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

To find a formula for the function b(t) that models the test data, we assume a sinusoidal function of the form b(t) = A*cos(wt + phi) + C. Using the given data points, we calculate the values of A, w, phi, and C. The formula for b(t) is b(t) ≈ 0.6*cos(pi/25*t + 11pi/5) + 0.44. For the first minute of the test, the breath volumes are approximately 0.44, 0.6, 0.68, 0.87, 1.04, 1.17, 1.26, 1.30, 1.30, 1.26, 1.17 liters.

Step-by-step explanation:

To find a formula for the function b(t) that models the test data, we need to use a sinusoidal function. Let's assume the equation is of the form b(t) = A*cos(wt + phi) + C, where A is the amplitude, w is the frequency, phi is the phase shift, and C is the vertical shift. Given that the smallest volume per breath is 0.6 liters at 5 seconds, and the largest volume per breath is 1.8 liters at 55 seconds, we can use these points to determine the values of A, w, and phi. The amplitude A is half the difference between the maximum and minimum values, so A = (1.8 - 0.6)/2 = 0.6. The frequency w can be found using the formula w = 2*pi / T, where T is the period. The period is the time between the maximum values, which is 55 - 5 = 50 seconds, so w = 2*pi / 50 = pi/25. The phase shift phi is the value of wt when the maximum value occurs, so wt = phi = pi/25 * 55 = 11pi/5. Therefore, the formula for b(t) is b(t) = 0.6*cos(pi/25*t + 11pi/5) + C. To find the value of C, we can use the fact that the smallest volume per breath occurs 5 seconds into the test, so b(5) = 0.6*cos(pi/25*5 + 11pi/5) + C = 0.6. Solving for C, we get C = 0.6 - 0.6*cos(pi/25*5 + 11pi/5) = 0.6 - 0.6*cos(pi/5 + 11pi/5) = 0.6 - 0.6*cos(12pi/5) ≈ 0.44. Therefore, the formula for b(t) is b(t) ≈ 0.6*cos(pi/25*t + 11pi/5) + 0.44.

To find the breath volumes during the first minute of the test, we can evaluate b(t) for values of t from 0 to 60 seconds. Plugging in these values, we get:

b(0) ≈ 0.6*cos(0 + 11pi/5) + 0.44 ≈ 0.44 liters

b(5) ≈ 0.6*cos(pi/25*5 + 11pi/5) + 0.44 ≈ 0.6 liters

b(10) ≈ 0.6*cos(pi/25*10 + 11pi/5) + 0.44 ≈ 0.68 liters

b(15) ≈ 0.6*cos(pi/25*15 + 11pi/5) + 0.44 ≈ 0.87 liters

b(20) ≈ 0.6*cos(pi/25*20 + 11pi/5) + 0.44 ≈ 1.04 liters

b(25) ≈ 0.6*cos(pi/25*25 + 11pi/5) + 0.44 ≈ 1.17 liters

b(30) ≈ 0.6*cos(pi/25*30 + 11pi/5) + 0.44 ≈ 1.26 liters

b(35) ≈ 0.6*cos(pi/25*35 + 11pi/5) + 0.44 ≈ 1.30 liters

b(40) ≈ 0.6*cos(pi/25*40 + 11pi/5) + 0.44 ≈ 1.30 liters

b(45) ≈ 0.6*cos(pi/25*45 + 11pi/5) + 0.44 ≈ 1.26 liters

b(50) ≈ 0.6*cos(pi/25*50 + 11pi/5) + 0.44 ≈ 1.17 liters

b(55) ≈ 0.6*cos(pi/25*55 + 11pi/5) + 0.44 ≈ 1.04 liters

b(60) ≈ 0.6*cos(pi/25*60 + 11pi/5) + 0.44 ≈ 0.87 liters

So, the breath volumes during the first minute of the test are approximately:

0.44, 0.6, 0.68, 0.87, 1.04, 1.17, 1.26, 1.30, 1.30, 1.26, 1.17 liters.