course Mth 164 I am looking for some help on the proofs. I understand the procedure completely and can understand it if the complete process is infront of me but I have a problem with the different variations and figuring out which one will work. ?Y???????????Q???assignment #005005. The graph keeps going and going. Thru what coordinates on the t axis does it pass etc
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19:09:43 Goals for this Assignment include but are not limited to the following: 1. Construct a table of the values of y = A sin(x) for a given value of A, construct a graph of a single cycle of this function, then extending the graph forward and/or backward for any specified number of complete cycles. 2. Given a function y = A sin(theta) with theta given as a function of x, construct a y vs. x graph of a single cycle of this function, then extending the graph forward and/or backward for any specified number of complete cycles. 3. Interpret the function and graph corresponding to Goal 2 in terms of angular motion on a circle of appropriate radius. Click once more on Next Question/Answer for a note on Previous Assignments.
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RESPONSE --> self critique assessment:
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19:09:49 Previous Assignments: Be sure you have completed Assignment 3 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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19:16:08 `q001. If the red ant keeps traveling around the circle (e.g., assume an immortal though mindless red ant that has been traveling around the circle since before time began and will continue until time ends, whatever that means, traveling continually without pause or change in speed) then its y coordinate will go from 0 to 1 to 0 to -1 to 0, then to 1 to 0 to -1 to 0 again and again and again, endlessly repeating the cycle. A graph of y vs. theta will therefore alternates peaks at 1 and valleys at -1. By how much does theta change from one peak of the graph to the next?
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RESPONSE --> The amount of change in theta from one peak of the graph to the next would be dependent on how fast this red ant was moving I would think. If the ant was moving at the standard speed of a standard sine graph then theta would change by 2pi each cycle. confidence assessment: 1
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19:17:07 The peak of the graph corresponds to the angular position pi/2, at the 'top' of the circle. In order for the graph to get from one peak to the next, the ant must travel from the top of the circle all the way around until it reaches the top once more. The angular displacement corresponding to a circuit around the circle is 2 pi. The angle theta on the graph therefore changes by 2 pi between peaks.
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RESPONSE --> I was not sure of the speed, which would be pi/2 self critique assessment: 3
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19:27:13 `q002. The graph of the basic sine function is centered on the theta axis, running in both the positive and negative directions, never ending, always repeating with a period of 2 pi. The graph represents never-ending motion around the unit circle with angular velocity 1, extending back forever, extending for forever. That is, extending forever into the past and forever into the future. How many complete cycles of the sine function will be completed between theta = -100 and theta = 100?
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RESPONSE --> My first initiative would be to find the total number of ""spaces"" between -100 and 100, this would be 200 I assume. Since this number is not in terms of pi I think to divide 200 by pi to get the number of pis so to speak. Like dividing 10 by 5 to get 2 (the number of fives in 2) 200/pi = 63.66198 number of pis in 200. If a cycle is completed every 2 pi, then divide this number by 2, to get 31.83099 which tells me that there are 31 completed cycles of 2pi between theta = -100 and theta = 100. confidence assessment: 1
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19:28:08 Between theta = -100 and theta = +100 the change is 100 - (-100) = 200. Since the complete cycle is completed as theta changes by 2 pi, the graph will complete 200 / (2 pi) = 100 / pi = 31.8 cycles, approximately. This corresponds to 31 complete cycles. However there is be another way of answering this question. If a complete cycle is defined as the cycle from y values 0 to 1 to 0 to -1 and back to 0, then between theta = 0 and theta = 100 there are 15.9 cycles, or 15 complete cycles, and moving to the left, between theta = 0 and theta = -100 there are another 15 complete cycles. This totals only 30 complete cycles. The answer to the question therefore depends on just where cycles are considered to begin. See Figure 25. If you count 'valleys' from the far left to the far right you find that there are 32 'valleys' so that the function completes 31 cycles between theta = -100 and theta = 100. However if you count complete cycles from the origin moving to the right you find only 15, and the same number from the origin moving to the left, so in that sense there appear to be only 30 cycles.
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RESPONSE --> self critique assessment: 3
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19:33:52 `q003. The graph of y = sin(3 t) represents the y coordinate of motion around the arc of the unit circle at 3 units per second. This graph of y vs. t can, just as in the preceding problems, continue forever into the past and into the future. What will be the distance between the peaks of this graph?
