Qa 2

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course Mth 164

February 8 1:11 AM

Question: `q001. Note that this assignment has 9 activities.

If the red ant (review preceding qa if necessary) moves at an angular velocity of pi/6 radians every second, starting from the standard initial point, then

what will be its angular position at the end of each of the first 12 seconds?

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Your solution:

You would change by pi/6 every second to each position so it would be pi/6, 2pi/6 or pi/3, 3pi/6 or pi/2, 4pi/6 or 2pi/3, 5pi/6, 6pi/6 or pi, 7pi/6, 8pi/6 or

4pi/3, 9pi/6 or 3pi/2, 10pi/6 or 5pi/3, 11pi/6 and 12pi/6 or 2pi.

confidence rating #$&*:

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3

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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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.

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Your solution:

pi/6, 2pi/6 or pi/3, 3pi/6 or pi/2, 4pi/6 or 2pi/3, 5pi/6, 6pi/6 or pi, 7pi/6, 8pi/6 or 4pi/3, 9pi/6 or 3pi/2, 10pi/6 or 5pi/3, 11pi/6 and 12pi/6 or 2pi.

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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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)??

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Your solution:

The 2/3 point would be between pi/2 and pi so the 2/3 points would either be 2pi/3 or 5pi/6. The 2/3 point would be 5pi/6.

The 1/3 point between pi/2 and pi would be 2pi/3.

The points between pi and 3pi/2 are 7pi/6 and 4pi/3. 4pi/3 would be at the 2/3 position and 7pi/6 would be at the 1/3 position.

The points between 3pi/2 and 2pi would be 1/3 at 5pi/3 and 2/3 at 11pi/6.

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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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?

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Your solution:

The ant would be moving pi/4 radians per second so the position would be pi/4, 2pi/4 or pi/2, 3pi/4, 4pi/4 or pi, 5pi/4, 6pi/4 or 3pi/2, 7pi/4 and 8pi/4 or

2pi.

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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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.

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Your solution:

pi/4, 2pi/4 or pi/2, 3pi/4, 4pi/4 or pi, 5pi/4, 6pi/4 or 3pi/2, 7pi/4 and 8pi/4 or 2pi.

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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Question: `q006. Sketch a unit circle (i.e., a circle of radius 1) 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)?

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Your solution:

1/2 point between (0,1) pi/2 and (-1,0) pi would be 3pi/4

1/2 point between (0,-1) 3pi/2 and (1,0) 2pi would be 7pi/4

1/2 point between(0,-1) 3pi/2 and (0,-1) 3pi/2 that is the same positions so there is no halfway point.

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

I think the last question wasn't stated right because it shows different in the given solution.

@&

You are correct.

*@

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Self-critique Rating:

3

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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.

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Your solution:

If the ant is moving pi/3 radians every second starting at pi/3 then you would move pi/3 radians to the next position which would be 2pi/3 then pi, 4pi/3

then 5pi/3 and then 2pi.

@&

You have listed 5 new positions.

The ant would be at position 2 pi after 5 seconds.

In another second it would be at position 7 pi / 3.

Note that this position is coterminal with the original pi/3 position.

*@

confidence rating #$&*:

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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.

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Self-critique (if necessary):

2

I am confused. If I am moving pi/3 radians around a circle and my positions are as below:

pi/3 x

pi/2

2pi/3 x

5pi/6

pi x

7pi/6

4pi/3 x

3pi/2

5pi/3 x

11pi/6

2pi x

pi/6

??? How did the solution give 7pi/3???

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Self-critique Rating:

3

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Question: `q008. Where is the angular position 7 pi/3 located?

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Your solution:

I don't know where 7pi/3 is on the circle. I thought it stopped at 2pi.

@&

The ant could certainly keep moving around the circle. There is no need for it to stop at the 2 pi position. This would in fact be unlikely, since the ant probably has no awareness of where on the circle it is.

*@

confidence rating #$&*:

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1

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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.

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Self-critique (if necessary):

???Where is the figure that the given solution is refering to???

@&

7 pi / 3 is the position pi/3 radians past the 2 pi position.

It is therefore at the same point on the circle as at the start.

*@

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Self-critique Rating:

1

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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.

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Your solution:

Moving by pi/4 the positions would be pi/3 + pi/4=7pi/12, pi/3 + 2 pi/4=5pi/12, pi/3 + 3 pi/4 =13pi/12, pi/3 + 4 pi/4=4pi/12, pi/3 + 5 pi/4=19pi/12, pi/3 + 6

pi/4=11pi/6, pi/3 + 7 pi/4=25pi/12 and pi/3 + 8 pi/4= 7pi/3.

@&

Good.

*@

confidence rating #$&*:

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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.

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Self-critique (if necessary):

OK

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Self-critique Rating:

3

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Question: `q010. Starting at angular position pi/4 and moving at pi/3 radians / second what will be the resulting positions after each of the first 6

seconds?

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Your solution:

pi/4+pi/3=7pi/12

pi4+2pi/3=11pi/12

pi4+3pi/3=5pi/4

pi4+4pi/3=19pi/12

pi4+5pi/3=23pi/12

pi4+6pi/3=9pi/4

confidence rating #$&*:

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3

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Question: `q011. 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.

So after 12 seconds we will have moved through an arc of 12 pi / 6 radians. Since 12 pi / 6 reduces to 2 pi, we will have moved through an arc of 2 pi

radians, and we will be back at our starting point.

If we continue to move around the circle for one more second we will have moved, in 13 seconds, through a total angle of 13 pi / 6 radians, and we will be at

the same point on the circle as when we had moved through pi/6 radians.

Through what total angle will we have moved by the end of each of the next 4 seconds, and at what previously visited point on the circle will we be located

at the end of each?

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Your solution:

14pi/6 or 7pi/3 which is pi/3; 15pi/6 which is pi/2; 16pi/6 or8pi/3 which is 2pi/3; 17pi/6 which is 5pi/6.

confidence rating #$&*:

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Question: `q012. This is an optional challenge question. If we start from angular position 0 and move through 7 pi / 4 radians every second, through what

total angle will we have moved and where on the circle will we be at the end of each of the first 4 seconds?

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Your solution:

o would equal to 12pi/12, so each second would be:

12pi/12 + 7pi/4radians = 11pi/4 radians

12pi/12 + 2*7pi/4 radians= 9pi/2

12pi/12 + 3*7pi/4 radians= 25pi/4

12pi/12 + 4*7pi/4 radians = 8pi

We will have completed 4 full circles and would be at 48pi/12.

confidence rating #$&*:

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3

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&#This looks good. See my notes. Let me know if you have any questions. &#