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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.

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2 Answers 2

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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.

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    $\begingroup$ Just realized that myself, feel stupid myself now. Thanks though. $\endgroup$ Feb 26, 2020 at 15:24
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    $\begingroup$ Always happy to make people feel stupid! $\endgroup$
    – user247327
    Feb 26, 2020 at 15:25
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You can choose an arbitrarily small step say 0.1

$$ x= \pm(.1,.2..3,.4,.5,.. 0.8,0.9,1)$$ $$ y= +(.01,.04,.09,.16,.25, ... 0.64,0.81, 1.0) $$

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  • $\begingroup$ That way I could get the interval where $t$ takes values, but couldn't be a methed to achieve this more rapidly? $\endgroup$ Feb 26, 2020 at 15:09
  • $\begingroup$ 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? $\endgroup$ Feb 26, 2020 at 15:14
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    $\begingroup$ 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. $\endgroup$
    – Narasimham
    Feb 26, 2020 at 15:24

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