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RESPONSE --> The distance between the peaks of the graph y = sin (3t) will be 2pi/3, the peaks will occur three times as quickly as the original sine graph so the peaks will be 2pi/3 distanced. confidence assessment: 2
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19:34:24 The distance between peaks corresponds to the time required for the ant to complete a cycle around the circle. The ant moves around the unit circle, which has circumference 2 pi. At 3 units per second the time required is 2 pi / 3 seconds. The peaks of the graph will therefore be separated by 2 pi / 3 units along the t axis.
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RESPONSE --> self critique assessment: 3
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19:35:15 `q004. The graphs of y = sin(3 t) and y = sin(t) both go through the origin and both with positive slope. Which graph goes through the origin with a greater slope?
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RESPONSE --> The graph of y = sin (3t) goes through the origin with a greater slope. confidence assessment: 6
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19:35:32 As shown in Figure 41, the graph of y = sin(3t) has the same y values but the horizontal distance between peaks is 1/3 that of the y = sin(t) graph. This compression of the graph triples the slopes, so the graph of y = sin(3t) has three times the slope at corresponding points compared to the graph of y = sin(t). The graphs do correspond at the origin, so the slope of the y = sin(3t) is three times as great at the origin as the slope of the graph of y = sin(t).
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RESPONSE --> self critique assessment: 3
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19:41:08 `q005. Through what coordinates on the t axis does the graph of y = sin(t - pi/3) pass? Through what points on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin?
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RESPONSE --> The graph of y = sin (t-pi/3) goes through the t axis at pi/3, 4pi/3, and 7pi/3 and does not pass through the origin. confidence assessment: 2
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19:42:29 The graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at tthe circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t - pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we see that t = theta + pi/3 takes values t = 0 + pi/3, pi + pi/3, 2 pi + pi/3, etc.. These values simplify to give us t = pi/3, 4 pi/3, 7 pi/3, ... . A graph is shown in Figure 91. Note that only the values t = pi/3 and t = 4 pi/3 lie between 0 and 2 pi.
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RESPONSE --> I said 7pi/3 because it was part of the complete cycle, I overlooked that the directions were between 0 and 2pi on the t axis. self critique assessment: 3
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19:47:08 `q006. Through what coordinates on the t axis does the graph of y = sin(t + pi/3) pass? Through what point on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin?
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RESPONSE --> The graph of y = sin ( t+pi/3) crosses through the t axis at 2pi/3 and 5pi/3 in the interval 0<=t<=2pi and it does not pass through the origin confidence assessment: 3
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19:47:32 The graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t + pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we see that t = theta - pi/3 takes values t = 0 - pi/3, pi - pi/3, 2 pi - pi/3, etc.. These values simplify to give us t = -pi/3, 2 pi/3, 5 pi/3, 8 pi/3, ... . The graph is shown in Figure 91. Only the values t = 2 pi/3 and t = 5 pi/3 lie between 0 and 2 pi.
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RESPONSE --> self critique assessment: 3
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19:49:52 `q007. What are the first four t coordinates at which the graph of y = sin(t - pi/3) passes through the t axis, for t > 0?
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RESPONSE --> The first four t coordinates at which the graph of y = sin(t-pi/3) crosses through the t axis for t > 0 are pi/3, 4pi/3, 7pi/3 and 10pi/3. confidence assessment: 3
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19:50:41 The graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t - pi/3 takes values 0, pi, 2 pi, 3 pi, 4 pi, ..., we see that t = theta + pi/3 takes values t = 0 + pi/3, pi + pi/3, 2 pi + pi/3, 0 + 4 pi / 3. These values simplify to give us t = pi/3, 4 pi/3, 7 pi/3, 10 pi/3, ... . These are the first four positive values of t for which the graph passes through the t axis.
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RESPONSE --> self critique assessment: 3
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19:53:12 `q008. What are the first four positive t coordinates through which the graph of y = sin(t + pi/3) passes through the t axis, for t > 0?
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RESPONSE --> The first four positive t coordinates through which the graph of y = sin (t+pi/3) passes through the t axis for t> 0 are 2pi/3, 5pi/3, 8pi/3 and 11pi/3. confidence assessment: 3
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19:58:40 The graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t + pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we see that t = theta - pi/3 takes values t = 0 - pi/3, pi - pi/3, 2 pi - pi/3, 3 pi - pi/3, 4 pi - pi/3, etc.. These values simplify to give us t = -pi/3, 2 pi/3, 5 pi/3, 8 pi/3, 11 pi/3, ... . The first four positive values of t for which the graph passes through the t axis are 2 pi/3, 5 pi/3, 8 pi/3 and 11 pi/3. The corresponding graph is shown in Figure 9.
