course Mth 164 004. Extending to circle of radius 3Precalculus II
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20:26:42 `q001. Now we have a circle of radius 3. An angular position of 1 radian again corresponds to an arc displacement equal to the radius of the circle. Which point on the circle in the picture corresponds to the angular position of 1 radian?
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RESPONSE --> I'm too sure on this one. I understand that the radius is 3 so around the whole circle the radius is the same. The point on the circle would have to be B. confidence assessment: 2
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20:28:05 The distance along the arc will be equal to the radius at point b. So the angular position of one radian occurs at point b. We see that when the circle is scaled up by a factor of 3, the radius becomes 3 times as great so that the necessary displacement along the arc becomes 3 times as great. Note that the 1-radian angle therefore makes the same angle as for a circle of radius 1. The radius of the circle doesn't matter.
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RESPONSE --> I see what is going on here. Also I see that the radius becomes 3 times as great along the arc. self critique assessment: 3
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20:29:07 `q002. On the circle of radius 3 what arc distance will correspond to an angle of pi/6?
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RESPONSE --> It would become 3 times the arc distance. You would have 3*pi/6=pi/2. confidence assessment: 2
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20:29:55 On a circle of radius 1 the arc distance pi/6 corresponds to an arc displacement of pi/6 units. When the circle is scaled up to radius 3 the arc distance will become three times as great, scaling up to 3 * pi/6 = pi/2 units.
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RESPONSE --> I see what is being done here. The arc is three times greater and you have to multiply the 3 times pi/6 in order to get pi/2 units. self critique assessment: 3
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20:30:54 `q003. If the red ant is moving along a circle of radius 3 at a speed of 2 units per second, then what is its angular velocity--i.e., its the rate in radians / second at which its angular position changes?
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RESPONSE --> It would be 2/3 radians/second. confidence assessment: 2
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20:34:41 `q004. If the red ant is moving along at angular velocity 5 radians/second on a circle of radius 3, what is its speed?
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RESPONSE --> I messed up here a little bit. I hit the wrong button. 5/3 radians/second confidence assessment: 2
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20:35:54 Each radian on a circle of radius 3 corresponds to 3 units of distance. Therefore 5 radians corresponds to 5 * 3 = 15 units of distance and 5 radians/second corresponds to a speed of 15 units per second.
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RESPONSE --> I see here that you have to multiply the 5*3 to get 15. You would then have 5 radians/second in order to get a speed of 15 units per second. self critique assessment: 3
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20:46:22 `q005. Figure 17 shows a circle of radius 3 superimposed on a grid with .3 unit between gridmarks in both x and y directions. Verify that this grid does indeed correspond to a circle of radius 3. Estimate the y coordinate of each of the points whose angular positions correspond to 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, pi, 7 pi/6, 4 pi/3, 3 pi/2, 5 pi/3, 11 pi/6 and 2 pi.
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RESPONSE --> At y-coordinate 0 you owould have angles 0, pi, and 2pi. Position 2/pi the angle would be -3. At pi/6 the point would be about (2.6, 1.5). At pi/3 the point is (1.5, .87) At 2pi/3 the point is 2.6. At 4pi/3 and 5pi/3 the point is -2.6 At 7pi/6 and 11pi/6 the point would be -1.5. confidence assessment: 2
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20:50:18 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/2 the y coordinate is equal to the radius 3 of the circle; at 3 pi/2 the y coordinate is -3. At angular position pi/6 the point on the circle appears to be close to (2.7,1.5); the x coordinate is actually a bit less than 2.7, perhaps 2.6, so perhaps the coordinates of the point are (2.6, 1.5). Any estimate close to these would be reasonable. The y coordinate of the pi/6 point is therefore 1.5. The coordinates of the pi/3 point are (1.5, .87), just the reverse of those of the pi/6 point; so the y coordinate of the pi/3 point is approximately 2.6. The 2 pi/3 point will also have y coordinate approximately 2.6, while the 4 pi/3 and 5 pi/3 points will have y coordinates approximately -2.6. The 5 pi/6 point will have y coordinate 1.5, while the 7 pi/6 and 11 pi/6 points will have y coordinate -1.5.
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RESPONSE --> I see that at y-coordinate 0 the angles include 0, pi and 2 pi. At angular position pi/2 the y coordinate is equal to the radius 3 of the circle at 3 pi/2 the y coordinate is -3. At angular position pi/6 the point on the circle appears to be close to (2.7,1.5); the x coordinate is actually a bit less than 2.7, perhaps 2.6, so perhaps the coordinates of the point are (2.6, 1.5). Sometimes it is hard to get it rigt on the point. It could either be 2.7 or 2.6. The y-coordinate is pi/6. self critique assessment: 3
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20:52:41 `q006. The y coordinates of the unit-circle positions 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 are 0, .5, .87, 1, .87, .5, 0, -.5, -.87, -1, -.87, -.5, 0. What should be the corresponding y coordinates of the points lying at these angular positions on the circle of radius 3? Are these coordinates consistent with those you obtained in the preceding problem?
