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21:21:30 Goals for this Assignment include but are not limited to the following:
1. Construct a table of the values of y = sin(x) for a complete cycle of this function, with x equal to multiples of pi/6 or pi/4, and using the table construct a graph of one cycle of y = sin(x ). 2. Given a function y = sin(theta) with theta given as a function of x, construct a table of the values of y = sin(theta) for a complete cycle of this function with theta equal to multiples of pi/6 or pi/4, then determine the x value corresponding to each value of theta. Using a table of y vs. x construct a graph of one cycle of y = sin(theta) in terms of the given function theta of x, clearly labeling the x axis for each quarter-cycle of the function. 3. Interpret the function and graph corresponding to Goal 2 in terms of angular motion on a unit circle. Click once more on Next Question/Answer for a note on Previous Assignments.......!!!!!!!!...................................
RESPONSE -->
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21:21:40
Previous Assignments: Be sure you have completed Assignment 1 as instructed under the Assts link on the homepage at 164.106.222.236 and submitted the result of the Query and q_a_ from that Assignment.......!!!!!!!!...................................
RESPONSE --> Ok
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21:22:21 `q001. Note that this assignment has 15 activities.
Figure 93 shows the angular positions which are multiples of pi/4 superimposed on a grid indicating the scale relative to the x y coordinate system. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi.......!!!!!!!!...................................
RESPONSE --> Ok, I may have done this assignment
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21:22:30 The angular positions of the points coinciding with the positive and negative x axes all have y coordinate 0; these angles include 0, pi and 2 pi. At angular position pi/4 the point of circle appears to be close to (.7,.7), perhaps a little beyond at (.71,.71) or even at (.72,.72). Any of these estimates would be reasonable. Note for reference that, to two decimal places the coordinate so are in fact (.71,.71). To 3 decimal places the coordinates are (.707, .707), and the completely accurate coordinates are (`sqrt(2)/2, `sqrt(2) / 2).
The y coordinate of the pi/4 point is therefore .71. The y coordinate of the 3 pi/4 point is the same, while the y coordinates of the 5 pi/4 and 7 pi/4 points are -.71.......!!!!!!!!...................................
RESPONSE --> ok
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21:27:46 `q002. Figure 31 shows the angular positions which are multiples of pi/6 superimposed on a grid indicating the scale relative to the x y coordinate system. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi.
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RESPONSE --> 0, 0 pi/6, 1/2 or .5 pi/3,sqrt3/2 or .87 pi/2 ,1 , 2 pi/3, sqrt3/2 or .87 5 pi/6, 1/2 or .5 pi, 0 7 pi/6,- 1/2 or -.5 4 pi/3, -sqrt3/2 or -.87 3 pi/2, -1 5 pi/3, -sqrt3/2 or -.87 11 pi/6 ,- 1/2 or -.5 2 pi, 0
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21:29:11 The angular positions of the points coinciding with the positive and negative x axes all have y coordinate 0; these angles include 0, pi and 2 pi. At angular position pi/6 the point on the circle appears to be close to (.9,.5); the x coordinate is actually a bit less than .9, perhaps .87, so perhaps the coordinates of the point are (.87, .5). Any estimate close to these would be reasonable. Note for reference that the estimate (.87, .50) is indeed accurate to 2 decimal places. The completely accurate coordinates are (`sqrt(3)/2, 1/2).
The y coordinate of the pi/6 point is therefore .5. The coordinates of the pi/3 point are (.5, .87), just the reverse of those of the pi/6 point; so the y coordinate of the pi/3 point is approximately .87. The 2 pi/3 point will also have y coordinate approximately .87, while the 4 pi/3 and 5 pi/3 points will have y coordinates approximately -.87. The 5 pi/6 point will have y coordinate .5, while the 7 pi/6 and 11 pi/6 points will have y coordinate -.5.......!!!!!!!!...................................
RESPONSE --> ok,
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21:32:46 `q003. Make a table of y coordinate vs. angular position for points which lie on the unit circle at angular positions theta which are multiples of pi/4 with 0 <= theta <= 2 pi (i.e., 0, pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, 2 pi). You may use 2-significant-figure approximations for this exercise.
Sketch a graph of the y coordinate vs. angular position. Give your table and describe the graph.......!!!!!!!!...................................
