On a summer day in Cape Cod, the depth of the water at a dock was 4ft at low tide at 2:00 AM. At high tide 5 hours later, the heist of the water at the dock rose to 14 feet. Write a cosine function to model the height of the water at the dock x hours after the day began at midnight

Question

On a summer day in Cape Cod, the depth of the water at a dock was 4ft at low tide at 2:00 AM. At high tide 5 hours later, the heist of the water at the dock rose to 14 feet. Write a cosine function to model the height of the water at the dock x hours after the day began at midnight

Answer 1

To model the height of the water at the dock as a function of time, we can use a cosine function because it exhibits periodic behavior.

Let's start by considering the midline of the cosine function, which represents the average water level throughout the day. Since the depth of the water at low tide is 4 feet, the midline would be at a height of (4+14)/2 = 9 feet.

Next, we need to determine the amplitude of the cosine function, which represents half the distance between the maximum and minimum water levels. In this case, the maximum water level is 14 feet and the minimum water level is 4 feet. Therefore, the amplitude is (14-4)/2 = 5 feet.

Lastly, we need to determine the period of the cosine function, which represents the time it takes for the water level to complete one full cycle. In this case, the period is 12 hours because high tide occurs 6 hours after low tide and completes another cycle 12 hours later.

Putting it all together, the cosine function that models the height of the water at the dock x hours after the day began at midnight is:

h(x) = A * cos(2π/P * (x - x0)) + M

where:
A = amplitude = 5 feet
P = period = 12 hours
x0 = phase shift (time at which the cosine function starts) = 2 hours after midnight
M = midline = 9 feet

Therefore, the cosine function to model the height of the water at the dock is:

h(x) = 5 * cos(2π/12 * (x - 2)) + 9

Answer 2

amplitude: (14-4)/2 = 5

center line: (14+4)/2 = 9
so, we can start with

y = 9cos(kx)

the period is 5+5 = 10 hours, so 2?/k = 10, making k=?/5

y = 9cos(?/5 x)

The high tide was at 7:00, rather than midnight, so

y = 9cos(?/5 (x-7))
or
y = 9cos(?/5 x - 7?/5)

see

http://www.wolframalpha.com/input/?i=9cos(%CF%80%2F5+(x-7))

Answer 3

Ummh, the amplitude is 5

We also need a vertical shift, as well as a phase shift.

I had
y = 5cos (?/5(x + 3)) + 9

test: let x =2 , (2:00 am)
y = 5cos ? + 9 = 5(-1)+9 = 4 ft,
let x = 7, (5 hours later, the max of 14)
y = 5cos (2?) + 9
= 5(1) + 9 = 14 ft

http://www.wolframalpha.com/input/?i=5+cos(1%2F5+%CF%80+(x+%2B+3))+%2B+9

Answer 4

Huh. Good thing we got a second opinion. Ahem. As I was saying,

amplitude: (14-4)/2 = 5
center line: (14+4)/2 = 9
so, we can start with

y = 9+5cos(kx)
y = 9+5cos(π/5 x)
y = 9+5cos(π/5 (x-7))

As for the shift, (x-7), (x+3)
tomato, tomahto. Since the period is 10, it's the same.