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RESPONSE --> You would multiply all of the values by 3. 0, .5, .87, 1, .87, .5, 0, -.5, -.87, -1, -.87, -.5, 0. You would then get 0, 1.5, 2.61, 3, -3, -2.61, -1.5, 0. confidence assessment: 2
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20:53:43 On a radius 3 circle the y coordinates would each be 3 times as great. The coordinates would therefore be obtained by multiplying the values 0, .5, .87, 1, .87, .5, 0, -.5, -.87, -1, -.87, -.5, 0 each by 3, obtaining 0, 1.5, 2.61, 3, 2.61, 1.5, 0, -1.5, -2.61, -3, -2.61, -1.5, 0. These values should be close, within .1 or so, of the estimates you made for this circle in the preceding problem.
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RESPONSE --> I understand how this is done. You just need to multiply the values by 3 in order to get the answer. Sometimes the answer will vary. self critique assessment: 3
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20:54:34 `q007. The exact y coordinates of the unit-circle positions 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 are 0, .5, sqrt(3) / 2, 1, sqrt(3) / 2, .5, 0, -.5, -sqrt(3) / 2, -1, -sqrt(3) / 2, -.5, 0. What should be the corresponding y coordinates of the points lying at these angular positions on the circle of radius 3? Are these coordinates consistent with those you obtained in the preceding problem?
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RESPONSE --> You would multiply 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 are 0, .5, sqrt(3) / 2, 1, sqrt(3) / 2, .5, 0, -.5, -sqrt(3) / 2, -1, -sqrt(3) / 2, -.5, 0 by three in order to get the points. confidence assessment: 3
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20:55:16 On a radius 3 circle the y coordinates would each be 3 times as great. The coordinates would therefore be obtained by multiplying the values 0, .5, sqrt(3) / 2, 1, sqrt(3) / 2, .5, 0, -.5, -sqrt(3) / 2, -1, -sqrt(3) / 2, -.5, 0 each by 3, obtaining 0, 1.5, 3 sqrt(3) / 2, 3, 3 sqrt(3) / 2, 1.5, 0, -3 sqrt(3) / 2, -3 sqrt(3) / 2, -3, -2.61, -1.5, 0.
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RESPONSE --> I see how it will be three times greater and in order to get that you would have to multiply everything by 3. self critique assessment: 3
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20:56:41 `q008. Sketch a graph of the y coordinate obtained for a circle of radius 3 in the preceding problem vs. the anglular position theta.
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RESPONSE --> I think that the maximun and minimum values would be 3 and -3. This is done with a y=sin(theta) formula. confidence assessment: 2
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20:58:15 Your graph should be as shown in Figure 54. This graph as the same description as a graph of y = sin(theta) vs. theta, except that the slopes are all 3 times as great and the maximum and minimum values are 3 and -3, instead of 1 and -1.
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RESPONSE --> I see how it looks like figure 54. The y=sin(theta) vs theata. I left the verus part out. I see that it is 3 times greater ad the maximum and minimum values are 3 and -3. self critique assessment: 3
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21:00:40 `q009. If the red ant starts on the circle of radius 3, at position pi/3 radians, and proceeds at pi/3 radians per second then what will be its angular position after 1, 2, 3, 4, 5 and 6 seconds? What will be the y coordinates at these points? Make a table and sketch a graph of the y coordinate vs. the time t. Describe the graph of y position vs. clock time.
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RESPONSE --> The angular positions are 1, 2, 3,4 5, 6 are 2pi/3, pi, 4pi/3, 5pi/3, 2pi and 7pi/3. The y-coordinates would have to be at sqrt(3). confidence assessment: 2
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21:03:37 The angular positions at t = 1, 2, 3, 4, 5 and 6 are 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi and 7 pi/3. The corresponding y coordinates are 3 * sqrt(3) / 2, 0, -3 * sqrt(3) / 2, -3 * sqrt(3) / 2, 0 and 3 * sqrt(3) / 2. If you just graph the corresponding points you will miss the fact that the graph also passes through y coordinates 3 and -3; from what you have seen about these functions in should be clear why this happens, and it should be clear that to make the graph accurate you must show this behavior. See these points plotted in red in Figure 45, with the t = 0, 2, 4, 6 values of theta indicated on the graph. The graph therefore runs through its complete cycle between t = 0 and t = 6, starting at the point (0, 3 * sqrt(3) / 2), or approximately (0, 2.6), reaching its peak value of 3 between this point and (1, 3 * sqrt(3) / 2), or approximately (1, 2.6), then reaching the x axis at t = 3 as indicated by the point (2, 0) before descending to (3, -3 * sqrt(3) / 2) or approximately (3, -2.6), then through a low point where y = -3 before again rising to (4, -3 * sqrt(3) / 2) then to (5, 0) and completing its cycle at (6, 3 * sqrt(3) / 2). This graph is shown in Figure 86.
