Rectangular to cylindrical conversion

  1. Mar 2, 2006 #1
    Hi =)
    I was given this problem on a test:
    a vector A = 2yi - Zj +3xk, was given in rectangular (cartesian) coordinates and I had to convert it to cylindrical coords. What I did to solve it was this:

    1) A = 2rsin(theta)i - zj + 3rcos(theta)k

    2) partial derivatives
    a) d/dr = 2sin(theta)i + 3cos(theta)j
    b) d/d(theta) = 2rcos(theta)i - 3rsin(theta)k
    c) z = k = 3rcos(theta)k

    3) dot product (initial vector A with each of the partial derivatives)
    a)*A = 4rsin^2(theta) + 9rcos^2(theta) = r
    b)*A = 4(r^2)sin(theta)cos(theta) - 9(r^2)cos(theta)sin(theta) = -5(r^2)sin(theta)cos(theta) = theta
    c) Z = K = 3rcos(theta)

    the equations in bold are my final answers. I would appreciate any feedback on what I did (if it is right or wrong). Many thanks!
     
    Last edited: Mar 2, 2006
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  3. Mar 2, 2006 #2

    HallsofIvy

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    Your statment of the problem is not very clear. First you say "I was given this problem on a test:
    a vector A = 2yi - Zj +3xk, was given in rectangular (cartesian) coordinates and I had to convert it to cylindrical coords. What I did to solve it was this:"
    and you have the answer to that correct.

    Then you say "differentiations". Was that an additional part of the problem? I presume you mean "find the partial derivatives".

    You have the partial derivatives with respect to r almost correct (it is "j" rather than "k") and [itex]\theta[/itex] right- though you should say "[itex]\frac{\partial A}{\partial r}[/itex]" (or dA/dr if you'd rather not use LaTex) rather than just "d/dr",

    Unfortunately, "z = k = 3rcos(theta)k" makes no sense at all. Even if I assume the "z" on the left was really "dA/dz", surely you know that k is not 3rcos(theta)k! I have no idea what you meant here.
    The only place z appears in A is in "-zj". dA/dz= -j.

    Finally, you have "dot products". Dot products of what vectors? A with what? Or is it the partial derivatives? I think you mean the dot product of A with each of its partial derivatives.

    You have
    The first is right but that is certainly not equal to r! What happened to the 4 and 9?
    Assuming you mean [itex]A\dot\frac{\partial A}{\partial \theta}[/itex] that is correct.
    Since you got the derivative of A with respect to z wrong above, this is wrong.
     
  4. Mar 2, 2006 #3
    I'm sorry you're right, I was not very clear on how I proceeded. But i did indeed take the partial derivatives with respect to r and (theta), I didn't take the partialwith respect to "z" because z=z when converting between cartesian and cylindrical. The dot products were that of the inicial vector A with each of the partial derivatives that I got.

    according to my teacher, the answer to this problem is:
    (2rsin(theta)cos(theta) - zsin(theta), 2rsin^2(theta) - zcos(theta), 1 + 3rcos(theta))

    I'm confused, because that's definately not the answer I got. Is my answer (in the first post) wrong or right? Thanks guys!
     
  5. Mar 3, 2006 #4
    .........?
     
  6. Mar 3, 2006 #5
    I am just going to write this out a little big more clearly

    Given:
    [tex]A = 2y \hat{i} - Z\hat{j} +3x\hat{k}[/tex]

    Solve:

    a.) Convert it into a cylindrical coordinate system

    b.) find the partial derivative of A w.r.t. [tex]r[/tex], [tex]\theta[/tex] and [tex]z[/tex].

    c.) Find the dot product (initial vector A with each of the partial derivatives)

    Your Solutions were:
    a.) A = 2rsin(theta)i - zj + 3rcos(theta)k

    That seems right, although (i,j,k) are unit vectors in cartesian coordinate systems, and typically cylindrical coordinate systems are in terms of [tex] (\hat{r}, hat{\theta}, \hat{z}) [/tex]. I am not 100% sure if it is proper or not to append the (i,j,k) to it or not. I suppose it is ok.

    2) partial derivatives
    a) d/dr = 2sin(theta)i + 3cos(theta)j
    b) d/d(theta) = 2rcos(theta)i - 3rsin(theta)k
    c) z = k = 3rcos(theta)k

    I think part a looks ok, as does part b. Part c is wrong, becuase z does not equal k. it should read:

    [tex] \frac {\partial A}{\partial z} = -j [/tex]
     
    Last edited: Mar 3, 2006
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