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Math Help - converting from spherical coordinates

  1. #1
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    converting from spherical coordinates

    Hey everyone, I really need some help on this one, I just can seem to find a connection anywhere between formulas for each coordinate system.

    Convert \phi = \frac{\pi}{6} from spherical coordinates in to both Cartesian coordinates and cylindrical coordinates.

    Can anyone help me at all?

    Thanks again.
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  2. #2
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    Is it as simple as just plugging in phi=pi/6 into the relavant forumula relating spherical and cylindrical, and spherical and cartesian coordinates? And leaving as it is, with say rho and theta still in the equation?
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  3. #3
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    spherical coordinates refers to an angle inscribed on a circle of unit length. So the polar coordinates og your point will be (1,pi/6). Now we know that x^2+y^2=1 and tan^-1(x/y)=pi/6. solving both of them we get your point in cartesian system as (+-sqrt(3)/2,+-1/2). hope your are comfortable with my notations. i have no latex installed and having some problem with uploading bmp files as well.
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  4. #4
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    As for cylindrical coordinates is it the hyperbolic or elliptical system?? if that is the case then simply solve for the equation and tanh^-1(y/x)=pi/6. i suppose that would suffice.
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  5. #5
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    Thanks Pulock2009, but we haven't done the hyperbolic system yet. Can anyone help me with the cylindrical coordinates?
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  6. #6
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    Quote Originally Posted by Pulock2009 View Post
    spherical coordinates refers to an angle inscribed on a circle of unit length. So the polar coordinates og your point will be (1,pi/6). Now we know that x^2+y^2=1 and tan^-1(x/y)=pi/6. solving both of them we get your point in cartesian system as (+-sqrt(3)/2,+-1/2). hope your are comfortable with my notations. i have no latex installed and having some problem with uploading bmp files as well.
    And isn't Cartesian coordinates supposed to be (x,y,z), I don't think we can find all 3 of these when we are only given phi.
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  7. #7
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    The aren't asking for a specific point, but for the equation satisfied by all points that for which [tex]\phi= \frac{\pi}{6}.

    The connection between Cartesian (x, y, z) coordinates and Spherical ( \theta, \phi, \rho) coordinates is
    x= \rho sin(\phi)cos(\theta)
    y= \rho sin(\phi)sin(\theta)
    z= \rho cos(\phi).

    With \phi= \pi/6 then sin(\phi)= \frac{1}{2} and cos(\phi)= \frac{\sqrt{3}}{2} so that
    x= \frac{\rho}{2}cos(\theta)
    y= \frac{\rho}{2}sin(\theta)
    z= \rho\frac{\sqrt{3}}2}

    Then x^2+ y^2= \frac{\rho^2}{4}(cos^2(\theta)+ sin^2(\theta)= \frac{\rho^2}{4} and
    z^2= \frac{3\rho^2}{4}.

    So z^2= 3(x^2+ y^2). That's the equation of the cone of all points satifying \phi= \frac{\pi}{6}.

    In cylindrical coordinates, x^2+ y^2= r^2 so that equation is z^2= 3r^2 in cylindrical coordinates.
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  8. #8
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    in your case z will always be zero because we are dealing with 2-D. polar coordinates generally refers to 2-D.
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