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Rectangular to cylindrical conversion 
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#1
Mar206, 01:07 AM

P: 35

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! 


#2
Mar206, 05:52 AM

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PF Gold
P: 39,553

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 


#3
Mar206, 08:54 AM

P: 35

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! 


#4
Mar306, 01:16 AM

P: 35

Rectangular to cylindrical conversion
.........?



#5
Mar306, 01:49 AM

P: 4,777

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] 


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