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Another algebra and trig question (maybe not though)

  1. Mar 17, 2006 #1
    I have [itex] u(r,\theta)=r \cos \theta + r^{-1}\cos \theta [/itex] and then am changing coordinates [itex] u(x,y)=|x|+\frac{|x|}{x^2+y^2} [/itex]. I then have to compute the vector [itex] \vec v = (u_x,u_y) [/itex].

    The thing is, is the derivative of [itex] |x| [/itex] is [itex] sign\,\,x [/itex]. I don't think we should be dealing with this function. So I'm not sure how I should be dropping the absolute value to get to: [itex] u(x,y)=x+\frac{x}{x^2+y^2} [/itex].

    I'm thinking it is from the conversion to polar to cartesian... but I'm not sure. Maybe it is within the definition of [itex] \tan^{-1} [/itex]... but again I'm not sure. Well I'm sick of throwing random darts, so if anyone has any suggestions, I'm all ears :)
    Please note that [itex] u(x,y) [/itex] was my conversion from polar coordiantes to cartesian, so this could also be a problem, and might be incorrect.

    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 17, 2006 #2


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    Consider the polar point (1, [itex] \pi [/itex]) this should be the same as (-1,0) in Cartesian coordinates.

    These points should yield the same result for your function. Do they?

    Consider your rectangular function both with and without the abs, which gives the correct result?
  4. Mar 17, 2006 #3


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    Consider that [itex] u(r,\theta)=r \cos \theta + r^{-1}\cos \theta = r \cos \theta + \frac{r\cos \theta}{r^2} [/itex] and you can get it from there, correct?
  5. Mar 18, 2006 #4
    Interesting. That makes sense.
    [tex] u_{polar}(1,\pi) = \cos \pi + \cos \pi =-2 [/tex]

    [tex] u_{cart}(-1,0) = |-1| + \frac{|-1|}{1}=2 \neq -2 [/tex]

    So I should drop the abs functions for [itex] u(x,y) [/tex]. I'm curious. How did you know to think of this as a counter example?

    I wish I understood what you are conveying in the algebraic rewrite of the expression. I'm sorry but I don't understand.
  6. Mar 18, 2006 #5


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    Use the transformation equations

    Recall that

    [tex]x=r\cos \theta, \, y=r\sin \theta,\mbox{ and }x^2+y^2=r^2[/tex]​

    (even though we won't use the second equation) so that

    [tex] u(r,\theta)=r \cos \theta + r^{-1}\cos \theta = r \cos \theta + \frac{r\cos \theta}{r^2} = x + \frac{x}{x^2+y^2}[/tex] ​
  7. Mar 18, 2006 #6
    Oh snap!
    I didn't see what you were showing before. That's cool. Thanks man :)

    hehe, I don't know how I didn't see that!
  8. Mar 18, 2006 #7


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    Sure, glad to help: so post another.
  9. Mar 18, 2006 #8
    lol, will do. I have another homework problem that is bugging me, but I need to get some sleep. It's been a fun st patricks day, and I'm not really in the proper mind to be working on it right now :)
  10. Mar 18, 2006 #9


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    Frog pad,
    It is a very good practice on this type of problem to pick a easy to verify point. We needed some thing that would yield a negitive result, thus cos [itex] \pi [/itex].

    Looks like every thing is happy now! :smile:
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