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17:37:23 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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17:37:30
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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17:42:22 `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 haven't seen the graph but i know what you want... Assuming its the unit circe with r=1 then the following values are : 0, sqrt2/2 , 1, sqrt2/2, 0 , -Sqrt2/2 , -1 , -sqrt2/2 and 0 all respectively...
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17:43:19 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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17:47:09 `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 --> Again assuming unit circle with r =1, the folowing coordinate for y is : 0 , 1/2 , sqrt3/2 , 1 , sqrt3/2 , 1/2 , 0 , - 1/2 , -sqrt3/2 , -1 , -sqrt3/2 , -1/2 , 0 all respective to the list opposite...
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17:47:30 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/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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18:15:47 `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 --> Table y Vs Theta (0,0) (pi/4, sqrt2/2) (Pi/2, 1) (3pi/4 , sqrt2/2) (pi , 0) (5pi/4 , -sqrt2/2) (3pi/2 , -1) (7pi/4 , -sqrt2/2)
The graph should be that of the Sine function...................................................
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18:16:44 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 --> Nice language used there !!
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18:25:02 `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 --> The motion of the graph is determined by the angular speed "omega". As the point starts off from point (0,0) it is travelling in an anti-clockwise direction on the unit circle in an ascending fashion. I' m not sure I understand the statement at a decreasing rate, since surely the omega is constant, but I would agree that as it approaches the point (pi/2, 1) it will appear to decrease prior to the apex or maximum. Then descent at an increasing rate.
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18:34:28 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 --> I understand your explaination and it is excellent. I can see why it moves at smaller and smaller increaments of y as it approaches the y=1 coordinate. The term decreasing rate while not unfamiliar with, I got mixed up with the omega of the function. Whether it moves faster or slower the graph will always act in the same manner and I should not have got these mixed up.. Sorry !! With the explanation given one should not have to draw out to see what is happening..
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18:44:51 `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 I was visualising these in my head maybe I should draw them out next time...
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18:45:43 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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18:49:58 `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 --> If I understand correctly, you are asking one to put the values pi/6, pi/3 etc in for the x coordinate which should not be done.
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18:55:24 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 --> Well, as you can see, I totally misunderstood your question, but understand what you are requiring I have made note of the figures and will look them up later...
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19:20:47 `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 --> x y=sin (2*x) pi/6 1/2 pi/4 1 pi/3 1/2 pi/2 0 2pi/3 1/2 3pi/4 -1 5pi/6 0 pi 0 7pi/6 1/2 5pi/4 -1 4pi/3 0 3pi/2 0 5pi/3 1/2 7pi/4 1 11pi/4 1
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19:31:18 pi / 12 doesn't reduce. 2 pi/12 reduces to pi/6. 3 pi/12 reduces to pi/4. 4 pi/12 reduces to pi/3. 5 pi/12 doesn't reduce. 6 pi/12 reduces to pi/2. 7 pi/12 doesn't reduce 8 pi/12 reduces to 2 pi/3 9 pi/12 reduces to 3 pi/4 10 pi/12 reduces to 5 pi/6 11 pi/12 doesn't reduce 12 pi/12 reduces to pi
Doubling these values and taking the sines we obtain the following table: x 2x sin(2x) 0 0 0.0 pi / 12 pi/6 0.5 pi/6 pi/3 0.87 pi/4 pi/2 1.0 pi/3 2 pi/3 0.87 5 pi/12 5 pi/6 0.5 pi/2 pi 0.0 7 pi/12 7 pi/6 -0.5 2 pi/3 4 pi/3 -0.87 3 pi/4 3 pi/2 -1.0 5 pi/6 5 pi/3 -0.87 11 pi/12 11 pi/6 -0.5 pi/2 pi -0.0......!!!!!!!!...................................
RESPONSE --> Ok, What I thought you wanted me to do to was superimpose the uiit circle broken into eight on top of the unit circle broken into twelve sectors .. I'm sure you noticed this from my break down above.. I also may have got confused and given you a mixture of cosine with sine . This is what I get for not taking my time and trying to do it in my head and not fully on paper..."Very sorry"... But I see the point!!
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19:33:43 `q008. Given the table of values obtained in the preceding problem, sketch a graph of y vs. x. Describe your graph. By how much does x change as the function sin(2x) goes through its complete cycle, and how does this compare with a graph of y = sin(x)?
