Assignment 4 qa

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course Mth 164

Question:  `q001.  In the preceding assignment we saw how to model the sine function using a circle of radius 1. 

Now we consider 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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Your solution: 

 point b because we know that one radian is slightly less that 60 degrees.

 

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Given Solution: 

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 affect the picture; the radius simply determines the scale at which the picture is interpreted.

 

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

 

 

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Your solution: 

 

3*pi/6= pi/2 

 

 

confidence rating #$&*3

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Given Solution: 

`aOn 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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Question:  `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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Your solution: 

 2/3 rads/sec

 

 

confidence rating #$&*3

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Given Solution: 

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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Question:  `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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Your solution: 

  

 

 5 rads/sec with a radius of 3 therefore:

5*3= 15 moves in distance per second

 

confidence rating #$&*3

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Given Solution: 

`aEach 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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Question:  `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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Your solution: 

 

 0 0

pi/6 1/2

pi/3 sqrt(3)/2

pi/2 1

2pi/3 sqrt(3)/2

5pi/6 1 /2

pi 0

7pi/6 -1/2

4pi/3 -sqrt(3)/2

3pi/2 -1

5pi/3 -sqrt(3)/2

11pi/6 -1/2

2pi 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/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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Question:  `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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Your solution: 

 It must be 3 times the value there since we have a radius of 3.

So therefore we get:

 0, 1.5, 2.61, 3, 2.61, 1.5, 0, -1.5, -2.61, -3, -2.61, -1.5, 0.

 

confidence rating #$&*3

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Given Solution: 

`aOn 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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Question:  `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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Your solution: 

 

 

 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.

 

Yes that are consistent with the values I obtained in problem 6. 

 

confidence rating #$&*3

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Given Solution: 

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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Question:  `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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Your solution: 

 

The graph is very similar to the ones we have made using sin(theta) and theta. However, this had a radius of 3 so everything is 3 times what it was before. 

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Given Solution: 

`aYour 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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Question:  `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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Your solution: 

 3 * sqrt(3) / 2, 0, -3 * sqrt(3) / 2, -3 * sqrt(3) / 2, 0 and 3 * sqrt(3) / 2

 (0, 2.6)

 (1, 2.6)

(2, 0)

(3, -2.6)

(4, -2.6)

(5, 0)

(6, 2.6)

the graph goes up for the first second then comes back down crossing through zero at 2 then reaching its low point at 3 seconds then crossing through zero again at 5.

confidence rating #$&*3

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Given Solution: 

`aThe 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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Question:  `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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Your solution: 

 

 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         3pi/4                 .71               2.1

 8/3            pi                    0                 0

11/3          5pi/4                -.71              -2.1

14/3          3pi/2                -1                 -1

17/3          7pi/4                -.71              -2.1

20/3           2 pi                   0                 0

 

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Given Solution: 

`aUsing 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."

&#Good work. Let me know if you have questions. &#

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