To prove the given identity, [tex]\(\frac{\partial r}{\partial x} = \frac{\partial x}{\partial r}\)[/tex], we start by understanding the relationship between Cartesian coordinates [tex]\((x, y)\)[/tex] and polar coordinates [tex]\((r, \theta)\)[/tex].
The relationships are given by:
[tex]\[ x = r \cos \theta \][/tex]
[tex]\[ y = r \sin \theta \][/tex]
To proceed, we first express [tex]\( r \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
### Step 1: Compute [tex]\(\frac{\partial r}{\partial x}\)[/tex]
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
Taking the partial derivative of [tex]\( r \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} (\sqrt{x^2 + y^2}) \][/tex]
Using the chain rule:
[tex]\[ \frac{\partial r}{\partial x} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \][/tex]
[tex]\[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \][/tex]
Since [tex]\( r = \sqrt{x^2 + y^2} \)[/tex], we can rewrite:
[tex]\[ \frac{\partial r}{\partial x} = \frac{x}{r} \][/tex]
### Step 2: Compute [tex]\(\frac{\partial x}{\partial r}\)[/tex]
From the initial relation:
[tex]\[ x = r \cos \theta \][/tex]
Taking the partial derivative of [tex]\( x \)[/tex] with respect to [tex]\( r \)[/tex]:
[tex]\[ \frac{\partial x}{\partial r} = \frac{\partial}{\partial r} (r \cos \theta) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial x}{\partial r} = \cos \theta \][/tex]
### Step 3: Evaluate [tex]\(\cos \theta\)[/tex]
We know:
[tex]\[ \cos \theta = \frac{x}{r} \][/tex]
### Step 4: Compare the results
From Step 1:
[tex]\[ \frac{\partial r}{\partial x} = \frac{x}{r} \][/tex]
From Step 2:
[tex]\[ \frac{\partial x}{\partial r} = \cos \theta = \frac{x}{r} \][/tex]
We observe that both partial derivatives are equal:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial x}{\partial r} = \frac{x}{r} \][/tex]
Thus, we have proved that:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial x}{\partial r} \][/tex]
