Answer:
[tex]y(t)= 6-3cos(\dfrac{2\pi}{14}t )[/tex]
Step-by-step explanation:
The function that could model this periodic phenomenon will be of the form
[tex]y(t) = y_0+Acos(wt)[/tex]
The tide varies between 3ft and 9ft, which means its amplitude [tex]A[/tex] is
[tex]A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}[/tex]
and its midline [tex]y_0[/tex] is
[tex]y_o=3+3 \\\\\boxed{y_o= 6ft}[/tex].
Furthermore, since at [tex]t=0[/tex] the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is [tex]-cos(\omega t)[/tex].
The period of the full cycle is 14 hours, which means
[tex]\omega t =2\pi[/tex]
[tex]\omega (t+14)= 4\pi[/tex]
giving us
[tex]\boxed{\omega = \dfrac{2\pi}{14}.}[/tex]
With all of the values of the variables in place, the function modeling the situation now becomes
[tex]\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}[/tex]
