course mth164

Assignment #10 last one

??????????assignment #010

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

010.

Precalculus II

07-11-2007

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10:34:13

Query problem 8.1.24 polar coordinates of (-3, 4`pi)

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RESPONSE -->

ok

confidence assessment:

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10:36:03

Describe the position of the given point on the polar coordinate axes.

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RESPONSE -->

The given point since it is negative is in the direction opposite the terminal side. The distance from the vertex is 3 and the angle is 4pi.

confidence assessment: 3

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10:38:27

Give other possible polar coordinates for the same point and describe in terms of the graph how you obtained these coordinates.

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RESPONSE -->

(-3,2pi), (-3,6pi)

confidence assessment: 2

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10:38:43

Query problem 8.1.42 rect coord of (-3.1, 182 deg)

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RESPONSE -->

ok

confidence assessment:

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10:40:50

What are the rectangular coordinates of the given point?

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RESPONSE -->

x=-3.1cos(182) = 3.098

y = -3.1sin(182) = .1082

(3.098,.1082)

confidence assessment: 3

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10:41:12

Explain how you obtained the rectangular coordinates of this point.

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RESPONSE -->

I used the formulas x=rcos(theta) and x=rsin(theta)

confidence assessment: 3

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10:43:54

Query problem 8.1.54 polar coordinates of (-.8, -2.1)

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RESPONSE -->

(2.25,1.21) (angle in radians)

or

(2.25,69.1) (angle in degrees)

confidence assessment: 3

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10:45:02

Give the polar coordinates of the given point.

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RESPONSE -->

(2.25,1.21) (angle in radians)

or

(2.25,69.1) (angle in degrees)

r was found by sqrt((-.8)^2+(-2.1)^2)

theta was tan ^-1(-2.1/-.8)

confidence assessment: 3

This point is in the third quadrant so the angle would be pi + 1.21 rad, or 69.1 deg + 180 deg.

When the x component is negative the angle is in the second or third quadrant; the range of the arctan is the fourth and first quadarnt so when x is negative you need to add pi rad or 180 deg to arctan(y/x).

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10:45:17

Explain how you obtained each of these polar coordinates.

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RESPONSE -->

I explained in the previous answer.

confidence assessment:

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10:45:36

Query problem 8.1.60 write y^2 = 2 x using polar coordinates.

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RESPONSE -->

ok

confidence assessment:

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10:48:02

What is the polar coordinate form of the equation y^2 = 2x?

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RESPONSE -->

rsin^2(theta) = cos(theta)

confidence assessment: 2

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10:49:28

Explain how you obtained this equation.

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RESPONSE -->

replace y with rsin(theta) and x with rcos(theta)

and get (rsin(theta))^2=2(rcos(theta))

simplified

r^2sin^2(theta) = 2rcos(theta)

divide by r

rsin^2(theta)=2cos(theta)

Note: I think I forgot to include the 2 in the right side of the equation in my previous answer

confidence assessment: 3

Good. The following includes an additional solution and extra step or two of simplification:

** We can rearrange the equation to give the form

y^2 - 2 x = 0. Then since y = r sin (theta) and x = r cos (theta) we have

(r^2 sin ^2 theta) - ( 2 r cos (theta) = 0.

We can factor out r to get

r [r sin^2(theta) - 2 cos(theta) ] = 0,

which is equivalent to

r = 0 or r sin^2(theta) - 2 cos(theta) = 0.

The latter form can be solved for r. We get

r = 2 cos(theta) / sin^2(theta).

This form is convenient for graphing. **

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10:52:24

Query problem 8.1.72 rect coord form of r = 3 / (3 - cos(`theta))

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RESPONSE -->

ok

confidence assessment:

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10:52:45

What is the rectangular coordinate form of the given equation?

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RESPONSE -->

3sqrt(x^2+y^2)-x=3

confidence assessment: 3

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10:54:07

Explain how you obtained this equation.

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RESPONSE -->

multiply both sides of the equation by 3-cos(theta) and get 3r-rcos(theta)=3

substitute for r sqrt(x^2+y^2) and for rcos(theta) x and get

3sqrt(x^2+y^2)-x=3

confidence assessment: 3

Good. If you write your equation as

3sqrt(x^2+y^2)=3 + x

and square both sides you get

3(x^2 + y^2) = x^2 + 6 x + 9, which simplifies to

2 x^2 + y^2 - 6x - 9 = 0.

Completing the square on 2 x^2 - 6x we get

2( x^2 - 3 x + 9/4 - 9/4 ) + y^2 = 9 so

2 ( x+3/2)^2 - 9/2 + y^2 = 9 so

2 ( x + 3/2)^2 + y^2 = 9/2 so

(x+3/2)^2 / (9/4) + y^2 / (9/2) = 1.

