Laplace Operator in Polar Coordinates: Steps & Solutions

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Homework Help Overview

The discussion revolves around the Laplace operator in polar coordinates, specifically how to express it in terms of the variables r and Θ. The original poster attempts to differentiate the operator after converting Cartesian coordinates to polar coordinates but encounters difficulties with the differentiation process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of differentiating the Laplace operator after converting coordinates, with some questioning the correctness of the differentiation steps taken. There is mention of the need for a nested chain rule and product rule in the differentiation process.

Discussion Status

Some participants are seeking clarification on the differentiation steps, particularly regarding the application of the chain rule in multivariable calculus. There is an ongoing exploration of the correct approach to differentiate the expressions derived from polar coordinates.

Contextual Notes

Participants note the complexity of the differentiation process and the potential for confusion due to the nested chain rule and product rule. There is an acknowledgment of the original poster's struggle with the differentiation, as well as the constraints of time due to other academic commitments.

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Homework Statement



The Laplace operator Δ is defined by: Δ=
c65dd028c1c9fb80a8288ca893e949da.png


Show in polar coordinates r and Θ, that the Laplace operator takes the following form:

http://upload.wikimedia.org/wikipedia/en/math/0/7/a/07a878276cffd0c680f3f827204aba24.png

Homework Equations



x=rcos(Θ), y=rsin(Θ), r ≥ 0, Θ ∈ [0,2∏]

The Attempt at a Solution



It seems simple; convert the x and y in terms of polar coordinates, and differentiate twice. However, when i attempt to differentiate twice, in respect to r and theta, my r disappears and the trig function simply reverts.
 
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Can you show your work where you differentiate the first time?
 
I like Serena said:
Can you show your work where you differentiate the first time?

Sorry for the delayed response, been studying for other finals.

Here is where i am stuck.

For the x partial derivative:

\partial / \partialx= -rsinθ+cosθ

For the y partial derivative:

\partial / \partialy= sinθ+rcosθ

I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?
 
tsamocki said:
For the x partial derivative:

\partial / \partialx= -rsinθ+cosθ

What did you differentiate? That seems to be missing.
And how did you get this result?


tsamocki said:
I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?

Yes, the chain rule for multivariable differentiation is:
{\partial \over \partial x}f(u(x,y), v(x,y))={\partial f \over \partial u}{\partial u\over \partial x}+{\partial f \over \partial v}{\partial v \over \partial x}

You should do something similar to f(x(r,θ), y(r,θ))...
 

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