#$&* course M 277 1-13 9 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aWe see visually that the point a lies at an arc distance less than the radius of the circle. We also see that the point c lies at an arc distance that is clearly greater than the radius of the circle. The only possible candidate for a 1 radian angle, which must lie at an arc distance equal to one radius, is therefore point b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. If the first ant moves at a constant speed, moving through 1 radian every second, then approximately how long, to the nearest second, do you think it will take for the ant to move along the arc to the point where the circle meets the negative x-axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If it takes 1sec to get to pt b, then it should take approx. 3sec to travel to pt where circle meets the neg x axis. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aVisual examination, perhaps accompanied by a quick sketch, shows that it takes approximately 3 arcs each of one radian to get from the positive x-axis to the negative x-axis when moving along the arc of the circle. In figure 37 the points b, c and d lie at approximately 1, 2 and 3 radians. Remember that each radian corresponds to an arc distance equal to the radius of the circle. At 1 radian / second it will take about 3 seconds to move the approximately 3 radians to the negative x axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So it would take a little more than 3sec but not much more correct.
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Given Solution: `aAfter 1 second the angular position would be 1/2 radian, which would correspond to point a. Note that after 2 seconds the angular position would be 1 radian, corresponding to point b, and after three seconds the angular position would be 3 * 1/2 radian = 3/2 radian and the ant would be at position c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Would it not be at the pt directly in between pt b and c(approx. 90deg mark) at 3sec and at pt c at 4sec?????????
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Given Solution: `aThe circumference of the circle is 2 pi r, where r is the radius of the circle. This is the distance traveled by the ant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That is a much better answer and totally understand it, if we were measuring in radians it would be a little more than 6radians all the way around correct. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. As we just saw the distance around the circle is its circumference 2 pi r, where r is the radius. Through how many radians would the ant travel from the initial point, where the circle meets the positive x-axis, if the motion was in the counterclockwise direction and ended at the original point after having completed one trip around the circle. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Approx. -6radians because moving counterclockwise right. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAn arc displacement of r corresponds to an arc distance of 1 radian on the circle. Arc distances of 2, 3, 4, ... time the radius would correspond to 2, 3, 4, ... radians of arc. That is, arc distance of r, 2r, 3r, 4r, ... correspond to 1, 2, 3, 4, ... radians of arc. We understand by these examples that if we divide the arc distance by the radius, we will get the number of radians of angular distance. The arc distance around the circle is 2 pi r, which therefore corresponds to 2 pi r / r = 2 pi radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand, so for anything related to measuring arc distance wrt the radius is the same constant value in radians. Like if something traveled the dist. 4*radius around the arc of the circle it has also traveled 4radians around the arc. This is related to the fact that a radian is a measurement where the arc dist is equal to the radius. ??????So when moving clockwise or counterclockwise around arc of circle it does not affect whether the measurement is pos or neg. Like for this question we were moving counterclockwise but solution was not -2`pi radians just 2`pi radians.?????????????????
