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The rate of growth of a fish population was model…

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Problem 83 Hard Difficulty

Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 s. The maximum rate of air flow into the lungs is about 0.5 L/s. This explains, in part, why the function $ f(t) = \frac{1}{2} \sin (2\pi t/5) $ has often been used to model the rate of air flow into the lungs. Use this model to find the volume of inhaled air in the lungs at time $ t $.

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Video Transcript

here, we're going to find a model for the volume of inhaled air into the lungs at time T. And what we are given is a function for the rate of airflow into the lungs at time. T. So I've written it out here on our left and I've got it plotted on the right, and so to find the volume of air inside the lungs, we're gonna have to use an integral. So why an integral? Well, if we say this is time T and we want to find the volume of air in the lungs at time T, what we can do is add up how much air has been flowing into the lungs each time. And so this will just be the area under the curve or the integral. So what we need to do, I guess I can I'll keep that shaded actually. Yeah, there we go. So that's what we're looking for. So our function, our model, which I'll call VF T V for volume, it's going to be the integral From 0 to T. Sorry, your T of F F T GT. And so now to compute this, we can just take the anti derivative of our function here. So the anti derivative of Sine was going to give us negative coastline. So it's going to be 1/2 negative co sign of the same input to piety over five. And now we have to use the chain rule what we're doing it backwards since we're integrating. So the derivative of our inside The derivative of to pity over five is just 2.45. We have to divide by it. So let's multiply by 5/2 pi. We do this so that if we were to take the derivative of all this of this expression here, you should get back to your FFT FFT. So This is all going to be evaluated from 0 to T. So let's combine like terms and then evaluate. So the Constance We're going to take out to the front. So in front we have a negative five over four pi and then inside we have co sign of two pi. And so our variables just T. So we're gonna keep T in there. We're doing essentially is plugging in T. For T. Like this tea is going in to this team now we're going to do it again. But this time something in zero. So it's gonna be minus co sign and when you plug in the zero for T everything just goes to zero. So it's just just one. So if we clean up our like terms and I'm going to distribute this negative here to both of these terms. So what we're going to get is 5/4 pi Course in of zero is just one and we're Distributing negative. So we have one minus co sign two pi T over five. Close parentheses and this is our model there we go

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

James StewartCalculus: Early Transcendentals

Grace He

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

James Stewart

Calculus: Early Transcendentals