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Note by Raja Metronetizen
3 years, 10 months ago

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Disregard my previous comment, this can be proved. We will begin with (ii), (though it doesn't really matter) because we will use the result of (ii) to prove (i).

$$(x)^2 + (y)^2 = r^2$$ $$\Rightarrow$$ $$\frac{\partial} {\partial x}$$$$((x)^2 + (y)^2)$$ = $$(2r)(\frac{\partial r} {\partial x})$$

$$\Rightarrow$$ $$2x = (2r)(\frac{\partial r} {\partial x})$$

$$\Rightarrow$$ $$\frac{x} {r} = \frac {\partial r} {\partial x}$$

Note that $$cos(\theta)=\frac{x} {r}$$,

Hence $$cos(\theta)=\frac {\partial r} {\partial x}$$

$$x=rcos(\theta)$$ $$\Rightarrow$$ $$\frac {\partial x} {\partial r} = \frac {\partial (rcos(\theta))} {\partial r}$$

$$\Rightarrow$$ $$\frac{\partial x} {\partial r} = \frac {\partial r} {\partial x} = cos(\theta)$$

This proves (ii). Now for (i):

$$x=rcos(\theta)$$ $$\Rightarrow$$ $$\frac {\partial x} {\partial \theta} = \frac {\partial} {\partial \theta} (rcos(\theta))$$ = $$-rsin(\theta)$$

$$\Rightarrow$$ $$(\frac {1} {r})\frac{\partial x} {\partial \theta}= -sin(\theta)$$

$$x=rcos(\theta)$$ $$\Rightarrow$$ $$\frac{\partial x} {\partial x}= \frac{\partial} {\partial x} (rcos(\theta)$$

$$\Rightarrow$$ $$1= cos(\theta)\frac{\partial r} {\partial x} -rsin(\theta) \frac {\partial \theta} {\partial x}$$

In (ii) we showed that: $$\frac{\partial r} {\partial x} = cos(\theta)$$, hence:

$$1=cos^2(\theta) - rsin(\theta)\frac {\partial \theta} {\partial x}$$

$$\Rightarrow$$ $$\frac{1-cos^2(\theta)} {-rsin(\theta)}=\frac{\partial \theta} {\partial x}$$

$$\Rightarrow$$ $$\frac {-sin(\theta)} {r} = \frac {\partial \theta} {\partial x}$$

$$\Rightarrow$$ $$r\frac{\partial \theta} {\partial x} = -sin(\theta) = \frac {1} {r} \frac {\partial x} {\partial \theta}$$

This proves (i). Q.E.D. · 3 years ago