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Partial differentiation problem, multiple variables (chain rule?)

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data

    if z = x2 + 2y2 , x = r cos θ , y = r sin θ , find the partial derivative

    [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex]

    2. Relevant equations

    z = x2 + 2y2
    x = r cos θ
    y = r sin θ

    3. The attempt at a solution

    The textbook says that the equation should be re-written to include only the variables θ and x, and then differentiated with respect to θ.

    Substituting y = r sin θ :

    z = x2 + 2r2 sin2 θ

    then [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2sin θ cos θ

    However the solutions in the book give

    [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = 4r2 tan θ

    What am I missing here?

    Thanks in advance.
     
  2. jcsd
  3. May 14, 2014 #2

    pasmith

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    You need to differentiate [itex]y^2 = r^2 \sin^2 \theta[/itex] using the product rule: [itex]r[/itex] is not independent of [itex]\theta[/itex], since [tex]
    r = \frac{x}{\cos \theta}[/tex]
     
    Last edited: May 14, 2014
  4. May 14, 2014 #3
    Thanks! I used the product rule to differentiate, but I still think I'm missing something. My working is as follows:

    z = x2 + 2r2 sin2 θ

    [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] = [itex]\frac{\partial}{\partial \theta}2r^{2}sin^{2}\theta[/itex]

    [itex]= 4r\frac{\partial r}{\partial \theta}sin^{2}\theta + 2r^{2}2sin\theta cos\theta[/itex] (chain rule for dr/dθ)

    since [itex] r = \frac{x}{cos \theta}[/itex], [itex]\frac{\partial r}{\partial \theta} = \frac{-x sin \theta}{cos^{2} \theta}[/itex]

    giving [itex] -4r \frac{x sin \theta}{cos^{2} \theta} sin^{2} \theta + 4r^{2} sin \theta cos \theta [/itex]

    I can replace x with r / cos θ, but I don't see how it reduces to the given solution of 4r2 tan θ.

    Thanks again
     
  5. May 14, 2014 #4

    Ray Vickson

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    You are missing the fact that ##x## is held constant. One way to do it is:
    [tex] dz = 2 x\, dx + 4 y \,dy\\
    dx = \cos(\theta) \, dr - r \sin(\theta)\, d \theta\\
    dy = \sin(\theta)\, dr + r \cos(\theta)\, d \theta[/tex]
    But ##dx = 0 \Longrightarrow dr = r \tan(\theta) \, d \theta##, so
    [tex] dy = r \sin(\theta)\tan(\theta)\, d \theta + r \cos(\theta) \,d \theta
    = r \left( \frac{\sin^2(\theta)}{\cos(\theta)} + \cos(\theta)\right)\, d \theta
    = \frac{r}{\cos(\theta)}\, d \theta [/tex]
    Thus
    [tex] dz = 4 y dy = 4 r \sin(\theta) (r/\cos(\theta)) \,d \theta = 4 r^2 \tan(\theta) \, d \theta [/tex]
    The partial ##(\partial z/\partial \theta)_{x} ## is the coefficient of ##d \theta## in the above.
     
  6. May 14, 2014 #5

    pasmith

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    The derivative of [itex]u^{-1}[/itex] with respect to [itex]u[/itex] is [itex]-u^{-2}[/itex]. The derivative of [itex]\cos \theta[/itex] with respect to [itex]\theta[/itex] is [itex]-\sin \theta[/itex]. The two minus signs cancel.

    You can't, but you can replace [itex]x[/itex] with [itex]r \cos \theta[/itex] and do some trigonometric simplifications; the first step is to express everything in terms of sines and cosines.
     
  7. May 14, 2014 #6
    Try saying z = x2 + 2[itex]\frac{x^2}{cos^2θ}[/itex]sin2θ = x2(1 + 2tan2θ)

    Now with this for z you can perform [itex]\left(\frac{\partial z}{\partial \theta}\right)_{x}[/itex] quite easily.

    Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.
     
  8. May 14, 2014 #7
    That was a typo, my bad. So after sorting the minus sign, I'm left with essentially what I had before, but I can't see anyway of reducing [itex] 4r^{2} tan \theta sin^{2} \theta + 4r^{2} sin \theta cos \theta[/itex] to [itex] 4r^{2} tan \theta[/itex] without ending up with a huge mess.

    Sorry for being dense :P
     
  9. May 14, 2014 #8
    Figured it out. Thanks guys. Turns out I laid it out the way jaytech said, but didn't use the product rule properly so I abandoned that method and tried it another way, which lead to that whole mess. Whoops!

    Thanks to everyone who helped!
     
  10. May 14, 2014 #9
    You should try the steps I previously mentioned. Then reflect on why it works..
     
  11. May 14, 2014 #10

    pasmith

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    [tex]\sin \theta \cos \theta = \tan\theta \cos^2 \theta[/tex]
     
  12. May 14, 2014 #11
    I did use the steps you suggested, jaytech. As for why it works, I imagine that I should be able to reach the solution from any starting point, with proper application of the chain rule/product rule, but arranging it as you suggested means I can skip over a lengthy simplification process after the operation. Any other insight you care to offer? :)
     
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