure in her muscle increases by an amount that is proportional to the difference in 2.4 and the partial pressure of nitrogen in her muscle at the beginning of that minute. a. Write a dynamic equation with initial condition to describe the \(\mathrm{N}_{2}\) partial pressure in her muscle. b. Your dynamic equation should have a proportionality constant. Assume that constant to be 0.067. Write a solution to your dynamic equation. c. At what time will the \(\mathrm{N}_{2}\) partial pressure be \(1.6 ?\) d. What is the half-life of \(N_{2}\) partial pressure in the muscle, with the value of \(K=0.067 ?\)

Step-by-Step Solution

Answer

a) \( \frac{dP}{dt} = 0.067(2.4 - P(t)) \) with \( P(0) = 0.8 \). b) \( P(t) = -1.6e^{-0.067t} + 2.4 \). c) \( t \approx 10.35 \) minutes. d) Half-life is \( \approx 10.35 \) min.

Step 1: Understand the Initial Problem

We need to find a dynamic equation that models the nitrogen partial pressure changes in a scuba diver's muscle. The change in partial pressure per minute is proportional to the difference between a constant partial pressure of 2.4 atm in her blood and the current muscle partial pressure.

Step 2: Define the Dynamic Equation

Let \( P(t) \) be the nitrogen partial pressure in the muscle at time \( t \). The rate of change of nitrogen partial pressure can be modeled by the differential equation:\[ \frac{dP}{dt} = K (2.4 - P(t)) \]where \( K = 0.067 \). The initial condition is given by \( P(0) = 0.8 \).

Step 3: Solve the Differential Equation

This is a first-order linear differential equation with a standard solution form:\[ P(t) = C e^{-Kt} + A \]where \( A \) is the equilibrium pressure, 2.4 atm, and \( C \) is a constant determined by initial conditions. Applying the initial condition \( P(0) = 0.8 \), we get:\[ 0.8 = C e^{0} + 2.4 \]\[ 0.8 = C + 2.4 \]\[ C = 0.8 - 2.4 = -1.6 \]Thus, the solution is:\[ P(t) = -1.6 e^{-0.067t} + 2.4 \]

Step 4: Finding the Time When Pressure is 1.6 atm

We need to find \( t \) when \( P(t) = 1.6 \):\[ 1.6 = -1.6 e^{-0.067t} + 2.4 \]Rearranging gives:\[ -0.8 = -1.6 e^{-0.067t} \]\[ 0.5 = e^{-0.067t} \]Taking the natural log of both sides, we get:\[ -0.067t = \ln(0.5) \]\[ t = \frac{\ln(0.5)}{-0.067} \]Calculating numerically, \( t \approx 10.35 \) minutes.

Step 5: Determining the Half-Life of the Pressure

The half-life is the time it takes for \( P(t) \) to reach the midpoint between its initial value and equilibrium, that is, when \( P(t) = 1.6 \) from the initial condition of 0.8 to equilibrium at 2.4. We previously found it takes approximately 10.35 minutes, which matches our half-life calculation. Thus, the half-life is approximately 10.35 minutes.

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is an equation involving the derivatives of a function and the function itself, usually expressing a rate of change in terms of time. The general form looks like this:

• \( \frac{dP}{dt} + aP = b \)

where \(a\) and \(b\) are constants, and \(P\) is the function of time, \(t\). In our exercise, the nitrogen partial pressure \(P(t)\) changes as the scuba diver remains at depth. The change is directly influenced by the difference between the equilibrium pressure in her blood and the current pressure in her muscle. The specific equation derived is:

• \( \frac{dP}{dt} = K (2.4 - P(t)) \)

This demonstrates a relationship based on how far \(P(t)\) is from equilibrium. Starting with an initial value, it progresses towards stability, reflecting real-world systems like chemical reactions or population models. Understanding such equations lays the groundwork for modeling various dynamic systems.

Proportionality Constant

The proportionality constant, denoted as \(K\), plays a crucial role in the differential equation, expressing how fast a change occurs relative to the difference from equilibrium. In our context, \(K = 0.067\) provides the scaling factor for the rate at which the nitrogen partial pressure in the diver’s muscle adjusts to the equilibrium pressure. This means each minute, the pressure change is scaled by \(0.067\) of whatever pressure difference exists.

• Think of \(K\) as a speedometer for change, depicting how swiftly or lazily the state shifts towards its desired state.

If \(K\) were larger, the adjustment would occur more rapidly; smaller \(K\), slower. Grasping the concept of a proportionality constant is pivotal for predicting the rate of change in processes influenced by other factors.
Whether it's a chemical concentration varying with time, or population dynamics, once we understand \(K\), we can predict how quickly equilibrium might be achieved.

Equilibrium Pressure

Equilibrium pressure in this scenario refers to the stable nitrogen pressure a diver eventually reaches in their muscle when external conditions remain constant. It acts like a target pressure, the muscle pressure aspires to match while at 30 meters depth. At equilibrium, the pressures equalize, and no net change occurs in \(P(t)\). At an equilibrium point:

• \( \frac{dP}{dt} = 0 \)
• \( P(t) = 2.4 \, \text{atm} \)

This 2.4 atm pressure serves as our equilibrium reference. It signifies that when the muscle pressure equals the blood pressure at depth, the nitrogen pressures are balanced, signifying no further net transfer of nitrogen. Comprehensive understanding of equilibrium is essential in various fields including thermodynamics and fluid mechanics, as all dynamic processes seek some form of equilibrium.

Half-life Calculation

Half-life is a term borrowed from radioactive decay but commonly used in processes involving exponential changes, representing the time for a quantity to reach half of its initial deviation from equilibrium. In our case, it measures how quickly \(P(t)\) approaches halfway to its final equilibrium value. In solving for half-life, we calculated:

• The time \(t\) when \(P(t) = 1.6\), halfway from 0.8 atm to 2.4 atm.
• The formula: \( t = \frac{\ln(0.5)}{-0.067} \)
• The result was approximately 10.35 minutes.

This calculation provides practical insight into the dynamics of change. It highlights how quickly the muscle pressure transitions in the water environment the diver is exposed to, underscoring the importance of half-life in making informed predictions about dynamic systems governed by first-order linear differential equations.