Question

Cheyne-Stokes Respiration, a respiratory ailment, causes the volume per breath to increase and decrease as a function of time, and this change can be modeled by a sinusoidal function. A machine is used to record a plot of volume per breath versus time (in seconds). Let b(t) be a function of time (t) that tells us the volume (in liters) of a breath that starts at time t. During a test, the smallest volume per breath is 0.6 liters and this occurs 5 seconds into the test. The largest volume per breath is 1.8 liters and this first occurs for a breath beginning 55 seconds into the test.

a) Find a formula for the function b(t) whose graph will model the test data for this patient.
b) If the patient begins a breath every 5 seconds, what are the breath volumes during the first minute of the test?

Answer 1

The sinusoidal function to model Cheyne-Stokes Respiration volume per breath over time is represented by the function b(t) = 0.6 sin((2π/(55 - 5))(t - 5)) + 1.2. To find the volumes of breath at each 5 second interval during the first minute, we substitute these time values into the function.

Step-by-step explanation:

We are tasked with modeling Cheyne-Stokes Respiration with a sinusoidal function representing the volume per breath over time. To create this model, we will use the given test data that indicates the smallest volume of 0.6 liters occurs at t=5 seconds, and the largest volume of 1.8 liters occurs at t=55 seconds.

Part A: Finding a Formula for b(t)

Let's denote the sinusoidal function as b(t) = A sin(B(t - C)) + D, where:

• A represents the amplitude
• B determines the period of the function
• C zeroes the horizontal shift
• D determines the vertical shift

The amplitude can be calculated as half the difference between the maximum and minimum volumes, which leads to A = (1.8 - 0.6) / 2 = 0.6 liters.

The vertical shift D is found by averaging the maximum and minimum volumes, giving D = (1.8 + 0.6) / 2 = 1.2 liters.

The breath cycle is completed by the time the largest volume occurs again. Since we only have data for the first occurrence at t=55 seconds, we'll assume the period is twice the difference in time between the first minimum and maximum volumes: B = 2π / (55 - 5).

The horizontal shift C corresponds to when the first minimum volume occurs, which is at t=5 seconds. Therefore, C=5.

The complete formula for b(t) is thus: b(t) = 0.6 sin((2π/(55 - 5))(t - 5)) + 1.2

Part B: Breath Volumes During the First Minute

To determine the volumes for each breath during the first minute, we find the value for b(t) at multiples of 5 seconds up to one minute. These time instances are t=0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, and 55 seconds.