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?

Guest Jul 23, 2019

#1**0 **

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

ElectricPavlov Jul 23, 2019

#1**0 **

Best Answer

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

ElectricPavlov Jul 23, 2019