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RESPONSE --> I wasn't too sure about the graph so I didn't want to jump ahead until I was sure about it. I see how to go about t = 1, 2, 3, 4, 5 and 6 are 2 pi/3, pi, 4 pi/3, 5 pi/3, 2 pi and 7 pi/3. The corresponding y coordinates are 3 * sqrt(3) / 2, 0, -3 * sqrt(3) / 2, -3 * sqrt(3) / 2, 0 and 3 * sqrt(3) / 2. I see where Figure 45, with the t = 0, 2, 4, 6 values of theta indicated on the graph. Also how the graph runs through its complete cycle between t = 0 and t = 6, starting at the point (0, 3 * sqrt(3) / 2), or approximately (0, 2.6), reaching its peak value of 3 between this point and (1, 3 * sqrt(3) / 2), or approximately (1, 2.6), then reaching the x axis at t = 3 as indicated by the point (2, 0) before descending to (3, -3 * sqrt(3) / 2) or approximately (3, -2.6), then through a low point where y = -3 before again rising to (4, -3 * sqrt(3) / 2) then to (5, 0) and completing its cycle at (6, 3 * sqrt(3) / 2). self critique assessment: 3
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21:16:50 `q010. Make a table for the graph of y = 3 sin(pi/4 * t + pi/3), using theta = 0, pi/4, pi/2, etc., and plot the corresponding graph. Describe your graph.
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RESPONSE --> sin (theta)0, pi/4, pi/2, 3pi/4, pi, 5pi/4, 3pi/2, 7pi/4, 2pi and for 3*sin(theta) is 0, .71, 0, .71, 0, -.71, -1, -.71, 0. confidence assessment: 1
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21:21:30 Using columns for t, theta and sin(theta) as we have done before, and an additional column for 3 * sin(theta) we obtain the following initial table: t theta = pi/4 t + pi/3 sin(theta) 3 * sin(theta) 0 0 pi/4 .71 pi/2 0 3 pi/4 .71 pi 0 5 pi/4 -.71 3 pi/2 -1 7 pi/4 -.71 2 pi 0 0 The solution to pi/4 t + pi/3 = theta is obtained by first adding -pi/3 to both sides to obtain pi/4 t = theta - pi/3, then multiplying both sides by 4 / pi to obtain t = 4 / pi * theta - 4 / pi * pi/3, and finally simplifying to get t = 4 / pi * theta - 4/3. Substituting in the given values of theta we obtain t values -4/3, -1/3, 2/3, 5/3, 8/3, 11/3, 14/3, 17/3, 20/3. We also multiply the values of sin(theta) by 3 to get the values of 3 sin(theta): t theta = pi/4 t + pi/3 sin(theta) 3 * sin(theta) -4/3 0 0 0 -1/3 pi/4 .71 2.1 2/3 pi/2 0 0 5/3 3 pi/4 .71 2.1 8/3 pi 0 0 11/3 5 pi/4 -.71 -2.1 14/3 3 pi/2 -1 -1 17/3 7 pi/4 -.71 -2.1 20/3 2 pi 0 0 Since theta = pi/4 t + pi/3, if we graph the final column vs. the first we have the graph of y = 3 sin(pi/4 t + pi/3) vs. t. This graph is shown in Figure 19, with red dots indicating points corresponding to rows of the table. The graph is as in the figure (fig 3 sin(pi/4 * t + pi/3)), with a standard cycle running from t = -4/3 to t = 20/3. During this cycle y goes from 0 to 3 to 0 to -3 to 0. The duration of the cycle is 20/3 - (-4/3) = 24/3.
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RESPONSE --> Okay, I didn't go into much depth because I was starting to get off track. I see that you have to have t theta = pi/4 t + pi/3 sin(theta) 3 * sin(theta) 0 0 pi/4 .71 pi/2 0 3 pi/4 .71 pi 0 5 pi/4 -.71 3 pi/2 -1 7 pi/4 -.71 2 pi 0 0 Then you have pi/4 t + pi/3 = theta is obtained by first adding -pi/3 to both sides to obtain by pi/4 t = theta - pi/3, then multiplying both sides by 4 / pi to obtain t = 4 / pi * theta - 4 / pi * pi/3, and finally simplifying to get t = 4 / pi * theta - 4/3. Then you have to substitute the values of theta you have t values -4/3, -1/3, 2/3, 5/3, 8/3, 11/3, 14/3, 17/3, 20/3. Then you have to multiply the values of sin(theta) by 3 to get the values of 3 sin(theta): t theta = pi/4 t + pi/3 sin(theta) 3 * sin(theta) -4/3 0 0 0 -1/3 pi/4 .71 2.1 2/3 pi/2 0 0 5/3 3 pi/4 .71 2.1 8/3 pi 0 0 11/3 5 pi/4 -.71 -2.1 14/3 3 pi/2 -1 -1 17/3 7 pi/4 -.71 -2.1 20/3 2 pi 0 0 self critique assessment: 3
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21:21:37 Complete Assignment 3, 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 --> Completed self critique assessment: 3
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