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RESPONSE --> self critique assessment: 3
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20:02:22 `q009. Through what coordinates on the t axis does the graph of y = sin(3t + pi/3) pass? Through what point on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin?
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RESPONSE --> The graph of y = sin ( 3t + pi/3 ) passes through 0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, and 2pi between the interval 0<=t<=2pi and it does cross through the origin. confidence assessment: 2
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20:07:43 The graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(3t + pi/3) then when theta = 3t + pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we solve for t in order to see that t = theta/3 - pi/9 takes values t = 0/3 - pi/9, pi/3 - pi/9, 2 pi/3 - pi/9, 3 pi/3 - pi/9, 4 pi/3 - pi/9, ..., n pi/3 - pi/9, ..., where n = 0, 1, 2, 3, ... . These values simplify to give us t = -pi/9, 2 pi/9, 5 pi/9, 8 pi/9, 11 pi/9, ... . If we list these values starting with the first positive value 2 pi/9, continuing as long as t < 2 pi, we obtain values 2 pi/9, 5 pi/9, 8 pi/9, 11 pi/9, 14 pi/9, 17 pi/9. Since 2 pi = 18 pi/9 any subsequent solution will be greater than 2 pi so is not included. The corresponding graph is shown if Fig 66.
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RESPONSE --> I just was not thinking on this one. The graph's frequency is increased by three and then shifted to the left pi/3 or t = theta/3 - pi/9. self critique assessment: 2
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20:07:53 Complete Assignment 4, including Class Notes, text problems and Web-based problems as specified on the Assts page. When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_.
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RESPONSE --> self critique assessment:
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???????????? assignment #007 007. Tangent Function Precalculus II 02-12-2007
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20:09:15 Goals for this Assignment include but are not limited to the following: 1. Understand and be able to explain how the vertical asymptotes of the graph of the tangent function occur. 2. Using exact values construct y vs. x graphs of y = A sin(theta) or y = A cos(theta), where theta is given as a function of x. Click once more on Next Question/Answer for a note on Previous Assignments.
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RESPONSE --> self critique assessment:
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20:09:20 Previous Assignments: Be sure you have completed Assignment 5 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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20:13:51 `q001. The tangent function is defined in terms of the unit circle (a circle of radius 1 centered at the origin): For an angular position theta the tangent of theta is the ratio y / x of the y coordinate to the x coordinate at the corresponding point on the circle. What are the values of the tangent(theta) for theta = 0, pi/6, pi/4, and pi/3? Sketch a graph of tan(theta) vs. theta for 0 <= theta <= pi/3. Are the slopes of the graph increasing or decreasing.
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RESPONSE --> The slope of the graph of tan (theta) seems to be increasing at an increasing rate. confidence assessment: 2
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20:14:17 The unit-circle points corresponding to the given angles are (1,0), (sqrt(3)/2, 1/2), (sqrt(2)/2, sqrt(2)/2) and (1/2, sqrt(3)/2). So the values of the tangent, each calculated as y / x, are as follows: tan(0) = 0/1 = 0, tan(pi/6) = 1/2 / (sqrt(3)/2) = 1/2 * 2/sqrt(3) = 1/sqrt(3) = sqrt(3)/3, tan(pi/4) = sqrt(2)/2 / (sqrt(2)/2) = 1 and tan(pi/3) = sqrt(3)/2 / (1/2) = sqrt(3). The corresponding graph is shown in Figure 73. The slopes of the graph between 0 and pi/6, between pi/6 and pi/4, and between pi/4 and pi/3 are, respectively: (sqrt(3)/3 - 0)/ (pi/6 - 0) = 6 sqrt(3) / pi = 1.10, (1 - sqrt(3)/3)/(pi/4-pi/6) = 1.62 and (sqrt(3) - 1)/(pi/3 - pi/4) = 2.80. The slopes are increasing, slowly at first, then more quickly.
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RESPONSE --> self critique assessment: 3
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20:26:52 `q002. Figure 52 indicates the unit circle positions corresponding to the angles which are multiples of pi/18. The grid shows intervals of .5. Estimate the x coordinates of the first-quadrant points then make a table of tangent (theta) vs. theta for the data from 0 to 8 pi/18. Give your values and use them to extend the graph of tan(theta) through theta = 8 pi/18.
