Help with Partial Derivatives and Implicit Differentiation

  • Thread starter end3r7
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  • #1
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The problem is as follows:

Cartesian and polar coordinates are related by the formulas
[tex]x = r\cos\theta[/tex]
[tex]y = r\sin\theta[/tex]
Determine [tex]\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}[/tex]. Differentiate the equations above implcitly adn then solve the resulting system of 4 equations in the four unknowns [tex]\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}[/tex].

Now, this seems a whole lot harder than just differentiating [tex]r^2 = x^2 + y^2[/tex], but I did anyway and the 4 equations are:

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

[tex]\frac{\partial x}{\partial y} = \frac{\partial r}{\partial y}\cos\theta - \sin\theta\frac{\partial\theta}{\partial y}r[/tex]

[tex]\frac{\partial y}{\partial x} = \frac{\partial r}{\partial x}\sin\theta + \cos\theta\frac{\partial\theta}{\partial x}r[/tex]

[tex]\frac{\partial y}{\partial y} = \frac{\partial r}{\partial y}\sin\theta + \cos\theta\frac{\partial\theta}{\partial y}r[/tex]

Now, if I try to solve for [tex]\frac{\partial r}{\partial x}[/tex], I know what the answer should be by differentiating [tex]r^2 = x^2 + y^2[/tex], and it's [tex]\frac xr[/tex] or [tex]\cos\theta[/tex], however I can't seem to get that from combining the 4 equations.

What am I doing wrong?

Thanks in advance.
 
Last edited:

Answers and Replies

  • #2
0rthodontist
Science Advisor
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Use the fact that dx/dx = 1 and dx/dy = 0, etc.
 
  • #3
168
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I did... it still doesn't give me [tex]\cos\theta[/tex] as an answer.

For example, to calculate dr/dx
using eq 1
I get [tex]1 = \frac{\partial r}{\partial x}\cos\theta[/tex]
using eq 3
[tex]0 = \frac{\partial r}{\partial x}\sin\theta[/tex]

If I use both separatedly or try to combine them, I don't get the same answer as deriving r^2 = x^2 + y^2
 
  • #4
0rthodontist
Science Advisor
1,230
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You seem to have cut off the end parts of those two equations. Try multiplying eq. 1 by cos theta / sin theta and adding it to eq. 3.
 
  • #5
168
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Why mutliply times tangent will help?
I cut them off because I multiplied teh other stuff by zero.
I can solve it using the Jacobian, but I just want to know the logic behind it.
 
  • #6
0rthodontist
Science Advisor
1,230
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You multiply eq. 1 by cos theta/sin theta so that when you add it to eq. 3, you cancel the terms involving dtheta/dx.

When you say you cut them off because you "multiplied the other stuff by 0" you can't just multiply one term of an equation by 0 and assume it will still be correct. The stuff you cut off was not in general zero.
 
  • #7
168
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Oh ****, you are right. Thanks a lot man, really appreciate it! =D
 

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