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I'm asked to show the following four derivatives from the polar coordinate conversions. ($x = r\cos \theta$, $y=r\sin \theta$, and $r^2=x^2+y^2$)

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

I've shown the fist two relatively easy, but I'm not sure how to show $$\frac{\partial \theta}{\partial x} = \frac{-\sin \theta}{r} \ , \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}$$

I know how to do this for $$\frac{\partial x}{\partial \theta} \ \ \text{and} \ \ \frac{\partial y}{\partial \theta}$$

Any help would be appreciated, thank you.

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    $\begingroup$ $\dfrac{\partial x}{\partial r} = \cos \theta$ $\endgroup$ – Nosrati Feb 2 '17 at 3:50
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The Jacobian matrix for the map $\phi:(x,y)\mapsto(r\cos\theta,r\sin\theta)$ is $$J_\phi=\pmatrix{{\partial x\over\partial r}&{\partial x\over\partial\theta}\\{\partial y\over\partial r}&{\partial y\over\partial\theta}}=\pmatrix{\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta}.$$ Inverting this matrix yields the Jacobian of the inverse map:$$J_\phi^{-1}=\frac1r\pmatrix{r\cos\theta&r\sin\theta\\-\sin\theta&\cos\theta}=\pmatrix{\cos\theta&\sin\theta\\-{\sin\theta\over r}&{\cos\theta\over r}}=J_{\phi^{-1}}=\pmatrix{{\partial r\over\partial x}&{\partial r\over\partial y}\\{\partial\theta\over\partial x}&{\partial\theta\over\partial y}}.$$

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Since $\theta=\theta(x, y)=\tan^{-1}\frac{y}{x}$, then \begin{align} \frac{\partial \theta}{\partial x} = \frac{-y/x^2}{1+y^2/x^2}= \frac{-y}{x^2+y^2} = -\frac{r\sin\theta}{r^2}. \end{align}

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Note that we have

$$\tan(\theta)=y/x \tag 1$$

Therefore, taking the partial derivative with respect to $x$ on both sides of $(1)$ reveals

$$\begin{align}\frac{\partial \tan(\theta)}{\partial x}&=\sec^2(\theta)\frac{\partial \theta}{\partial x}\\\\ &=-y/x^2 \end{align}$$

where upon solving for $\frac{\partial \theta}{\partial x}$ and using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ yields

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

Similarly, taking the partial derivative with respect to $y$ on both sides of $(1)$ reveals

$$\begin{align}\frac{\partial \tan(\theta)}{\partial y}&=\sec^2(\theta)\frac{\partial \theta}{\partial y}\\\\ &=1/x \end{align}$$

where upon solving for $\frac{\partial \theta}{\partial y}$ and using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ yields

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

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