Question

When a person is breathing normally the amount of air in their lawns varies sinusoidally. When full Karen’s lungs hold 2.8 L of air when empty her Lawrence hold 0.6 L of air her brother starts timing her breathing at T equals two seconds she has exhaled completely and at T equals five seconds she has completely inhaled. Create an equation for this

Answer 1

The sinusoidal equation modeling the volume of air in Karen's lungs as a function of time is f(T) = 1.1 sin((π/3)(T - 2)) + 1.7, with a maximum capacity of 2.8 L and a minimum of 0.6 L.

Step-by-step explanation:

To determine the equation for the sinusoidal variation of air in the lungs during normal breathing, we can use the information about the maximum and minimum lung capacities and the timing of the breath cycle provided in the question. Karen's lungs hold a maximum of 2.8 L of air when full and a minimum of 0.6 L of air when empty. We are given that at T = 2 seconds, she has exhaled completely (minimum), and at T = 5 seconds, she has completely inhaled (maximum).

To find the amplitude, we subtract the minimum capacity from the maximum capacity and divide by 2:

Amplitude (A) = (2.8 - 0.6) / 2 = 1.1 L

The vertical shift (D) is the average of the maximum and minimum volumes, which is:

Vertical Shift (D) = (2.8 + 0.6) / 2 = 1.7 L

Additionally, the time it takes to complete one full breathing cycle can be found by considering that she exhales completely at T = 2 seconds and inhales completely at T = 5 seconds, indicating that a half-cycle (from minimum to maximum) is 3 seconds long. Thus, a full cycle is 6 seconds. The period (B) of the sinusoidal function is the full cycle time:

Period (B) = 6 seconds

To get the sinusoidal function in the form f(T) = A sin(B(T - C)) + D, where C represents the horizontal shift, we consider the minimum point at T = 2 seconds. Since the sinusoidal curve for breathing starts at its lowest point, which corresponds to a sine function at -π/2, we adjust for this starting phase. The value of B is calculated by using B = 2π/Period, which in this case is 2π/6. After finding B, we can write the function as:

f(T) = 1.1 sin((π/3)(T - 2)) + 1.7

Answer 2

`A(t) = 2.2\sin ((t - 2)\pi )/(6) + 0.6`

Step-by-step explanation:

Let the function of quantity in the lung of air be A(t)

So
`A(t) \alpha \sin ((t - \alpha )/(k) )`

so, A(t) = Amax sin t + b

A(t) = 2.8t⇒ max

A(t) = 0.6t ⇒ min

max value of A(t) occur when sin(t) = 1

and min value of A(t) = 0

So b = 0.6

and A(max) = 2.2

`A(t) = 2.2\sin ((t))/(k) + 0.6`

at t = 2 sec volume of a is 0.6

So function reduce to

`A(t) = 2.2\sin ((t - 2))/(k) + 0.6`

and t = 5 max value of volume is represent

so,

`\sin (t - \alpha )/(k) = 1`

`(t - 2)/(k) = (\pi )/(2)` when t = 5

`(6)/(\pi ) = k`

so the equation becomes

`A(t) = 2.2\sin ((t - 2)\pi )/(6) + 0.6`