#$&* course M 277 1-18 12 005. The graph keeps going and going. Thru what coordinates on the t axis does it pass etcGoals for this Assignment include but are not limited to the following:
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Given Solution: `aThe peak of the graph corresponds to the angular position pi/2, at the 'top' of the circle. In order for the graph to get from one peak to the next, the ant must travel from the top of the circle all the way around until it reaches the top once more. The angular displacement corresponding to a circuit around the circle is 2 pi. The angle theta on the graph therefore changes by 2 pi between peaks. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see you meant from maximum pt to maximum pt, not max pt to min pt. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. The graph of the basic sine function is centered on the theta axis, running in both the positive and negative directions, never ending, always repeating with a period of 2 pi. The graph represents never-ending motion around the unit circle with angular velocity 1, extending back forever, extending for forever. That is, extending forever into the past and forever into the future. How many complete cycles of the sine function will be completed between theta = -100 and theta = 100? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100/`pi = approx. 32 cycles, because dist. travled on `theta’s axis is 200(100 in pos direction and 100 in neg direction) with every 2`pi travled equal to 1 cycle. 200/2`pi = 100/`pi = 31.831 approx confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Understood ------------------------------------------------ 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: dist in peaks would be 2`pi/3 because object is moving 3 times as fast so it will take 1/3 the time to complete cycle. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance between peaks corresponds to the time required for the ant to complete a cycle around the circle. The ant moves around the unit circle, which has circumference 2 pi. At 3 units per second the time required is 2 pi / 3 seconds. The peaks of the graph will therefore be separated by 2 pi / 3 units along the t axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. The graphs of y = sin(3 t) and y = sin(t) both go through the origin and both with positive slope. Which graph goes through the origin with a greater slope? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=sin(3t) would have to because its moving at a greater speed. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: No does not pass through origin passes through pt `pi-`pi/3 = 2`pi/3 and pt 2`pi - `pi/3 = 5`pi/3 because zeros are at `pi and 2`pi and we are subtracting `pi/3 the fn is the same to y=sin(t) in every other regard so it’s the same fn just shifted to the left by `pi/3( goes for everything zero’s, max’s and min’s pts) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at tthe circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t - pi/3 takes values 0, pi, 2 pi, 3 pi, ..., we see that t = theta + pi/3 takes values t = 0 + pi/3, pi + pi/3, 2 pi + pi/3, etc.. These values simplify to give us t = pi/3, 4 pi/3, 7 pi/3, ... . A graph is shown in Figure 91. Note that only the values t = pi/3 and t = 4 pi/3 lie between 0 and 2 pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Rushing I put graph shifted to left but I had the right idea only fn is shifted to right instead. Solution understood ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. Through what coordinates on the t axis does the graph of y = sin(t + pi/3) pass? Through what point on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No does not pass through origin and the solution I stated in previous question applies to this equation with graph being shifted to left instead of right. confidence rating #$&*: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, 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): The graph above applies to previous question correct and not this one????? I’m pretty sure it does and if so solution is understood. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `pi/3, 4`pi/3, 7`pi/3, 10`pi/3. All are `pi/3 greater then when fn for sin(`theta) usually passes through origin. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe graph of y = sin(theta) passes through the horizontal axis when the reference circle position is at the rightmost or leftmost point, one of the points for the circle meets the x axis. This occurs repeatedly, at angular positions theta = 0, pi, 2 pi, 3 pi, 4 pi, etc.; and also at angular positions theta = -pi, -2 pi, -3 pi, etc.. If we are considering the graph of y = sin(t - pi/3) then when theta = t - pi/3 takes values 0, pi, 2 pi, 3 pi, 4 pi, ..., we see that t = theta + pi/3 takes values t = 0 + pi/3, pi + pi/3, 2 pi + pi/3, 0 + 4 pi / 3. These values simplify to give us t = pi/3, 4 pi/3, 7 pi/3, 10 pi/3, ... . These are the first four positive values of t for which the graph passes through the t axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. What are the first four positive t coordinates through which the graph of y = sin(t + pi/3) passes through the t axis, for t > 0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2`pi/3, 5`pi/3, 8`pi/3, 11`pi/3. Same logic as previous problem only -`pi/3 shift from normal sin fn zeros. confidence rating #$&*: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): I noticed you asked for fist 4 pos almost got me with -`pi/3 because when thinking of normal graph and 0 being 1st time t axis is crossed you want to count -`pi/3 1st. ------------------------------------------------ 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: Yes does pass thru origin, because fn has 3 times the slope it will complete 3 cycles where it normally would only complete 1. After we see how fn is acting as 3t it is easy to see that it takes `pi/3 to travel from either max or min back to t axis, so will pass through origin only will be headed in neg direction. Then passes through pts `pi/3, 2`pi/3, `pi, 4`pi/3, 5`pi/3, 2`pi. So it is easy to see sin(3t + `pi/3) passes through origin and completes 3 cycles between 0<=t<=2`pi. confidence rating #$&*:3 ********************************************* Question: `q009. Through what coordinates on the t axis does the graph of y = sin(3t + pi/3) pass? Through what point on the t axis does the graph pass in the interval 0 <= t <= 2 pi? Does this graph pass through the origin? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes does pass thru origin, because fn has 3 times the slope it will complete 3 cycles where it normally would only complete 1. After we see how fn is acting as 3t it is easy to see that it takes `pi/3 to travel from either max or min back to t axis, so will pass through origin only will be headed in neg direction. Then passes through pts `pi/3, 2`pi/3, `pi, 4`pi/3, 5`pi/3, 2`pi. So it is easy to see sin(3t + `pi/3) passes through origin and completes 3 cycles between 0<=t<=2`pi. confidence rating #$&*:3 #(*!