Our cycle of normal breathing takes place every 5 seconds. Velocity of air flow, y, measured in liters per second, after x seconds is modeled by : y=0.6sin(2Ĉ/5)x. Velocity of air flow is positive when we inhale.
Within each breathing cycle, when are we inhaling 0.3 liters per second? Round to the nearest tenth of a second.
There is no need to post the same question twice.
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To find the time when the velocity of air flow, y, is equal to 0.3 liters per second, we can set up the equation:
0.3 = 0.6sin(2π/5)x
First, let's isolate the sine term:
sin(2π/5)x = 0.3/0.6
Simplifying the right side:
sin(2π/5)x = 0.5
To find the value of x, we need to take the inverse sine (sin^(-1)), also known as arcsine, of both sides:
sin^(-1)(sin(2π/5)x) = sin^(-1)(0.5)
Now, using the inverse sine function, we have:
2π/5 x = π/6 + 2πn or 4π/6 + 2πn
Where n is an integer representing the number of cycles completed.
To find x, we divide both sides of the equation by 2π/5:
x = (π/6 + 2πn)/(2π/5) or (4π/6 + 2πn)/(2π/5)
Simplifying further:
x = (5/2) * (π/6 + 2πn) or (5/2) * (4π/6 + 2πn)
x = (5π/12 + 10πn/2) or (10π/12 + 10πn/2)
Now, to find the time within each breathing cycle when the velocity is 0.3 liters per second, we need to find the fractional part of x.
Since the cycle of normal breathing takes place every 5 seconds, we can simplify the expression further:
x = (5π/12 + 5πn) or (10π/12 + 5πn)
To calculate within each breathing cycle, we need to consider the single period between 0 and 5 seconds.
For the first expression:
x = 5π/12 + 5πn when 0 ≤ x ≤ 5
For the second expression:
x = 10π/12 + 5πn when 0 ≤ x ≤ 5
Now, we can numerically solve for x within the interval [0, 5] to find when the velocity is 0.3 liters per second. Rounding to the nearest tenth of a second, we can use a graphing calculator or software to calculate the value of x.