qa assignment 4

course Mth 164

?????R???????assignment #004004. Extending to circle of radius 3

Precalculus II

02-22-2009

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15:58:42

`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 confused about which picture you're referring to. Is it the one we drew in a previous assignment, or a figure that we should open ? I tried to answer the question without knowing which but can't.

confidence assessment: 0

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15:59:10

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 -->

self critique assessment:

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15:59:39

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 -->

The answer doesn't make sense without the correct picture to refer to.

self critique assessment: 1

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16:00:32

`q002. On the circle of radius 3 what arc distance will correspond to an angle of pi/6?

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RESPONSE -->

3(pi/6) = pi/2

confidence assessment: 1

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16:00:44

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 -->

self critique assessment:

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16:01:12

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 just remembered from the last question that the distance would be 3 times as great

self critique assessment: 2

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16:02:06

`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 -->

3*2 = 6 radians/sec

confidence assessment: 1

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16:03:49

Since 3 units corresponds to one radian, 2 units corresponds to 2/3 radian, and 2 units per second will correspond to 2/3 radian/second.

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RESPONSE -->

I thought that the radius of 3 would make the ant move 3 times as fast. Do you multiply for distance and divide for angular velocity ?

self critique assessment: 1

It's very difficult to remember what to multiply or divide by what--way too easy to get confused.

Best to reason it out from the meaning of a radian:

A radian is the angle for which arc distance matches the radius.

On a circle of radius 3, a radian of angle would correspond to 3 units along the arc.

So 2 units along the arc corresponds to less than a radian. Specifically, 2 units along the arc correspond to 2/3 of a radian.

Since the object moves at 2 units per second, it therefore moves at 2/3 of a radian per second.

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16:04:44

`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 -->

5*3 = 15 rad/sec.

confidence assessment: 1

15 units / second; radians measure angle, not arc distance

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16:04:57

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 -->

self critique assessment:

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16:05:18

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 -->

my assumption must have been correct

self critique assessment: 2

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16:33:19

`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 -->

angular position y-coord.

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pi/6 1 .5

pi/3 2.7

pi/2 3

2pi/3 2.7

5pi/6 1.5

pi 0

7pi/6 -1.5

4pi/3 -2.7

3pi/2 -3

5pi/3 -2.7

11pi/6 -1.5

2pi 0

confidence assessment: 2

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16:33:53

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 -->

self critique assessment:

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16:34:19

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 -->

It helps when you find the right file figure !

self critique assessment: 2

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16:36:14

`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 -->

yes - for example >5*3 = 1.5

confidence assessment: 2

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16:36:29

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 -->

self critique assessment: 0

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19:52:32

`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 -->

unit-circle position y-coord.w/radius 3

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0 0

pi/6 1.5

pi/3 2.6

pi/2 3

2pi/3 2.6

5pi/6 1.5

pi 0

7pi/6 -1.5

4pi/3 -2.6

3pi/2 -3

5pi/3 -2.6

11pi/6 -1.5

2pi 0

yes

confidence assessment: 2

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19:53:06

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 -->

self critique assessment: 0

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20:00:31

`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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confidence assessment: 2

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20:06:22

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 -->

figure 54 is y = 3sin(x). Is this correct ? I sketched a circle with radius 3 and marked the y-coord., angular position, and angle for each value in the table. For example, pi/6 = 30 deg., y = 1.5

self critique assessment: 2

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20:12:53

`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 -->

t pos y-coord.

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0 pi/3 2.6

1 2pi/3 2.6

2 7pi/6 -1.5

3 3pi/2 -3

4 11pi/6 -1.5

5 pi/6 1.5

6 pi/2 3

confidence assessment: 2

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20:18:14

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 was trying to plot this on the circle, moving 60 deg. each time to correspond to pi/3

self critique assessment: 1

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21:49:47

`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 -->

t theta x-coord. y-coord.

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1 pi/4 .87 3

2 pi/4 .52 3

3 pi/4 .52 3

4 pi/4 .39 3

As time increases, the x-coordinate decreases

confidence assessment: 1

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21:57:46

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 -->

I used y=3sin(pi/4*t +pi/3) for my graph and compared the x-coord. at times 1-4 I thought that was what I was supposed to do. I'm not sure how that relates to this

self critique assessment: 1

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You should review the 'open' qa's posted for Assignments 3 and 4, and I recommend you use the 'open' qa's rather than the program. I've just completed the task of inserting all the pictures into the 'open' qa's. The tables in these two assignments are better aligned in those documents. I've also inserted additional explanations of the process of setting up the tables, to help clarify the process.

The basic idea is that you start with a listing the columns for theta and sin(theta), using the standard values for theta.

Then for example if theta = pi / 4 t + pi / 3, you solve for t and substitute your theta values to get the corresponding t values.