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Thread: Sketching polar graph

  1. #1
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    Sketching polar graph

    Sketch the curve $\displaystyle r = 2cos4\theta$

    Attempt:

    when $\displaystyle \theta$ = 0, r = 2

    when $\displaystyle \theta$ = $\displaystyle \frac{\pi}{6}$, r = -1

    when $\displaystyle \theta$ = $\displaystyle \frac{\pi}{4}$, r = -2

    The graph I got was a circle (not ment to look like an ellipse in the drawing), but the correct answer was a flower
    Attached Thumbnails Attached Thumbnails Sketching polar graph-polar.jpg  
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Sketch the curve $\displaystyle r = 2cos4\theta$

    Attempt:

    when $\displaystyle \theta$ = 0, r = 2

    when $\displaystyle \theta$ = $\displaystyle \frac{\pi}{6}$, r = -1

    when $\displaystyle \theta$ = $\displaystyle \frac{\pi}{4}$, r = -2

    The graph I got was a circle (not ment to look like an ellipse in the drawing), but the correct answer was a flower
    You need more points.

    Hint: Find the points where r=0.
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  3. #3
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    Hello, SyNtHeSiS!

    $\displaystyle \text{Sketch the curve: }\;r \:=\: 2\cos4\theta$

    We are expected to recognize this as a "rose curve", a flower-shaped curve.
    . . [But it looks more like a daisy.]

    There are two basic forms: .$\displaystyle \begin{Bmatrix}r &=& a\sin n\theta \\ r &=& a \cos n\theta \end{Bmatrix}$

    $\displaystyle \,a$ is the length of a "petal".

    If $\displaystyle \,n$ is odd, there are $\displaystyle \,n$ petals.
    If $\displaystyle \,n$ is even, there are $\displaystyle 2n$ petals.
    . . The petals are equally spaced about the center.


    We have: .$\displaystyle r \:=\:2\cos4\theta$

    Since $\displaystyle n = 4$ (even), there are 8 petals, spaced 45 degrees apart.
    Each petal is 2 units long.


    Now where are the petals?
    . . When is $\displaystyle r = 2$ ?

    We have: .$\displaystyle 2\cos4\theta \:=\:2 \quad\Rightarrow\quad \cos4\theta \:=\:1 \quad\Rightarrow\quad 4\theta \:=\:\cos^{\text{-}1}(1)$

    . . . . . . . . $\displaystyle 4\theta \:=\:0^o \quad\Rightarrow\quad \theta\:=\:0^o$

    Hence, the "first" petal is on the $\displaystyle 0^o$-line (positive $\displaystyle \,x$-axis).


    You should be able to sketch the graph now.

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