# Thread: Model of a form

1. ## Model of a form

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.

2. Originally Posted by christenc05
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.
I think the information is not complete. Give us the time of the high tide (when the level of the water is at its highest), or the time of the low tide (at the lowest water level).

3. This is all the information that I have. Nothing else is given.

4. 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

5. Originally Posted by christenc05
This is all the information that I have. Nothing else is given.
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 ----**

---------------
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.

-------------------------
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.