precalculus question

Question

The average depth of water at the end of a dock is 6 feet. This varies 2 feet in both directions with the tide. Suppose there is a high tide at 4AM. If the tide goes from low to high every 6 hours, write a sinusoidal equation describing the depth of the water as a function of time with t=0 corresponding to 12AM. What is the depth of the water at 8AM?

Answer 1

If the tide goes from low to high every 6 hours, then it makes a complete cycle (low-to-high and then high-to-low) every 12 hours meaning that the period is 12 hours.

If the average is 6 feet then that is the midline or vertical shift.

If it varies by 2 feet in each direction, then the amplitude is 2 feet.

Since we are given information about a high tide, and since a cosine function starts at a high point and goes down from there, we can model this as a cosine function.

If we are starting (t = 0) at midnight and there is a high tide at 4AM, then the horizontal shift (also known as phase shift) is 4 hours to the right.

The standard cosine equation looks like this:

A = 2

To find B, use T = 2π/B or B = 2π/T meaning B = 2π/12 = π/6

C = 4

D = 6

So depth as a function of time is:

d(t) = 2cos[π/6(t - 4)] + 6

To find the depth at 8AM (which is 8 hours after midnight) plug in t = 8:

d(8) = 2cos[π/6(8 - 4)] + 6

d(8) = 2cos[π/6(4)] + 6

d(8) = 2cos[2π/3] + 6

d(8) = 2(-0.5) + 6

d(8) = -1 + 6

d(8) = 5

So the water is 5 feet deep at 8:00 AM