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For a differentiable function: $f:\mathbb{R}^n\rightarrow\mathbb{R}$

prove that:

$$\int_{C_{\boldsymbol{pq}}}{\nabla f}\cdot\mathrm{d}\boldsymbol{r}=f(\boldsymbol{q})-f(\boldsymbol{p})$$

where $C$ is any differentiable path from $\boldsymbol{p}$ to $\boldsymbol{q}$

Then: What can you say about $\oint_C{\nabla f\cdot\mathrm{d}\boldsymbol{r}}$ for a closed curve $C$, explain that claim.

My attempt (more thoughts)

f being differentiable is a way of saying it is continuous (differentiability implies continuity), while I can't formally state continuity in multiple dimensions, the definition of partial derivative is enough to tell me a proof wouldn't be difficult and the idea is largely trivial.

Now consider n=2, a surface, a continuous surface, to get from a height a to a height b is a conservative field because the difference between "the uppy bits" and "the downy bits" is constant no matter how you go from a to b, I should say perhaps "the amount of upness is constant" but that's equally bad. I don't like using words like uppy.

This is not a real analysis question, it is a calculus question, so I need not work with definitions (I need not formally state the above basically)

I'm not quite sure where to go from here. I am shocked that differentiability is sufficient for a conservative field (sufficient is all the question asks, it is not nessasary because of things like the field around an electric point charge, that is (using the same thoughts) conservative, hence electrical potential, although it does say any differentiable path.... hmmm)

I'm quite comfortable with the dot product and the idea of a scalar field (in this case) I'm just not sure how I'd go about proving it.

If anyone wants to bring Analysis into this I'd love that, unfortunately I am away from my books currently and it's been a while.

PS: How do I underline? Thanks

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  • $\begingroup$ You're looking for \nabla (that is, $\nabla$). You also want \cdot (that is, $\cdot$) for dot products. What is it you need to underline? $\endgroup$
    – Muphrid
    Aug 29, 2013 at 22:45
  • $\begingroup$ @Muphrid any vectors in there, I used bold because I wasn't sure what it was. $\endgroup$
    – Alec Teal
    Aug 29, 2013 at 22:52
  • $\begingroup$ Please note that the continuous-differentiability of $f$ is enough for $\nabla f$ to be conservative. It is false that any continuously-differentiable vector field is conservative. $\endgroup$
    – Jonathan Y.
    Aug 29, 2013 at 22:53
  • $\begingroup$ @JonathanY. please put some of that into an answer! I am still quite blank sadly. $\endgroup$
    – Alec Teal
    Aug 29, 2013 at 22:54
  • $\begingroup$ @Muphrid I'm sorry for being rude! Thanks for answering by the way, I just ignored the rest of the comment and answered the underline part, I am very grateful. $\endgroup$
    – Alec Teal
    Aug 29, 2013 at 22:57

1 Answer 1

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Suppose $r: [a,b] \rightarrow C_{pq}$, with $r(a)=p$ and $r(b)=q$.

We have $\nabla f(r(t))\cdot r'(t)= D_t f(r(t))$. So, $$ \int_{C_{pq}} \nabla f ds=\int_a^b \nabla f(r(t))\cdot r'(t) dt = \int_a^b D_t f(r(t)) dt=f(r(b))-f(r(a))=f(q)-f(p) $$

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  • $\begingroup$ I should have seen that (honest) thanks, lovely and short. $\endgroup$
    – Alec Teal
    Aug 29, 2013 at 22:59
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    $\begingroup$ I'll add that we're assuming $r$ is $C^1$, but the result is easily generalized to the case that it fails to be so at a finite number of points. $\endgroup$
    – Jonathan Y.
    Aug 29, 2013 at 23:00

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