Problem 25 - Thomas Calculus

Question

Cardiac output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min}\) . At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min} .\) If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min.}\) Your cardiac output can be calculated with the formula $$ y=\frac{Q}{D} $$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{m} 1 / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L}\) $$ y=\frac{233 \mathrm{ml} / \mathrm{min}}{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min} $$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

Step-by-Step Solution

Answer

The cardiac output is increasing by approximately 0.277 L/min.

Step 1: Understanding the Given Formula

We are provided with the cardiac output formula: \( y = \frac{Q}{D} \). Here, \( Q \) is the amount of carbon dioxide exhaled per minute, given as 233 ml/min, and \( D \) is the difference in CO2 concentration, initially 41 ml/L. We also know that \( D \) is decreasing at a rate of 2 ml/L per minute, and we want to find out how this change affects \( y \), the cardiac output.

Step 2: Using Related Rates

Since \( D \) is changing, we should find the rate of change of cardiac output \( \frac{dy}{dt} \). We will differentiate both sides of \( y = \frac{Q}{D} \) with respect to time \( t \), keeping in mind that \( Q \) is constant. The relationship is: \[ \frac{dy}{dt} = -\frac{Q}{D^2} \times \frac{dD}{dt}. \]

Step 3: Substitute Known Values

Substitute the known values into the derivative formula: \( Q = 233 \), \( D = 41 \), and \( \frac{dD}{dt} = -2 \) (since \( D \) is decreasing). Calculate \( \frac{dy}{dt} \) using: \[ \frac{dy}{dt} = -\frac{233}{41^2} \times (-2) = \frac{233 \times 2}{41^2}. \]

Step 4: Perform Calculations

Calculate \( 41^2 = 1681 \). Now, continue with: \[ \frac{dy}{dt} = \frac{466}{1681}. \] Perform the division to get the numerical value of \( \frac{dy}{dt} \).

Step 5: Interpret the Result

Since \( \frac{dy}{dt} = \frac{466}{1681} \approx 0.277 \) L/min, the cardiac output is increasing. This is because as \( D \) decreases, since \( D \) is in the denominator, \( y \) increases. The cardiac output increases by approximately 0.277 L/min.

Key Concepts

Related Rates

Related rates involve the study of how quantities change with respect to time. In our scenario, the cardiac output calculation is impacted by the change in the concentration difference of carbon dioxide (CO2) in the blood. As the difference (D) changes over time, it influences the cardiac output (y). This relationship is mathematically modeled as a function that relates these changing variables.
To find how quickly the cardiac output rate (\(\frac{dy}{dt}\)) changes as D changes, we employ the concept of related rates. We differentiate both sides of the equation \(y = \frac{Q}{D}\) with respect to time. This allows us to determine the instantaneous rate of change, showing how the rate of cardiac output adapts to alterations in blood CO2 concentration over time.

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how functions change. When applied to real-world problems like cardiac output, it helps calculate rates of change.
The formula for cardiac output is \(y = \frac{Q}{D}\). Q remains constant, whereas D reduces over time. Differentiating this equation with respect to time, using the quotient rule of derivatives, we get:
\[\frac{dy}{dt} = -\frac{Q}{D^2} \times \frac{dD}{dt}\]
This shows how y changes as D changes over time. By plugging in the known values for Q, D, and \(\frac{dD}{dt}\), we compute the precise rate at which cardiac output is changing.
Through differentiation, we track the dynamic relationship between the variables, capturing the exact adjustment in cardiac output in response to physiological shifts in CO2 concentration.

Mathematical Modeling

Mathematical modeling is the use of mathematical expressions to represent real-world phenomena. It is crucial in physiology to predict and analyze biological processes, such as cardiac output.
In this instance, the model \(y = \frac{Q}{D}\) represents cardiac output as a function of exhaled CO2 quantity and the difference in CO2 concentration. This expression models how cardiac output can be estimated based on measurable quantities (\(Q\) and \(D\)). Using this model, we can explore scenarios like what happens if D changes over time.
By incorporating the rates of change through differentiation, the model evolves to reflect the dynamic nature of physiological processes, providing valuable insights into cardiac function under varying conditions. This transforms static measurements into a tool for predicting outcomes and understanding how heart efficiency adapts to changes in respiratory function.

Physiology

Physiology is the study of how biological systems function. In cardiac physiology, understanding cardiac output is vital as it determines how effectively the heart is delivering blood and oxygen to the body.
The formula \(y = \frac{Q}{D}\) helps illuminate this function by connecting oxygen delivery to CO2 management. Q represents how much CO2 is expelled, giving insight into metabolic activity, while D highlights the CO2 concentration difference, a marker of respiratory efficiency.
When D decreases, the heart's ability to expel CO2 efficiently and circulate blood quickly increases. This indicates the body’s responsive mechanism to maintain equilibrium and meet physiological demands. Physiological insights garnered from these calculations are crucial in medical diagnostics, patient monitoring, and establishing baselines for normal and stress conditions, such as rest or extreme physical activity.