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How to prove this corollary in line integral?


Dhamnekar Win,odd

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  • 2 weeks later...

If f is in the direction of r'(t) then it is in the direction of increasing r(t). So, the directions of f(t) and r'(t) will also be same.

 Definition of dot product:

image.png.337af54b0f08a78de2bb2e5dc0d147c6.png

And ds =|| r'(t)|| ⇒ [math]\sqrt{x'(t)^2 + y'(t)^2}dt [/math] so, work = force × distance = [math] \displaystyle\int_C f\cdot dr= \displaystyle\int_C \left\vert\vert{f} \right\vert\vert ds[/math]  

 

Edited by Dhamnekar Win,odd
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  • 1 month later...
On 7/4/2022 at 3:45 PM, Dhamnekar Win,odd said:

If f is in the direction of r'(t) then it is in the direction of increasing r(t). So, the directions of f(t) and r'(t) will also be same.

Exactly!

They key fact is your observation that,

\[ \frac{\boldsymbol{f}\left(\boldsymbol{r}\left(t\right)\right)}{\left\Vert \boldsymbol{f}\left(\boldsymbol{r}\left(t\right)\right)\right\Vert }=\frac{\boldsymbol{r}'\left(t\right)}{\left\Vert \boldsymbol{r}'\left(t\right)\right\Vert } \]

You only have to dot-multiply by \( d\boldsymbol{r}\left(t\right)=\boldsymbol{r}'\left(t\right)dt \), remember the definition of the norm, and you're there.

I hope that helps.

Edited by joigus
Latex editing
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