Results 1 to 8 of 8

Thread: find the cylindrical , rectangular, spherical coordinates

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    123

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    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 $\displaystyle z = (0.5 \, cm/min)(2 \, min) = 1 cm$.

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

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

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

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

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

    So
    $\displaystyle (x, y, z) = (0, 10, 1)$

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

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

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

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

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

    -Dan
    Last edited by topsquark; Jan 9th 2007 at 01:18 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    Posts
    123
    Hi topsquark,

    Thank you very much for your reply.

    for part a) why r = 10?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2006
    Posts
    123
    Hi topsquark,

    i see. Also , in part a) why thetasub0 = pi/2?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    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 $\displaystyle \theta_0 = \pi /2 \, rad$.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2006
    Posts
    123
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Oct 30th 2010, 12:48 PM
  2. Replies: 0
    Last Post: Mar 29th 2009, 06:53 AM
  3. cylindrical or spherical coordinates problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 23rd 2008, 09:22 AM
  4. Replies: 4
    Last Post: Nov 24th 2007, 08:28 AM
  5. Replies: 1
    Last Post: Nov 20th 2007, 06:57 PM

Search Tags


/mathhelpforum @mathhelpforum