At a certain point on the beach, a post sticks out of the sand, it's top being 76 cm above the beach. The depth of the water at the post varies sinusoidally with time due to the motion of the tides. The depth d is modeled by the equation d=40+60cos(PI/2(t-2)) where t is the time in hours since midnight.

a. Sketch a graph of the sinusoid curve without using a calculator.
b. What is the earliest time of day at which the water level is just at the top of the post?
c. At the time you calculated in part b, is the post just going under water or just emerging from the water? Explain.
d. When d is negative, the tide is completely out and there is no water at the post. Between what times is the entire post out of the water?

what don't you get? You have the equation for d(t). Just do the math. For the graph, see

http://www.wolframalpha.com/input/?i=40%2B60cos%28%28pi%2F2%29%28t-2%29%29

For the rest, just solve for t under the given conditions:

(b) 40+60cos(PI/2(t-2)) = 76
(c) is d(t) on its way up, or down?
(d) 40+60cos(PI/2(t-2)) < 0

a. To sketch the graph of the sinusoid curve, we can start by determining the key characteristics of the equation:

Amplitude: The amplitude of the wave is the coefficient of the cosine function, which in this case is 60.
Vertical shift: The vertical shift is given by the constant term in the equation, which is 40.
Period: The period of the wave can be calculated by finding the time it takes for the cosine function to complete a full cycle. In this case, the coefficient of (t - 2) is π/2, so the period is 2π/(π/2) = 4 hours.
Phase shift: The phase shift of the wave is given by the constant term inside the cosine function, which is -2.

Now, let's draw the graph:
- Start by drawing the x-axis representing time and the y-axis representing the water depth.
- Mark the vertical shift of 40 cm on the y-axis.
- The amplitude of 60 cm will determine the maximum and minimum heights of the wave above and below the vertical shift.
- Since the period is 4 hours, mark the x-axis accordingly, going from 0 to 4 and dividing it into equal intervals.
- The phase shift of -2 hours means that the wave starts 2 hours after midnight.
- Begin by plotting the starting point at (0,40) and then continue plotting points using the amplitude and period to determine the height at different times. Connect the points smoothly to complete the curve.

b. To find the earliest time of day when the water level is just at the top of the post, we need to find when the depth of the water, d, equals the height of the post, which is 76 cm.

Set the equation equal to 76 and solve for t:
40 + 60cos(π/2(t-2)) = 76
60cos(π/2(t-2)) = 36
cos(π/2(t-2)) = 36/60
cos(π/2(t-2)) = 3/5

To find the earliest time, we need to find the smallest positive value of t that satisfies this equation. We can use the inverse cosine function (arccos) to find the angle.

arccos(3/5) ≈ 0.9273

Since cosine is positive in the first and fourth quadrants, we have two possible solutions:
π/2(t-2) ≈ 0.9273
t-2 ≈ 0.9273/(π/2)
t-2 ≈ 0.589

And
π/2(t-2) ≈ -0.9273
t-2 ≈ -0.9273/(π/2)
t-2 ≈ -0.589

Solving both equations, we get:
t ≈ 2 + 0.589 ≈ 2.589 hours
t ≈ 2 - 0.589 ≈ 1.411 hours

Therefore, the earliest time when the water level is just at the top of the post is approximately 1.411 hours, which is around 1:24 AM.

c. To determine if the post is just going under water or just emerging from the water at the time calculated in part b, we need to examine the behavior of the graph.

At t = 1.411 hours, the depth of the water is equal to the height of the post, which means the water level is touching the top of the post. Since the height continues to increase after this point, it means that the post is just emerging from the water at this time.

d. When the depth, d, is negative, it means the tide is out, and there is no water at the post. To find the times when the entire post is out of the water, we need to determine when d is less than or equal to zero.

40 + 60cos(π/2(t-2)) ≤ 0

Let's solve for t:
60cos(π/2(t-2)) ≤ -40
cos(π/2(t-2)) ≤ -2/3

The range of values for which the cosine function is less than or equal to -2/3 is 2πn + π/6 ≤ π/2(t-2) ≤ 2πn + 5π/6, where n is an integer.