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RESPONSE --> My x coordinate estimates are as follows 4.9, 4.7, 4.4, 3.8, 3.2, 2.5, 2.2 and 0.8 respectively for pi/18, pi/8, pi/6, 2pi/9, 5pi/18, pi/3, 7pi/18, and 4pi/9. This results in tangent found from theta/x will be about 0.0356, 0.0743, 0.1190, 0.1837, 0.2727, 0.4189, 0.5553, 1.7453 respectively. confidence assessment: 1
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20:30:01 The angles 3 pi/18 = pi/6 and 6 pi/18 = pi/3 are known to be approximately sqrt(3)/3 = .58 and sqrt(3) = 1.73, respectively. The values at the remaining points can be estimated more or less accurately. The actual values, to 2 significant figures, strarting with angle pi/18, are .18, .36, .58, .84, 1.19, 1.7, 2.7 and 5.7. The rapidly increasing slopes as theta exceeds pi/3 are apparent from these numbers. The extended graph shown in Figure 97 depicts only the results for angles through 7 pi/18; the figure would have to be twice as high to include the point (8 pi/18, 5.7), or in reduced for (4 pi/9, 5.6). The colored line segments just below the x axis indicate the multiples of pi/18.
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RESPONSE --> I seem to have been mistaken on the increments of the graph. self critique assessment: 2
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20:32:53 `q003. Sketch a series of points on the unit circle approaching the pi/2 position. As we approach closer and closer to pi/2, what happens to the y coordinate? What happens to the x coordinate? Does the y coordinate approach a limiting value? Does the x coordinate approach a limiting value?
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RESPONSE --> Approaching pi/2, the y coordinate grows closer and closer together, there is less of a difference between two of the coordinates but rising closer and closer to 1 or the radius of the circle. The x coordinates grow farther apart from the previous but reach 0. The y coordinate approaches the limiting value of 1, the radius of the circle, as does x with the same value. confidence assessment: 2
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20:33:20 Figure 48 shows a series of points on the unit circle approaching the pi/2 position. It should be clear that the y coordinate approaches 1 and the x coordinate approaches 0.
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RESPONSE --> self critique assessment: 3
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20:37:41 `q004. As the angle approaches pi/2 from the first quadrant the y coordinate approaches 1 and the x coordinate approaches 0, as we saw in the last problem. What happens to the ratio 1/x as x take values 0.1, 0.01, 0.001, 0.0001? What happens as x continues to approach 0? Is there a limit to how large 1/x can get?
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RESPONSE --> The ratio 1/x grows larger the smaller the value of x is. As x continues to approach 0, the ratio continues to grow larger at an increasing rate. There is no limit that is evident for the large values of 1/x. confidence assessment: 2
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20:37:51 If x = 0.1, 0.01, 0.001 and 0.0001 we see that 1/x = 10, 100, 1000 and 10,000. There is no limit to how large 1/x can get; we can make it as large as we wish by choosing x small enough.
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RESPONSE --> self critique assessment: 3
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20:42:49 `q005. What happens to the tangent of theta as theta approaches pi/2?
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RESPONSE --> The tangent of theta increases at an increasing rate as theta approaches pi/2. tan 0 = 0 tan pi/6 = 0.5774 tan pi/2 = 1 tan pi/3 = 1.7321 tan pi/2 = undefined. confidence assessment: 2
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20:43:11 Since the x coordinate approaches zero and the y coordinate approaches 1, it follows that tan(theta) = y / x gets larger and larger, without bound. We say that this quantity approaches infinity.
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RESPONSE --> self critique assessment: 3
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20:44:54 `q006. As we have seen the value of tangent (theta) exceeds all bounds as theta approaches pi/2 from within the first quadrant. How do we extend the graph of tangent (theta) vs. theta to cover the domain 0 <= theta <= pi/2?
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RESPONSE --> There is going to have to be a asymptote at multiples of pi/2. confidence assessment: 1
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20:45:24 Preceding solutions and graphs showed how the slopes of the graph increased as we moved from theta = 0 to theta = 8 pi / 18 (which reduces to 4 pi / 9, just a little less than pi/2). The slopes will continue to increase, and the ratio y / x will increase without bound so that the values of the function increase without bound as theta approaches pi / 2. The graph therefore forms a vertical asymptote at theta = pi/2.
