Question

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?

Answer 1

Answer:

The "independent variable" is the "time" (in hours) that is plotted on the "x-axis".  Note the "time" (in hours) can be "manipulated" in the sense that is it "chosen" by the humans who are measuring the data (e.g. when to start, how many hours, and at what intervals.

The "dependent variable" is the measurement (in feet) above or below the sea level, and this is plotted on the "y-axis".  These values cannot be "manipulated" in the sense that one cannot "choose" what value or measurement the tide level would be at any particular time.

Answer 2

Data

x = 0; y = midline + 6 feet  

x = 6; y = midline - 6 feet

x = 12; y = midline + 6 feet  

x = 18; y = midline - 6 feet

Part B

amplitude (A) = 6 feet (distance, in feet, from maximum to midline)

period (T) = 12 hours (distance, in hours, between two maximums)

frequency (f) = 1/T = 0.083 hours^-1

midline = average sea level

vertical shift (B) = average sea level

phase shift $\phi$ = 0 (maximum occurs at x = 0)  

PartC

General formula  

$x(t) = -A \times cos(\frac{2 \pi t}{T} + \phi) + B$

Replacing

$x(t) = -6 \times cos(\frac{2 \pi t}{12}) + average \; sea \; level$

Part E

$x(t) = -6 \times cos(\frac{2 \pi 93}{12}) + average \; sea \; level$

x(t) = average sea level