I recently came across this statement which I was told was true. $$\oint_C{f\nabla f}\cdot d\mathbf{r}=0$$ Can anyone provide a proof of why this is the case? Is there any way to show that $$f\nabla f=\nabla g$$ for some function $g$?
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With no restriction on $f$?!? – Eric Towers Mar 20 '14 at 21:08
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2Of course you can check that $f \nabla f = \frac{1}{2} \nabla (f^{2})$. – Sangchul Lee Mar 20 '14 at 21:09
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No restriction besides $f$ being a scalar function $\mathbb{R}^2\rightarrow\mathbb{R}$ – George1811 Mar 20 '14 at 21:10
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@sos440 How would I go about doing that? – George1811 Mar 20 '14 at 21:13
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$$\nabla(f^2) = \frac \partial {\partial x} (f^2) \hat\imath + \frac \partial {\partial y} (f^2) \hat\jmath + \frac \partial {\partial z} (f^2) \hat k$$ Apply the chain rule? – Klaas van Aarsen Mar 20 '14 at 21:26
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Won't $\nabla (f^2)=2f$? I'm not very good at multivariate calculus – George1811 Mar 20 '14 at 21:35
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So make it regular calculus. Suppose we have a function g(x). What is $\frac d {dx}(g(x)^2)$? – Klaas van Aarsen Mar 20 '14 at 21:40
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AH I see, god sake it's so blatant. I was thinking that the derivative of $f$ was $1$ which it's clearly not. Thank you very much :) – George1811 Mar 20 '14 at 21:44
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Following up on sos440, applying Stokes gives us: $$\oint f \nabla f \cdot d\mathbf r = \frac 1 2 \oint \nabla (f^2) \cdot d\mathbf r = \frac 1 2 \iint \nabla \times \nabla(f^2) \cdot d\mathbf S$$ One of the non-trivial rules for $\nabla$ is that for any $g$: $$\nabla \times \nabla g = 0$$
Klaas van Aarsen
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Using the product rule for the curl of a vector field multiplied by a scalar field,
$$\nabla\times(f\nabla f)=\nabla f \times \nabla f + f\nabla\times(\nabla f)=0.$$
(Cross product of a vector with itself is identically zero, and so is the curl of a gradient field.)
Applying Stokes's theorem,
$$\oint_C{f\nabla f}\cdot d\mathbf{r}=\int_Sd\mathbf{a}\cdot\nabla\times(f\nabla f)=\int_Sd\mathbf{a}\cdot\mathbf{0}=0.$$
David H
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