Rectangular to cylindrical conversion

  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
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,550
    Staff Emeritus
    Science Advisor

    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. 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. .........?
     
  6. 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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