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. $\frac{dr}{dt}=0$ at $t=0$. But geometrically it shows the curve (parabola) is smooth at $(0,0)$. Why is this so?
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2It is only an arc of parabola. – Bernard Aug 15 '18 at 09:08
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@Bernard What's the difference? – Arthur Aug 15 '18 at 09:08
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1It 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 '18 at 09:12
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1That explains why $\mathrm dr/\mathrm dt=0$. – Bernard Aug 15 '18 at 09:12
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To expand @Bernard's comment it is only the right half of the parabola since $t^2$ is always positive. – Jeff Mar 20 '22 at 20:48
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A curve is non-smooth at a point if every parametrisation is either not differentiable or has $\frac{dr}{dt} = 0$ 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.
Arthur
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