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Let $\vec{n}$ be the out of unit normal vector on the curve $\varGamma$,and the $\varGamma:x^2+y^2=R^2$.define$$r=\sqrt{x^2+y^2}.$$Calculate$$\oint_{\varGamma}\dfrac{\partial\ln r}{\partial{\vec{n}}}\mathrm{d}s.$$ I made the following attempts,Known that $\vec{n}=(\dfrac{x}{R},\dfrac{y}{R})$.So$$\dfrac{\partial\ln r}{\partial{\vec{n}}}=\dfrac{\partial\ln r}{\partial{x}}\cdot\dfrac{x}{R}+\dfrac{\partial\ln r}{\partial{y}}\cdot\dfrac{y}{R}$$ But I don't understand why $$\dfrac{\partial\ln r}{\partial\vec{n}}=\dfrac{d\ln r}{d r}$$ Can you explain it to me, thank you.

Hilbert1994
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    Note the normal vector in this case is $\vec{r}$. So when you are finding the rate of change of $f(r) = \ln r$ in the direction of the unit normal, it means you are finding the rate of change of $f(r)$ in the direction of $\vec{r}$. Does that explain? – Math Lover Jan 16 '21 at 06:43
  • @Math Lover Thank you,but why the normal vector is$\vec{r}$ not $\vec{n}$? – Hilbert1994 Jan 16 '21 at 07:28
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    $\vec{n}$ is a general notation for a unit normal vector. For a circle centered at the origin, an outward normal vector will be from the origin pointing out radially. So the normal vector is $\vec{r} = x \hat{i} + y \hat{j}$ where $(x, y)$ is on the circle. When it comes to unit normal it can be written as $\frac{\vec{r}}{r}$. But the point you should focus on is that you are trying to find rate of change of $f(r)$ in the direction of $\vec{n}$ which is radially outward vector. – Math Lover Jan 16 '21 at 07:37
  • @Math Lover I think I know what you mean. Thank you very much! – Hilbert1994 Jan 16 '21 at 07:40
  • @Math Lover it means that the $\vec{n}=\frac{\vec{r}}{r}$ in this case? – Hilbert1994 Jan 16 '21 at 07:43
  • Yes that is correct. – Math Lover Jan 16 '21 at 07:46
  • @Math Lover Can you help me with this problem, thank you https://math.stackexchange.com/questions/3987320/how-to-prove-the-following-inequality-with-trigonometric-function – Hilbert1994 Jan 16 '21 at 07:51

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Note that $R$ belongs to $\Gamma$.

At first the function $\frac{\partial \ln r}{\partial \vec n}$ is considered independently of $\Gamma$.

Now, interpreting $\frac{\partial \ln r}{\partial \vec n}$ as the derivative of $\ln r$ in the direction of $\vec n = \left( \frac x{\color{blue}{r}}\:\: \frac y{\color{blue}{r}} \right)$ you get

\begin{eqnarray*} \frac{\partial \ln r}{\partial \vec n} & = & \frac 12 \left(\frac{\partial \ln (x^2+y^2)}{\partial x}\cdot\frac xr + \frac{\partial \ln (x^2+y^2)}{\partial y}\cdot\frac yr \right) \\ & = & \frac 12 \left(\frac {2x}{r^2}\cdot\frac xr + \frac {2y}{r^2}\cdot\frac yr\right) \\ & = & \frac 1r\\ & = & \frac{d\ln r}{dr} \end{eqnarray*}

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