RESPONSE --> (0, 0) (pi/4, sqrt2/2) (pi/2, 1) (3 pi/4,sqrt2/2) (pi,0) (5 pi/4,-sqrt2/2) ( 3 pi/2, -1) (7 pi/4,-sqrt2/2) (2 pi,0)
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21:33:48 The table is
theta y coordinate 0 0.0 pi/4 0.71 pi/2 1.0 3 pi/4 0.71 pi 0.0 5 pi/4 -0.71 3 pi/2 -1.0 7 pi/4 -0.71 2 pi 0.0. We note that the slope from (0,0) to (pi/4,.71) is greater than that from (pi/4,.71) to (pi/2,1), so is apparent that between (0,0) and (pi/2,1) the graph is increasing at a decreasing rate. We also observe that the maximum point occurs at (pi/2,1) and the minimum at (3 pi/2,-1). The graph starts at (0,0) where it has a positive slope and increases at a decrasing rate until it reaches the point (pi/2,1), at which the graph becomes for an instant horizontal and after which the graph begins decreasing at an increasing rate until it passes through the theta-axis at (pi, 0). It continues decreasing but now at a decreasing rate until reaching the point (3 pi/2,1), for the graph becomes for an instant horizontal. The graph then begins increasing at an increasing rate until it again reaches the theta-axis at (2 pi, 0).......!!!!!!!!...................................
RESPONSE --> ok
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21:41:22 `q004. In terms of the motion of the point on the unit circle, why is it that the graph between theta = 0 and pi/2 increases? Why is that that the graph increases at a decreasing rate?
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RESPONSE --> As the point travels in an anti clockwise direction from (0,0) to (pi/2,1) in at first increases rapidly and then prior to reaching (0,Pi/2) it decreases to make its turn downward. It's almost like the graph knowing that it has to slow down in oder to round the curve thus, obaying the laws of physics, much like you or I must slow a car down to make a dangerous turn. I understand that the graph at first is covering more ground in the vertical direction but then the slope is falling off rapidly as it reaches the maximum.
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21:43:02 As we move along the unit circle from the theta = 0 to the theta = pi/2 position it is clear that the y coordinate increases from 0 to 1. At the beginning of this motion the arc of the circle take us mostly in the y direction, so that the y coordinate changes quickly. However by the time we get near the theta = pi/2 position at the 'top' of the circle the arc is carrying us mostly in the x direction, with very little change in y.
If we continue this reasoning we see why as we move through the second quadrant from theta = pi/2 to theta = pi the y coordinate decreases slowly at first then more and more rapidly, reflecting the way the graph decreases at an increasing rate. Then as we move through the third quadrant from theta = pi to theta = 3 pi/2 the y coordinate continues decreasing, but at a decreasing rate until we reach the minimum y = -1 at theta = 3 pi/2 before begin beginning to increase an increasing rate as we move through the fourth quadrant. If you did not get this answer, and if you did not draw a sketch of the circle and trace the motion around the circle, then you should do so now and do your best to understand the explanation in terms of your picture. You should also document in the notes whether you have understood this explanation.......!!!!!!!!...................................
RESPONSE --> Undersatnd this completely !!
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21:50:48 `q005. The table and graph of the preceding problems describe the sine function between = 0 and theta = pi/2. The sine function can be defined as follows:
The sine of the angle theta is the y coordinate of the point lying at angular position theta on a unit circle centered at the origin. We write y = sin(theta) to indicate the value of this function at angular position theta. Make note also of the definition of the cosine function: The cosine of the angle theta is the x coordinate of the point lying at angular position theta on a unit circle centered at the origin. We write x = cos(theta) to indicate the value of this function at angular position theta. We can also the line tangent function to be tan(theta) = y / x. Since for the unit circle sin(theta) and cos(theta) are respectively y and x, it should be clear that tan(theta) = sin(theta) / cos(theta). Give the following values: sin(pi/6), sin(11 pi/6), sin(3 pi/4), sin(4 pi/3), cos(pi/3), cos(7 pi/6).......!!!!!!!!...................................
