!Understanding Partial Derivatives of Coordinate Transformation

[onchoa]

Hi Everyone,

I was studying coordinate transformation and I came across this equation, that I couldn't understand how it came up. Let me put it this way:
x = rcosθ

Then if I want to express the partial derivative (of any thing) with respect to x, what would be the expression? i.e. ∂/∂x=?
According to the text, answer would be
∂/∂x=cosθ*∂/∂r - (sinθ/r)*∂/∂θ

Please explain to me how to come up with this expression.

Thnaks

 

[sweet springs]

Hi.

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}$$

$$r^2=x^2+y^2$$

$$2rdr=2xdx+2ydy$$

$$\frac{\partial r}{\partial x}=\frac{x}{r}=cos \theta$$

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

$$\frac{d\theta}{cos^2 \theta}=-\frac{ydx}{x^2}+\frac{dy}{x}$$

$$\frac{\partial \theta}{\partial x}=-\frac{ cos^2 \theta y} {x^2} = -\frac{sin \theta} {r}$$

I hope these calculations are helpful.

 

[HallsofIvy]

The chain rule:
[tex]\frac{\partial F}{\partial x}= \frac{\partial F}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial F}{\partial \theta}\frac{\partial \theta}{\partial x}[/tex]

But since there are two variable, r and [itex]\theta[/itex], just "[itex]x= r cos(\theta)[/itex]" is not enough. You also need [itex]y= r sin(\theta)[/itex]. Then we can write [itex]r= (x^2+ y^2)^{1/2}[/itex] and [itex]\theta= arctan(y/x)[/itex] so that
[tex]\frac{dr}{dx}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{\sqrt{x^2+ y^2}}= \frac{rcos(\theta)}{r}= cos(\theta)[/tex]
and
[tex]\frac{d\theta}{dx}= \frac{1}{1+ y^2/x^2}\left(-\frac{y}{x^2}\right)= \frac{-y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r}sin(\theta)[/tex]

So that
[tex]\frac{\partial F}{\partial x}= cos(\theta)\frac{\partial F}{\partial r}- \frac{sin(\theta)}{r}\frac{\partial F}{\partial \theta}[/tex]