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Rectangular to cylindrical conversion

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itzela
#1
Mar2-06, 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!
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HallsofIvy
#2
Mar2-06, 05:52 AM
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P: 39,565
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
a) *A = 4rsin^2(theta) + 9rcos^2(theta) = r
The first is right but that is certainly not equal to r! What happened to the 4 and 9?
b)*A = 4(r^2)sin(theta)cos(theta) - 9(r^2)cos(theta)sin(theta) = -5(r^2)sin(theta)cos(theta) = theta
Assuming you mean [itex]A\dot\frac{\partial A}{\partial \theta}[/itex] that is correct.
c) Z = K = 3rcos(theta)
Since you got the derivative of A with respect to z wrong above, this is wrong.
itzela
#3
Mar2-06, 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!

itzela
#4
Mar3-06, 01:16 AM
P: 35
Rectangular to cylindrical conversion

.........?
Cyrus
#5
Mar3-06, 01:49 AM
Cyrus's Avatar
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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