Is $y=x^2$ smooth at origin?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite
1












if $r(t)=t^2 i + t^4 j$, then it is a parabola $y = x^2$. It satisfies the condition of non smooth curve i.e. $fracdrdt=0$ at $t=0$. But geometrically it shows the curve (parabola) is smooth at $(0,0)$. Why is this so?




conic-sections




share|cite|improve this question


















  • 1




    It is only an arc of parabola.
    – Bernard
    Aug 15 at 9:08










  • @Bernard What's the difference?
    – Arthur
    Aug 15 at 9:08






  • 1




    It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
    – Fakemistake
    Aug 15 at 9:12










  • That explains why $mathrm dr/mathrm dt=0$.
    – Bernard
    Aug 15 at 9:12















up vote
0
down vote

favorite
1












if $r(t)=t^2 i + t^4 j$, then it is a parabola $y = x^2$. It satisfies the condition of non smooth curve i.e. $fracdrdt=0$ at $t=0$. But geometrically it shows the curve (parabola) is smooth at $(0,0)$. Why is this so?




conic-sections




share|cite|improve this question


















  • 1




    It is only an arc of parabola.
    – Bernard
    Aug 15 at 9:08










  • @Bernard What's the difference?
    – Arthur
    Aug 15 at 9:08






  • 1




    It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
    – Fakemistake
    Aug 15 at 9:12










  • That explains why $mathrm dr/mathrm dt=0$.
    – Bernard
    Aug 15 at 9:12













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





if $r(t)=t^2 i + t^4 j$, then it is a parabola $y = x^2$. It satisfies the condition of non smooth curve i.e. $fracdrdt=0$ at $t=0$. But geometrically it shows the curve (parabola) is smooth at $(0,0)$. Why is this so?




conic-sections




share|cite|improve this question














if $r(t)=t^2 i + t^4 j$, then it is a parabola $y = x^2$. It satisfies the condition of non smooth curve i.e. $fracdrdt=0$ at $t=0$. But geometrically it shows the curve (parabola) is smooth at $(0,0)$. Why is this so?





conic-sections





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 9:04









Cornman

2,61721128




2,61721128










asked Aug 15 at 9:00









Sumit

1




1







  • 1




    It is only an arc of parabola.
    – Bernard
    Aug 15 at 9:08










  • @Bernard What's the difference?
    – Arthur
    Aug 15 at 9:08






  • 1




    It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
    – Fakemistake
    Aug 15 at 9:12










  • That explains why $mathrm dr/mathrm dt=0$.
    – Bernard
    Aug 15 at 9:12













  • 1




    It is only an arc of parabola.
    – Bernard
    Aug 15 at 9:08










  • @Bernard What's the difference?
    – Arthur
    Aug 15 at 9:08






  • 1




    It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
    – Fakemistake
    Aug 15 at 9:12










  • That explains why $mathrm dr/mathrm dt=0$.
    – Bernard
    Aug 15 at 9:12








1




1




It is only an arc of parabola.
– Bernard
Aug 15 at 9:08




It is only an arc of parabola.
– Bernard
Aug 15 at 9:08












@Bernard What's the difference?
– Arthur
Aug 15 at 9:08




@Bernard What's the difference?
– Arthur
Aug 15 at 9:08




1




1




It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
– Fakemistake
Aug 15 at 9:12




It depends on the parametrization you have: Take for example $(t,t^2)$ and $(t^2,t^4)$ and calculate the derivative. It is $(1,2t)$ and $(2t,4t^3)$. The first is never equal to zero, but the second...
– Fakemistake
Aug 15 at 9:12












That explains why $mathrm dr/mathrm dt=0$.
– Bernard
Aug 15 at 9:12





That explains why $mathrm dr/mathrm dt=0$.
– Bernard
Aug 15 at 9:12











1 Answer
1






active

oldest

votes

















up vote
3
down vote













A curve is non-smooth at a point if every parametrisation is either not differentiable or has $fracdrdt = 0$ at that point. It's not enough that you have found one such parametrisation.



