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Thread: Find the cartesian form

  1. #1
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    Find the cartesian form

    Find the cartesian form of r^2=sec(2theta)

    I know I have to use conversion formulas x=rcos(theta) and y=rsin(theta) but it's not clear to me how. I've been having trouble with this kind of problem and figure I should get help before I run into more trouble down the line. :s Would appreciate help with this.
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  2. #2
    MHF Contributor chisigma's Avatar
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    You have an implicit function in the form $\displaystyle f(r, \theta) = r^{2} - \frac{1}{\cos 2 \theta}=0$ that can be 'transformed' is an implicit function in the form $\displaystyle f(x,y)=0$ setting $\displaystyle r^{2}= x^{2} + y^{2}$ and $\displaystyle \theta = \tan^{-1} \frac{y}{x}$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by mcsquared View Post
    Find the cartesian form of r^2=sec(2theta)

    I know I have to use conversion formulas x=rcos(theta) and y=rsin(theta) but it's not clear to me how. I've been having trouble with this kind of problem and figure I should get help before I run into more trouble down the line. :s Would appreciate help with this.
    This is the same as

    $\displaystyle r^2\cos^2(2\theta)=1 \iff r^2(2\cos^2(\theta)-1)=1 \iff 2(r\cos(\theta))^2-r^2=1$

    Now just use what you wrote above
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  4. #4
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    Helpful equations:

    $\displaystyle \displaystyle r^2 = x^2 + y^2$

    $\displaystyle \displaystyle \sec{\theta} = \frac{1}{\cos{\theta}}$

    $\displaystyle \displaystyle \cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}$

    $\displaystyle \displaystyle x = r\cos{\theta} \equiv \cos{\theta} = \frac{x}{r} = \frac{x}{\sqrt{x^2 + y^2}}$

    $\displaystyle \displaystyle y = r\sin{\theta} \equiv \sin{\theta} = \frac{y}{r} = \frac{y}{\sqrt{x^2 + y^2}}$.
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