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Math Help - spherical coordinates

  1. #1
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    Exclamation spherical coordinates

    does anyone know how to plot (6, 11pi/12, 0) ? i tried transforming it to rectangular coordinates and got (0,0,1/2) is this right?

    also, for cylindrical coordinates, how do you plot (3, pi/4, 2)? i'm not sure if i did that one right.

    greatly appreciated
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  2. #2
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    cylindrical coordinates are : (r,\theta, z) right?

    spherical coordinates are: (\theta, \phi, \rho) right?



    so in your second question the r = 3, theta = pi/4 and z = 2

    r and theta are on the xy plane. You're looking at a point that has a radius 3 and an angle of 45 degrees measured from the x axis. So that's roughly (\sqrt{1.5},\sqrt{1.5}) on the x-y plane. Then your Z is your height.

    So you just have a point 2 spaces above that.

    About your first question if you could be a little clearer, it seems like it should also be cylindrical but you named the thread as "spherical" coordinates.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    (6, 11pi/12, 0)


    in rect coord x= psin(phi)cos(theta) =0
    y= psin(phi)sin(theta) = 0
    z= pcos(phi) = 6cos(0) = 6

    For (3, pi/4, 2) See attachment
    Attached Thumbnails Attached Thumbnails spherical coordinates-cldrcal.jpg  
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  4. #4
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    whoa, what'd you use to generate that image?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    spherical are listed ( p,theta,phi)

    disappoiningly I just used paintbrush
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  6. #6
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    whoa thanks! so in your picture is the point the part in towards the bottom right?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails spherical coordinates-cldrcal2.jpg  
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  8. #8
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    yay thank you thats where i had it i think i'm finally getting the hang of this!
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