Why does the chain rule apply to partial derivatives in cylindrical coordinates?

[ViktigLemma]

I find this passage $$\frac{\partial}{\partial x} = \cos(\phi)\frac{\partial}{\partial \rho } - \frac{\sin(\phi)}{\rho}\frac{\partial}{\partial \phi}$$ difficult to understand.

My teacher wrote this as an explanation:

$$\frac{\partial V}{\partial x} = \frac{\partial\rho}{\partial x}\frac{\partial V}{\partial \rho} + \frac{\partial\phi}{\partial x}\frac{\partial V}{\partial \phi} + \frac{\partial z}{\partial x}\frac{\partial V}{\partial z}$$ *

And then inserting for $$\rho$$ and $$\phi$$, which will yield a correct result.

What I don't understand is how * can be correct? To me it seems that the right side of the equation is equal to $$3\frac{\partial V}{\partial x}$$

Please enlighten me.

 

[ViktigLemma]

There, I fixed it :)

 

[HallsofIvy]

Could you tell us WHY you think there should be a "3" in that?

$$\frac{\partial V}{\partial x} = \frac{\partial\rho}{\partial x}\frac{\partial V}{\partial \rho} + \frac{\partial\phi}{\partial x}\frac{\partial V}{\partial \phi} + \frac{\partial z}{\partial x}\frac{\partial V}{\partial z}$$
is simply the chain rule!
That would be 3 times dV/dx only if all those were equal to dV/dx separately and they are not!

Take a simple example. Suppose V(u,w)= u²+ 2w² and
that u= 3x-1, v= 2x+3.
One way of finding dV/dx would be to substitute. V(x)= (3x-1)²+ 2(2x+3)²= 9x²- 6x+ 1+ 2(4x²+ 12x+ 9)= 9x²- 6x+ 1+ 8x²+ 24x+ 18= 17x²+ 18x+ 19.

dV/dx= 34x- 30.

But a simpler way is to use the chain rule: $\frac{dV}{dx}= \frac{dV}{du}\frac{du}{dx}+ \frac{dV}{dw}\frac{dw}{dx}$= 2u(3)+ 4w(2)= 6(3x-1)+ 8(2x-3)= 18x- 6+ 16x- 24= 34x- 30 just as before. dV/dx is that sum, not the individual parts.

 

[Gamma]

$$\frac{\partial}{\partial x} = \cos(\phi)\frac{\partial}{\partial \rho } - \frac{\sin(\phi)}{\rho}\frac{\partial}{\partial \phi}$$

I find this passage difficult to understand.

If you understand how chain rule works just as explained by HallsofIvy,

this is not so difficult.
Lets say $$V=V(\rho, \phi, z)$$

$$\frac{\partial V}{\partial x} = \frac{\partial V}{\partial \rho}\frac{\partial\rho}{\partial x} +\frac{\partial V}{\partial \phi} \frac{\partial\phi}{\partial x}+ \frac{\partial V}{\partial z}\frac{\partial z}{\partial x}$$



Transformation from cylindrical to cartisian coordinates,
$$x= \rho cos \phi ,\\ y= \rho sin \phi ,\\ z= z$$

$$\rho = \sqrt(x^2 +y^2)$$

$$tan \phi = \frac{y}{x}$$

From here,

$$\frac{\partial\rho}{\partial x}= \frac{x}{\rho}$$

$$\frac{\partial\phi}{\partial x}= \frac{-sin\phi}{\rho}$$

$$\frac{\partial z}{\partial x}= 0$$


Substituting you get your result.

 

[ViktigLemma]

Thank you both very much! I was thinking in the wrong paths entirely. To explain:

What I figured, not being well versed in this kind of manipulation, was this:


$$\frac{\partial V}{\partial \phi}\frac{\partial \phi}{\partial x} = \frac{\partial V}{\partial x}\frac{\partial \phi}{\partial \phi} = \frac{\partial V}{\partial x}$$

Because I remember doing something like that once.


Thanks again for your answers :)

 

[vincentchan]

$$\frac{\partial V}{\partial \phi}\frac{\partial \phi}{\partial x} = \frac{\partial V}{\partial x}\frac{\partial \phi}{\partial \phi} = \frac{\partial V}{\partial x}$$

this is not valid in general, you can only do this in a very special case, i.e.

$$V = V(\phi(x))$$

 

[ViktigLemma]

Yes that's the chain rule doing its magic I guess :D