This is the equation of an ellipse centered at (-3/2, 0) with semi-axes 3/2 in the x direction and 3 sqrt(2) / 2 in the y direction.

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10:54:21

Query problem 8.2.10 graph r = 2 sin(`theta).

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RESPONSE -->

ok

confidence assessment:

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10:56:26

Describe your graph of r = 2 sin(`theta).

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RESPONSE -->

circle with center at (0,1) and radius 1

confidence assessment: 3

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10:58:03

Explain how you obtained your graph.

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RESPONSE -->

First I converted to rectangular coordinates by multiplying by r, substituting for r^2=x^2+y^2 and for rsin(theta) = y

Giving

x^2+y^2=2y

Subtracting 2y and completing the square gives

x^2 + (y-1)^2 = 1 which is a circle with center (0,1) and radius 1

confidence assessment: 3

Good. Compare with the following:

When `theta = 0 or `pi, r = 0 and the graph point coincides with the origin.

If `theta = `pi/2, r = 2. If `theta = `pi/4, r = `sqrt(2). So as `theta increases from 0 to `pi/2, r increases from 0 to 2 and the graph follows an arc (actually the arc of a circle) in the first quadrant from the origin to (2, `pi/2).

As `theta moves from `pi/2 to `pi the graph follows an arc in the second quadrant which leads back to the origin.

As `theta goes from `pi to 3 `pi/2, angles in the third quadrant, r becomes negative since sin(`theta) is negative. At 3 `pi / 2, r will be -2 and the point (-2, 3 `pi / 2) coincides with (2, `pi/2). The graph follows the same arc as before in the first quadrant.

As `theta goes from 3 `pi / 2 to 2 `pi, r will remain negative, which places the graph along the same second-quadrant arc as before.

Thus the graph will consist of a circle sitting on the x axis at the origin, having radius 1.

** Conversion to rectangular coordinates also yields the same circle:

we can multiply both sides by r to get

r ^2 = 2 r sin (theta).

x ^2 + y ^2 = 2 y

x ^2 + ( y ^2 -2 y ) = 0

x ^ 2 + ( y ^2 - 2 y + 1 ) = 1

x ^2 + ( y -1 ) ^2 = 1.

This circle has center (0, 1) with radius 1. This is the same as the circle described above. **

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10:58:30

Was it possible to use symmetry in any way to obtain your graph, and if so how did you use it?

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RESPONSE -->

I did not use symmetry. I used algebra and conic sections.

confidence assessment: 2

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10:58:59

Query problem 8.2.36 graph r=2+4 cos `theta

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RESPONSE -->

ok

confidence assessment:

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11:06:34

Explain how you obtained your graph.

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RESPONSE -->

First test for polar axis symmetry by replacing theta with -theta and it passes.

Test for line theta=pi/2, by replacing theta with pi-theta, test fails.

Test for the pole by replacing r with -r and fails,.

since a

Graph is on the xaxis extending in the positive direction to 6 with an inner loop extending to 2.

confidence assessment: 2

You probably understand the following details already, but in case they are helpful I'm including them here:

** This graph is symmetric with respect to the pole, since replacing theta by -theta doesn't affect the equation (since cos(-theta) = cos(theta) ).

At theta goes from 0 to pi/2, cos(theta) goes from 1 to 0 so that r will go from 6 to 2.

Then when theta = pi/3 we have cos(theta) = -1/2 so that r = 0. As theta goes from pi/2 to pi/3, then, r goes from 2 to 0. To this point the graph forms half of a heart-shaped figure lying above the x axis.

Between theta = pi/3 and theta = 2 pi / 3 the value of cos(theta) will go from -1/2 to -1 to -1/2, so that r will go from 0 to -2 and back to 0. The corresponding points will move from the origin to the 4th quadrant as theta goes past pi / 3, reaching the point 2 units along the pole when theta = pi then moving into the first quadrant, again reaching the origin when theta = 2 pi/3. These values will form an elongated loop inside the heart-shaped figure.

From theta = 2 pi / 3 thru theta = 3 pi / 2 and on to theta = 2 pi the values of r will go from 0 to 2 then to 6, forming the lower half of the heart-shaped figure. **

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11:06:51

Was it possible to use symmetry in any way to obtain your graph, and if so how did you use it?

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RESPONSE -->

Yes, it was symmetric to the polar axis

confidence assessment: 3

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11:07:21

Comm on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

The polar graphs were tricky when you weren't able to convert them readily to rectangular coordinates.

confidence assessment:

.................................................

"

&#

Good responses. See my notes and let me know if you have questions. &#

course mth164

Assignment #10 last one

??????????assignment #010

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

010.

Precalculus II

07-11-2007

......!!!!!!!!...................................

10:34:13

Query problem 8.1.24 polar coordinates of (-3, 4`pi)

......!!!!!!!!...................................

