At an ocean depth of 20 meters, a buoy bobs up and then down 2 meters from the ocean's depth. Four seconds pass from the time the buoy is at its highest point to when it is at its lowest point. Assume at x = 0, the buoy is at normal ocean depth.
Use the sine tool to graph the function. The first point must be on the mid-line and the second point must be a maximum or minimum value on the graph closest to the first point.
To graph the given function using the sine tool, we need to understand how the properties of the sine function correspond to the given situation.
The general form of the sine function is given by: y = A * sin(B(x - C)) + D
Where:
A represents the amplitude of the graph, which is the maximum distance the graph moves from the mid-line.
B represents the period of the function, which is the distance between two consecutive maximum or minimum points.
C represents the horizontal shift of the graph.
D represents the vertical shift of the graph.
In this case, the given function representing the buoy's motion can be written as: y = 2 * sin(B(x - C)) + D
The buoy bobs up and down 2 meters from the ocean's depth, so the amplitude is 2. Therefore, A = 2.
The period of the function is the time it takes for the buoy to complete one full cycle, including going up and down. We know that it takes 4 seconds for the buoy to go from the highest to the lowest point, so the period is 4 seconds. Therefore, B = 2π / 4 = π/2.
Since at x = 0, the buoy is at the normal ocean depth, its mid-line is at y = 0. Therefore, D = 0.
Using the given information, we can find the value of C. We know that the first point must be on the mid-line, so (0, 0) is a point on the graph. This gives us the equation: 0 = 2 * sin(B(0 - C)) + 0
Simplifying the equation, we find: sin(-BC) = 0
For sin(-BC) to be equal to 0, -BC must be equal to 0, π, 2π, 3π, etc. This means that BC must be equal to 0, π/2, π, 3π/2, etc.
Since B = π/2, we get: (π/2)C = 0, π/2, π, 3π/2, etc.
So, C can be 0, 1, 2, 3, etc., depending on the value of π/2.
To summarize:
A = 2 (amplitude)
B = π/2 (period)
C = 0, 1, 2, 3, ... (horizontal shift)
D = 0 (vertical shift)
Now, you can plot the points on the graph using the values of A, B, C, and D. Start with the first point on the mid-line at (0, 0) and the second point at the closest maximum or minimum value to the first point. Repeat this process to draw the entire graph by finding consecutive maximum or minimum points using the period and horizontal shift.
To graph the function that describes the buoy's motion using the sine tool, we need to start by determining the period, amplitude, and mid-line.
The period is the time it takes for the buoy to complete one full cycle of motion. In this case, since the buoy takes 4 seconds from its highest point to its lowest point, the period is 4 seconds.
The amplitude is the maximum distance the buoy moves up or down from the mid-line. Here, the buoy bobs up and down 2 meters from its normal ocean depth of 20 meters. Thus, the amplitude is 2 meters.
The mid-line is the average of the highest and lowest points the buoy reaches. Since the buoy's normal ocean depth is at x = 0, the mid-line is at a height of 20 meters.
Now let's plot the first two points based on the given information:
1. The first point must be on the mid-line, which is at a height of 20 meters. So the first point is (0, 20).
2. The second point must be a maximum or minimum value on the graph closest to the first point. Since the buoy bobs up and down, the second point will be either a maximum or minimum value. From the information given, the buoy bobs up and down 2 meters from the mid-line. Therefore, the second point will be a maximum or minimum value at 22 or 18 meters, respectively.
Let's plot the second point as (t, y), where t is the horizontal distance from the first point in seconds and y is the height of the buoy from the mid-line in meters.
If the second point is a maximum, then (t, y) = (1, 22). If the second point is a minimum, then (t, y) = (1, 18).
Plotting these points will give you a sine wave graph that describes the buoy's motion at an ocean depth of 20 meters.
y = 20 + 2 sin (2 pi t/T)
that is 0 of the sin function at t = 0
max to min is half a period t = T/2 (t = T/4 to t = 3 T/4) so T = 8 seconds
y = 20 + 2 sin ( pi t/4)
you seem to be using x for time so
y = 20 + 2 sin (pi x/4)