#$&* course Mth 164 12:16 pm2/9/15
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Given Solution: `aThe peak of the graph corresponds to the angular position pi/2, at the 'top' of the circle. In order for the graph to get from one peak to the next, the ant must travel from the top of the circle all the way around until it reaches the top once more. The angular displacement corresponding to a circuit around the circle is 2 pi. The angle theta on the graph therefore changes by 2 pi between peaks. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I misunderstood the question but I see now that the intervals are complete cycles and they are in intervals of 2 pi ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q002. The graph of the basic sine function is centered on the theta axis, running in both the positive and negative directions, never ending, always repeating with a period of 2 pi. The graph represents never-ending motion around the unit circle with angular velocity 1, extending back forever, extending for forever. That is, extending forever into the past and forever into the future. How many complete cycles of the sine function will be completed between theta = -100 and theta = 100? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: IF similar to the previous question and is moving at theta intervals of 2 pi then I believe I would find this by dividing the total theta intervals by a single interval of 2pi from -100 to 100 there is a difference of 200 which divided by 2 pi becomes 100/pi. I’m not sure that this would be the method to find this solution though, the answer doesn’t seem correct to me. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aBetween 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was surprised to find that my solution was correct (although missing the simplifying to an approximate amount). I was satisfied to see that I understood the concepts behind the problem more than I gave myself credit for. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe this would the total 2 pi divided by 3 since the complete cycle of 2 pi is completed in 3 units per second. So 2 pi / 3 between peaks. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance between peaks corresponds to the time required for the ant to complete a cycle around the circle. The ant moves around the unit circle, which has circumference 2 pi. At 3 units per second the time required is 2 pi / 3 seconds. The peaks of the graph will therefore be separated by 2 pi / 3 units along the t axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. The graphs of y = sin(3 t) and y = sin(t) both go through the origin and both with positive slope. Which graph goes through the origin with a greater slope? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe the original graph would have the greater slope based on the idea that the graph y = sin(3 t) completes a cycle in less time compared to the original graph. This would cause a shorter distance between peaks and other points and therefore I would think there would be a steeper incline for the graph. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAs 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Sometimes I feel like I have trouble with the idea of understanding more or less slope. I think if I had drawn a good graph for each of these I would have been able to see and calculate the slope to see the greater one. Sometimes I think steep will mean less slope since it’s headed in the vertical direction and I think that is where my erroneous reasoning came from. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For points 0 <= t <=2 pi being pi/6, pi/3, pi/2, 2pi/3 and so on we get the following values for y = sin(t - pi/3) Then I get the values: T (t - pi/3) Pi/ 6 -pi/6 Pi /3 0 Pi/2 Pi/6 2pi/3 Pi/3 5pi/6 Pi/2 Pi 2pi/3 7pi/6 5pi/6 4pi/3 Pi 3pi/2 7pi/6 5pi/3 4pi/3 11pi/6 3pi/2 2pi 5pi/3 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was confused about the given statement 0 <= t <= 2 pi. When I first read the question I thought it said the values for t were 0 <= t <= 2 pi/. I should have paid more attention I thought it was strange that it seemed somewhat out of context for the given assignment. I am still a little confused about the given answer though. How did the problem become (t + pi /3) instead of (t - pi /3)???? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m afraid I don’t understand. ???Do we go from positive to negative and vice versa because of even and odd functions???? ------------------------------------------------ Self-critique Rating: 1
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I’m understanding what is happening here. T - pi /3 equals theta and taking the values of theta we get the following values for T. Not the values for y. So when theta equal the given values it results in the following T values: Theta T 0 pi / 3 Pi 4 pi /3 2 pi 7 pi/3 3 pi 10pi / 3 4 pi 13 pi/3 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When Theta = t + pi/3 T value 0 -pi/3 Pi 2 pi/3 2 pi 5pi /3 3 pi 8 pi / 3 4 pi 11pi /3 So 2 pi/3, 5 pi/3, 8 pi / 3 and 11 pi /3 are the 1st four positive values. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It is relieving that I am starting the see the process here. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When theta = 3 t + pi / 3 Theta t 0 - pi / 9 Pi 2 pi / 9 2 pi 5 pi / 9 3 pi 8 pi / 9 4 pi 11 pi / 9 5 pi 14 pi / 9 6 pi 17 pi / 9 I think 0<= t <= 2 pi would be a wide range since given this scenario 2 pi would be equal to 18 pi / 9 which we almost reach at t = 17 pi /9. This range excludes - pi /9 since it would be below 0 but includes the rest of the given list. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(3t + pi/3) then when theta = 3t + pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we solve for t in order to see that t = theta/3 - pi/9 takes values t = 0/3 - pi/9, pi/3 - pi/9, 2 pi/3 - pi/9, 3 pi/3 - pi/9, 4 pi/3 - pi/9, ..., n pi/3 - pi/9, ..., where n = 0, 1, 2, 3, ... . These values simplify to give us t = -pi/9, 2 pi/9, 5 pi/9, 8 pi/9, 11 pi/9, ... . If we list these values starting with the first positive value 2 pi/9, continuing as long as t < 2 pi, we obtain values 2 pi/9, 5 pi/9, 8 pi/9, 11 pi/9, 14 pi/9, 17 pi/9. Since 2 pi = 18 pi/9 any subsequent solution will be greater than 2 pi so is not included. The corresponding graph is shown if Fig 66. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. Through what coordinates on the t axis does the graph of y = sin(3t + pi/3) pass? Through what point on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When theta = 3 t + pi / 3 Theta t 0 - pi / 9 Pi 2 pi / 9 2 pi 5 pi / 9 3 pi 8 pi / 9 4 pi 11 pi / 9 5 pi 14 pi / 9 6 pi 17 pi / 9 I think 0<= t <= 2 pi would be a wide range since given this scenario 2 pi would be equal to 18 pi / 9 which we almost reach at t = 17 pi /9. This range excludes - pi /9 since it would be below 0 but includes the rest of the given list. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!