course MTH174 ?? ???W€??????assignment #003003. The Sine Function
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09:30:02 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.
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RESPONSE --> self critique assessment: 3
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09:30:11 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.
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RESPONSE --> self critique assessment:
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09:34:48 `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.
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RESPONSE --> 0, = 0 pi/4, = 0.71 pi/2, = 1 3 pi/4, = 0.71 pi, = 0 5 pi/4, = -0.71 3 pi/2, = -1 7 pi/4, = -0.71 2 pi, = 0 confidence assessment: 3
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09:35:22 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.
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RESPONSE --> correct self critique assessment: 3
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09:40:12 `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, pi/6, = 0 pi/3, = 0.87 pi/2, = 1 2 pi/3, = 0.87 5 pi/6, = 0.5 pi, = 0 7 pi/6, = -0.5 4 pi/3, = -0.87 3 pi/2, = 1 5 pi/3, = -0.87 11 pi/6 = -0.5 2 pi, = 0 confidence assessment: 3
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09:41:35 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.
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RESPONSE --> correct self critique assessment: 3
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23:17:12 `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.
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RESPONSE --> 0, =0 pi/4, 0.71 pi/2, = 1 3 pi/4, 0.71 pi, = 0 5 pi/4, = -0.71 3 pi/2, = -1 7 pi/4, = -0.71 2 pi =0 my graph resembles a sine wave confidence assessment: 3
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23:17:40 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).
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RESPONSE --> self critique assessment: 3
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23:19:40 `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 --> because the y coodnate rises, the graph increases at a decreasing rate because the cirle levels off at the top confidence assessment: 3
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23:20:02 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.
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RESPONSE --> self critique assessment: 3
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