The volume V, in liters, of air in the lungs during a two-second respiratory cycle is approximated by the model
V = 0.1729t + 0.1522t^2 − 0.0374t^3,
where t is the time in seconds. Approximate the average volume of air in the lungs during one cycle. (Round your answer to four decimal places.)
To find the average volume of air in the lungs during one cycle, we need to calculate the net change in volume over the entire cycle and then divide it by the time of one complete cycle.
To do this, we need to find the volume at the beginning and end of one cycle.
Let's assume that the beginning of one cycle is at t = 0 seconds, and the end of one cycle is at t = 2 seconds.
So, we need to find V(0) and V(2).
Substituting t = 0 into the equation for V, we have:
V(0) = 0.1729(0) + 0.1522(0)^2 − 0.0374(0)^3
V(0) = 0
Substituting t = 2 into the equation for V, we have:
V(2) = 0.1729(2) + 0.1522(2)^2 − 0.0374(2)^3
V(2) = 0.3458 + 0.1522(4) − 0.0374(8)
V(2) = 0.3458 + 0.6088 − 0.2992
V(2) = 0.6554
Now that we have the initial and final volumes, we can calculate the average volume by finding the net change in volume and dividing it by the time of one complete cycle.
Net change in volume = V(2) - V(0) = 0.6554 - 0 = 0.6554
Average volume = Net change in volume / Time of one complete cycle
Average volume = 0.6554 / 2
Therefore, the average volume of air in the lungs during one cycle is approximately 0.3277 liters.
To approximate the average volume of air in the lungs during one cycle, we need to find the average value of the volume function over the interval of one cycle.
The volume function is given by V = 0.1729t + 0.1522t^2 - 0.0374t^3. We can calculate the average value of this function over one cycle using definite integration.
To find the average value of a function f(x) over an interval [a, b], we can use the following formula:
average value = (1 / (b - a)) * ∫[a to b] f(x) dx
In this case, the interval is [0, 2] seconds, as it represents one cycle.
Using this information, we can now calculate the average value of the volume function over one cycle.
First, let's evaluate the integral:
∫[0 to 2] (0.1729t + 0.1522t^2 - 0.0374t^3) dt
To integrate each term, we can use the power rule of integration:
For the term 0.1729t, the integral is (0.1729/2)t^2 = 0.08645t^2
For the term 0.1522t^2, the integral is (0.1522/3)t^3 = 0.050733t^3
For the term -0.0374t^3, the integral is (-0.0374/4)t^4 = -0.00935t^4
Now, let's compute the integral of the volume function over one cycle:
∫[0 to 2] (0.08645t^2 + 0.050733t^3 - 0.00935t^4) dt
Evaluating this integral gives us:
(0.08645/3)(2^3 - 0^3) + (0.050733/4)(2^4 - 0^4) - (0.00935/5)(2^5 - 0^5)
Simplifying further, we get:
(0.08645/3)(8) + (0.050733/4)(16) - (0.00935/5)(32)
= 0.231267 + 0.203332 - 0.11904
= 0.315559
Finally, to find the average volume over one cycle, we divide this result by the duration of one cycle, which is 2 seconds:
average volume = 0.315559 / 2
≈ 0.1578 liters
Therefore, the approximate average volume of air in the lungs during one cycle is 0.1578 liters.
That would just be
(V(2)-V(0))/(2-0) = 0.6554/2 = 0.3277
Hmmm. I misspoke. The average value is
∫[0,2] V(t) dt
--------------------
2-0
= 0.3010