How Is the Parabolic Path y=x^2 Related to the Line Integral Calculation?

[gl0ck]

Homework Statement

Homework Equations

Trued only 1st question..
Unfortunately I lost my notes about this and cannot find anything relevant to this.
I think,
∫_cF.dr = ∫_cF.dr/dt dt ..
also dr/dt isn't it = ∂x/∂ti +∂y/∂tj+∂z/∂tk
Also it seems that C is with parabolic shape?
Can someone tell me what is the relationship with y=x^2 and 't'
Also if there is just dot product of F and ∂x/∂ti +∂y/∂tj+∂z/∂tk and boundaries of integral 0 and 1
why he have this y=x^2 from (0,0) to (1,1) ?
I think I miss something general here..

Thanks

 

[voko]

$t$ is the parameter of the curve, so that the curve can be described by $x = f(t), y = g(t), z = h(t)$. In your case, you can simply let $x = t$.

 

[pasmith]

Homework Statement

Homework Equations

Trued only 1st question..
Unfortunately I lost my notes about this and cannot find anything relevant to this.
I think,
∫_cF.dr = ∫_cF.dr/dt dt ..
also dr/dt isn't it = ∂x/∂ti +∂y/∂tj+∂z/∂tk
Also it seems that C is with parabolic shape?
Can someone tell me what is the relationship with y=x^2 and 't'

The principle is that C can be parametrized by any \mathbf{r}(t) = (x(t),y(t)) such that \mathbf{r}(0) = (0,0), \mathbf{r}(1) = (1,1), together with \|d\mathbf{r}/dt\| > 0 for all t \in (0,1) and y(t) = (x(t))^2 for all t \in (0,1). Subject to those constraints the integral doesn't depend on the particular choice of \mathbf{r}(t).

Can you think of a simple choice for x(t) and y(t)?

 

[gl0ck]

I got it it was simpler that I thought .. Thanks for the time :)
https://www.dropbox.com/s/3az4t1r6sj52rkw/IMG_20140415_224810.jpg