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RESPONSE --> vertical asymptotes more specifically. self critique assessment: 3
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20:53:27 `q007. What are the values of tan(theta) as theta changes from -8 pi/18 to 0 in increments of pi/18? Use your estimates of the coordinates of the first-quadrant points as a basis for your estimates of the values of the tangent function. Extend your graph of the function so that it is graphed for -8 pi / 18 <= theta <= 8 pi / 18, then show what happens as theta approaches -pi/2.
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RESPONSE --> the tangent of theta takes values of -5.67, -2.75, -1.73, -1.19, -0.84, -0.58, -0.36, -0.18, 0 for the theta values from -8pi/18 to 0 in increments of pi/18. As theta approaches -pi/2, the tangent values increase and increase but never reach the value of -pi/2 because it is a vertical asymoptote. confidence assessment: 2
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20:53:45 Figure 71 shows the multiples of pi/18 from -8 pi/18 to 0. It is clear that for all values of theta the x coordinates are positive and for all values of theta except 0 the y coordinates negative, which make the values of tan(theta) = y / x negative. Otherwise the values are the same as those obtained for theta = 0 to 8 pi/18. The actual values, to 2 significant figures, strarting with angle -pi/18, are -.18, -.36, -.58, -.84, -1.19, -1.7, -2.7 and -5.7. The graph is shown in Figure 8. Note how the graph now approaches the vertical line theta = -pi/2 as an asymptote. The values of y = tan(theta) for -pi/2 < theta < pi/2 combines the graph just obtained with the graph obtained in the preceding exercise, and is shown in Figure 97. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |.
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RESPONSE --> self critique assessment: 3
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20:55:11 `q008. What happens to the value of the tangent function as the angle approaches pi/2 through second-quadrant angles?
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RESPONSE --> The value of the tangent function again increase seemingly to infinity and pi/2 is a vertical asymptote with using second-quadrant angles. confidence assessment: 2
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20:56:23 In the second quadrant, as we approach theta = pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches +1. Thus tan(theta) = y / x will be negative, with the magnitude of the quantity approaching infinity. We say that tan(theta) approaches -infinity. The graph will be asymptotic to the negative portion of the line theta = pi/2, as indicated in Figure 69, which combines this portion of the graph with the existing graph of the full cycle of the tangent function. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |.
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RESPONSE --> self critique assessment: 3
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20:56:59 `q009. What happens to the value of the tangent function as the angle approaches 3 pi/2 through third-quadrant angles?
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RESPONSE --> Again, it approaches infinity seemingly and never touches 3pi/2 as it is a vertical asymptote. confidence assessment:
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20:57:15 In the third quadrant, as we approach theta = 3 pi/2 on the circle the x coordinates are negative as the approach 0 while the y value approaches -1. Thus tan(theta) = y / x will be positive, with the magnitude of the quantity approaching infinity. We see that tan(theta) approaches +infinity. The graph will be asymptotic to the positive portion of the line theta = 3 pi/2, as in Figure 30. Note that the graph as shown is terminated at y values exceeding | tan( 7 pi / 18) |.
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RESPONSE --> self critique assessment: 3
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20:59:07 `q010. Sketch a graph of the tangent function from theta = -pi/2 through theta = 3 pi / 2.
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RESPONSE --> There is a vertical asymptote at -pi/2, pi/2, and 3pi/2. The graph of tangent crosses the x axis at the origin and pi. To the left of the x axis intersection the graphs decrease at an increasing rate to the left and tothe right of the x axis intersection, the graphs increase at an increasing rate to the right. confidence assessment: 3
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20:59:27 The graph is shown in Figure 30. Note how the tangent function goes through its complete cycle of values, from -infinity to +infinity, between theta = -pi/2 and theta = pi/2, then again between theta = pi/2 and theta = 3 pi/2. The horizontal 'distance' corresponding to a complete cycle is thus seen to be pi.
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RESPONSE --> self critique assessment: 3
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21:00:59 `q011. Sketch a graph of y = tan ( 2 x ), showing two complete cycles. What is the length of a cycle?
