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Math Help - Help with polar coordinates

  1. #1
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    Help with polar coordinates

    Consider a curve given in polar coordinates by the equation
    r = sec(theta +pi/4).
    Find the equation of the tangent line (in Cartesian coordinates) at the point where  theta= 0.
    Simplify your answer as much as possible.
    Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by khuezy View Post
    Consider a curve given in polar coordinates by the equation
    r = sec(theta +pi/4).
    Find the equation of the tangent line (in Cartesian coordinates) at the point where  theta= 0.
    Simplify your answer as much as possible.
    Thanks.
    We have that r=\frac{1}{\cos\left(\theta+\frac{\pi}{4}\right)} Now a little trig will lead you to r=\frac{\sqrt{2}}{\cos(\theta)-\sin(\theta)}, or 1=\frac{\sqrt{2}}{r\cos(\theta)-r\sin(\theta)}\implies{1=\frac{\sqrt{2}}{x-y}}

    I think you can take it from there.
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  3. #3
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    ?

    how did u do that trig?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by khuezy View Post
    how did u do that trig?
    Using the two facts \cos\left(A+B\right)=\cos\left(A\right)\cos\left(B  \right)-\sin\left(A\right)\sin\left(B\right) and \sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi  }{4}\right)=\frac{\sqrt{2}}{2}

    So

    \begin{aligned}\cos\left(x+\frac{\pi}{4}\right)&=\  cos(x)\cos\left(\frac{\pi}{4}\right)-\sin(x)\sin\left(\frac{\pi}{4}\right)\\<br />
&=\frac{\sqrt{2}}{2}\cos(x)-\frac{\sqrt{2}}{2}\sin(x)\\<br />
&\implies\frac{1}{\cos\left(x+\frac{\pi}{4}\right)  }\\<br />
&=\frac{1}{\frac{\sqrt{2}}{2}\cos(x)-\frac{\sqrt{2}}{2}\sin(x)}\\<br />
&=\frac{\sqrt{2}}{\cos(x)-\sin(x)}\end{aligned}
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  5. #5
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    awesome

    thanks, how did u get the root2/2 to go on top and the /2 to go away?
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