Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:
The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.
So far so good.
But then they write:
However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$:
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$
And thus:
$$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$
It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t – \cos (2t), y=2\sin t – \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.
However my professor said that this is wrong ("because the derivative is not zero" – indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined – "$0/0$").
So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.
Thanks.
Short version of the question: can a parametric curve have a horizontal tangent line at a singular point?
I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).
Best Answer
Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.
We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).
In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.
Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.
Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.
Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.
In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.
The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),
$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$
The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,
$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$
Again, we obtain an intersection number $\ge 2$, which means tangency.
What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".
$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.