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Let $c$ be a curve given by $x=f(t)$ and $y=g(t)$ with $t\in \mathbb{R}$. Lets assume $f$ and $g$ are $C^1$. I am trying to learn how to analyze the smoothness of $c$ in the cases where $f'(t_0) = 0$ and $g'(t_0) = 0$.

For example, if we have $x=t^3$ and $y=t^2$, then $t_0=0$. In this case, $\dfrac{dx}{dy} = \dfrac{f'(t)}{g'(t)}=\dfrac{3t}{2}$ which goes to $0$ as $t\to 0$. My notes say the curve is not smooth because $dy/dt$ changes sign at $t=0$ and I can't see precisely why (why the sign changing implies is not smooth).

Here is another example, if $x=t \sin t$ and $y=t^3$ then $f'(t) = \sin t + t \cos t$ and $g'(t) = 3t^2$. Both vanishe at $t=0$. Now we have $$\lim _{t \rightarrow 0} \frac{d y}{d x}=\lim _{t \rightarrow 0} \frac{3 t^{2}}{\sin t+t \cos t}=\lim _{t \rightarrow 0} \frac{6 t}{2 \cos t-t \sin t}=0$$

but again since $dx/dt$ changes sign at $t=0$, the curve is not smooth at $t=0$. And I still don't see the relation.

Also, i noted that when my notes compute $dx/dy$ they study if $y'(t)$ changes sign, and if they compute $dy/dx$, they study if $x'(t)$ changes sign.

What I imagine is, okay imagine we are computing the derivative wrt $y$, $dx/dy$. Then we study $dy/dt$ to see if it changes suddenly of sign. That would imply $dx/dy$ is not continuous where $dy/dt$ changed sign since we are studying the derivative with respect to $y$. But i fail trying to precise this with math terms and i am not sure at all if this is correct.

Odestheory12
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  • Does this question help? (Not marking as a duplicate for now.) – Andrew D. Hwang Dec 24 '21 at 12:31
  • @AndrewD.Hwang it doesn't. In fact, not sure if that is true. Consider $\gamma(t) = (t \sin{t}, t^{3})$ which should have a continuous extension at $t=0$ according to your post (if I understood it correctly) but it actually doesn't (it has a cusp). – Odestheory12 Dec 24 '21 at 15:08
  • That curve does have cusp (and its tangent vector does not extend continuously to $0$), because $t\sin t = t^2 + O(t^3)$ at $t = 0$. – Andrew D. Hwang Dec 24 '21 at 15:28
  • I do know it has a cusp but I can't follow your explanation. What has to do $t\sin t = t^2 + O(t^3)$ with having a cusp at $t=0$? The tangent vector is $T(t) = \dfrac{(t \sin{t}, t^3)}{\sqrt{9t^4 + (\sin{t}+t\cos{t})^2}}$ – Odestheory12 Dec 24 '21 at 15:48
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    We have (n.b. the numerator) $$T(t)=\frac{(\sin t + t\cos t, 3t^2)}{\sqrt{9t^4 + (\sin t + t\cos t)^{2}}}.$$Factoring $\sqrt{t^2}=|t|$ (n.b., not $t$) from the denominator and canceling gives the first component as $\frac{((\sin t)/t + \cos t)}{\sqrt{9t^2 + ((\sin t)/t + \cos t)^{2}}}, t/|t|$, which does not extend continuously to $0$ because the first factor is $1$ in the limit at $t = 0$. – Andrew D. Hwang Dec 24 '21 at 16:32
  • Right. Your characterization is pretty attractive but can we link it with what I am asking on this question. i.e Let $\gamma(t)$ be a curve given by $x = f(t)$, $y = g(t)$, and lets assume that $f,g$ are $C^1$. Suppose there is a $t_0$ such that $f'(t_0) = g'(t_0)=0.$ If $dy/dx$ (resp $dx/dy$) goes to $0$ as $t\to t_0$ and $g'(t)$ doesn't change sign in $t_0$ (resp $f'(t_0)$), is $\gamma(t)$ regular? – Odestheory12 Dec 24 '21 at 16:46
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    This paper solved my question: https://www.researchgate.net/publication/222470960_Detecting_Cusps_and_Inflection_Points_in_Curves – Odestheory12 Dec 24 '21 at 17:58

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