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Math Help - identify the curve

  1. #1
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    identify the curve

    identify the curve by finding a Cartesian equation for the curve.

    r=3sin(theta)

    how do you get a circle of radius 3/2 centered at (0, 3/2)?

    i got
    r^2=3rsin(theta)
    x^2+y^2=3y

    did i do something wrong already? or do i just not know how to go to the next step or something?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    identify the curve by finding a Cartesian equation for the curve.

    r=3sin(theta)

    how do you get a circle of radius 3/2 centered at (0, 3/2)?

    i got
    r^2=3rsin(theta)
    x^2+y^2=3y

    did i do something wrong already? or do i just not know how to go to the next step or something?
    you are right so far, now complete the square
    x^2 + y^2 = 3y
    => x^2 + y^2 - 3y = 0
    => x^2 + y^2 - 3y + (-3/2)^2 - (-3/2)^2 = 0
    => x^2 + y^2 - 3y + (-3/2)^2 = (-3/2)^2
    => x^2 + (y - 3/2)^2 = 9/4
    => x^2 + (y - 3/2)^2 = (3/2)^2

    this is a circle with center (0, 3/2) and radius 3/2
    Last edited by Jhevon; April 5th 2007 at 01:25 PM.
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  3. #3
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    dang.....can you tell me how you came up with the numbers to complete the square?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    dang.....can you tell me how you came up with the numbers to complete the square?
    ok, here's how to complete the square:
    a quadratic in a variable x (and the coefficient of x^2 has to be 1) is a complete square if the lone constant is the square of half the coeffcient of x. so if we have no constant, to complete the square we simply add half the coefficient of x (and since we are adding something that was not there before, we have to subtract it again). once we do that, we can just contract with the x and the new constant and square that group. let's use the question i just did as an example. let's forget about the x^2 that was there for the moment. we have

    y^2 - 3y = 0
    the coefficient of y^2 is 1, so we are good with that condition. now for this to be a complete square, we must have a constant that is the square of half the coefficient of y, namely, (half of -3)^2. so we add that

    y^2 - 3y + (-3/2)^2 = 0

    however, we just added a constant, which means we changed the value of the function, to keep the function the same we have to subtract what we added (or you can add the same thing on the other side of the equal sign, it amounts to the same thing). so we get:

    y^2 - 3y + (-3/2)^2 - (-3/2)^2 = 0

    y^2 - 3y + (-3/2)^2 = (-3/2)^2

    now we contract using y and the new constant and we square the group:

    (y - 3/2)^2 = (-3/2)^2

    now we simplify, a radius can't be negative, so we square the number to get it positive, and rewrite it as a square.


    (y - 3/2)^2 = 9/4

    (y - 3/2)^2 = (3/2)^2

    see Completing the square for more info
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  5. #5
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    thanks! aha...i thought that was pretty helpful. i never really learned how to complete the square... i know that a few of my profs have gone over it before, i just dont pay attention =T i just find it hard to learn things sitting down in class.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    thanks! aha...i thought that was pretty helpful. i never really learned how to complete the square... i know that a few of my profs have gone over it before, i just dont pay attention =T i just find it hard to learn things sitting down in class.
    So you understood everything?

    why not practice completing the square, it is a very important technique to learn. post some problems and solutions on this site so we can guide you
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