Partial derivative question involving polar coordinates

Question

Given the plane polar coordinates x = r cos (θ), y = r sin (θ), show that ∂r/∂x*∂x/∂r + ∂θ/∂x*∂x/∂θ = 1

The solution I have says Xr = cos(θ) and Rx = cos(θ).

but I don't understand how you can't divide x =rcos(θ) by cos(θ). and then Rx would be 1/cos(θ)? Any help would be appreciated.

Answer 1

Notice, we have $x=r\cos\theta$ $$\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}$$ $$\frac{\partial x}{\partial r}=\cos\theta$$ $$\frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r}$$ $$\frac{\partial x}{\partial \theta}=-r\sin\theta$$ Now, set the corresponding values as follows $$\frac{\partial r}{\partial x}\cdot \frac{\partial x}{\partial r}+\frac{\partial \theta}{\partial x}\cdot \frac{\partial x}{\partial \theta}$$$$=\left(\frac{1}{\cos\theta}\right)\left(\cos \theta\right)+\left(-\frac{\sin\theta}{r}\right)\left(-r\sin\theta\right)=\cos^2\theta+\sin^2\theta=1$$

Answer 2

From what you said $x=r\cos \theta$ and $y=r\sin \theta$. Also in the reverse direction $r=\sqrt{x^2+y^2}$ and $\theta = \tan^{-1}(y/x)$. Now First of all if $\theta=\theta(x,y)$, then $$ -\frac{y}{x^2}=\frac{\partial}{\partial x}\tan\theta=(1+\tan^2\theta)\frac{\partial \theta}{\partial x}=\frac{x^2+y^2}{x^2}\frac{\partial \theta}{\partial x}\Longrightarrow \frac{\partial \theta}{\partial x}=-\frac{y}{r^2}=\frac{-\sin\theta}{r} $$

$$ \begin{cases} \frac{\partial x}{\partial r}=\cos\theta\\ \frac{\partial x}{\partial \theta}=-r\sin\theta\\ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r} = \cos\theta\\ \frac{\partial \theta}{\partial x} = -\frac{\sin\theta}{r} \end{cases} $$

Therefore $$ \frac{\partial x}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial x}{\partial \theta}\frac{\partial \theta}{\partial x}= \cos^2\theta +\sin^2\theta=1 $$

Answer 3

Let $\phi: (r,\theta)\mapsto(x,y)$ be the polar to Cartesian map. We have the two Jacobian matrices $$ J_\phi=\pmatrix{{\partial x\over\partial r} & {\partial x\over\partial\theta} \\ {\partial y\over\partial r} & {\partial y\over\partial\theta}} \\ J_{\phi^{-1}}=\pmatrix{{\partial r\over\partial x} & {\partial r\over\partial y} \\ {\partial\theta\over\partial x} & {\partial\theta\over\partial y}} $$ The given expression is the upper-left corner of the product $J_\phi J_{\phi^{-1}}=J_\phi J_\phi^{-1}=I$, so it must be equal to $1$.