The motion of the tide is periodic, such that the maximum of minimum
height are reached at regular intervals.
The correct responses are;
Reasons:
Height of the first high tide = 13.25 ft.
Time of first high tide = 4:30 am = 4.5 hours
Height of first low tide = 1.88 feet
Time of second high tide = 5:00 pm
a) The equation that models the tide is found as follows; sine function
The general form of the equation is f(x) = a·sin(B·(x - C)) + D
The period of the tide, T = 5:00 pm - 4:30 am = 12.5 hours
[tex]\displaystyle B = \frac{2 \cdot \pi}{T} = \frac{2 \cdot \pi}{12} = \mathbf{\frac{\pi}{6}}[/tex]
[tex]\displaystyle \mathrm{The \ amplitude \ of \ the \ tide, } \ a = \frac{13.25 \ ft. - 1.88 \ ft. }{2} = \mathbf{5.685}[/tex]
The amplitude, a = 5.685 ft.
D = The vertical shift = Height of midline = 1.88 + a = 1.88 + 5.685 = 7.565
D = 7.565 ft.
At x = 4.3, f(x) = 13.25, therefore;
[tex]\displaystyle 13.25 = 5.685 \times sin \left(\frac{\pi}{6} \cdot \left(4.30 - C\right)\right) + 7.565[/tex]
[tex]\displaystyle sin \left(\frac{\pi}{6} \cdot \left(4.30 - C\right)\right) = \frac{13.25 - 7.565}{ 5.685 }[/tex]
[tex]\displaystyle C = \mathbf{4.5 - \frac{6}{\pi} \times arcsine \left(\frac{13.25 - 7.565}{ 5.685 }\right)}[/tex]
Therefore;
C = 1.5
Which gives the equation modelling the tide as follows;
b) At 8 am, we have, x = 8, which gives;
[tex]\displaystyle f(8) = 5.685 \times sin \left(\frac{\pi}{6} \cdot \left(8 - 1.5\right)\right) + 7.565 \approx 6.094[/tex]
The height of the tide at 8 a.m., f(8) ≈ 6.0964 ft.
c) The vertical shift, D = 7.565 feet
The vertical shift represents the height of the midline above the ground level or x-axis.
d) The amplitude, a = 5.685 feet
The amplitude represents the maximum distances the tide oscillates above and below the midline.
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