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Math Help - converting polar equation to cartesian equation

  1. #1
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    converting polar equation to cartesian equation

    With this problem it ask to convert r=-3 cos(theta) from polar form to cartesian

    This is what I get
    <br /> <br />
\begin{array}{l}<br />
 r =  - 3\sin (\theta ) \\ <br />
 (x - \frac{3}{2})(x - \frac{3}{2}) + (\frac{{ - 3}}{2})^2  + y^2  = 0 + (\frac{{ - 3}}{2})^2  \\ <br />
 (x - \frac{3}{2})^2  + (\frac{{ - 3}}{2})^2  + y^2  = (\frac{{ - 3}}{2})^2  \\ <br />
 \end{array}<br /> <br />

    However the answer given is as below. What have I done wrong?
    <br /> <br />
 (x - \frac{3}{2})^2  + y^2  = (\frac{{ - 3}}{2})^2  \\ <br /> <br />
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  2. #2
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    Hello,

    If the polar coordinate is defined as:
    x=r cos(theta)
    y=r sin(theta)

    then,
    r=\sqrt{x^2+y^2},
    cos(theta)=x/r,
    sin(theta)=y/r.

    By substituton,
    r=-3 cos(theta) becomes
    r=-3x/r that is, r^2=-3x
    so that x^2+y^2=-3x.
    ((x+3/2)^2+y^2=(3/2)^2.)

    Please adjust the above argument to the appropriate polar coordinate you are using...

    Bye.
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  3. #3
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    Sorry could someone clarify my answer please.
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  4. #4
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    Hello,

    Quote Originally Posted by Craka View Post
    Sorry could someone clarify my answer please.
    Sorry I couldn't figure out what you had written.
    Would you explain how you got these?
    (If you are asked to convert r=-3 cos(theta), why begin with r=-3sin(theta)? Where does the second line come from?)


    Besides, there are several ways to define coordinates.
    What is the definition (in your book) of the polar coordinates?

    Bye.
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  5. #5
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    Hello, Craka!

    Convert r\:=\:-3\cos\theta to cartesian form.

    \text{Multiply by }r\!:\;\;\underbrace{r^2} \;\;=\;-3\underbrace{r\cos\theta}

    . . . . . . . . . x^2+y^2 \;=\quad\; -3x


    Then: . x^2 + 3x + y^2 \;=\;0

    . . x^2 + 3x + {\color{blue}\frac{9}{4}} + y^2 \;=\;{\color{blue}\frac{9}{4}}

    . . . . \boxed{\left(x + \frac{3}{2}\right)^2 + y^2 \;=\;\frac{9}{4}}


    This is a circle with center \left(-\frac{3}{2},\:0\right) and radius \frac{3}{2}

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