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
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.
5 answers
the answer above is straight up wrong.
wrong !!!!!
completely wrong
This is completely wrong not what my teacher was looking for.