The average depth of the water at the end of the dock is 6 feet. High tide occurs at 6 am and low tide occurs at 6 pm. The depth of the water at high tide is 9 feet. The information can be graphed using a sinusoidal function. Your boat needs at least 4.5 feet of water to come into the dock. What times of the day is the water going to be too low to come into the dock?

model this as depth of water:

D=6 -3sin wt w=PI/12 and t is in Universal 24 hr time (6am is 6, 6PM is 18)

so when is D<4.5
4.5=6-3sin(PI*t/12)
sin(PI*t/12)=1/2
PI*t/12=(30 deg)=PI*t/6
t=2, so by symettry, at low tide +- 2hrs, the boat wont float. That is 4Pm to 8PM

Wow thank you so much :)

To find the times of the day when the water is too low to come into the dock, we need to determine the equation of the sinusoidal function that represents the water depth.

Let's start by defining the variables:
- h: high tide depth (9 feet)
- l: low tide depth (unknown)
- x: time in hours since high tide
- d: average depth at the end of the dock (6 feet)
- a: amplitude (difference between high and low tide depths divided by 2)
- b: frequency (12 hours per cycle)

Since the water depth oscillates between high and low tide depths, we can use the formula for a sinusoidal function:
d = a * sin(b * x) + (h + l) / 2

Given that the average depth (d) is 6 feet, the high tide depth (h) is 9 feet, and the amplitude (a) is (h - l) / 2, we can substitute these values into the equation:
6 = [(9 - l) / 2] * sin[(2π / 12) * x] + (9 + l) / 2

To solve for the low tide depth (l), we can rearrange the equation:
[(9 - l) / 2] * sin[(2π / 12) * x] + (9 + l) / 2 = 6

Multiply both sides by 2 to eliminate the fraction:
(9 - l) * sin[(2π / 12) * x] + 9 + l = 12

Next, isolate the term with sin[(2π / 12) * x]:
(9 - l) * sin[(2π / 12) * x] = 12 - (9 + l)

Simplify the right-hand side:
(9 - l) * sin[(2π / 12) * x] = 3 - l

Now, divide both sides by (9 - l):
sin[(2π / 12) * x] = (3 - l) / (9 - l)

To solve for x, we can take the inverse sine of both sides:
(2π / 12) * x = arcsin[(3 - l) / (9 - l)]

Finally, divide both sides by (2π / 12) to isolate x:
x = 12 * arcsin[(3 - l) / (9 - l)] / 2π

To find the times when the water is too low for the boat to come into the dock, we need to substitute x back into the time equation:
time = 6 am + x hours

Given that the boat needs at least 4.5 feet of water to come into the dock, we can set the equation for the water depth equal to 4.5 feet and solve for the times (x) when this occurs.

Please note that finding the exact values for x and converting them to times requires numerical calculations.

To determine the times of the day when the water will be too low for the boat to come into the dock, let's analyze the given information and create a sinusoidal function to represent the water depth.

First, we identify the important data points:
- Average depth: 6 feet
- High tide depth: 9 feet
- Low tide depth: unknown
- Range of water depth: 9 feet - unknown
- Period of the tide: 12 hours (from 6 am to 6 pm)

To create a sinusoidal function, we need to find the amplitude, period, and vertical shift.

Amplitude:
The amplitude is half the difference between the maximum and minimum values. In this case, the average depth is the mid-point, so the amplitude is (9 - 6) / 2 = 1.5 feet.

Period:
The period represents the length of one complete cycle. In this case, it is 12 hours since high tide and low tide occur every 12 hours.

Vertical Shift:
Since the average depth is 6 feet, the vertical shift is 6.

Now, we can construct the sinusoidal function in the form:
depth(t) = A * sin(B * (t - C)) + D

where:
A = amplitude = 1.5
B = 2π / period = 2π / 12 = π/6
C = phase shift = 0 (since high tide occurs at 6 am)
D = vertical shift = 6

Substituting these values into the function, we get:
depth(t) = 1.5 * sin((π/6) * (t - 0)) + 6

Next, we need to determine the times when the depth falls below the minimum requirement of 4.5 feet.

depth(t) < 4.5

1.5 * sin((π/6) * (t - 0)) + 6 < 4.5
1.5 * sin((π/6) * (t - 0)) < -1.5
sin((π/6) * (t - 0)) < -1

To solve this inequality, we need to find the values of "t" that satisfy it. Since sin(theta) can only be between -1 and 1, there are no solutions to this inequality. Therefore, there are no times during the day when the water is too low for the boat to come into the dock.

Therefore, the water will never be too low for the boat to come into the dock during the given 12-hour period (6 am to 6 pm).