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http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

That is, the conversions of the vectors, as well as the conversions of the unit vectors.

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i mean with proofs because i already read that website page

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i mean with proofs because i already read that website page

Hmm, well, you can basically read them directly off the drawing of the coordinates.

Is there any one in particular for which you would like an explanation?

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in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates

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in the definition of unit vectors how did we get the cartesian coordinates in terms of cylinderical coordinates

Do you mean:

[tex]

\begin{matrix}

\mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\

\mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi} \\

\mathbf{\hat z} & = & \mathbf{\hat z}

\end{matrix}

[/tex]

?

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yes please because we know how the opposite happens

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yes please because we know how the opposite happens

The only interesting ones are [itex]\mathbf{\hat x}[/itex] and [itex]\mathbf{\hat y}[/itex].

So we're actually talking about 2-dimensional polar coordinates.

Here's one way to derive the unit vectors.

[itex]\boldsymbol{\hat \rho}[/itex] corresponds in this case with the vector from the origin to a point on the unit circle at angle [itex]\phi[/itex].

The vector [itex]\boldsymbol{\hat \rho}[/itex] can be seen as the result of a rotation of [itex]\mathbf{\hat x}[/itex] over an angle of [itex]\phi[/itex].

Or in reverse, the vector [itex]\mathbf{\hat x}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \rho}[/itex] over an angle of [itex]-\phi[/itex].

Similarly the vector [itex]\mathbf{\hat y}[/itex] is the result of a rotation of [itex]\boldsymbol{\hat \phi}[/itex] over an angle of [itex]-\phi[/itex].

We need the rotation matrix for an angle [itex]\phi[/itex] is to get what we want:

[tex]R = \begin{pmatrix}\cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}[/tex]

Multiply [itex]R[/itex] with [itex]\boldsymbol{\hat \rho}[/itex] and [itex]\boldsymbol{\hat \phi}[/itex] and the result rolls out:

[tex]

\begin{matrix}

\mathbf{\hat x} & = & \cos\phi\boldsymbol{\hat \rho}-\sin\phi\boldsymbol{\hat\phi} \\

\mathbf{\hat y} & = & \sin\phi\boldsymbol{\hat \rho}+\cos\phi\boldsymbol{\hat\phi}

\end{matrix}

[/tex]

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How about this:

Apparently you already know:

[tex]

\begin{matrix}

\boldsymbol{\hat \rho} & = & \cos\phi\mathbf{\hat x}+\sin\phi \mathbf{\hat y}\\

\boldsymbol{\hat\phi} & = & -\sin\phi\mathbf{\hat x}+\cos\phi \mathbf{\hat y}

\end{matrix}

[/tex]

Which is:

[tex]\begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix} = \begin{pmatrix}

\cos \phi & \sin \phi \\

-\sin \phi & \cos \phi \end{pmatrix}

\begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix}

[/tex]

The inverse is:

[tex]\begin{pmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \end{pmatrix} = \begin{pmatrix}

\cos \phi & -\sin \phi \\

\sin \phi & \cos \phi \end{pmatrix}

\begin{pmatrix} \boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \end{pmatrix}

[/tex]

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ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh now i get it okay okay thank you alotttttttttttttttttttt

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