You nailed it, except for one step in your arithmetic near the end. See my note.
Excellent analysis.
Problem Number 4
Assuming that high tides occur every 12.5 hours, give a model of the form y = A sin(`omega t - `phi) + c for height of the tide vs. clock time if at 2: 30 a.m. the water level is 13 feet, and 6.25 hours later the level is 11 feet. Assume that t is the time since midnight.
Sir,
On this question I see that the high tide is at 2:30am and low tide at 8:55 am .
The A Vlue is therefore (13-10)/2 = 1.5 …( c ) = (13+10)/2=11.5 and Omega= 2pi/12.5 or 4pi/25
If y= 13 at t=2:30.…. Then : ( 2pi/12.5*2.5-Phi ) =pi/2, because the Sin theta =1 occurs at pi/2
Therefore:
( 2pi/12.5 * 2.5 -phi) =pi/2,
( 5pi/12.5-phi)=pi/2
(5pi/12.5-pi/2) =phi
(10pi/25 - 12.5pi/25)=phi
(-2.5pi/25)=phi
phi= -.1pi
Therefore, y = A sin(`omega t - `phi) + c
Y=1.5sin(4pi/25 * t +.1pi) +11.5
Phase shift + phi/omega = -.1pi/(4pi/25)=-1.25
-.1pi/(4pi/25) = -.625, half of your -1.25.
If you then plot this it works.
Your function was shifted a bit over 1/2-hour to the left.
So,
Y=1.5sin(4pi/25(t+1.25)+11.5
When I graphed this it does not look right.
Can you advise..