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Transformation from cylindrical 2 cartisian

  • Thread starter Thread Man
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  • #1
it's not homework but it's something I can't make any sense with it

why when we transform from cylindrical 2 cartisian or the inverse we take the unit vector in our

consideration and transform it also not just transform the function and relate it to the other

unit vectors ??.

for example : the vector (A= 5 ar + 3π/2 aΦ )

we can say that : r^2 = x^2 + y^2 that's mean that : 25 = x^2 +y^2

and also : x = r cos Φ which means : x= 5 cos 3π/2 so: x = 0

and : y = r sin Φ which means : y= -5


so that our vector will be directly A= -5 ay

but if we did unit vector transformation : ar = cosΦ ax + sinΦ ay
aΦ = -sinΦ ax + cosΦ ay

we will get that : A= -5 ay + π/2 ax

please, I'm so confused in this part and cannot think in it anymore so I just wanna clearance...
 

Answers and Replies

  • #2
vela
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it's not homework but it's something I can't make any sense with it

why when we transform from cylindrical to cartesian or the inverse do we take the unit vector in our consideration and transform it also not just transform the function and relate it to the other unit vectors?

for example: the vector (A= 5 ar + 3π/2 aΦ )

we can say that : r^2 = x^2 + y^2 that's mean that : 25 = x^2 +y^2

and also : x = r cos Φ which means : x= 5 cos 3π/2 so: x = 0

and : y = r sin Φ which means : y= -5


so that our vector will be directly A= -5 ay

but if we did unit vector transformation : ar = cosΦ ax + sinΦ ay
aΦ = -sinΦ ax + cosΦ ay

we will get that : A= -5 ay + π/2 ax

please, I'm so confused in this part and cannot think in it anymore so I just wanna clearance...
Your confusion arises because there are two vectors involved. One vector is r=(x,y), which is the point to which the second vector A=(Ax,Ay) is assigned. With r, you have the usual relations:

[tex]r^2 = x^2+y^2[/tex]
[tex]\tan \theta = y/x[/tex]

At that point, you have the unit vectors

[tex]\hat{r} = \cos\theta\hat{x} + \sin\theta\hat{y}[/tex]
[tex]\hat{\theta} = \sin\theta\hat{x} - \cos\theta\hat{y}[/tex]

and vector A can be expressed in terms of these vectors:

[tex]\vec{A}=A_x\hat{x}+A_y\hat{y}=A_r\hat{r}+A_\theta\hat{\theta}[/tex]

Note that A has its own "r" and "θ":

[tex]r_a^2 = A_x^2+A_y^2[/tex]
[tex]\tan \theta_A = A_y/A_x[/tex]

and these quantities aren't the same as Ar and Aθ.

This topic came up in another thread yesterday. You might find helpful to read post 6, in which I gave an example.

https://www.physicsforums.com/showthread.php?t=386337
 
  • #3
thanks vela 4 explanation

it was really useful ,,
 

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