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RESPONSE --> Since I did such a royal job bollixing up the last problem, I 'll pass here and wait to print of a copy..
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19:33:50 Your graph should pass through the origin (0,0) with a positive slope. It will have a peak at x = pi/4, pass back through the x-axis at x = pi/2, reach a minimum at x = 3 pi/4 and return to the x-axis at x = pi.
More detail: The graph of y vs. x passes through the origin (0,0) with a positive slope and increases at a decreasing rate until it reaches a maximum value at the x = pi/4 point (pi/4,1). The graph is horizontal for an instant and then begins decreasing at an increasing rate, again reaching the x-axis at the x = pi/2 point (pi/2,0). The graph continues decreasing, but now at a decreasing rate until a reaches its minimum value at the x = 3 pi/4 point (3 pi/4,0). The graph is horizontal for an instant then begins increasing an increasing rate, finally reaching the point (pi, 0). The graph goes through its complete cycle as x goes from 0 to pi. A graph of y = sin(x), by contrast, would go through a complete cycle as x changes from 0 to 2 pi. We see that placing the 2 in front of x has caused the graph to go through its cycle twice as fast. Note that the values at multiples of the function at 0, pi/4, pi/2, 3 pi/4 and 2 pi are clearly seen on the graph. Note in Figure 3 how the increments of pi/12 are labeled between 0 and pi/4. You should complete the labeling of the remaining points on your sketch.......!!!!!!!!...................................
RESPONSE --> ok
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19:52:24 `q009. Now consider the function y = sin(theta) = sin(3x). What values must x take so that theta = 3x can take the values 0, pi/6, pi/3, pi/2, ... ?
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RESPONSE --> x y=sin(3*x)
0 0 pi/6 1 pi/3 0 pi/2 -1 2pi/3 0 5pi/6 1 pi 0 7pi/6 1 4pi/3 0 3pi/2 1 ( found by 3*(3pi/2)= 9pi/2 ) 5pi/3 0 11pi/6 0 2pi 0.................................................
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19:54:40 If theta = 3x then x = theta / 3. So if theta = 3x takes values 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 2 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi then x takes values 0, 1/3 * pi/6, 1/3 * pi/3, 1/3 * pi/2, 1/3 * 2 pi/3, 1/3 * 5 pi/6, 1/3 * pi, 1/3 * 7 pi/6, 1/3 * 2 pi/3, 1/3 * 3 pi/2, 1/3 * 5 pi/3, 1/3 * 11 pi/6, 1/3 * 2 pi, or 0, pi/18, pi/9, pi/6, 2 pi/9, 5 pi/18, pi/3, 7 pi/18, 4 pi/9, pi/2, 5 pi/9, 11 pi/18, and 2 pi/3.
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RESPONSE --> Ok and another royal job I,ve done.....
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20:15:17 `q010. Make a table consisting of 3 columns, one for x, one for theta and one for sin(theta). Fill in the column for theta with the values 0, pi/6, pi/3, ... which are multiples of pi/6, for 0 <= theta <= 2 pi. Fill in the column under sin(theta) with the corresponding values of the sine function. Now change the heading of the theta column to 'theta = sin(3x)' and the heading of the sin(theta) column to 'sin(theta) = sin(3x)'. Fill in the x column with those values of x which correspond to the second-column values of theta = 3x. The give the first, fifth and seventh rows of your table.
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RESPONSE --> Theta=sin(3x) Sin theta=sin(3x) x Thet Sin theta Ist row = pi/2.........pi/6............. 1/2 5th row = 15pi/6...... 5pi/6................. 1/2 7th row = 21pi/6.... .7pi/6................ .-1/2
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20:20:54 The table originally reads as follows:
x theta sin(theta) 0 0 0.0 0 pi/6 0.5 0 pi/3 0.87 0 pi/2 1.0 0 2 pi/3 0.87 0 5 pi/6 0.5 0 pi 0.0 0 7 pi/6 -0.5 0 4 pi/3 -0.87 0 3 pi/2 -1.0 0 5 pi/3 -0.87 0 11 pi/6 -0.5 0 2 pi -0.0 After inserting the values for x and changing column headings the table is x theta = 3x sin(3x) 0 0 0.0 pi/18 pi/6 0.5 pi/9 pi/3 0.87 pi/6 pi/2 1.0 2 pi/9 2 pi/3 0.87 5 pi/18 5 pi/6 0.5 pi/3 pi 0.0 7 pi/18 7 pi/6 -0.5 4 pi/9 4 pi/3 -0.87 pi/2 3 pi/2 -1.0 5 pi/9 5 pi/3 -0.87 11 pi/18 11 pi/6 -0.5 2 pi/3 2 pi -0.0......!!!!!!!!...................................
