Assuming that high tides occur every 12.5 hours, give a model of the form y=Asin(`omega t -`phi)+c for height of the tide vs. clock time if at 4:30am the water level is 13 feet, and 6.25 hours later the level is 10 feet. Assume that t is the time since midnight.
Assuming that high tides occur every 12.5 hours, give a model of the form y=Asin(`omega t -`phi)+c for height of the tide vs. clock time if at 4:30am the water level is 13 feet, and 6.25 hours later the level is 10 feet. Assume that t is the time since midnight.
13 = Asin(4.5omega + phi)
10 = Asin(10.75omega + phi)
A = high tide = 13
4.5omega + phi = pi/2
phi = pi/2 - 4.5omega
10/13 = sin(6.25omega + pi/2)
from that you can work out omega easily with basic trig
hope that helps
I see. Okay.
Assuming that high tides occur every 12.5 hours, give a model of the form y=Asin(`omega t -`phi)+c for height of the tide vs. clock time if at 4:30am the water level is 13 feet, and 6.25 hours later the level is 10 feet. Assume that t is the time since midnight.
So the period , or one cycle, is 12.5 hrs.
So half-cycle = 12.5 /2 = 6.25 hr
In a basic sine curve, at (1/4)cycle, y = maximum and then half-cycle later, y = minimum.
"...at 4:30am the water level is 13 feet, and 6.25 hours later the level is 10 feet..."
That means at 4:30 AM, y = 13ft , and later on, after 6.25 hours, or half-cycle, y = 10ft.
Meaning, at 4.5 hrs after midnight, y = max, or there is a hight tide
and at (4.5 +6.25) = 10.75 hrs after midnight, y = minimum, or there is a low tide.
y = Asin(`omega t -`phi)+c ----(i)
For less typing or confusion, we change that to
y = A*sin(Bt -C ) +D -----(ii)
A = amplitude = (max y - min y)/2 = (13 -10)/2 = 1.5ft ----**
D = vertical shift = (min y + amplitude) = 10 +1.5 = 11.5 ft. ----**
Period = (2pi)/B
B = (2pi)/period) = (2pi)/12.5 = pi/6.25 ----**
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For the C:
(Bt -C) = B(t -C/B)....where C/B = phase angle or horizontal shift.
It is given, or implied, that max y is at 4.5 hrs from midnight. This max y must be at (1/4)cycle or (1/4)period.
(1/4)period = 12.5/4 = 3.125 hrs.
That means, the relocated sine curve has a y=0, or the relocated sine curve started, at 3.125 hrs before 4:30 AM.
Meaning, there is a horizontal shift of (4.5 -3.125) = 1.375 hrs.
Meaning, the relocated curve has a y=0, or the curve started, at 1.375 hrs after midnight.
So,
C/B = 1.375 hrs.
Then,
(Bt -C) = B[t -C/B] = (pi/6.25){t - 1.375) = (pi/6.25)t -0.22pi ----**
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Therefore, the model, finally, is
y = 1.5sin[(pi/6.25)t -0.22pi] +11.5 --------in feet, answer.
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Check when t = 4.5 hrs
y must be 13 ft.
y = 1.5sin[(pi/6.25)(4.5) -0.22pi] +11.5
y = 13 ft-----so, OK..
Check when t = 10.75 hrs
y must be 10 ft.
y = 1.5sin[(pi/6.25)(10.75) -0.22pi] +11.5
y = 10 ft-----so, OK.