If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
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 13
Asst # 13
04-01-2002
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20:43:50
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**** Query problem 9.4.12 graph of hyperbola with vertices at (-4,0) and (4,0), asymptote y = 2x **** Give the equation of the specified hyperbola.
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Your solution:
Confidence Assessment:
Given Solution:
** The equation of a hyperbola centered at the origin is
x^2 / a^2 - y^2 / b^2 = 1,
where (a, 0) and (-a, 0) are the vertices and y = +-b / a * x are the asymptotes.
In the present case we have a = 4 and y = 2 x is an asymptote.
It follows that b / a = 2 so that b = 2 a = 2 * 4 = 8.
Thus the equation is
x^2 / 4^2 - y^2 / 8^2 = 1 or
x^2 / 16 - y^2 / 64 = 1. **
Self-critique (if necessary):
Self-critique Rating:
Explain how you know whether the equation is x^2 / a^2 - y^2 / b^2 = 1, x^2 / a^2 + y^2 / b^2 = 1 or -x^2 / a^2 + y^2 / b^2 = 1.
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13:18:39
Your solution:
Confidence Assessment:
Given Solution:
** The point (4, 0) wouldn't make sense in an equation of the form -x^2 / a^2 + y^2 / b^2 = 1, since for this point -x^2 / a^2 would be negative and y^2 / b^2 would be positive, making the left-hand side negative while the right-hand side is 1.
For large x and y the value of x^2 / a^2 + y^2 / b^2 would be large and couldn't be equal to 1.
The form x^2 / a^2 - y^2 / b^2 = 1 makes sense for (4, 0) and also for large x and y, since the difference x^2 / a^2 - y^2 / b^2 could indeed equal 1 even for large x and y. **
Self-critique (if necessary):
Self-critique Rating:
**** Query problem 9.4.26 graph of hyperbola defined by rectangle of width 4, centered at the origin, lines y = + - 2x passing through diagonals, opening right and left. **** Give the equation of the specified hyperbola.
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Your solution:
Confidence Assessment:
Given Solution:
** The width of the rectangle is 4 so its vertical sides are at x = -2 and x = 2.
Since the parabola opens to the right the vertices are on the right and left sides of the rectangle, at (-2, 0) and (2, 0).
The lines y = +- 2 x are the asymptotes.
The form of the equation is x^2 / a^2 - y^2 / b^2 = 1, with vertices (a, 0) and (-a, 0) and asymptotes y = +- b / a * x.
It follows that a = 2 (vertices at (-a,0) and (a,0)) and b / a = 2 (since y = +-2 x = +- b / a * x).
We get b = 2 a = 2 * 2 = 4 so the equation is
x^2 / 2^2 - y^2 / 4^2 = 1 or
x^2 / 4 - y^2 / 16 = 1. **
20:59:30
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 9.6.6 conic defined by a r = 6 / (8 + 2 sin(`theta)). **** What conic is defined by the given polar equation?
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21:00:25
Your solution:
Confidence Assessment:
Given Solution:
Since e < 1 the conic is an elipse.
** Good. For reference two derivations are included below: **
** Derivation of forms of a conic:
If the directrix D oriented perpendicular to the polar axis lies at distance p in the negative direction from that axis, then the distance from point P = (r, `theta) to the directrix will be measured in the direction parallel to the polar axis. r cos(`theta) is the displacement of P with respect to the polar axis do this distance is
dist(D,P) = r cos(`theta) + p.
If the focus is at the pole then dist(F,P) = distance from pole to P = r.
The ratio of the latter distance to the former is the eccentricity so we have:
r / [ r cos(`theta) + p ] = e
r = e r cos(`theta) + e p
r - e r cos(`theta) = e p
r = e p / (1 - e cos(`theta) ).
For directrix p units above pole parallel to polar axis we have for point P
dist(D, P) = r sin(`theta) - p
dist(F, P) = r so
r / [ r sin(`theta) - p ] = e
r = e r sin(`theta) - e p
etc.
r = e p / (e r sin`theta - 1) **
** We can convert to rectangular coordinates. We can rearrange the equation to get
8 r + 2 r sin(`theta) = 6. Since r sin(`theta) = y and r = `sqrt(x^2+y^2) we have
8 `sqrt(x^2+y^2) + 2 y = 6. Getting the `sqrt on one side we have
`sqrt(x^2 + y^2) = (6-2y)/8. Squaring both sides we have
x^2 + y^2 = (36 - 24 y + 4y^2) / 64 so that
64 x^2 + 60 y^2 + 24 y = 36.
If we complete the square on y we do end up with a positive number on the right-hand side, so we indeed have an ellipse. **
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21:00:26
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Self-critique (if necessary):
Self-critique Rating:
**** How do you know that the conic is the one you indicated?
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21:02:19
Your solution:
Confidence Assessment:
Given Solution:
STUDENT SOLUTION: After dividing both numerator and denominator by 8, I get r = (3/4) / (1+ 1/4sin('theta)) so I know that e = 1/4 which is less than 1. This results in an elipse.
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21:02:20
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Self-critique (if necessary):
Self-critique Rating:
**** What is the position of the directrix, if one exists?
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21:03:05
Your solution:
Confidence Assessment:
Given Solution:
STUDENT ANSWER: The directrix is parallel to the polar axis at a distance of p = 3 above the pole.
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21:03:06
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 9.6.16 identify and graph r(2-cos(`theta)) = 2. **** Describe in detail the graph you obtain from the given equation.
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21:07:33
Your solution:
Confidence Assessment:
Given Solution:
The conic is an ellipse with e = 1/2.
** Good.
The equation rearranges to give you r = 2 / (2 - cos(`theta) );
dividing numerator and denominator by 2 you get r = 1 / (1 - 1/2 cos(`theta) ).
So e = 1/2 and since ep = 1/2 p = 1 we have p = 2. **
We can check for plausibility: for theta values
0, pi/2, pi and 3 pi/2 the r values are respectively 2, 1, 2/3 and 1, so the
points in polar coordinates are
(2, 0 radians), (1, pi/2 radians), (2/3, pi radians) and (1, 3 pi/2 radians).
These points are easily graphed, and are thus seen to correspond to (x, y) coordinates (2, 0), (0, 1), (-2/3, 0) and (0, -1).
These points could not lie on a parabola; they are clearly compatible with an ellipse.
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21:07:34
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Self-critique (if necessary):
Self-critique Rating:
**** Query problem 9.6.28 convert to rectangular coordinates r(2-cos(`theta)) = 2. **** What is the rectangular coordinate form of the given equation, and how did you obtained it a?
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21:12:21
Your solution:
Confidence Assessment:
Given Solution:
r ( 2- cos ('theta)) = 2
2r - r cos('theta) = 2 distribute the r
2r = 2 + r cos('theta) rearrange
4r^2 = ( 2 + r cos('theta))^2 square both sides
4 (x^2 + y^2) = (2 + x)^2 use transformations
4x^2 + 4y^2 = x^2 + 4x +4 distribute
4y^2 = -3x^2 + 4x +4 isolate the y
** Keeping the equation in the standard rectangular form of a conic we have
3 x^2 + 4 y^2 - 4 x - 4 = 0.
From this form it should evident that completing the square on x will give you the standard equation of an ellipse. **
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Self-critique (if necessary):
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