RESPONSE --> Sir, On the second chart row # 2 should it not be for (x )= 3*(pi/6) = 3/1*pi/6 = 3pi/6 = pi/2 etc,etc...and if not what am I donig wrong..
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20:21:44 `q011. Sketch the graph corresponding to your table for sin(3x) vs. x. Does the sine function go through a complete cycle? By how much does x change as the sine function goes through its complete cycle?
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RESPONSE --> I'm not sure I'll have to wait untill I finish and print off you numbers...
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20:25:26 Your graph should pass through the origin (0,0) with a positive slope. It will have a peak at x = pi/6, pass back through the x-axis at x = pi/3, reach a minimum at x = pi/2 and return to the x-axis at x = 2 pi/3.
More detail: The graph of y vs. x passes through the origin (0,0) with a positive slope and increases at a decreasing rate until it reaches a maximum value at the x = pi/6 point (pi/6,1). The graph is horizontal for an instant and then begins decreasing at an increasing rate, again reaching the x-axis at the x = pi/3 point (pi/3,0). The graph continues decreasing, but now at a decreasing rate until a reaches its minimum value at the x = pi/2 point (pi/2,0). The graph is horizontal for an instant then begins increasing an increasing rate, finally reaching the point (2 pi/3, 0). The graph goes through its complete cycle as x goes from 0 to 2 pi/3. A graph of y = sin(x), by contrast, would go through a complete cycle as x changes from 0 to 2 pi. We see that placing the 3 in front of x has caused the graph to go through its cycle three times as fast.......!!!!!!!!...................................
RESPONSE --> ok, so you have increased the angular frequency "omega" which tightens the peak to peak distance..
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20:26:40 `q012. For the function y = sin(3x), what inequality in the variable x corresponds to the inequality 0 <= theta <= 2 pi?
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RESPONSE --> I have no idea..Sorry !!
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20:33:09 If theta = 3x then the inequality 0 <= theta <= 2 pi becomes
0 <=3x <= 2 pi. If we multiply through by 1/3 we have 1/3 * 0 <= 1/3 * 3x <= 1/3 * 2 pi, or 0 <= x <= 2 pi/3. In the preceding problem our graph when through a complete cycle between x = 0 and x = 2 pi/3. This precisely correspond to the inequality we just obtained.......!!!!!!!!...................................
RESPONSE --> I'm not sure why your multiplying through by 1/3. I can see theta=3*x which results in x= theta/3 so should I see it as 0/3<=theta/3 <=2pi/3 in other-words put the whole inequality over three...
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20:35:40 `q013. For y = sin(theta) = sin(2x - 2 pi/3), what values must x take so that theta = 2x - pi/3 will take the values 0, pi/6, pi/3, pi/2, ... ?
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RESPONSE --> This is I can not do at present...I do not understand
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20:42:06 If theta = 2x - pi/3 then 2 x = theta + 2 pi/3 and x = theta/2 + pi/6. So if theta = 2x - pi/3 takes values 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 2 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi then x = theta/2 + pi/6 takes values
x values: 0 + pi/6, pi/12 + pi/6, pi/3 + pi/6, pi/4 + pi/6,pi/3 + pi/6, 5 pi/12 + pi/6, pi/2 + pi/6, 7 pi/12 + pi/6,2 pi/6 + pi/6,3 pi/4 + pi/6,5 pi/6 + pi/6,11 pi/12 + pi/6,pi + pi/6, which are added in the usual manner and reduce to added and reduced x values: pi/6, pi/3, 5 pi/12, pi/2, 7 pi/12, 2 pi/3, 3 pi/4, 5 pi/6, 11 pi/12, pi, 13 pi/12, 7 pi/6.......!!!!!!!!...................................