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Given Solution: `aThe unit circle has radius 1 and is centered at the origin, so the circle meets the positive x-axis 1 unit from the origin at (x, y) = (1,0). Similarly the circle meets the positive y-axis at the 'top' of the circle, 1 unit from the origin at (x, y) = (0,1); the circle meets the negative x-axis at (-1, 0); and the circle meets the negative y-axis at (0,-1). Figure 84 shows these points on the unit circle. Note that in this figure the small dots are located at increments of .1 unit in the x and y directions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Without looking at Figure 84, sketch a picture of the unit circle, complete with labeled points where the circle meets the x and y axes. Indicate the arc from the standard initial point, where the circle meets the x-axis, to the point where the circle meets the positive y axis. Describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Im not sure but I believe you are talking about the same pic we just saw correct, but Quad #1 arc would be highlighted. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour sketch should show the x and y axes and a circle of radius 1, with the points (1,0), (0, 1), (-1, 0) and (0, -1) where the circle meets the coordinate axes labeled. The arc will run along the first quadrant of the circle from (1,0) to (0,1). Your figure should match figure 84. You should be able to quickly draw this picture any time you need it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought thats what the question was asking, got it. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q008. How many radians of angular displacement correspond to the arc displacement from the standard initial point, where the circle meets the x-axis, to the point where the circle meets the positive y axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Little more than 1.5radians or because we know there is 2`pi radians in a circle we could take a Ό of that measurement to be more precise which would be (`pi/2)radians. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe trip around the entire circle, which corresponds to an angular displacement of 2 pi radians, corresponds to a trip from the initial point to the point where the circle meets the positive y-axis (i.e., the point (0,1)), then from this point to the point where the circle meets the negative x-axis (i.e., the point (-1,0)), then from this point to the point where the circle meets the negative y-axis (i.e., the point (0,-1)), then from this point back to the point where the circle meets the positive x-axis (i.e., the point (1,0)). Because of the symmetry of the circle, the arc corresponding to each of these displacements is the same. The arc from (1,0) to (0,1) is 1/4 of the 2 pi radian angular displacement around the entire circle, so its angular displacement is 2 pi/4 = pi/2 radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Recall some of this material but am going to work thru all of it anyway. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q009. We have just seen that the angular position of the (1,0) point is 0 and the angular position of the (0,1) point is pi/2. What are the angular positions of the (-1,0) and (0,-1) points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `pi and 3`pi/2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThese points are reached after successive angular displacements of pi/2. The (-1,0) point is reached from the pi/2 position by an additional angular displacement of pi/2, which puts it at angular position pi. The (0,-1) point is reached after another angular displacement of pi/2, which puts it at pi + pi/2 = 2 pi/2 + pi/2 = 3 pi/2. Note that still another angular displacement of pi/2 puts us back at the initial point, whose angular position is 0. This shows that the initial point has angular position 0, or angular position 3 pi/2 + pi/2 = 4 pi/2 = 2 pi, consistent with what we already know. You should label your picture with these angular positions pi/2, pi, 3 pi/2 and 2 pi specified at the appropriate points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. What is the angular displacement from the standard initial point of the point halfway along the arc of the circle from (1,0) to (0,1)? Note that you should begin with a sketch of the circle and of the arc specified here. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `pi/4, half of `pi/2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a(1,0) is the point at which the circle meets the positive x-axis and (0,1) is the point at which the circle meets the positive y-axis. The trip along the arc of the circle from (1,0) to (0,1) will move along the first-quadrant arc from angular position 0 to angular position pi/2. Halfway along this arc, the angular position will be 1/2 * pi/2 = pi/4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. What will be the angular positions of the arc points halfway between the (0,1) and (-1,0) points of the circle? What will be the angular positions of the arc points halfway between the (-1,0) and (0,-1) points of the circle? What will be the angular positions of the arc points halfway between the (0,-1) and (1,0) points