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 4 AM. If the tide goes from low to high every 6 hours, write a cosine function d(t) describing the depth of the water as a function of time with t = 4 corresponding to 4 AM. At what two times within one cycle is the tide at a depth of 5 feet?

by the way,stick to radians, not degrees :)

To write a cosine function describing the depth of water as a function of time, we can use the formula:

d(t) = A * cos(Bt - C) + D,

where A is the amplitude, B is the frequency, C is the phase shift, and D is the vertical shift.

Given that the average depth is 6 feet, and it varies 2 feet in both directions with the tide, we can determine the amplitude as half of the variation, which is 2/2 = 1 foot. Therefore, A = 1.

The frequency is the number of cycles that occur within a given time span, so if the tide goes from low to high every 6 hours, the frequency is 1 cycle per 6 hours or 1 cycle per 360 minutes. Therefore, B = 2π/360 = π/180.

The phase shift represents the starting point of the function in relation to time. Since the high tide occurs at 4 AM, which corresponds to t = 4, the phase shift is 0 (no shift). Therefore, C = 0.

Since the average depth is 6 feet, the vertical shift is also 6 feet. Therefore, D = 6.

Putting it all together, the cosine function d(t) describing the depth of the water as a function of time is:

d(t) = cos((π/180)t) + 6.

To find the two times within one cycle when the tide is at a depth of 5 feet, we need to solve the equation d(t) = 5.

5 = cos((π/180)t) + 6.

Subtracting 6 from both sides, we get:

-1 = cos((π/180)t).

Taking the inverse cosine (cos^(-1)) of both sides, we obtain:

(π/180)t = cos^(-1)(-1).

The inverse cosine of -1 is π, so we have:

(π/180)t = π.

Simplifying, we get:

t = 180.

Therefore, within one cycle, the tide is at a depth of 5 feet at t = 180 minutes or 3 hours after the high tide at 4 AM.

To find the second time, we need to find another angle θ such that cos(θ) = -1. The cosine of θ is -1 at θ = π, but this has already been accounted for. Therefore, we need to add another half cycle to find the second time.

Adding another half cycle, we get:

t = 180 + 180 = 360.

Therefore, within one cycle, the tide is at a depth of 5 feet at t = 360 minutes or 6 hours after the high tide at 4 AM.

To find the cosine function that describes the depth of the water as a function of time, we need to consider the given information:

1. The average depth of water at the end of the dock is 6 feet.
2. The depth varies 2 feet in both directions with the tide.

The cosine function that describes the depth of the water as a function of time can be represented as follows:

d(t) = A * cos(B(t - C)) + D

Where:
- A represents the amplitude of the function, which is half of the variation, in this case: 2 feet.
- B represents the frequency of the function, which depends on how often the tide repeats its cycle. Since the tide goes from low to high every 6 hours, the frequency can be calculated as 2π/6.
- C represents the phase shift of the function, which is defined as the time at which the tide is at its maximum or minimum value. Since the high tide is at 4 AM, the phase shift is 4.
- D represents the vertical shift of the function, which is the average depth of the water, in this case: 6 feet.

Substituting the given values into the equation, we get:

d(t) = 2 * cos((2π/6)(t - 4)) + 6

To find the two times within one cycle when the tide is at a depth of 5 feet, we can set the function equal to 5 and solve for t:

5 = 2 * cos((2π/6)(t - 4)) + 6

Subtracting 6 from both sides:

-1 = 2 * cos((2π/6)(t - 4))

Dividing by 2:

-1/2 = cos((2π/6)(t - 4))

Now we can take the inverse cosine (arccos) of both sides to find the values of (2π/6)(t - 4):

arccos(-1/2) = (2π/6)(t - 4)

Using the unit circle, we know that the angle whose cosine is -1/2 is 2π/3. So:

2π/3 = (2π/6)(t - 4)

Simplifying:

2π/3 = (π/3)(t - 4)

Multiplying both sides by 3/π:

2 = t - 4

Adding 4 to both sides:

t = 6

So, within one cycle, the tide reaches a depth of 5 feet twice - once at 6 AM and once at 6 PM.

T = period = 12 hr

d(t) = 6 + 2 cos (2 pi t /12 - phase )
max at t = 4, so when t = 4 we want :
(2 pi t /12 - phase ) = 0
because cos 0 = 1, the max
2 pi * 4/12 = phase
phase = 8 pi/12 = 2 pi/3
so
d(t) = 6 + 2 cos ( pi t /6 - 2 pi/3 )
so put in 5 for d
-1/2 = cos ( pi t /6 - 2 pi/3 )
cosine of what 2 angles = -1/2 ?