RESPONSE --> Sir, where you say, If theta = 2x - pi/3 then 2 x = theta + 2 pi/3 and x = theta/2 + pi/6. I would have solved Theta=2x-pi/3 2x =Theta + pi/3 x=(Theta+pi/3)/2 I'm kind of lost here..
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20:42:33 `q014. Make a table consisting of 3 columns, one for x, one for theta and one for sin(theta). Fill in the column for theta with the values 0, pi/6, pi/3, ... which are multiples of pi/6, for 0 <= theta <= 2 pi. Fill in the column under sin(theta) with the corresponding values of the sine function. Now change the heading of the theta column to 'theta = sin(2x - pi/3)' and the heading of the sin(theta) column to 'sin(theta) = sin(2x - pi/3)'. Fill in the x column with those values of x which correspond to the second-column values of theta = 2x - pi/3. Give the first, fifth and seventh rows of your table.
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RESPONSE --> I'm going to just click on here..
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20:42:36 Our first table is the same as before, as it will always be:
x theta sin(theta) 0 0.0 pi/6 0.5 pi/3 0.87 pi/2 1.0 2 pi/3 0.87 5 pi/6 0.5 pi 0.0 7 pi/6 -0.5 4 pi/3 -0.87 3 pi/2 -1.0 5 pi/3 -0.87 11 pi/6 -0.5 2 pi -0.0 Our second table is as follows: x theta = 2x - pi/3 sin(2x-pi/3) pi/6 0 0.0 3 pi/12 pi/6 0.5 pi/3 pi/3 0.87 5 pi/12 pi/2 1.0 pi/2 2 pi/3 0.87 7 pi/12 5 pi/6 0.5 2 pi/3 pi 0.0 3 pi/4 7 pi/6 -0.5 5 pi/6 4 pi/3 -0.87 11 pi/12 3 pi/2 -1.0 pi 5 pi/3 -0.87 13 pi/12 11 pi/6 -0.5 7 pi/6 2 pi -0.0 The second table indicates that the function y = sin(2x - pi/3) goes through a complete cycle for x values running from pi/3 to 5 pi/3, with y running from 0 to 1 to 0 to -1 to 0.......!!!!!!!!...................................
RESPONSE -->
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20:42:39 `q015. Sketch the graph corresponding to your table for sin(2x - pi/3) vs. x. Does the sine function go through a complete cycle? By how much does x change as the sine function goes through its complete cycle?
For the function y = sin(2x - pi/3), what inequality in the variable x corresponds to the inequality 0 <= theta <= 2 pi?......!!!!!!!!...................................
RESPONSE -->
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20:42:46 If theta = 2x - pi/3 then the inequality 0 <= theta <= 2 pi becomes
0 <=2x - pi/3 <= 2 pi. If we add pi/3 to both sides we get pi/3 <= 2x <= 2 pi + pi/3. If we then multiply through by 1/2 we have 1/2 * pi/3 <= 1/2 * 2x <= 1/2 * 2 pi + 1/2 * pi/3, or pi/6 <= x <= 7 pi/6. In the preceding problem our graph when through a complete cycle between x = pi/6 and x = 7 pi/6. This precisely corresponds to the inequality we just obtained. A graph of y = sin(2x - pi/3) vs. x is shown in Figure 43. This graph goes through its cycle in an x 'distance' of pi, between pi/6 and 7 pi/6. In this it is similar to the graph of y = sin(2x), which also requires an x 'distance' of pi. It differs in that the graph is 'shifted' pi/6 units to the right of that graph.......!!!!!!!!...................................
RESPONSE -->
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20:48:56
Complete Assignment 2 Includes Class Notes #3 (Class Notes are accessed under the Lectures button at the top of the page and are included on the CDs starting with CD #1). Text Section 5.2 and Section 5.3, 'Blue' Problems (i.e., problems whose numbers are highlighted in blue) and odd multiples of 3 in text and the Web version of Ch 5 Problems Section 5.2 and 5.3 (use the link in the Assts page to access the problems). When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.......!!!!!!!!...................................
RESPONSE --> Sir, I failed miserably on this qa.. I was able to finish all of section 5.2 and 5.3 questions and enjoyed them.. I am not sure how I should feel about this performance..I do realise that the qa is a learning tool but this was intimadating, partly because I was misunderstanding the question.. I will await you comments, print off and study them hard.. Regards Slan...
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