of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3`pi/4, 5`pi/4, 7`pi/4 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aHalfway between the (0,1) point, which corresponds to the the the position pi/2, and the (-1,0) point, which corresponds to angular position pi, will be the point lying at angular position pi/2 + pi/4 = 2 pi / 4 + pi / 4 = (2 pi + pi)/4 = 3 pi / 4. Halfway between the (-1,0) point, which corresponds to the the position pi,and the (0,-1) point, which corresponds to angular position 3 pi / 2, will be the point lying at angular position pi + pi/4 = 4 pi / 4 + pi / 4 = (4 pi + pi)/4 = 5 pi / 4. Halfway between the (0,-1) point, which corresponds to the the position 3 pi/2, and the (-1,0) point, which corresponds to angular position 2 pi, will be the point lying at angular position 3 pi/2 + pi/4 =62 pi / 4 + pi / 4 = (6 pi + pi)/4 = 7 pi / 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. What is the angular position of the point lying 1/3 of the way along the arc of the circle between the points (1,0) and (0,1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `pi/6, or 1/3 of `pi/2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe arc from (1,0) to (0,1) corresponds to an angular displacement of pi/2. One-third of the arc corresponds to an angular displacement of 1/3 * pi/2 = pi/6. The angular position of the specified point is therefore pi/6. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe arc from (1,0) to (0,1) corresponds to an angular displacement of pi/2. One-third of the arc corresponds to an angular displacement of 1/3 * pi/2 = pi/6. The angular position of the specified point is therefore pi/6. "
#$&* course M 277 1-13 9 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The angular position changes by pi/6 radians every second. Starting at angular position 0, the angular positions at t = 1, 2, 3, 4, ..., 12 will be pi/6, 2 pi/6, 3 pi/6, 4 pi/6, 5 pi/6, 6 pi/6, 7 pi/6, 8 pi/6, 9 pi/6, 10 pi/6, 11 pi/6, and 12 pi/6. You might have reduced these fractions the lowest terms, which is good. In any case this will be done in the next problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Reduce the fractions pi/6, 2 pi/6, 3 pi/6, 4 pi/6, 5 pi/6, 6 pi/6, 7 pi/6, 8 pi/6, 9 pi/6, 10 pi/6, 11 pi/6, and 12 pi/6 representing the angular positions in the last problem to lowest terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Already done in previous problem confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe reduced fractions are pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q003. Sketch a circle centered at the origin of an x-y coordinate system, depicting the angular positions pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi. What are the angular positions of the following points: The point 2/3 of the way along the arc between (0,1) and (-1,0) The point 1/3 of the way along the arc from (0, 1) to (-1,0) The points 1/3 and 2/3 of the way along the arc from (-1,0) to (0,-1) The points 1/3 and 2/3 of the way along the arc from (0, -1) to (0,1)?? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5`pi/6, 2`pi/3, 7`pi/6 and 4`pi/3, 5`pi/3 and 11`pi/6 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe points lying 1/3 and 2/3 of the way along the arc between the points (0,1) and (-1,0) are at angular positions 2 pi/3 and 5 pi/6; the point 2/3 of the way between these points is at angular position 5 pi/6. The points lying 1/3 and 2/3 of the way along the arc between the points (-1,0) and (0,1) are at angular positions 7 pi/6 and 4 pi/3. The points lying 1/3 and 2/3 of the way along the arc between the points (0,-1) and (1,0) are at angular positions 5 pi/3 and 11 pi/6. Note that you should be able to quickly sketch and label this circle, which depicts the angles which are multiples of pi/6, whenever you need it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. If the red ant moves at an angular velocity of pi/4 radians every second then what will be its angular position at the end of each of the first 8 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `pi/4, `pi/2, 3`pi/4, `pi, 5`pi/4, 3`pi/2, 7`pi/4, 2`pi confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe angular position changes by pi/4 radians every second. Starting at angular position 0, the angular positions will be pi/4, 2 pi/4, 3 pi/4, 4 pi/4, 5 pi/4, 6 pi/4, 7 pi/4, and 8 pi/4. You might have reduced these fractions the lowest terms, which is good.In any case this will be done in the next problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Reduce the fractions pi/4, 2 pi/4, 3 pi/4, 4 pi/4, 5 pi/4, 6 pi/4, 7 pi/4, and 8 pi/4 representing the angular positions in the last problem to lowest terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Already done confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe reduced fractions are pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Sketch a circle centered at the origin of an x-y coordinate system, depicting the angular positions pi/4, pi/2, 3 pi/4, pi, 5 pi/4, 3 pi/2, 7 pi/4, and 2 pi. What are the angular positions of the following points: The point 1/2 of the way along the arc between (0,1) and (-1,0) The point 1/2 of the way along the arc from (0, -1) to (1,0) The point 1/2 of the way along the arc from (0,-1) to (0, -1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3`pi/4, 7`pi/4, 3`pi/2??? confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe point lying 1/2 of the way along the arc between the points (0,1) and (-1,0) (the topmost and leftmost points of the circle) is at angular position 3 pi/4. The point lying 1/2 of the way along the arc between the points (0,-1) and (1,0) is at angular position 7 pi/4. The point lying 1/2 of the way along the arc between the points (-1,0) and (0,-1) is at angular position 5 pi/4. These angles are shown in Figure 21. Note that the degree equivalents of the angles are also given. