Please help A question about partial differentiation

In summary, the conversation discusses solving for the gradient operator in polar coordinates and differentiating in those coordinates. The vector notation and polar unit vectors are also mentioned. One person thanks the other for their help and mentions using different coordinates.
  • #1
tony_engin
45
0
hi all!
I know how to solve the part (i) but not part (ii).
Could anyone help?
 

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  • #2
Note that in vector notation, we have, from (i):
[tex]\nabla{u}=\frac{\partial{u}}{\partial{x}}\vec{i}+\frac{\partial{u}}{\partial{y}}\vec{j}=\frac{\partial{u}}{\partial{r}}\vec{i}_{r}+\frac{\partial{u}}{r\partial\theta}\vec{i}_{\theta}[/tex]
when the polar unit vectors are:
[tex]\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
Thus, the gradient operator has the form, in polar coordinates:
[tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}[/tex]
In agreement thus far?
 
  • #3
Leave vectors alone and simply differentiate

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

[tex] \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)[/tex]

[tex]=\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right) [/tex]

and the same way for "y"...

Daniel.
 
Last edited:
  • #4
Oh! Thank you very much for your reply!
But I just manage to get the correct answer!
Thank you anayway! ^^
 
  • #5
So that's the laplacian in polar plane coordinates.BTW,it's customarily to denote these coordinates by [itex] \left(\rho,\theta\right) [/itex] and physicists prefer [itex] \left(\rho,\phi\right) [/itex].

Daniel.
 

Related to Please help A question about partial differentiation

What is partial differentiation?

Partial differentiation is a mathematical technique used to find the rate of change of a function with respect to one of its variables while holding all other variables constant.

Why is partial differentiation useful?

Partial differentiation allows us to analyze complex functions by breaking them down into simpler parts. It is particularly useful in fields such as physics and economics where multiple variables are involved.

What is the difference between partial differentiation and ordinary differentiation?

Ordinary differentiation involves finding the rate of change of a function with respect to a single variable, while partial differentiation involves finding the rate of change with respect to one variable while holding all others constant.

How do I perform partial differentiation?

To perform partial differentiation, you must take the derivative of the function with respect to the variable of interest while treating all other variables as constants. This can be done using the standard rules of differentiation.

What are some real-world applications of partial differentiation?

Partial differentiation is used in various fields of science and engineering, including physics, economics, and chemistry. It is particularly useful in analyzing systems with multiple variables, such as in optimization problems or in determining the rate of change of a physical quantity.

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