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Math Help - find the cylindrical , rectangular, spherical coordinates

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    Question find the cylindrical , rectangular, spherical coordinates

    A right circular cylinder of radius 10cm spinning at 3 revolutions per minute about the z-axis. At time t=0s, a bug at the point (0,10,0) begins walking straight up the face of the cylinder at the rate of 0.5cm/min.

    a) find the cylindrical coordinates of the bug after 2 min.

    b) Find the rectangular coordinates of the bug after 2 min.

    c) find the spherical coordinates of the bug after 2 min.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jenny20 View Post
    A right circular cylinder of radius 10cm spinning at 3 revolutions per minute about the z-axis. At time t=0s, a bug at the point (0,10,0) begins walking straight up the face of the cylinder at the rate of 0.5cm/min.

    a) find the cylindrical coordinates of the bug after 2 min.

    b) Find the rectangular coordinates of the bug after 2 min.

    c) find the spherical coordinates of the bug after 2 min.
    The linear and rotational motions are independent, so after 2 min the bug has a z coordinate of z = (0.5 \, cm/min)(2 \, min) = 1 cm.

    Rotationally speaking, the bug starts on the y-axis at 10 cm ( \theta_0 = \pi/2 \, rad) and spins for 2 min. The angular speed is constant, so
    \theta = \theta _0 + \omega t

    \theta = \pi/2 + (3 \cdot 2\pi \, rad/min)(2 \, min) = 25 \pi /2 \, rad ==> \theta = \pi /2 \, rad

    So in cylindrical coordinates the bug is at
    (r, \theta, z) = (10, \pi /2, 1)
    after 2 minutes.

    In rectangular coordinates we have
    x = r cos( \theta )
    and
    y = r sin( \theta )

    So
    x = 10 cos( \pi /2 ) = 0 \, cm
    y = 10 sin( \pi / 2 ) = 10 \, cm

    So
    (x, y, z) = (0, 10, 1)

    Using rectangular coordinates as the starting point, we have for spherical coordinates:
    x = \rho cos( \phi ) sin( \theta ) <-- Not the same \theta as before!
    y = \rho sin( \phi ) sin( \theta )
    z = \rho cos( \theta )

    So
    \rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{101}

    Now, \phi would normally be tan^{-1} \left ( \frac{y}{x} \right ) but since x is 0 this cannot be calculated directly. Suffice it to say that \phi needs to be such that tan( \phi ) is positive and infinite. Thus \phi = \pi /2 \, rad. (The other way to see this is that \phi in spherical coordinates is the same as \theta in cylindrical coordinates.)

    \theta = cos^{-1} \left ( \frac{z}{\rho} \right ) = cos^{-1} \left ( \frac{1}{\sqrt{101}} \right ) \approx 1.47113 \, rad (As far as I know there is no exact expression of this.)

    So in spherical coordinates we have:
    (\rho, \phi, \theta ) = (\sqrt{101}, \pi /2, 1.47113 )

    -Dan
    Last edited by topsquark; January 9th 2007 at 01:18 PM.
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    Hi topsquark,

    Thank you very much for your reply.

    for part a) why r = 10?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jenny20 View Post
    Hi topsquark,

    Thank you very much for your reply.

    for part a) why r = 10?
    Because the bug is on a spinning cylinder of radius 10 cm. I'm assuming it stays on the cylinder.

    -Dan
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    Hi topsquark,

    i see. Also , in part a) why thetasub0 = pi/2?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jenny20 View Post
    Hi topsquark,

    i see. Also , in part a) why thetasub0 = pi/2?
    The bug starts at (0, 10, 0) in Cartesian coordinates. That means the bug is on the y-axis, which corresponds to \theta_0 = \pi /2 \, rad.

    -Dan
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    Hi topsquark,

    Thank you very much. I got the question now.

    You have a typo in part c):

    cos phi = z/rho = 1/sqrt(101)
    so phi approximately equal to 1.47 radians.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jenny20 View Post
    Hi topsquark,

    Thank you very much. I got the question now.

    You have a typo in part c):

    cos phi = z/rho = 1/sqrt(101)
    so phi approximately equal to 1.47 radians.
    Heh. Sorry about that. I've fixed it in my original post.

    -Dan
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