How to prove this corollary in line integral?

[Dhamnekar Win,odd]

. Show that if f points in the same direction as r'(t) at each point r(t) along a smooth curve C, then [math]\displaystyle\int_C f\cdot dr = \displaystyle\int_C ||f|| ds  ?[/math] 

 

How can I prove this corollary in line integral?

[Dhamnekar Win,odd]

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:

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 July 4, 2022 by Dhamnekar Win,odd

[joigus]

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 August 11, 2022 by joigus
Latex editing