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Math Help - rectangular to polar conversions

  1. #1
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    Exclamation rectangular to polar conversions

    I need to convert y=x^2 to polar form preferably with r in terms of theta. So far I have:
    y=x^2
    x^2-y=0
    (r^2cos^2θ)-(rsinθ)=0
    after that I get lost, help me please
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shadowsiren View Post
    I need to convert y=x^2 to polar form preferably with r in terms of theta. So far I have:
    y=x^2
    x^2-y=0
    (r^2cos^2θ)-(rsinθ)=0
    after that I get lost, help me please
    you were on the right track. now to get it into a nice-looking form, we solve for r. Here
    Attached Thumbnails Attached Thumbnails rectangular to polar conversions-polar.gif  
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shadowsiren View Post
    I need to convert y=x^2 to polar form preferably with r in terms of theta. So far I have:
    y=x^2
    x^2-y=0
    (r^2cos^2θ)-(rsinθ)=0
    after that I get lost, help me please
    Going from what you did:

    ... (r^2cos^2θ)-(rsinθ)=0
    => rcos^2θ - sinθ = 0 .............divided both sides by r
    => rcos^2θ = sinθ ..................add sinθ to both sides
    => r = sinθ/cos^2θ .................divide both sides by cos^2θ
    => r = sinθsec^2θ ..................change 1/cos^2θ to sec^2θ by trig identity
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  4. #4
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    Thanks 4 the help.
    Could someone explain to me how to convert (x^2)-(y^2)=25 to polar?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shadowsiren View Post
    Thanks 4 the help.
    Could someone explain to me how to convert (x^2)-(y^2)=25 to polar?
    You should start a new question in a new thread.

    Let t = theta for convenience.

    x = r*cos(t)
    y = r*sin(t)

    So
    (x^2)-(y^2)=25

    (r*cos(t))^2 - (r*sin(t))^2 = 25

    r^2*cos^2(t) - r^2*sin^2(t) = 25

    r^2*(cos^2(t) - sin^2(t)) = 25

    r^2 * cos(2t) = 25

    -Dan
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  6. #6
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    Hello, shadowsiren!

    Convert y = x▓ to polar form.

    We have: . . . . . . x▓ .= .y

    Substitute: .r▓Ěcos▓θ .= .rĚsinθ

    Divide by r: . rĚcos▓θ .= .sinθ

    Divide by cos▓θ: . . r .= .sinθ/cos▓θ

    Therefore: . . . . . . r .= .secθĚtanθ

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    We have: . . . . . . . . . . .x▓ - y▓ .= .25

    Substitute: .(rĚcosθ)▓ - (rĚsinθ)▓ .= .25

    Then: . . . . . r▓Ěcos▓θ - r▓Ěsin▓θ .= .25

    Factor: . . . . . r▓(cos▓θ - sin▓θ) .= .25

    and we have: . . . . . . r▓Ěcos2θ .= .25

    Therefore: . . . . . . . . . . . . .r▓ .= .25Ěsec2θ

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