Answer :
To prove the given partial derivative relationships, we will utilize the definitions of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in terms of [tex]\(r\)[/tex] and [tex]\(\theta\)[/tex], and use chain rule techniques. The equations given are:
[tex]\[x = r \cos \theta\][/tex]
[tex]\[y = r \sin \theta\][/tex]
### (i) Proving that [tex]\(\frac{\partial r}{\partial x} = \frac{\partial x}{\partial r}\)[/tex]
First, we express [tex]\(r\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[r = \sqrt{x^2 + y^2}\][/tex]
Now, let's find [tex]\(\frac{\partial r}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2 + y^2} \][/tex]
Considering [tex]\(y\)[/tex] as a constant while differentiating with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} \][/tex]
Next, let's find [tex]\(\frac{\partial x}{\partial r}\)[/tex]:
[tex]\[ x = r \cos \theta \][/tex]
Considering [tex]\(\theta\)[/tex] as a constant while differentiating with respect to [tex]\(r\)[/tex]:
[tex]\[ \frac{\partial x}{\partial r} = \cos \theta \][/tex]
Given [tex]\(x = r \cos \theta\)[/tex], we can rewrite [tex]\(\cos \theta\)[/tex] as:
[tex]\[ \cos \theta = \frac{x}{r} \][/tex]
So,
[tex]\[ \frac{\partial x}{\partial r} = \frac{x}{r} \][/tex]
Therefore, we have:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial x}{\partial r} \][/tex]
This completes the proof for part (i).
### (ii) Proving that [tex]\(\frac{1}{r} \frac{\partial x}{\partial \theta} = r \frac{\partial \theta}{\partial x}\)[/tex]
First, let's find [tex]\(\frac{\partial x}{\partial \theta}\)[/tex]:
[tex]\[ x = r \cos \theta \][/tex]
Considering [tex]\(r\)[/tex] as a constant while differentiating with respect to [tex]\(\theta\)[/tex]:
[tex]\[ \frac{\partial x}{\partial \theta} = -r \sin \theta \][/tex]
Next, let's express [tex]\(\theta\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \theta = \tan^{-1}\left( \frac{y}{x} \right) \][/tex]
Now, let's find [tex]\(\frac{\partial \theta}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x} \tan^{-1}\left( \frac{y}{x} \right) \][/tex]
Considering [tex]\(y\)[/tex] as a constant while differentiating with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial \theta}{\partial x} = \frac{-y}{x^2 + y^2} = \frac{-y}{r^2} \][/tex]
So, let's combine the results:
[tex]\[ \frac{1}{r} \frac{\partial x}{\partial \theta} = \frac{1}{r} \left(-r \sin \theta\right) = -\sin \theta \][/tex]
And:
[tex]\[ r \frac{\partial \theta}{\partial x} = r \left(\frac{-y}{r^2}\right) = \frac{-y}{r} \][/tex]
Given [tex]\(y = r \sin \theta\)[/tex], we can rewrite:
[tex]\[ \frac{-y}{r} = - \sin \theta \][/tex]
So, we have:
[tex]\[ \frac{1}{r} \frac{\partial x}{\partial \theta} = r \frac{\partial \theta}{\partial x} = - \sin \theta \][/tex]
This completes the proof for part (ii).
Thus, both statements are proven as required.
[tex]\[x = r \cos \theta\][/tex]
[tex]\[y = r \sin \theta\][/tex]
### (i) Proving that [tex]\(\frac{\partial r}{\partial x} = \frac{\partial x}{\partial r}\)[/tex]
First, we express [tex]\(r\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[r = \sqrt{x^2 + y^2}\][/tex]
Now, let's find [tex]\(\frac{\partial r}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2 + y^2} \][/tex]
Considering [tex]\(y\)[/tex] as a constant while differentiating with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} \][/tex]
Next, let's find [tex]\(\frac{\partial x}{\partial r}\)[/tex]:
[tex]\[ x = r \cos \theta \][/tex]
Considering [tex]\(\theta\)[/tex] as a constant while differentiating with respect to [tex]\(r\)[/tex]:
[tex]\[ \frac{\partial x}{\partial r} = \cos \theta \][/tex]
Given [tex]\(x = r \cos \theta\)[/tex], we can rewrite [tex]\(\cos \theta\)[/tex] as:
[tex]\[ \cos \theta = \frac{x}{r} \][/tex]
So,
[tex]\[ \frac{\partial x}{\partial r} = \frac{x}{r} \][/tex]
Therefore, we have:
[tex]\[ \frac{\partial r}{\partial x} = \frac{\partial x}{\partial r} \][/tex]
This completes the proof for part (i).
### (ii) Proving that [tex]\(\frac{1}{r} \frac{\partial x}{\partial \theta} = r \frac{\partial \theta}{\partial x}\)[/tex]
First, let's find [tex]\(\frac{\partial x}{\partial \theta}\)[/tex]:
[tex]\[ x = r \cos \theta \][/tex]
Considering [tex]\(r\)[/tex] as a constant while differentiating with respect to [tex]\(\theta\)[/tex]:
[tex]\[ \frac{\partial x}{\partial \theta} = -r \sin \theta \][/tex]
Next, let's express [tex]\(\theta\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \theta = \tan^{-1}\left( \frac{y}{x} \right) \][/tex]
Now, let's find [tex]\(\frac{\partial \theta}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial \theta}{\partial x} = \frac{\partial}{\partial x} \tan^{-1}\left( \frac{y}{x} \right) \][/tex]
Considering [tex]\(y\)[/tex] as a constant while differentiating with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial \theta}{\partial x} = \frac{-y}{x^2 + y^2} = \frac{-y}{r^2} \][/tex]
So, let's combine the results:
[tex]\[ \frac{1}{r} \frac{\partial x}{\partial \theta} = \frac{1}{r} \left(-r \sin \theta\right) = -\sin \theta \][/tex]
And:
[tex]\[ r \frac{\partial \theta}{\partial x} = r \left(\frac{-y}{r^2}\right) = \frac{-y}{r} \][/tex]
Given [tex]\(y = r \sin \theta\)[/tex], we can rewrite:
[tex]\[ \frac{-y}{r} = - \sin \theta \][/tex]
So, we have:
[tex]\[ \frac{1}{r} \frac{\partial x}{\partial \theta} = r \frac{\partial \theta}{\partial x} = - \sin \theta \][/tex]
This completes the proof for part (ii).
Thus, both statements are proven as required.