RESPONSE --> sin(pi/6), = 1/2 sin(11 pi/6), -1/2 sin(3 pi/4), =sqrt2/2 sin(4 pi/3),=-sqrt3/2 cos(pi/3), 1/2 cos(7 pi/6) =-sqrt3/2
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21:51:11 sin(pi/6) is the y coordinate on the unit circle of the point at the pi/6 position. We have seen that this coordinate is .5.
sin(11 pi/6) is the y coordinate on the unit circle of the point at the 11 pi/6 position, which lies in the fourth quadrant and in angular displacement of pi/6 below the positive x-axis. We have seen that this coordinate is -.5. sin(3 pi/4) is the y coordinate on the unit circle of the point at the 3 pi/4 position. We have seen that this coordinate is .71. sin(4 pi/3) is the y coordinate on the unit circle of the point at the 4 pi/3 position, which lies in the third quadrant at angle pi/3 beyond the negative x axis. We have seen that this coordinate is -.87. cos(pi/3) is the x coordinate on the unit circle of the point at the pi/3 position, which lies in the first quadrant at angle pi/3 above the negative x axis. We have seen that this coordinate is .5. cos(7 pi/6) is the y coordinate on the unit circle of the point at the 7 pi/6 position, which lies in the third quadrant at angle pi/6 beyond the negative x axis. We have seen that this coordinate is -.87.......!!!!!!!!...................................
RESPONSE --> ok
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22:26:37 `q006. Suppose that the angle theta is equal to 2 x and that y = sine (theta). Given values of theta correspond to x = pi/6, pi/3, ..., pi, give the corresponding values of y = sin(theta).
Sketch a graph of y vs. x. Not y vs. theta but y vs. x. Do you think your graph is accurate?......!!!!!!!!...................................
RESPONSE --> What I did here was set Theta = 2x where x=theta/2. I made a table of theda with pi/6 intervals. Plugging in these theta I solved for (x) then using these values of (x) I plugged them into the function y = Sine (theta) The following are my results... (0,0), (pi/12, appro .26 ) , ( pi/6,1/2) ,( pi/4,sqrt2/2), (pi/3 , sqrt3/2) , (5pi/12,.9659) ,(pi/2,1) , (7pi/12, .9659) , (2pi/3,Sqrt3/2) , (pi/2,1), (5pi/6,1/2) , (11pi/12, .2588), (pi,0) To the question of do I think it is accurate, well to be honest, if my points are correct the graph starts off rising fast then decreases, kind of flattens out, then rises with a gental slope to the maximum dips down then back up to the max the falls off rapidly then decreases much like a waterslide fast and furious but tapers off ...I don't think this is the type of model one would like to arrive at..
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22:32:11 The angles are in increments of pi/6, so we have angles pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6 and pi.
If x = pi/6, then 2x = 2 * pi/6 = pi/3. If x = pi/3, then 2x = 2 * pi/3 = 2 pi/3. If x = pi/2, then 2x = 2 * pi/2 = pi. If x = 2 pi/3, then 2x = 2 * 2 pi/3 = 4 pi/3. If x = 5 pi/6, then 2x = 2 * 5 pi/6 = 5 pi/3. If x = pi, then 2x = 2 * pi/6 = 2 pi. The values of sin(2x) are therefore sin(pi/3) = .87 sin(2 pi/3) = .87 sin(pi) = 0 sin(4 pi/3) = -.87 sin(5 pi/3) = -.87 sin(2 pi) = 0. We can summarize this in a table as follows: x 2x sin(2x) 0 0 0.0 pi/6 pi/3 0.87 pi/3 2 pi/3 0.87 pi/2 pi 0 2 pi/3 4 pi/3 -0.87 5 pi/6 5 pi/3 -0.87 0 2 pi 0.0. Figures 93 and 77 depict the graphs of y = sin(theta) vs. theta and y = sin(2x) vs. x. Note also that the graph of y = sin(2x) continues through another complete cycle as x goes from 0 to 2 pi; the incremental x coordinates pi/4 and 3 pi / 4 are labeled for the first complete cycle.......!!!!!!!!...................................
RESPONSE --> I don't know how I misunderstood this..
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23:18:56 `q007. Now suppose that x = pi/12, 2 pi/12, 3 pi/12, 4 pi/12, etc.. Give the reduced form of each of these x values. Given x = pi/12, pi/6, pi/4, pi/3, 5 pi/6, ... what are the corresponding values of y = sin(2x)?
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RESPONSE --> Sir, In the last question I did not look at the referenced graphs.However, when I look at your table for values of x you are referencing (x) for theta, so I see that I worked in reverse by setting theta = 2x and then solving for x.. I feel the rest of my procedure was ok, but even so, while describing the graph to you, I new by looking at the function y=sin (2x) that the omega had merely doubled and I never followed up by asking myself the question, if the omega of the function has only changed and no shifting has occured, then why was my sketch so funky.. If I think of x as clock time, which I should have done, then It would have been apparent that the function was travelling twice as fast and still kept the sine wave..
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