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Need help in drawing a level curve. |
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| Oct10-11, 08:05 PM | #1 |
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Need help in drawing a level curve.
1. The problem statement, all variables and given/known data
i want to draw a level curve for the following but im having some trouble. x-y^2/x^2+y 2. Relevant equations 3. The attempt at a solution I know i'm supposed to set c = x-y^2/x^2+y, but i can't find a way to set y in terms of x and c. |
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| Oct10-11, 08:11 PM | #2 |
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| Oct10-11, 08:15 PM | #3 |
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Whoops, guess I should've added brackets.
[x-y^2]/[x^2+y] sorry bout that. |
| Oct10-11, 08:34 PM | #4 |
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Need help in drawing a level curve.
Much better.
Your equation is z = (x - y2)/(x2 + y) The level curves are curves in some plane, z = c. For example, in the plane z = 1, the level curve is the equation (x - y2)/(x2 + y) = 1. Keep in mind that y can't equal -x2. Multiply both sides by x2 + y. What equation do you get? |
| Oct10-11, 08:47 PM | #5 |
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c(x^2+y) = (x-y^2)?
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| Oct11-11, 12:01 AM | #6 |
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So if you move all of the terms to one side, what do you get?
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| Oct11-11, 12:37 AM | #7 |
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cx^2 + cy - x + y^2 = 0
so when c = 0 i get a parabola, but what about when c = 2 for example..? |
| Oct11-11, 06:52 AM | #8 |
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2x2 - x + y2 +2y = 0 2(x2 - (1/2)x) + y2 +2y = 0 2(x2 - 2 (1/4)x + (1/4)2 -1/16 ) + y2 +2y + 1 - 1 = 0 2(x - 1/4)2 + (y - 1)2 = 1 + 1/8 (x - 1/4)2/(9/16) + (y - 1)2/(9/8) = 1 Do similar for c in general. |
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