Question

Group Activity Cardiac Output In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurtzberg, 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 liters a minute. 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 $$$=\frac{Q}{D}$$ 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 \((\mathrm{mL} / \mathrm{L})\) 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?

Short answer

When the rate of decrease of D (concentration difference of CO2 between the blood to and from the lungs) is 2 units per minute while Q (volume of exhaled CO2) remains constant, the cardiac output is increasing at an approximate rate of \( \frac{466}{1681} \) liters per minute.

Step-by-step solution

Understanding the Given Variables and Formula

In this scenario, the cardiac output is given by the equation \( Y = \frac{Q}{D} \), where Y represents the cardiac output, Q is the volume of exhaled CO2 in a minute, and D signifies the difference in CO2 concentration in the blood pumped to the lungs and returning from the lungs. As given, initially, \( Q = 233 \) and \( D = 41 \). The rate at which D is decreasing is also given, which is 2 units per minute.

Differentiating the Formula

Since we're asked about the rate of change, we need to differentiate the formula to obtain dY/dt. Using the quotient rule (which states that the derivative of two functions f and g, f/g, is given as (g * df/dt - f * dg/dt)/g²), we have \( \frac{dY}{dt} = \frac{D\frac{dQ}{dt} - Q\frac{dD}{dt}}{D^2} \). In our case, dQ/dt = 0 (since Q is constant) and dD/dt = -2 (since D is decreasing by 2units/minute).

Substituting Values into the Derivative

Substituting the values into the expression for dY/dt, we obtain \( \frac{dY}{dt} \) = \( \frac{D * 0 - 233 * (-2)}{41^2} \) = \( \frac{466} {1681} \) L/minute.

Interpreting the Result

The rate of change of cardiac output is positive, which means that the cardiac output is increasing even as the difference in CO2 concentration in blood (D) is decreasing. This result might seem counter-intuitive, as one might think that if D decreases, the cardiac output should decrease as well. However, in the function Y = Q/D, as D decreases and Q remains constant, Y actually increases, according to mathematical rules.