Solving the inequalities, we get:
2πn + π/6 ≤ π/2(t-2) ≤ 2πn + 5π/6

Dividing by π/2, we have:
4n + 1/3 ≤ t - 2 ≤ 4n + 5/3

Adding 2 to each inequality, we get:
4n + 7/3 ≤ t ≤ 4n + 17/3

Therefore, the entire post is out of the water between the times of approximately 7/3 hours (2:14 AM) and 17/3 hours (5:54 AM), where n is an integer.

a. To sketch a graph of the sinusoid curve without using a calculator, we'll consider the general shape of the graph and key points.

The general form of a sinusoidal function is: d = A + B * cos(C(t - D)), where A represents the vertical shift, B represents the amplitude, C represents the horizontal compression/stretch, and D represents the horizontal shift.

In this case, the equation for the depth is given as: d = 40 + 60 * cos(π/2(t - 2))

Let's analyze the key components of the equation:
- A = 40: This represents the vertical shift of the graph. Since the top of the post is 76 cm above the beach, this indicates that the water level will vary between 40 cm below the top of the post and 116 cm above the top of the post.
- B = 60: This represents the amplitude of the graph. It determines the vertical range of the graph.
- C = π/2: This represents the horizontal compression/stretch. A smaller value of C would result in a slower oscillation, while a larger value would make it oscillate more quickly.
- D = 2: This represents the horizontal shift. It determines where the starting point of the graph is.

Now, let's apply this information to sketch the graph. Start by drawing a horizontal line at y = 40, representing the vertical shift. From there, you can make points above and below this line, considering the amplitude of 60 and the horizontal compression based on the frequency of the graph. Then, try to connect the points smoothly to form a sinusoidal curve.

b. To find the earliest time of day when the water level is just at the top of the post, we need to solve the equation for d = 76.

76 = 40 + 60 * cos(π/2(t - 2))

Rearranging the equation:

60 * cos(π/2(t - 2)) = 36

cos(π/2(t - 2)) = 36/60

cos(π/2(t - 2)) = 3/5

To find the nearest time when the cosine function equals 3/5, we need to refer to the unit circle and identify the corresponding angle for which the cosine is 3/5. The cosine of an angle corresponds to the x-coordinate on the unit circle. By trigonometry, we know that the cosine of an angle is adjacent over hypotenuse.

In this case, the adjacent side is 3 and the hypotenuse is 5 (since 3/5 = 3/5). Therefore, the corresponding angle to 3/5 is arccos(3/5).

t - 2 = arccos(3/5)

To isolate t, add 2 to both sides:

t = arccos(3/5) + 2

By evaluating arccos(3/5) and adding 2, you can calculate the earliest time of day when the water level is just at the top of the post.

c. To determine if the post is just going under the water or just emerging from the water at the calculated time, we need to consider the behavior of the cosine function.

Recall that the cosine function oscillates between -1 and 1. Since the maximum water level is 116 cm above the top of the post (A + B = 40 + 60 = 100), and the post is at 76 cm above the beach, it means that the post is not submerged underwater at any point.

Therefore, at the calculated time when the water level is just at the top of the post, the post is just emerging from the water.

d. To determine the times during which the entire post is out of the water (when d is negative), we need to find the values of t when the depth d is less than 0.

0 > 40 + 60 * cos(π/2(t - 2))

Rearranging the equation:

-40 > 60 * cos(π/2(t - 2))

Dividing by 60 (since the amplitude is 60):

-40/60 > cos(π/2(t - 2))

Simplifying:

-2/3 > cos(π/2(t - 2))

To find the values of t in which cos(π/2(t - 2)) is less than -2/3, we can refer to the unit circle. The cosine function is negative in the second and third quadrants. Therefore, we can find two intervals when t satisfies the equation.

Interval 1: 180° < π/2(t - 2) < 360°
Solve for t:

180° < π/2(t - 2) < 360°

By dividing by π/2 and adding 2, you can calculate the first interval.

Interval 2: 360° < π/2(t - 2) < 540°
Solve for t:

360° < π/2(t - 2) < 540°

Again, by dividing by π/2 and adding 2, you can calculate the second interval.

These intervals will give you the times during which the entire post is out of the water.