At a certain point on the beach, a post sticks out of the sand, its 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/6(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 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?
b.
76=40+60cos(PI/6 (t-2))
.6=cos(theta)
theta=arccos.6=.972 rad
.972=PI/6 (t-2)
solve for t.
c. evaluate dh/dt=-60*PI/6*sin(PI/6(t-2)
at the t in b. is it negative? then level is going down. Positive, increasing.
find times when h=0, then check between those times. your graph in a should help.
a. To sketch a graph of the sinusoid curve without using a calculator, we can follow these steps:
1. Start by identifying the key components of the equation:
- Amplitude: The amplitude of the sinusoid is the coefficient in front of the cosine term. In this case, it is 60.
- Period: The period of the sinusoid is given by 2π divided by the coefficient multiplying the variable inside the cosine function. In this case, it is 2π/(π/6) = 12.
- Vertical shift: The vertical shift is the constant term added to the cosine function. In this case, it is 40.
2. Plot the main points on the graph:
- At t = 2 (hours since midnight), the depth is at its maximum, 100 cm (40 + 60).
- At t = 5 (hours since midnight), the depth is at zero, the beach level.
- At t = 8 (hours since midnight), the depth is at its minimum, -20 cm (40 - 60).
3. Use the period to determine the distance between consecutive maximum or minimum points and draw the curve accordingly.
b. To find the earliest time of day when the water level is just at the top of the post, we need to set the equation equal to the height of the post, which is 76 cm. Thus, we have:
40 + 60cos(pi/6(t - 2)) = 76
To solve for t, we can isolate the cosine term:
60cos(pi/6(t - 2)) = 36
cos(pi/6(t - 2)) = 36/60
cos(pi/6(t - 2)) = 3/5
Now, we need to find the angle whose cosine is 3/5. Using the inverse cosine function (cos^-1), we get:
pi/6(t - 2) = cos^-1(3/5)
Solving for t, we have:
t - 2 = 6cos^-1(3/5)/pi
t = 2 + 6cos^-1(3/5)/pi
Using a calculator to evaluate cos^-1(3/5) (the inverse cosine of 3/5), we find it to be approximately 0.9273. Thus:
t ≈ 2 + 6(0.9273)/pi
t ≈ 2.354 hours since midnight
Therefore, the earliest time of day when the water level is just at the top of the post is approximately 2 hours and 21 minutes after midnight.
c. At the time calculated in part b (approximately 2.354 hours since midnight), the post is just emerging from the water. This is because the water level is equal to the height of the post (76 cm), indicating that the post is no longer submerged.
d. When d is negative, it means the tide is completely 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 find the values of t for which d is negative.
40 + 60cos(pi/6(t - 2)) < 0
cos(pi/6(t - 2)) < -2/3
To find the values of t satisfying this inequality, we can use the unit circle and the reference angle for which the cosine function is -2/3. However, since cosine is positive in quadrants I and IV, this inequality has no solutions.
Therefore, there is no time when the entire post is out of the water.