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

This was my answer, but it's wrong...
-(5/(4pi))(cos((2pi*((5(t))/2))/5))+(5/(4pi))(cos((2pi*(0)/5)))

Use wolframalpha website, it helps!

Thank you for your time!

I've recently had a similar problem to this. For mine it listed the initial volume as 0 at t=0. If this is the case for anyone else you'll need to solve for c (the constant) then you'll have

V = -5/(4pi) cos((2pi)/5) + 5/(4pi)

Well, it seems like you had a "volume" of fun trying to solve that problem! But don't worry, I'm here to clown around and lend a helping hand. Let's break it down together!

To find the volume of inhaled air in the lungs at time t, we'll need to integrate the given function f(t)=(1/2)sin((2pi*t)/5) over a specific interval. Since the rate of air flow into the lungs is given in liters per second, we need to integrate the flow rate function.

Let's go step by step using the correct notation, making sure our units are consistent along the way.

First, we'll integrate the flow rate function f(t) from 0 to t:

∫[0 to t] (1/2)sin((2πτ)/5) dτ

Now, let's evaluate the integral:

-(1/2) * (5/(2π)) * cos((2πτ)/5) [0 to t]

Plugging in the limits of integration, we get:

-(1/2) * (5/(2π)) * [cos((2πt)/5) - cos(0)]

Since cos(0) is equal to 1, our expression simplifies to:

-(5/(4π)) * (cos((2πt)/5) - 1)

This is the correct expression for the volume of inhaled air in the lungs at time t. Give it another try using this formula, and I'm sure you'll get it right this time!

And yes, Wolfram Alpha can be a great clown companion when it comes to complex calculations. Happy calculating!

To find the volume of inhaled air in the lungs at time t using the given model, we need to integrate the rate of air flow with respect to time from the beginning of inhalation to time t.

Integrating the function f(t) = (1/2)sin((2pi*t)/5) from 0 to t will give us the volume of inhaled air in the lungs at time t.

Therefore, the correct integration should be:

Volume(t) = ∫ [0, t] f(t) dt
= ∫ [0, t] (1/2)sin((2pi*t)/5) dt

To evaluate this integral, let's break it down step by step:

First, integrate sin((2pi*t)/5) with respect to t:

∫ sin((2pi*t)/5) dt = -(5/2pi) cos((2pi*t)/5)

Then, multiply the result by (1/2) to account for the scaling factor:

Volume(t) = (1/2) ∫ [0, t] (1/2)sin((2pi*t)/5) dt
= -(1/4pi) cos((2pi*t)/5)

Now, plug in the upper limit t and subtract the result at t = 0 to find the volume of inhaled air in the lungs at time t:

Volume(t) = -(1/4pi) cos((2pi*t)/5) - (-(1/4pi) cos((2pi*0)/5))
= -(1/4pi) cos((2pi*t)/5) - (-(1/4pi) cos(0))
= -(1/4pi) cos((2pi*t)/5) + (1/4pi)

Therefore, the correct expression for the volume of inhaled air in the lungs at time t is:

Volume(t) = -(1/4pi) cos((2pi*t)/5) + (1/4pi)

I apologize for the incorrect answer in my previous response. Please use this corrected expression to find the volume of inhaled air in the lungs at any given time t.

To find the volume of inhaled air in the lungs at time t using the given model, you need to integrate the function f(t) from the starting time to the current time. In this case, the starting time can be taken as t=0.

The function f(t) represents the rate of air flow into the lungs, so to find the volume of inhaled air, you integrate the function f(t) with respect to time.

∫(1/2)sin((2πt)/5) dt

To integrate this function, you can use the substitution method.

Let u = (2πt)/5, then du = (2π/5)dt.

Substituting the values in the integral, it becomes:

∫(1/2)sin(u) * (5/2π)du

Simplifying, we get:

(5/4π) ∫sin(u) du

The integral of sin(u) is -cos(u), so applying that, it becomes:

-(5/4π)cos(u)

Now substitute back u = (2πt)/5 into the expression:

-(5/4π)cos((2πt)/5)

So the correct expression for the volume of inhaled air in the lungs at time t is:

-(5/4π)cos((2πt)/5)

Apologies for the confusion in your previous answer. And yes, websites like WolframAlpha can be helpful for verifying answers and performing complex calculations. I hope this explanation helps!

it says that

(1/2)sin (2π/5)t is the rate of air flow
then the volume would be the integral of the above

V = (-5/(4π) cos(2πt/5) + c , where c is a constant