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RESPONSE --> The length of a cycle is shortened down to pi/2 as opposed to pi. confidence assessment: 3
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21:01:13 We begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x and relabel the graph as indicated in Figure 79, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x <= 3 pi/2 so that we have -pi/4 <= x <= 3 pi/4. This shows us that x goes through a complete cycle between -pi/4 and pi/4 and again between pi/4 and 3 pi/4. Each cycle has length pi/2. Note how the period of the function tan(2x) is 1/2 the period of the tan(x) function, which is pi.
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RESPONSE --> self critique assessment: 3
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21:45:58 `q011. Sketch a graph of y = tan ( 2 x + pi/3 ), showing two complete cycles. What is the length of a cycle?
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RESPONSE --> The graph of y = tan (2x+pi/3) has vertial asymptotes of -pi/12 and 5pi/12. The length of a cycle is pi/2 because of the 2x. The graph passes through the x axis at pi/6. confidence assessment: 2
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21:47:58 We begin by sketching a graph of y = tan(theta) for -pi/2 <= theta <= 3 pi/2, as shown in Figure 30. We then let theta = 2x + pi/3 and relabel the graph as indicated in Figure 62, with pi/2 <= theta <= 3 pi/2 replaced by -pi/2 <= 2x + pi/3 <= 3 pi/2 so that we have -5 pi/12 <= x <= 7 pi/12. This graph shows us that x goes through a complete cycle between -5 pi/12 and pi/12 and again between pi/12 and 7 pi/12. Each cycle has length pi/2. Note how the period of the function tan(2x + pi/3) is 1/2 the period of the tan(x) function, which is pi.
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RESPONSE --> I did the graph in reverse, I thought that the (2x+pi/3) the pi/3 part meant that you horizonally shifted the graph to the left pi/3. self critique assessment: 2
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21:48:16 Complete Assignments 6, 7, including Class Notes, text problems and Web-based problems as specified on the Assts page. When you have completed the entire assignment run the Query program. Submit SEND files from Query and q_a_. Note that Assignment 7 consists of a test covering Assignments 1-5.
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RESPONSE --> self critique assessment:
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???????????assignment #004 004. Precalculus II 02-12-2007
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21:50:04 query problem 5.6.54 3 cos(2x+`pi) find characteristics and graph using transformations
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RESPONSE --> confidence assessment:
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21:52:32 explain how you use transformations to construct the graph.
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RESPONSE --> The graph I did took the basic cosine graph, added a vertical stretch of factor three, reduced the cycle to pi and added a horizontal shift of pi. confidence assessment: 3
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21:57:20 Describe the resulting graph by given its period, its the maximum and minimum y values and its phase shift, and describe how its phase shift affects the graph
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RESPONSE --> The period is pi. The maximum y value is 3 and the minimum is -3. The phase shift is pi. The phase shift affects the graph by changing the y intercept to -3 instead of what would without the shift be 3. confidence assessment: 3
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21:57:39 query problem 5.6.60 2 cos(2`pix-4) find characteristics and graph using transformations
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RESPONSE --> confidence assessment:
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21:58:43 explain how you use transformations to construct the graph.
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RESPONSE --> The amplitude was changed to 2 and the period of the graph was changed to 1 with a phase shift of four to the right. confidence assessment: 3
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21:59:56 Describe the resulting graph by given its period, its the maximum and minimum y values and its phase shift, and describe how its phase shift affects the graph
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RESPONSE --> The period is 1. The maximum y value is 2 and the minimum is -2. The phase shift is four which does not really affect the placement of the graph, the y intercept is still 2 as the normal cosine (2x) would be. confidence assessment: 3
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22:02:48 query problem 6.1.24 1 - sin^2 x /( 1-cos x) = -cos x
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RESPONSE --> confidence assessment:
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22:07:31 give the steps in your solution
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RESPONSE --> 1-sin^2x/1-cosx = -cos x 1-sin^2x/1-cosx = 1 - 1-cos^2/1-cos x, use pythagorean identity = -cos^2x/1-cos x, ones subtract = -cosx/1, -cos x cancels out = -cos x, simplify. confidence assessment: 3
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22:20:22 query problem 6.1.48 sec x / (1 + sec x) = (1-cos x) / sin^2 x
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RESPONSE --> confidence assessment:
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22:20:50 give the steps in your solution
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RESPONSE --> This one stumped me. I tried solving both sides about 20 different ways and did not come up with anything. confidence assessment: 0
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22:21:48 Comm on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The proofs proved to be quite difficult. I believe practice is the only answer, I understand the concept and the procedure, just at times it is difficult to get from point a to point b. confidence assessment:
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