assignment 10

course mth164

assignment #010

010.

Precalculus II

12-03-2008

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20:21:06

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

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RESPONSE -->

-3cos4pi=x

-3sin4pi=y

x=-2.928

y=-.652

confidence assessment: 2

(-3, 4 pi) corresponds to 2 complete revolutions, corresponding to the angle 4 pi, which directs you along the positive x axis. However r = -3 indicates that we move 3 units in the opposite direction, so we'll end up 3 units to the left of the origin and on the x axis.
This point could also be described by the polar coordinates (3, pi) or (3, -pi), or (-3, 0).

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20:21:59

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

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RESPONSE -->

these are the given x and y values that show the polar coordinates in the 3rd quadrant

confidence assessment: 2

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20:24:15

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 -->

the other possible coordinates are the same but in the 1rst quadrant becuase they are angular. You use the given triangle is shown and you then take the numbers and crreate a figure and extend all lines to find the results

confidence assessment: 2

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20:29:07

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

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RESPONSE -->

this uses the same polar formula and then you find and create a rectangle instead of a triangle

3.09

.108

confidence assessment: 2

Good.

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20:29:40

What are the rectangular coordinates of the given point?

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RESPONSE -->

they are the furthermost point from the origin which is in the rectangle

confidence assessment: 2

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20:29:58

Explain how you obtained the rectangular coordinates of this point.

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RESPONSE -->

by using the polar coordinates and the deriving a rectangle

confidence assessment: 2

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20:31:13

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

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RESPONSE -->

if -2.1 is a degree

-.799=x

.029=y

confidence assessment: 2

The given point is in rectangular coordinates; you are to find its polar coordinates.
For the rectangular coordinates of (-.8, -2.1) to find the polar coordinates:
we know that r^2 = x^2 + y^2
so thus we know r = sqrt( (-.8) ^2 + (-2.1) ^2)
so r = sqrt( .64 + 4.41)
r = sqrt 5.05
r = 2.247.
To find theta we use tan (theta) = y / x
So tan(theta) = (-2.1)/ (-.8)
tan (theta) = 2.625 and x < 0 so
theta = 69.145 deg + 180 deg = 249.145 deg
or about 249 degrees.
Thus the polar coordinate would be as follows,
( 2.2, 249 deg).
In radians we have arctan(2.625) = 1.21; adding pi radians because x < 0 we get 4.35 rad.
So the coordinates are (2.2, 4.35), approx..

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20:31:36

Give the polar coordinates of the given point.

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RESPONSE -->

they are as were given prevoiusly

confidence assessment: 2

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20:32:11

Explain how you obtained each of these polar coordinates.

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RESPONSE -->

they are the rsintheta for y and for x rcostheta

confidence assessment: 2

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20:34:47

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

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RESPONSE -->

(y^2,1)=(x,x)

confidence assessment: 1

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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20:35:23

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

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RESPONSE -->

r,o=r,o

confidence assessment: 1

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20:37:27

Explain how you obtained this equation.

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RESPONSE -->

because the polar coordinate is a point on the line of a shape and it does not change the shape of it by being ^2

confidence assessment: 1

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20:39:08

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

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RESPONSE -->

r,1=cos-3/3,1

confidence assessment: 1

Multiply both sides of the equation by 3-cos(theta) to get

3r – r cos(theta)=3.

Substitute sqrt(x^2+y^2) for r and x for r cos(theta) to get

3sqrt(x^2+y^2)-x=3. Add x to both sides to obtain

3sqrt(x^2+y^2)=3 + x and square both sides:

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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20:39:41

What is the rectangular coordinate form of the given equation?

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RESPONSE -->

a rectangle in the 1rst quadrant

confidence assessment: 1

The equation is that of an ellipse. See my response above.

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20:40:02

Explain how you obtained this equation.

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RESPONSE -->

by using the rect coordinate and a graph

confidence assessment: 1

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20:41:19

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

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RESPONSE -->

ok it goes throught the origin and travles over and under the x axis

confidence assessment: 2

The graph never goes beneath the x axis.

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 closed arc in the upper half-plane, tangent to the x axis at the origin.

This description doesn’t prove that the graph is a circle, but it turns out to be a circle whose radius is 1. The easiest way to prove this is to convert the equation to rectangular coordinates, as follows:

We can multiply both sides by r to get
r ^2 = 2 r sin (theta).

Substituting x^2 + y^2 for r and y for r sin(theta) we have

x ^2 + y ^2 = 2 y
x ^2 + ( y ^2 -2 y ) = 0
x ^ 2 + ( y ^2 - 2 y + 1 – 1 ) = 0
x ^2 + ( y -1 ) ^2 - 1 = 0
x^2 + (y-1)^2 = 1.

This is the standard form of the equation of a circle with center (0, 1) and radius 1. This is the circle described above.

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20:41:42

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

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RESPONSE -->

already did previously

confidence assessment: 2

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20:42:46

Explain how you obtained your graph.

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RESPONSE -->

by graphing it because 2 times the sin of x will show all values the are y of anything

confidence assessment: 2

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20:43:28

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 because the function travels the same on both sides of the x axis

confidence assessment: 2

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20:44:53

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

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RESPONSE -->

the cos function is stretched and passes the y axis at 6 with a small faining and the frequency is lengthened substantially

confidence assessment: 2

This graph is symmetric with respect to the pole, since replacing theta by -theta doesn't affect the equation (this is so because 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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20:45:09

Explain how you obtained your graph.

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RESPONSE -->

by modifying the cos function

confidence assessment: 2

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20:45:25

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 because it repeats

confidence assessment: 2

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20:46:50

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

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RESPONSE -->

i used the degree mode to calculate the first few problems instead of the radians, also I realize that all things done in this section are tied together

confidence assessment: 2

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You need to go over the given solutions carefully, and I suggest you submit a copy of this document with detailed self-critiques so we can be sure you understand these concepts and procedures.