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There was a miss print in question when asking for 3rd position. Understand ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q007. If the red ant starts at angular position pi/3 and moves at an angular velocity of pi/3 radians every second then what will be its angular position at the end of each of the first 6 seconds? Reduce your fractions to lowest terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2`pi/3, `pi, 4`pi/3, 5`pi/3, 2`pi, `pi/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe angular position changes by pi/3 radians every second. Starting at angular position pi/3, the angular positions after successive seconds will be 2 pi/3, 3 pi/3, 4 pi/3, 5 pi/3, 6 pi/3 and 7 pi/3, which reduce to 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi and 7 pi/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So after 2pi position will still be 7`pi/3 instead of `pi/3?????? ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q008. Where is the angular position 7 pi/3 located? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: !!!!Nevermind!!!! `pi/3, just answered my own question. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf you have not done so you should refer to your figure showing the positions which are multiples of pi/6. On your picture you will see that the sequence of angular positions 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi, 7 pi/3 beginning in the first quadrant and moving through the second, third and fourth quadrants to the 2 pi position, then pi/3 beyond that to the 7 pi/3 position. The 7 pi/3 position is therefore identical to the pi/3 position. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Understood. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q009. If the red ant starts at angular position pi/3 and moves at an angular velocity of pi/4 radians every second then what will be its angular position at the end of each of the first 8 seconds? Reduce your fractions to lowest terms. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7`pi/12, 5`pi/6, 13`pi/12, 4`pi/3, 19`pi/12, 11`pi/6, 25`pi/12, 7`pi/3(init starting pt) Found by getting common denm.-->12 having starting pos=4`pi/12 and adding 3`pi/12 at the end of every time interval and reducing if possible. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe angular position changes by pi/4 radians every second. Starting at angular position pi/3, the angular positions after successive seconds will be pi/3 + pi/4, pi/3 + 2 pi/4, pi/3 + 3 pi/4, pi/3 + 4 pi/4, pi/3 + 5 pi/4, pi/3 + 6 pi/4, pi/3 + 7 pi/4 and pi/3 + 8 pi/4. These fractions must be added before being reduced to lowest terms. In each case the fractions are added by changing each to the common denominator 12. This is illustrated for pi/3 + 3 pi/4: We first multiply pi/3 by 4/4 and 3 pi/4 by 3/3, obtaining the fractions 4 pi/12 and 9 pi/12. So the sum pi/3 + 3 pi/4 becomes 4 pi/12 + 9 pi/12, which is equal to 13 pi/12. The fractions add up as follows: pi/3 + pi/4 = 7 pi/12, pi/3 + 2 pi/4 = 5 pi/6, pi/3 + 3 pi/4 = 13 pi/12, pi/3 + 4 pi/4 = 4 pi/3, pi/3 + 5 pi/4 = 19 pi/12, pi/3 + 6 pi/4 = 11 pi/6, pi/3 + 7 pi/4 = 25 pi/12 and pi/3 + 8 pi/4 = 7 pi / 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: "
#$&* course M 277 1-13 9 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe angular positions of the points coinciding with the positive and negative x axes all have y coordinate 0; these angles include 0, pi and 2 pi. At angular position pi/4 the point of circle appears to be close to (.7,.7), perhaps a little beyond at (.71,.71) or even at (.72,.72). Any of these estimates would be reasonable. Note for reference that, to two decimal places the coordinates are in fact (.71,.71). To 3 decimal places the coordinates are (.707, .707), and the completely accurate coordinates are (`sqrt(2)/2, `sqrt(2) / 2). The y coordinate of the pi/4 point is therefore .71. The y coordinate of the 3 pi/4 point is the same, while the y coordinates of the 5 pi/4 and 7 pi/4 points are -.71. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was approx. but understand solution. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Figure 31 shows the angular positions which are multiples of pi/6 superimposed on a grid indicating the scale relative to the x y coordinate system. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0, ½, 8.5/10, 1, 8.5/10, ½, 0, -1/2, -8.5/10, -1, -8.5/10, -1/2, 0 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So the measurement I used of 8.5/10 or .85 is exactly = sqrt(3)/2 correct.
