The pressure, P (in lbs/ft2), in a pipe varies over time. Three times an hour, the pressure oscillates from a low of 88 to a high of 254 and then back to a low of 88. The pressure at t=0 is 88. Find a possible cosine formula for P=f(t), where t is time in minutes.
the answer above is straight up wrong.
wrong !!!!!
completely wrong
This is completely wrong not what my teacher was looking for.
To find a possible cosine formula for P=f(t), we can use the given information about the pressure oscillations.
First, we need to determine the period of the oscillations. Since the pressure oscillates three times per hour, the time for one complete oscillation is 60 minutes / 3 = 20 minutes.
Next, we need to determine the midline (average value) of the oscillations. The midline is the value the pressure oscillates around. In this case, the midline is given as 88.
Now, let's determine the amplitude of the oscillations. The amplitude is the half the difference between the high and low values. In this case, the high value is 254 and the low value is 88, so the amplitude is (254 - 88) / 2 = 83.
Therefore, the possible cosine function for P=f(t) can be written as:
P = 83 * cos((2π / 20) * t) + 88
In this formula, the time variable t is in minutes, and the cosine function oscillates with a period of 20 minutes. The amplitude of the oscillations is 83, and the midline is 88.
from max to min = 254-88 = 166
so amplitude = 84
period = 20 min
period = 2π/k
20 = 2π/k
k = π/10
let p(t) = 83 cos π/10(t + d) + 171
when t = 0 we want p(0) = 88
88 = 83cos (π/10)cos(0 + d) + 171
-1 = cos (dπ/10)
I know cos π = -1
so dπ/10 = π
dπ = 10π
d = 10
so a possible equation is
P(t) = 83 cos (π/10)(t + 10) + 171
check:
at t = 0, P(0) = 83cos
= = 83(-1) + 171
= 88, check!
when t = 10 , (should get my max)
P(10) = 83cos(π/10)(20) + 171
= 83(cos2π) + 171 = 83 + 171 = 254
looks ok