There are other parametisations of the same curve, such as $r(t) = ti + t^2j$, which does have a well-defined, non-zero derivative everywhere, showing that the curve is smooth.






share|cite|improve this answer






















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2883372%2fis-y-x2-smooth-at-origin%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    A curve is non-smooth at a point if every parametrisation is either not differentiable or has $fracdrdt = 0$ at that point. It's not enough that you have found one such parametrisation.



    There are other parametisations of the same curve, such as $r(t) = ti + t^2j$, which does have a well-defined, non-zero derivative everywhere, showing that the curve is smooth.






    share|cite|improve this answer


























      up vote
      3
      down vote













      A curve is non-smooth at a point if every parametrisation is either not differentiable or has $fracdrdt = 0$ at that point. It's not enough that you have found one such parametrisation.



      There are other parametisations of the same curve, such as $r(t) = ti + t^2j$, which does have a well-defined, non-zero derivative everywhere, showing that the curve is smooth.






      share|cite|improve this answer
























        up vote
        3
        down vote










        up vote
        3
        down vote









        A curve is non-smooth at a point if every parametrisation is either not differentiable or has $fracdrdt = 0$ at that point. It's not enough that you have found one such parametrisation.



        There are other parametisations of the same curve, such as $r(t) = ti + t^2j$, which does have a well-defined, non-zero derivative everywhere, showing that the curve is smooth.






        share|cite|improve this answer














        A curve is non-smooth at a point if every parametrisation is either not differentiable or has $fracdrdt = 0$ at that point. It's not enough that you have found one such parametrisation.



        There are other parametisations of the same curve, such as $r(t) = ti + t^2j$, which does have a well-defined, non-zero derivative everywhere, showing that the curve is smooth.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 15 at 9:11

























        answered Aug 15 at 9:07









        Arthur

        99.8k793175




        99.8k793175






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2883372%2fis-y-x2-smooth-at-origin%23new-answer', 'question_page');

            );

            Post as a guest
































































            -

            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
            閱讀完整內容

            Propositional logic and tautologies

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Taken from P. Suppes "Introduction to logic" pp16 Starting from the sentence "If $P$ tautologically implies $Q$, then.." Question: Is it true that, at least within propositional logic, we are able to prove only tautologies or do I mis-understand something? If yes, is propositional logic special in some way? Is the notion of tautology meaningful only in propositional logic? logic propositional-calculus share | cite | improve this question edited Nov 29 '17 at 18:53 Mauro ALLEGRANZA 60.8k 4 46 105 asked Nov 29 '17 at 16:32 Alvin Lepik 2,448 9 20 add a comment  |  up vote 0 down vote favorite Taken from P. Suppes "Introduction to logic" pp16 Starting from the sentence "If $P$ tautologically implies $Q$, then.." Question: Is it true that, at least within propositional logic, we are able to prove only tautologi...
            閱讀完整內容

            Distribution of Stopped Wiener Process with Stochastic Volatility

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Let $(W_s)_s geq 0$ be a Wiener process and $tau$ be a random variable with an exponential distribution with parameter $lambda$. Suppose that $W$ and $tau$ are independent. In this question, we see that the distribution of the stopped Wiener process $W_tau$ corresponds to a Laplace distribution with scale parameter $fracsigmasqrt2lambda$ where $sigma$ is the instantaneous variance of the Wiener process. Suppose now, that the variance $sigma^2$ also follows a stochastic process such that we have: $$dS = sigma SdW_s \ dsigma^2 = alphasigma^2dt + xisigma^2dW_sigma$$ where $alpha$ and $xi$ are independent of $S$ and $dW_s$ and $dW_sigma$ are independent Wiener processes. My aim is to derive the distribution of $log S_tau$. First, if $barV_T$ denotes the mean variance over some time interval $[0,T]$ defined by $$barV_T = frac1T intlimits_0^Tsigma^2(t)dt,$$ it is easy to show (Lemma of Îto) that $$...
            閱讀完整內容