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Answer :
Answer:
[tex]y(t) = 6 -3cos(\frac{2\pi }{14} )t[/tex]
[tex]y(t) = 6 -3cos(\frac{2\pi }{7} )t[/tex]
[tex]y(t) = 6 - 6cos(\frac{2\pi }{14} ) t[/tex]
[tex]y(t) = 3- 6cos(\frac{2\pi }{7} )t[/tex]
Step-by-step explanation:
Given that,
Hudson Bay tides vary between [tex]3 ft[/tex] and [tex]9 ft[/tex].
Tide is at its lowest when [tex]t=0[/tex]
Completes a full cycle in 14 hours.
To find:- What is the amplitude, period, and midline of a function that would model this periodic phenomenon?
So, The periodic function of this model is
[tex]y(t) = y^{'} + Acos(\omega\ t)[/tex] ...................(1)
where, [tex]A- Amplitude of cycle[/tex]
[tex]\omega = Angular speed (in Radian.)[/tex]
Then putting the value in given Equation(1) we get,
Amplitude = [tex]\frac{9-3}{2} ft = 3ft[/tex]
[tex]y^{'} = (3+ 3 )ft = 6ft[/tex]
Now, At [tex]t=0 sec[/tex] it complete full cycle in [tex]14 hours.[/tex] [tex]-cos(\omega t)[/tex] because it is at lowest at t=0sec.
∵ [tex]\omega t= 2\pi[/tex]
[tex]\omega (t+14) = 2\pi[/tex]
∴ [tex]\omega = \frac{2\pi }{14}[/tex]
Hence [tex]y(t) = 6 -3cos(\frac{2\pi }{14} )t[/tex]
[tex]y(t) = 6 -3cos(\frac{2\pi }{7} )t[/tex]
[tex]y(t) = 6 - 6cos(\frac{2\pi }{14} ) t[/tex]
[tex]y(t) = 3- 6cos(\frac{2\pi }{7} )t[/tex]
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Rewritten by : Barada
The tidal phenomenon in Hudson Bay can be modeled with an amplitude of 3 feet, a period of 14 hours, and a midline of y = 6 feet.
Tide Modeling in Hudson Bay
To model the tidal phenomenon in Hudson Bay, we need to determine the amplitude, period, and midline of the periodic function:
Amplitude: The highest tide is 9 feet and the lowest is 3 feet. The amplitude is half the difference between the high and low tide: (9 - 3) / 2 = 3 feet.
Period: The tide completes a full cycle in 14 hours.
Midline: The midline is the average of the high and low tide: (9 + 3) / 2 = 6 feet.
Therefore, the correct model is: Amplitude = 3 feet, Period = 14 hours, and Midline = y = 6 feet.