RESPONSE -->

ok

confidence assessment:

.................................................

......!!!!!!!!...................................

10:36:03

Describe the position of the given point on the polar coordinate axes.

......!!!!!!!!...................................

RESPONSE -->

The given point since it is negative is in the direction opposite the terminal side. The distance from the vertex is 3 and the angle is 4pi.

confidence assessment: 3

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10:38:27

Give other possible polar coordinates for the same point and describe in terms of the graph how you obtained these coordinates.

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RESPONSE -->

(-3,2pi), (-3,6pi)

confidence assessment: 2

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10:38:43

Query problem 8.1.42 rect coord of (-3.1, 182 deg)

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RESPONSE -->

ok

confidence assessment:

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10:40:50

What are the rectangular coordinates of the given point?

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RESPONSE -->

x=-3.1cos(182) = 3.098

y = -3.1sin(182) = .1082

(3.098,.1082)

confidence assessment: 3

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10:41:12

Explain how you obtained the rectangular coordinates of this point.

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RESPONSE -->

I used the formulas x=rcos(theta) and x=rsin(theta)

confidence assessment: 3

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10:43:54

Query problem 8.1.54 polar coordinates of (-.8, -2.1)

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RESPONSE -->

(2.25,1.21) (angle in radians)

or

(2.25,69.1) (angle in degrees)

confidence assessment: 3

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10:45:02

Give the polar coordinates of the given point.

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RESPONSE -->

(2.25,1.21) (angle in radians)

or

(2.25,69.1) (angle in degrees)

r was found by sqrt((-.8)^2+(-2.1)^2)

theta was tan ^-1(-2.1/-.8)

confidence assessment: 3

This point is in the third quadrant so the angle would be pi + 1.21 rad, or 69.1 deg + 180 deg.

When the x component is negative the angle is in the second or third quadrant; the range of the arctan is the fourth and first quadarnt so when x is negative you need to add pi rad or 180 deg to arctan(y/x).

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10:45:17

Explain how you obtained each of these polar coordinates.

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RESPONSE -->

I explained in the previous answer.

confidence assessment:

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10:45:36

Query problem 8.1.60 write y^2 = 2 x using polar coordinates.

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RESPONSE -->

ok

confidence assessment:

.................................................

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10:48:02

What is the polar coordinate form of the equation y^2 = 2x?

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RESPONSE -->

rsin^2(theta) = cos(theta)

confidence assessment: 2

.................................................

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10:49:28

Explain how you obtained this equation.

......!!!!!!!!...................................

RESPONSE -->

replace y with rsin(theta) and x with rcos(theta)

and get (rsin(theta))^2=2(rcos(theta))

simplified

r^2sin^2(theta) = 2rcos(theta)

divide by r

rsin^2(theta)=2cos(theta)

Note: I think I forgot to include the 2 in the right side of the equation in my previous answer

confidence assessment: 3

Good. The following includes an additional solution and extra step or two of simplification:

** We can rearrange the equation to give the form

y^2 - 2 x = 0. Then since y = r sin (theta) and x = r cos (theta) we have

(r^2 sin ^2 theta) - ( 2 r cos (theta) = 0.

We can factor out r to get

r [r sin^2(theta) - 2 cos(theta) ] = 0,

which is equivalent to

r = 0 or r sin^2(theta) - 2 cos(theta) = 0.

The latter form can be solved for r. We get

r = 2 cos(theta) / sin^2(theta).

This form is convenient for graphing. **

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10:52:24

Query problem 8.1.72 rect coord form of r = 3 / (3 - cos(`theta))

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RESPONSE -->

ok

confidence assessment:

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10:52:45

What is the rectangular coordinate form of the given equation?

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RESPONSE -->

3sqrt(x^2+y^2)-x=3

confidence assessment: 3

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10:54:07

Explain how you obtained this equation.

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RESPONSE -->

multiply both sides of the equation by 3-cos(theta) and get 3r-rcos(theta)=3

substitute for r sqrt(x^2+y^2) and for rcos(theta) x and get

3sqrt(x^2+y^2)-x=3

confidence assessment: 3

Good. If you write your equation as

3sqrt(x^2+y^2)=3 + x

and square both sides you get

3(x^2 + y^2) = x^2 + 6 x + 9, which simplifies to

2 x^2 + y^2 - 6x - 9 = 0.

Completing the square on 2 x^2 - 6x we get

2( x^2 - 3 x + 9/4 - 9/4 ) + y^2 = 9 so

2 ( x+3/2)^2 - 9/2 + y^2 = 9 so

2 ( x + 3/2)^2 + y^2 = 9/2 so

(x+3/2)^2 / (9/4) + y^2 / (9/2) = 1.

