# Finding the parameter values of parametrized curve

I'm kinda confused on how to find the values that a parameter takes given two points of parametrized curve, this is the problem I have:

Parametrize $$y=x^2$$ in the interval of the two points $$(-1,1)$$ and $$(1,1)$$ find the values of the parameter where this function takes value.

My solution was this, the parametrizatition is $$x=t$$ and $$y=t^2$$, this results in the fuction $$\alpha =t \hat i+t^2 \hat j$$, but I don't know what are the values of $$t$$ in the interval $$(-1,1)\cup(1,1)$$ for the parametrized curve.

Any help would be awesome, thanks.

I'm not sure whether you are very confused or if you are just using poor notation. You talk about $$(-1, 1)\cup (1, 1)$$ as if (-1, 1) and (1, 1) were intervals but you had already said that (-1, 1) and (1, 1) are points, not intervals! You have x= t so the values for t are just the values for x, -1 to 1.
$$x= \pm(.1,.2..3,.4,.5,.. 0.8,0.9,1)$$ $$y= +(.01,.04,.09,.16,.25, ... 0.64,0.81, 1.0)$$
• That way I could get the interval where $t$ takes values, but couldn't be a methed to achieve this more rapidly? Feb 26, 2020 at 15:09
• Or could I just evaluate $\alpha =t \hat i+t^2 \hat j$ in the two points I'm given to find the interval where $t$ takes values? Feb 26, 2020 at 15:14
• You are given endpoints. You can parametrize with regard to $x$ or slope or arc length. Step interval is to be chosen based on a choice of one of these for example..that is all your choice. Feb 26, 2020 at 15:24