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I would like to know how could I change the coordinates to cilindrical coordinates of the following differential operator.

$y\frac{\partial f}{\partial x} + xy^2z^5\frac{\partial f}{\partial y} + x^3yz^2\frac{\partial f }{\partial z}$

EDIT: cilindrical rather than polar coordinates

Fireeyes
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  • You already know how to derive $x,y,z$ in polar (or, rather, spherical) coordinates. What does the chain rule imply for the derivatives? – Jonathan Y. Aug 21 '13 at 09:19
  • Well, it implies that the rate of change of x, y and z is given by the rate of change of r and theta. – Fireeyes Aug 21 '13 at 10:21
  • More precisely, how can we get $\frac{\partial f}{\partial x}$ in terms of $\frac{\partial f}{\partial r}$,$\frac{\partial f}{\partial\theta}$? – Jonathan Y. Aug 21 '13 at 10:30
  • Oh, I think i understand. First of all I express $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial\theta}$ in terms of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ and I get a system of two equations and solve it for $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ – Fireeyes Aug 21 '13 at 10:40
  • You could, but it would be easier to do it the other way around to begin with (since you have to know how to express $x,y,z$ in terms of $r,\theta,\phi$ anyway). – Jonathan Y. Aug 21 '13 at 10:42
  • $ \frac{\partial f}{\partial r} = \cos(\theta)\frac{\partial f}{\partial x} + \sin(\theta)\frac{\partial f}{\partial y} $ and $ \frac{\partial f}{\partial\theta} = -r\sin(\theta)\frac{\partial f}{\partial x} + r\cos(\theta)\frac{\partial f}{\partial y}$ – Fireeyes Aug 21 '13 at 10:45
  • I don't really know how to do it the other way around since I have no definition of $f$. – Fireeyes Aug 21 '13 at 10:47

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As you say, $$\left(\begin{matrix}\frac{\partial f}{\partial r}\\ \frac{\partial f}{\partial \theta}\end{matrix}\right) = \left(\begin{matrix}\cos\theta & \sin\theta\\ -r\sin\theta & r\cos\theta\end{matrix}\right) \left(\begin{matrix}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\end{matrix}\right)$$ implying that (where $r\neq 0$) $$\left(\begin{matrix}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\end{matrix}\right) = \left(\begin{matrix}\cos\theta & -\frac{\sin\theta}{r}\\ \sin\theta & \frac{\cos\theta}{r}\end{matrix}\right) \left(\begin{matrix}\frac{\partial f}{\partial r}\\ \frac{\partial f}{\partial \theta}\end{matrix}\right)$$

Jonathan Y.
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  • Oh, it was that simple, you have my greatest thanks! Also I leave $z$ and $\frac{\partial f}{\partial z}$ as they are since the change of variable implies that z = z. – Fireeyes Aug 21 '13 at 11:36