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Homework Help: Polar Partials

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Why are the following not contradictory?
    [tex]\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos{\theta}[/tex]


    [tex] r=\frac{x}{cos{\theta}} [/tex]
    [tex]\frac{\partial r}{\partial x}=\frac{1}{cos{\theta}}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I understand that we got these results from differentiating two separate relations, but I don't see why they're different, and seemingly contradictory.
  2. jcsd
  3. Nov 3, 2008 #2


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    Homework Helper
    Gold Member

    Hint: what is the definition of [itex]\theta[/itex]?...If [itex]\theta[/itex] has any explicit dependence on [itex]x[/itex], then is [tex]\frac{\partial}{\partial x} \left(\frac{x}{cos(\theta)} \right)[/tex] Really just [itex]\frac{1}{cos(\theta)}[/itex]?Don't you have to use the product rule?
  4. Nov 3, 2008 #3


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    They are contradictory because the second is simply wrong. Yes,
    [tex]r= x/cos(\theta)[/tex] but the derivative is wrong: [itex]\theta[/itex] is not a constant, it depends on x itself. A correct calculation would be
    [tex]\frac{\partial r}{\partial x}= \frac{cos(\theta)+ xsin(\theta)\frac{\partial \theta}{\partial x}}{cos^2(\theta}[/itex]
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