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Can the slope in polar coordinate be expressed as $\frac{dr}{rd\theta}$?

Feynman
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    You should add some context: What makes you think it can (or can't) be done like this, or what inspired you to ask the question? – pjs36 Mar 08 '17 at 00:52
  • For example, I have a function $V_n=r^nsinn\theta$, and its gradient is $\vec X=\nabla {V_n}=X_r\hat r+X_{\theta}\hat \theta$. So, the slope of the gradient in $(r,\theta)$ can be expressed as $k=\frac{dr}{rd\theta}=\frac{X_r}{X_{\theta}}$. Is it right? – Feynman Mar 08 '17 at 07:02

2 Answers2

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The slope of a polar function is

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}} = \frac{\frac{\mathrm{d}r}{\mathrm{d}\theta}\cdot\sin\theta + r\cdot\cos\theta}{\frac{\mathrm{d}r}{\mathrm{d}\theta}\cdot\cos\theta - r\cdot\sin\theta}$$

Dando18
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Let $x=r \cos (\theta)$ and $y=r \sin (\theta)$.

We have by the product rule,

$$\frac{dx}{d\theta}=-r\sin( \theta)+\frac{dr}{d\theta}\cos(\theta)$$

We also have,

$$\frac{dy}{d\theta}=r\cos (\theta)+\frac{dr}{d\theta}\sin( \theta)$$

So it follows by the chain rule,

$$\frac{dy}{dx}\frac{dx}{d\theta}=\frac{dy}{d\theta}$$

Solving for $\frac{dy}{dx}$ gives,

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

Substituting in the values we found we get,

$$\frac{dy}{dx}=\frac{r\cos (\theta)+\frac{dr}{d\theta}\sin( \theta)}{-r\sin( \theta)+\frac{dr}{d\theta}\cos(\theta)}$$

Ahmed S. Attaalla
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  • Hmm, in fact, I want to know the expression of slope in the coordinator $(r,\theta)$, but in $(x,y)$. – Feynman Mar 08 '17 at 06:41
  • Not sure I understand, may you elaborate @Feynman – Ahmed S. Attaalla Mar 08 '17 at 06:42
  • Hmm, in fact, I want to know the expression of slope in the coordinator $(r,\theta)$, but in $(x,y)$. For example, I have a function $V_n=r^nsinn\theta$, and its gradient is $\vec X=\nabla {V_n}=X_r\hat r+X_{\theta}\hat \theta$. So, the slope of the gradient in $(r,\theta)$ can be expressed as $k=\frac{dr}{rd\theta}=\frac{X_r}{X_{\theta}}$. Is it right? – Feynman Mar 08 '17 at 06:58
  • It still isn't clear to me what you mean. Please edit your question. I think your probably looking for something like this http://math.stackexchange.com/questions/586848/how-to-obtain-the-gradient-in-polar-coordinates. – Ahmed S. Attaalla Mar 08 '17 at 23:30