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Math Help - polar / parametrics question

  1. #1
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    polar / parametrics question

    xy - 2x = 0

    Convert the rectangular equation into polar form.
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  2. #2
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    Plug in x=rcos{\theta}, \;\ y=rsin{\theta}
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  3. #3
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    I get...

    r^2 cosx sinx - 2r cosx = 0

    what do i do next, since this answer doesn't match the one in the back fo the book.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    xy - 2x = 0

    Convert the rectangular equation into polar form.
    We first parametrize it

    x=r\cos(\theta)

    and y=r\sin(\theta)

    Inputting this we get

    (r\cos(\theta))(r\sin(\theta))-2r\cos(theta)=0

    This can be rewritten as
    \frac{r^2}{2}\sin(2\theta)-2r\cos(\theta)=0

    Or we could have factored out a r\cos(\theta)

    r\cos(\theta)(r\sin(\theta)-1)=0

    Or finally we can solve this using the quadratic equation

    If we let u=\cos(\theta)\Rightarrow\sqrt{1-u^2}=\sin(\theta)

    so we have

    r^2u\sqrt{1-u^2}-2ru

    Now using the quadratic equation we get

    r=\frac{-u\sqrt{1-u^2}\pm\sqrt{u^2(1-u^2)-4(-2u)}}{2}

    So r=\frac{-\cos(\theta)\sin(\theta)\pm\sqrt{\cos^2(\theta)\si  n^2(\theta)+8\cos(\theta)}}{2}
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  5. #5
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    The answer in the book is r^2= 2 sec(theta)csc(theta)

    Is there a simpler way of deriving this answer?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    The answer in the book is r^2= 2 sec(theta)csc(theta)

    Is there a simpler way of deriving this answer?
    This cannot be right

    r^2=2\sec(\theta)\csc(\theta)\Rightarrow{r\cos(\th  eta)r\sin(\theta)=2}

    Or xy=2
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  7. #7
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    I guess the answer in the book is wrong then.

    r^2= 4 / sin(theta)^2 is this the right answer?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    I guess the answer in the book is wrong then.

    r^2= 4 / sin(theta)^2 is this the right answer?
    r^2=\frac{4}{\sin^2(\theta)}\Rightarrow{(r\sin(\th  eta))(r\sin(\theta))=4}

    So y^2=4

    So no.
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  9. #9
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    Book's wrong

    xy-2x=0

    xy=2x

    y=2, x does not equal zero

    OR

    xy-2x=0

    t=theta

    x=r*cos(t), -pi < t <pi

    y=r*sin(t)

    xy-2x=r*sin(t)*r*cos(t)-2*r*cos(t)

    r*sin(t)*r*cos(t)-2*r*cos(t)=0

    r*sin(t)*r*cos(t)=2*r*cos(t)

    r*sin(t)*cos(t)=2*cos(t), r does not equal zero

    r=sec(t)*csc(t)*(2*cos(t))

    r=2*sec(t)*csc(t)*[cos(t)]

    maybe this is where the computation in the book went wrong, you can see it's the same except for a factor of cos(t). Simplifying we get,

    r=2*csc(t)

    r=2/sin(t), r does not equal zero, -pi < t <pi

    Even editors make mistakes
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wkrepelin View Post
    xy-2x=0

    xy=2x

    y=2, x does not equal zero

    OR

    xy-2x=0

    t=theta

    x=r*cos(t), -pi < t <pi

    y=r*sin(t)

    xy-2x=r*sin(t)*r*cos(t)-2*r*cos(t)

    r*sin(t)*r*cos(t)-2*r*cos(t)=0

    r*sin(t)*r*cos(t)=2*r*cos(t)

    r*sin(t)*cos(t)=2*cos(t), r does not equal zero

    r=sec(t)*csc(t)*(2*cos(t))

    r=2*sec(t)*csc(t)*[cos(t)]

    maybe this is where the computation in the book went wrong, you can see it's the same except for a factor of cos(t). Simplifying we get,

    r=2*csc(t)

    r=2/sin(t), r does not equal zero, -pi < t <pi

    Even editors make mistakes
    *Ahem* I think you have made some undue assumptions here friend.
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  11. #11
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    Am I missing something here?

    xy-2x=0
    x(y-2) = 0
    By null factor theorem, x=0 or y =2

    in polar

    \theta = n\pi, n\in\mathbb{Z}
    or
    r=0
    or
    r = 2cosec(\theta)
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badgerigar View Post
    xy-2x=0
    x(y-2) = 0
    By null factor theorem, x=0 or y =2

    in polar

    \theta = n\pi, n\in\mathbb{Z}
    or
    r=0
    or
    r = 2cosec(\theta)
    Convert back

    r=2\csc(\theta)\Rigtharrow{r\sin(\theta)=2}

    So y=2

    Which is not at all what the original problem was.
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  13. #13
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    Sorry, in my previous post
    should have been \theta = \frac{\pi}{2}+n\pi, which I think mathstud was trying to point out.
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