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Given Solution: `aThe 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Solution understood. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because the slope or arc dist travled in the beginning of the interval between 0 and `pi/2 is greater which then tappers off before slope actually equals 0 at `pi/2. Slope is in regards to rate of change in y wrt x. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Understood. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ½, -1/2, `sqrt(2)/2, -sqrt(3)/2, ½, -1/2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Rushing caused me to miss last calculation. Understood ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For x=`pi/12, sin(2x) = sin(`pi/6) = .5 For x=`pi/6, sin(2x) = sin(`pi/3) = .87 For x=`pi/4, sin(2x) = sin(`pi/2) = 1 For x=`pi/3, sin(2x) = sin(2`pi/3) = .87 For x=5`pi/12, sin(2x) = sin(5`pi/6) = .5 For x=`pi/2, sin(2x) = sin(`pi) = 0 For x=7`pi/12, sin(2x) = sin(7`pi/6) = -.5 For x=2`pi/3, sin(2x) = sin(4`pi/3) = -.87 For x=3`pi/4, sin(2x) = sin(3`pi/2) = -1 For x=`5pi/6, sin(2x) = sin(5`pi/3) = -.87 For x=11`pi/12, sin(2x) = sin(11`pi/6) = -.5 For x=`pi, sin(2x) = sin(2`pi) = 0 !!!Hope you can read I reduced and found y value in the same line!!!!!!!!!!! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x goes from 0 to `pi, y goes from 0 to 1 back to 0 to -1 then back to 0. Looks similar to first graph we looked at. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour 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. STUDENT QUESTION Why does it peak at pi/4? I thought Pi/4=sqrt2/2 for sine and pi/2= sine of 1. What idea am I missing? INSTRUCTOR RESPONSE The question is about sin(2x) vs. x, and the graph depicts sin(2x) vs. x. On your table you have a column for x and a column for 2 x. The peak occurs when 2 x = pi / 2. The x value for this line of the table is pi /4 (half the values of 2 x). The graph therefore peaks at x = pi / 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Sorry I misunderstood question in regards to values of x, but my graph was exactly the same as the one given. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `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, ... ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x must = 0, `pi/18, `pi/9, `pi/6, 2`pi/9, 5`pi/18, `pi/3, 7`pi/18, 4`pi/9, 3`pi/6, 10`pi/9, 2`pi/3 ???Im not sure what the sequence is but assume you want to 2`pi)???? confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I made that problem a lot harder than it needed to be I did not think to mult. everything by 1/3 and got everything by trial and error and missed a couple. I dont know why the most obvious stuff is so hard for me to see sometimes. Self-critique Rating:3 ********************************************* Question: `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. Then give the first, fifth and seventh rows of your table. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???Do you mean to change heading to `theta = 3x, instead of Now change the heading of the theta column to 'theta = sin(3x)????? X `theta=3x Sin(3x) 0 0 0 2`pi/9 2`pi/3 .87 `pi/3 `pi 0 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe table originally reads as follows: x theta sin(theta) 0 0.0 pi/6 0.5 pi/3 0.87 pi/2 1.0 2 pi/3 0.87 5 pi/6 0.5 pi 0.0 7 pi/6 -0.5 4 pi/3 -0.87 3 pi/2 -1.0 5 pi/3 -0.87 11 pi/6 -0.5 2 pi -0.0 After inserting the values for x and changing column headings the table is x theta = 3x sin(3x) 0 0 0.0 pi/18 pi/6 0.5 pi/9 pi/3 0.87 pi/6 pi/2 1.0 2 pi/9 2 pi/3 0.87 5 pi/18 5 pi/6 0.5 pi/3 pi 0.0 7 pi/18 7 pi/6 -0.5 4 pi/9 4 pi/3 -0.87 pi/2 3 pi/2 -1.0 5 pi/9 5 pi/3 -0.87 11 pi/18 11 pi/6 -0.5 2 pi/3 2 pi -0.0 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Same table I got. