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Thread: r=2cos theta graph

  1. #1
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    Question r=2cos theta graph

    Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kittycat View Post
    Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.
    Decide on a set of $\displaystyle \theta$ values, say,
    $\displaystyle \theta = 0, \pi/4, \pi/2, 3\pi/4, pi, 5pi/4, 3\pi/2, 7\pi/4$ rad

    Then calculate the r value for each $\displaystyle \theta$. They are
    $\displaystyle r = 2, \sqrt{2}, 0, -sqrt{2}, -2, -\sqrt{2}, 0, \sqrt{2}$ (respectively)

    What do you do with negative r values? Well, you make the r value positive and then add (or subtract) $\displaystyle \pi$ rad to the angle.

    So we wish to plot the set of points:
    $\displaystyle (r, \theta)$
    $\displaystyle (2, 0)$
    $\displaystyle (\sqrt{2}, \pi/4)$
    $\displaystyle (0, \pi/2)$
    $\displaystyle (\sqrt{2}, 5\pi/4)$
    $\displaystyle (2, 0)$ <-- $\displaystyle \pi + \pi = 2\pi \to 0$
    $\displaystyle (\sqrt{2}, \pi/4)$
    $\displaystyle (0, 3\pi/2)$
    $\displaystyle (\sqrt{2}, 7\pi/2)$

    Then you plot the points on polar coordinate graph paper and sketch in a smooth curve between the points. I get the graph below.

    -Dan
    Attached Thumbnails Attached Thumbnails r=2cos theta graph-circle.jpg  
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kittycat View Post
    Could you please tell me how to plot r=2cos theta ? Please show me the step of plotting this graph. Thanking in advance.
    Or you could do it the hard way:
    $\displaystyle r = \sqrt{x^2 + y^2}$

    $\displaystyle cos(\theta) = \frac{x}{\sqrt{x^2 + y^2}}$

    So your equation becomes:
    $\displaystyle \sqrt{x^2 + y^2} = 2 \frac{x}{\sqrt{x^2 + y^2}}$

    Multiplying both sides by $\displaystyle \sqrt{x^2 + y^2}$ gives:
    $\displaystyle x^2 + y^2 = 2x$

    $\displaystyle (x^2 - 2x) + y^2 = 0$

    $\displaystyle (x^2 - 2x + 1) + y^2 = 1$

    $\displaystyle (x - 1)^2 + y^2 = 1$

    which is a circle centered on (1, 0) with radius 1.

    -Dan
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  4. #4
    Bar0n janvdl's Avatar
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    Sorry, i saw you used polar functions, my apologies. I'll delete my posts now.
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