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Area of Polar Graphs

  1. Mar 1, 2005 #1
    I have the AP Calc test coming up in May, and polar graphs are going to be of somewhat importance.

    The area of a polar region is: [tex] A = \frac{1}{2}\int_{a}^{b}r^2d\theta[/tex]

    The only thing I don't understand is how to know your bounds.

    Example: Find the area of one loop of the polar equation [tex]r = \cos(4\theta)[/tex]

    Help please :cry:

    -------------
    Jameson

    P.S. I need to do this without a calculator
     
  2. jcsd
  3. Mar 1, 2005 #2
    A loop presumably takes place when the radius r starts at zero (the origin) and ends at zero without hitting zero anywhere in between. Using only this definition, you can easily see how many loops the graph of r = [tex]1-\theta^2[/tex] has and where [tex]\theta[/tex] hits zero. You know the behavior of cos(t) and you should know the behavior of A*cos(B*t) + C for arbitrary A,B, and C. Can you find the bounds using the definition of loop given above ?
    It's really just another way of talking about the area between roots of the equation.
     
    Last edited: Mar 1, 2005
  4. Mar 1, 2005 #3
    [tex]A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos(4\theta) ^{2}d\theta[/tex] ?

    Those are when the graph will hit zero... I think at least.

    Jameson
     
  5. Mar 1, 2005 #4

    dextercioby

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    How about plotting that [tex] \cos \ 4\theta [/tex]...?You'll figure out immediately when it starts to thake the same values again...Or you can dirrectly plot the polar diagram and find from the graph to what values do the ends of the loop correspond to.

    Daniel.

    P.S.Shouldn't the polar coord.be [itex] \rho,\varphi [/itex]...?
     
  6. Mar 1, 2005 #5
    He is restricted from using a calculator. Using consecutive roots to find loops is the simplest non-visual way.
     
  7. Mar 1, 2005 #6
    Yep, that's one loop. :)
     
  8. Mar 1, 2005 #7
    What I did (incorrectly I know) was [tex]\cos(4*\frac{\pi}{8}) = 0[/tex]

    So I started there... and the second bound was where that equaled 0 again. I don't know how to plot polar coordinates really. I'm kind of in the dark here. Can someone tell why my bounds are incorrect?

    EDIT: I didn't see you told me I had the right answer. Thank you very much for your help :smile:
     
  9. Mar 1, 2005 #8

    dextercioby

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    Who said anything about using a calculator...?Can't he plot BY HAND [tex] \cos \ 4\theta [/tex] ??:surprised: That's absurd...:yuck:

    Daniel.
     
  10. Mar 1, 2005 #9
    Actually I can't plot by hand. I still need to learn. Thank you hypermorphism for your help.
     
  11. Mar 1, 2005 #10
    Another way is to see that f(t) = cos(4t) shrinks the graph along the x axis by a factor of 4 so that cos(2pi) is mapped to f(pi/2). The rest is just sketching the graph so that cos(t) in the interval [0,2pi] is shrunk to the interval [0,pi/2]. :)
     
  12. Mar 1, 2005 #11

    dextercioby

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    You can't plot "cosine" by hand...?That's bad...

    Daniel.
     
  13. Mar 1, 2005 #12
    I can plot cosine by hand Daniel. That's basic trig. I can't plot the polar graph I mentioned.
     
  14. Mar 1, 2005 #13
    My bad. :D Your plotting is what I refer to as sketching. I've usually only heard plotting in reference to hard-copy plotters/software.
     
  15. Mar 1, 2005 #14

    dextercioby

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    I was spealking about "cosine of 4 theta" which is a strangled version of cosine...

    Daniel.
     
  16. Mar 1, 2005 #15
    Of course I can plot that.... You're right, that would be very bad if I couldn't.
     
  17. Mar 1, 2005 #16
    Try plotting it and its friends (A*cos(B*t) + C) using the Cartesian graphs as guides. After awhile you won't need the guides. It's good to have multiple views of the same mathematical object.
     
  18. Mar 1, 2005 #17

    dextercioby

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    Shift on the horizontal an and compression on both directions...Quite useless on other functions,pretty handy when it comes to parabolas or circular trig.functions.

    Daniel.
     
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