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http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
SOLUTIONS/COMMENTARY ON QUERY 10
Query problem 8.1.28 polar coordinates of (-3, 4`pi)
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10:38:28
Your solution:
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Given Solution:
** (-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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**** Query problem 8.1.42 rect coord of (-3.1, 182 deg)
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Your solution:
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Given Solution:
GOOD STUDENT SOLUTION:
since we know the formula for converting polar coordinates to rectangular
coordinates we can say that r=-3.1 and theta=182 deg. we can first find the
value of x by saying
x=r cos theta which gives
(-3.1) cos 182 deg. = 3.1.
We can then find y by saying y= r sin theta so y=(-3.1)sin 182 deg. using a
calculator we get an approx. value of .1081.
Thus the rectangular coordinates are (3.1, .1081), approx..
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Query problem 8.1.54 polar coordinates of (-.8, -2.1)
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18:50:13
Your solution:
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Given Solution:
STUDENT SOLUTION:
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.. **
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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**** Query problem 8.1.60 write y^2 = 2 x using polar coordinates.
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19:05:48
Your solution:
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Given Solution:
** 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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**** Query problem 8.1.52 exact polar coordinates of (-2, -2`sqrt(3))
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10:58:03
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Given Solution:
r^2=x^2+y^2 so
r^2=(-2)^2+(-2 sqrt(3))^2 = 4+12 = 16 so
so r=4.
tan theta= y/x = (-2 sqrt(3))/(-2) = sqrt(3).
This occurs for the basic angle theta = pi/3, and also for theta = pi/3 + pi = 4 pi/3.
The given point is in the third quadrant, so the angle is 4 pi/3.
The polar coordinates are therefore (4, 4 pi/3).
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**** Query problem 8.1.62 write 4 x^2 y = 1 using polar coordinates.
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11:05:39
Your solution:
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Given Solution:
GOOD STUDENT SOLUTION
we can first say that since x=r cos theta and y= r sin theta that 4(r cos
theta)^2(r sin theta)=1
we then have
4 r^3 cos theta sin theta=1
then using the double angle formula we have
2r^3(sin 2 theta)=1
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11:05:40
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**** Query problem 8.1.72 rect coord form of r = 3 / (3 -
cos(`theta))
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Your solution:
Confidence Assessment:
Given Solution:
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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**** Query problem 8.2.10 graph r = 2 sin(`theta).
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Your solution:
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Given Solution:
** 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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**** Was it possible to use symmetry in any way to obtain your graph,
and if so how did you use it?
Your solution:
Confidence Assessment:
Given Solution:
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11:54:29
yes the horizontal symmetry caused us to bound the graph at theta=pi/2 so r
has a maximum height of 2. and its minimum horizontal line is drawn at -2
this maximum and minimum happens alternatley at the intervals of pi/2 and
the negatives.
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11:54:30
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**** Query problem 8.2.36 graph r=2+4 cos `theta
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11:58:23
Your solution:
Confidence Assessment:
Given Solution:
** 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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11:58:24
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