Finding the angle at which a parametric curve intersects itself

Question

I have a parametric curve, given by $$ \gamma: \mathbb{R} \to \mathbb{R}: t \mapsto \begin{bmatrix} t^2-1 \\ t(t^2-1) \end{bmatrix} $$ which intersects itself at the origin $(0,0)$. I have to find the angle at which this curve intersects itself, so naturally I would use the formula $$ \cos \theta = \frac{\langle \dot{\gamma}_1(t_1)| \dot{\gamma}_2(t_2)\rangle}{||\dot{\gamma}_1(t_1)|| \,||\dot{\gamma}_2(t_2)||}, \quad \theta \in [0,\pi] $$ that describes the intersection angle between two parametric curves, where, in this particular case, $\dot{\gamma}_1(t_1) = \dot{\gamma}_2(t_2) = \dot{\gamma} (t)$. However, $\dot{\gamma} (t)$ is $$ \dot{\gamma} (t) = \begin{bmatrix} 2t \\ 3t^2 -1 \end{bmatrix} $$ and thus I get $$ \begin{align} \cos \theta &= \frac{\begin{bmatrix} 2t \\ 3t^2 -1 \end{bmatrix} \cdot \begin{bmatrix} 2t \\ 3t^2 -1 \end{bmatrix}}{||\begin{bmatrix} 2t \\ 3t^2 -1 \end{bmatrix}||\cdot||\begin{bmatrix} 2t \\ 3t^2 -1 \end{bmatrix}||} \\ &= \frac{(2t)^2 + (3t^2-1)^2}{\sqrt{(2t)^2 + (3t^2-1)^2} \cdot \sqrt{(2t)^2 + (3t^2-1)^2}} \\ &= 1 \end{align} $$ which gives an angle of $\theta = 0 \,\mathrm{rad}$ via $\arccos{1} = 0$. As we can see from WolframAlpha this can't be correct (my initial guess would've been $\pi/2$ rad). Where did I go wrong here? It's probably some stupid sloppy mistake...

Answer 1

You are calculating the angle using the same vector. Obviously, the angle between a vector and itself is $0$. Instead, you should look at which values of $t$ it happens $\gamma(t) = 0$. It's clear that curve $\gamma$ only touches $0$ twice, once for $t_1 = -1$ and then for $t_2 = 1$. You should use the formula that provides the angle for vectors $\dot{\gamma}(-1)$ and $\dot{\gamma}(1)$.

Answer 2

Another way to solve without vectors.

The parametrics have an easy elimination of parameter $t$ which gives the nodal cubic of cartesian equation $$y^2=(x+1)x^2$$ so the derivative is $$2ydy=(x^2+2x(x+1))dx\Rightarrow \frac{dy}{dx}=\frac{3x^2+2x}{2x\sqrt{x+1}}$$ The value of this derivative at $x=0$ has the form $\frac 00$ so applying l'Hôpital Rule we get the form $$\frac{6x+2}{2\sqrt{x+1}+\frac{x}{\sqrt{x+1}}}$$ which have the determined value $1$ when $x=0$ and clearly $-1$ for the other negative value of $y=\pm x\sqrt{x+1}$.

Thus the searched angle is $90^{\circ}$.