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q011. Sketch the graph corresponding to your table for sin(3x) vs. x. Does the sine function go through a complete cycle? By how much does x change as the sine function goes through its complete cycle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes sine fn goes thru complete cycle. Fn peaks at x=`pi/6, and has min. value at x= `pi/2. Very similar to previous prob. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. For the function y = sin(3x), what inequality in the variable x corresponds to the inequality 0 <= theta <= 2 pi? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0<=`theta<=2`pi/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If theta = 3x then the inequality 0 <= theta <= 2 pi becomes 0 <=3x <= 2 pi. If we multiply through by 1/3 we have 1/3 * 0 <= 1/3 * 3x <= 1/3 * 2 pi, or 0 <= x <= 2 pi/3. In the preceding problem our graph when through a complete cycle between x = 0 and x = 2 pi/3. This precisely correspond to the inequality we just obtained. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Not thinking I put `theta back into inequality instead of x. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q013. For y = sin(theta) = sin(2x - 2 pi/3), what values must x take so that theta = 2x - pi/3 will take the values 0, pi/6, pi/3, pi/2, ... ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1st solve for x, `theta= 2x-`pi/3 2x=`theta+`pi/3 x=(`theta+`pi/3)/2 = `theta/2 + `pi/6 Now we plug in values of `theta and solve for x x = 0/2 + `pi/6 = `pi/6, for `theta=0 x = (`pi/6)/2 + `pi/6 =`pi/12+ `pi/6 = `pi/4, for `theta=`pi/6 x = (`pi/3)/2 + `pi/6 =`pi/6+ `pi/6 = `pi/3, for `theta=`pi/3 x = (`pi/2)/2 + `pi/6 =`pi/4+ `pi/6 = 5`pi/12, for `theta=`pi/2 x = (2`pi/3)/2 + `pi/6 =`pi/3+ `pi/6 = `pi/2, for `theta=2`pi/3 x = (5`pi/6)/2 + `pi/6 =5`pi/12+ `pi/6 = 7`pi/12, for `theta=5`pi/6 x = (`pi)/2 + `pi/6 =`pi/2+ `pi/6 = 2`pi/3, for `theta=`pi x = (7`pi/6)/2 + `pi/6 =7`pi/12+ `pi/6 = 3`pi/4, for `theta=7`pi/6 x = (4`pi/3)/2 + `pi/6 =4`pi/6+ `pi/6 = 5`pi/6, for `theta=4`pi/3 x = (3`pi/2)/2 + `pi/6 =3`pi/4+ `pi/6 = 11`pi/12, for `theta=3`pi/2 x = (5`pi/3)/2 + `pi/6 =5`pi/6+ `pi/6 = `pi, for `theta=5`pi/3 x = (11`pi/6)/2 + `pi/6 =11`pi/12+ `pi/6 = 13`pi/12, for `theta=11`pi/6 x = (2`pi)/2 + `pi/6 =`pi+ `pi/6 = 7`pi/6, for `theta=2`pi confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Had the right approach but my logic is off some where ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same answer as previous question X 'theta = 2x - pi/3 sin(2x - pi/3) `pi/6 0 0 7`pi/12 2 pi/3 .87 3`pi/4 `pi 0 confidence rating #$&*::3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our first table is the same as before. We will always start the table for the sine function in the following manner, leaving our x column blank, and listing the theta and sin(theta) columns: 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 obtained by solving theta = 2 x - pi / 3 for x, and finding x for each of our listed values of theta. We get the following: 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 This table indicates that the function y = sin(2x - pi/3) goes through a complete cycle, in which y values run from 0 to 1 to 0 to -1 to 0, when the x values run from pi/3 to 5 pi/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Im not sure where Im getting mixed up but most of my values match yours but my 5th value is your 6th, my 7th is your 8th etc.
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Given Solution: 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. Our graph differs from that ofy = sin(2x) in that the graph is 'shifted' pi/6 units to the right of that graph. 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_. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Our graph differs from that ofy = sin(2x) in that the graph is 'shifted' pi/6 units to the right of that graph. 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_. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #(*!