This is the equation of an ellipse centered at (-3/2, 0) with semi-axes 3/2 in the x direction and 3 sqrt(2) / 2 in the y direction.

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10:54:21

Query problem 8.2.10 graph r = 2 sin(`theta).

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RESPONSE -->

ok

confidence assessment:

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10:56:26

Describe your graph of r = 2 sin(`theta).

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RESPONSE -->

circle with center at (0,1) and radius 1

confidence assessment: 3

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10:58:03

Explain how you obtained your graph.

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RESPONSE -->

First I converted to rectangular coordinates by multiplying by r, substituting for r^2=x^2+y^2 and for rsin(theta) = y

Giving

x^2+y^2=2y

Subtracting 2y and completing the square gives

x^2 + (y-1)^2 = 1 which is a circle with center (0,1) and radius 1

confidence assessment: 3

Good. Compare with the following:

When `theta = 0 or `pi, r = 0 and the graph point coincides with the origin.

If `theta = `pi/2, r = 2. If `theta = `pi/4, r = `sqrt(2). So as `theta increases from 0 to `pi/2, r increases from 0 to 2 and the graph follows an arc (actually the arc of a circle) in the first quadrant from the origin to (2, `pi/2).

As `theta moves from `pi/2 to `pi the graph follows an arc in the second quadrant which leads back to the origin.

As `theta goes from `pi to 3 `pi/2, angles in the third quadrant, r becomes negative since sin(`theta) is negative. At 3 `pi / 2, r will be -2 and the point (-2, 3 `pi / 2) coincides with (2, `pi/2). The graph follows the same arc as before in the first quadrant.

As `theta goes from 3 `pi / 2 to 2 `pi, r will remain negative, which places the graph along the same second-quadrant arc as before.

Thus the graph will consist of a circle sitting on the x axis at the origin, having radius 1.

** Conversion to rectangular coordinates also yields the same circle:

we can multiply both sides by r to get

r ^2 = 2 r sin (theta).

x ^2 + y ^2 = 2 y

x ^2 + ( y ^2 -2 y ) = 0

x ^ 2 + ( y ^2 - 2 y + 1 ) = 1

x ^2 + ( y -1 ) ^2 = 1.

This circle has center (0, 1) with radius 1. This is the same as the circle described above. **

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10:58:30

Was it possible to use symmetry in any way to obtain your graph, and if so how did you use it?

......!!!!!!!!...................................

RESPONSE -->

I did not use symmetry. I used algebra and conic sections.

confidence assessment: 2

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10:58:59

Query problem 8.2.36 graph r=2+4 cos `theta

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RESPONSE -->

ok

confidence assessment:

.................................................

......!!!!!!!!...................................

11:06:34

Explain how you obtained your graph.

......!!!!!!!!...................................

RESPONSE -->

First test for polar axis symmetry by replacing theta with -theta and it passes.

Test for line theta=pi/2, by replacing theta with pi-theta, test fails.

Test for the pole by replacing r with -r and fails,.

since a

Graph is on the xaxis extending in the positive direction to 6 with an inner loop extending to 2.

confidence assessment: 2

You probably understand the following details already, but in case they are helpful I'm including them here:

** This graph is symmetric with respect to the pole, since replacing theta by -theta doesn't affect the equation (since cos(-theta) = cos(theta) ).

At theta goes from 0 to pi/2, cos(theta) goes from 1 to 0 so that r will go from 6 to 2.

Then when theta = pi/3 we have cos(theta) = -1/2 so that r = 0. As theta goes from pi/2 to pi/3, then, r goes from 2 to 0. To this point the graph forms half of a heart-shaped figure lying above the x axis.

Between theta = pi/3 and theta = 2 pi / 3 the value of cos(theta) will go from -1/2 to -1 to -1/2, so that r will go from 0 to -2 and back to 0. The corresponding points will move from the origin to the 4th quadrant as theta goes past pi / 3, reaching the point 2 units along the pole when theta = pi then moving into the first quadrant, again reaching the origin when theta = 2 pi/3. These values will form an elongated loop inside the heart-shaped figure.

From theta = 2 pi / 3 thru theta = 3 pi / 2 and on to theta = 2 pi the values of r will go from 0 to 2 then to 6, forming the lower half of the heart-shaped figure. **

.................................................

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11:06:51

Was it possible to use symmetry in any way to obtain your graph, and if so how did you use it?

......!!!!!!!!...................................

RESPONSE -->

Yes, it was symmetric to the polar axis

confidence assessment: 3

.................................................

......!!!!!!!!...................................

11:07:21

Comm on any surprises or insights you experienced as a result of this assignment.

......!!!!!!!!...................................

RESPONSE -->

The polar graphs were tricky when you weren't able to convert them readily to rectangular coordinates.

confidence assessment:

.................................................

"

&#

Good responses. See my notes and let me know if you have questions. &#