query_asst_10

course Mth 164

010.

Precalculus II

04-19-2007

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

23:22:14

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

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

RESPONSE -->

confidence assessment:

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

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

23:23:21

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

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

RESPONSE -->

The point (-3, 4'pi) has an angle 'theta of 4'pi and an r value of -3. So it is at the angle 4pi, which is identical to 2pi and back -3 units so it actually lies on 'theta = pi.

confidence assessment: 3

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

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

23:24:25

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

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

RESPONSE -->

(3,pi)

(-3, 2pi)

(-3,0)

I found these by looking at the graph and figuring out the positions dealing with pi and 2pi and also 0.

confidence assessment: 3

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

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

23:24:30

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

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

RESPONSE -->

confidence assessment:

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

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

23:24:46

What are the rectangular coordinates of the given point?

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

RESPONSE -->

(3.098, 0.108)

confidence assessment: 3

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

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

23:25:44

Explain how you obtained the rectangular coordinates of this point.

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

RESPONSE -->

The set (-3.1, 182 deg) is given r is -3.1 and 'theta = 182 deg. it is known that x = rcos'theta so x = -3.1 cos 182 = 3.098 approx and y = r sin 'theta = -3.1 sin 182 deg = 0.108 approx.

confidence assessment: 3

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

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

23:25:51

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

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

RESPONSE -->

confidence assessment:

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

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

23:26:04

Give the polar coordinates of the given point.

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

RESPONSE -->

(2.25, 249.15deg)

confidence assessment: 3

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

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

23:27:46

Explain how you obtained each of these polar coordinates.

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

RESPONSE -->

The point (-.8, -2.1) is in the third quadrant so to find 'theta I used 'theta = 180 + tan ^(-1) (y/x) = 180 + tan ^(-1)(-2.1/-.8) = 249.15 deg.

Then I found r using the distance formula r = sqrt(x^2+y^2) = sqrt((-.8)^2+(-2.1)^2 = sqrt(5.05)= 2.25.

confidence assessment: 3

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

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

23:27:58

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

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

RESPONSE -->

confidence assessment:

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

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

23:28:12

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

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

RESPONSE -->

sin ^2'theta = 2 cos 'theta

confidence assessment: 1

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

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

23:28:54

Explain how you obtained this equation.

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

RESPONSE -->

y^2 would be equal to sin^2'theta and 2x equal to 2 cos 'theta since the y value is also the sin 'theta and x the cos 'theta.

confidence assessment: 1

You left out the factor r in the expressions for x and for y. You should have obtained r sin^2(theta) = 2 cos(theta), or equivalent.

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.

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

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

23:28:59

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

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

RESPONSE -->

confidence assessment:

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

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

23:29:15

What is the rectangular coordinate form of the given equation?

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

RESPONSE -->

I am not sure of how to go about solving this.

confidence assessment: 1

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

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

23:29:26

Explain how you obtained this equation.

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

RESPONSE -->

I multiplied both sides by r and then am stuck.

confidence assessment:

** Since r cos(`theta) = x, cos(`theta) = x / r = x / `sqrt(x^2 + y^2).

So x^2 + y^2 = 9 / (cos(`theta) - 3)^2 = 9 / ( 3 - x / `sqrt(x^2 + y^2)) ^ 2.

Multiplying both sides by common denominator:

(3 - x / `sqrt(x^2 + y^2) )^2 * (x^2 + y^2) = 9. Expanding the left-hand side we get

[ x^2 / (x^2 + y^2) - 6 x / `sqrt(x^2 + y^2) + 9 ] * ( x^2 + y^2) = 9 or

x^2 - 6 x `sqrt(x^2 + y^2) + 9(x^2 + y^2) = 9, or

10 x^2 + 9 y^2 - 6x `sqrt(x^2 + y^2) - 9 = 0. **

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

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

23:29:32

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

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

RESPONSE -->

confidence assessment:

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

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

23:29:54

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

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

RESPONSE -->

The graph of r = 2sin'theta is a circle with radius 1 and center at (0,1)

confidence assessment: 3

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

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

23:30:27

Explain how you obtained your graph.

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

RESPONSE -->

r = 2sin'theta is in the form r = 2asin'theta with a = 1. It is known that this is a circle with radius a and center at (0,a)

confidence assessment: 3

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

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

23:30:49

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 but the graph is symmetric about the y axis so this could be used.

confidence assessment: 3

You should be able to do this sort of problem analytically. See 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.

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

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

23:30:54

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

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

RESPONSE -->

confidence assessment:

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

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

23:31:48

Explain how you obtained your graph.

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

RESPONSE -->

I found the graph of r = 2+4cos'theta to be symmetric about the x axis so I found the values of r for angles 0<='theta<=pi which gave me a limacon with inner loop.

confidence assessment: 3

Good. More detail, for comparison and for reference:

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.

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

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

23:32:36

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

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

RESPONSE -->

The graph had polar symmetry so I found the r values for 0<='theta<=pi and reflected them across the x axis in order to complete the graph.

confidence assessment: 3

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

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

23:32:42

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

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

RESPONSE -->

confidence assessment:

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

query_asst_10

Your work on this assignment is good. See my notes and let me know if you have questions.