In one day, there are two high tides and two low tides in equally spaced intervals. The high tide is observed to be 6 feet above the average sea level. After 6 hours pass, the low tide occurs at 6 feet below the average sea level. In this task, you will model this occurrence using a trigonometric function by using x as a measurement of time. Assume the first high tide occurs at x = 0.
Part B
Determine these key features of the function that models the tide:
amplitude
period
frequency
midline
vertical shift
phase shift
Part C
Create a trigonometric function that models the ocean tide for a period of 12 hours.
Part E
What is the height of the tide after 93 hours?
+36916 Amplitude = (highest - lowest)/2 = 6 - -6 /2 = 12/2 = 6 feet
period = time for 1 cycle from high to low to high = 6 + 6 = 12 hours
frequency = 1/period = 1/12 cycles per hour
midline = 0 (midpoint between high and low)
Vertical shift = 0 the high and the low are equal in magnitude
Phase shift = 0 if using cosine function to model this or 90 degrees (pi/2) leftward if using sine
Usual cos period is 2pi
we want 1/12 of the 1/12 x 2pi = pi/6
6 cos ( (pi/6) *x) = height or 6 sin ((pi/6)( x) + pi/2))
6 cos (pi/6 x) when x = 93 6 cos (93pi/6) = 0 ft
Here is the graph:
https://www.desmos.com/calculator/3pv8ltv8xh
+36916 Amplitude = (highest - lowest)/2 = 6 - -6 /2 = 12/2 = 6 feet
period = time for 1 cycle from high to low to high = 6 + 6 = 12 hours
frequency = 1/period = 1/12 cycles per hour
midline = 0 (midpoint between high and low)
Vertical shift = 0 the high and the low are equal in magnitude
Phase shift = 0 if using cosine function to model this or 90 degrees (pi/2) leftward if using sine
Usual cos period is 2pi
we want 1/12 of the 1/12 x 2pi = pi/6
6 cos ( (pi/6) *x) = height or 6 sin ((pi/6)( x) + pi/2))
6 cos (pi/6 x) when x = 93 6 cos (93pi/6) = 0 ft
Here is the graph:
https://www.desmos.com/calculator/3pv8ltv8xh
