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Homework Help: Transformation from cylindrical 2 cartisian

  1. Mar 16, 2010 #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...
  2. jcsd
  3. Mar 16, 2010 #2


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    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:


    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.

  4. Mar 20, 2010 #3
    thanks vela 4 explanation

    it was really useful ,,
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