I'm working from do Carmo's Differential Geometry of Curves and Surfaces, 2ed. He tends to use language like"the curve $\alpha$ and its tangent line approach [some line] $L$" or "the curve $\alpha$ and its tangent have $L$ as an asymptote."
I know what it means for a curve to have a (linear) asymptote and I know what the tangent line is, but am not sure how to formalize a tangent line approaching some other line.
The problem I'm looking at is the folium of Descartes, exercise 1-3.5c.
$$
\alpha: (-1,\infty) \to {\mathbb R^2} \\
\alpha(t) = 3a \dfrac{t}{1+t^3} \cdot \left( 1,t \right) ,\quad a>0
$$
With the reverse orientation
$$
\beta: (-\infty,1) \to {\mathbb R^2} \\
\beta(t) = \alpha(-t) = 3a \dfrac{t}{1-t^3} \cdot (-1,t)
$$
So
$$
\beta'(t) = \dfrac{3a}{1-t^3} \left[ 3 \dfrac{t^3}{1-t^3} \cdot (-1,t)
+ (-1,2t) \right]
$$
I've already shown that $\beta$ asymptotically approaches the line $L$ given by $y+x+a=0$ and parametrized by
$$\ell(h) = (0,-a) + h \cdot (-1,1) = (-h, h-a)$$
but I've had no luck proving the same for the tangent line.
The line tangent to $\beta(t)$ is the trace of $T_t:{\mathbb R} \to {\mathbb R^2}$ with
\begin{align}
T_{t}(h) &= \beta(t) + h \cdot \beta'(t) \\
&= \dfrac{3a}{1-t^3} \left\lbrace t \cdot (-1,t) +
h \cdot \left[ 3 \dfrac{t^3}{1-t^3} \cdot (-1,t)
+ (-1,2t)
\right]
\right\rbrace
\end{align}
I've considered a few methods. Finding the shortest distance from $T_t[\mathbb{R}]$ to $L$ does not work because non-parallel lines always intersect in $\mathbb R^2$. I then considered the dot product $T_t'(h) \cdot \ell'(k) = \beta'(t) \cdot (-1,1)$ as $t \to 1$ to show that the lines are parallel, but it looks like that limit doesn't exist.
What am I missing here?
Also, is it not true that the tangent of any asymptotic curve $\gamma: (a,b) \to {\mathbb R^n}$ is always parallel to the asymptote as $t \to b$ ?
Best Answer
I can't tell you what Do Carmo meant, but here's one reason why $\beta'(t) \cdot \binom{-1}{1}$ shouldn't be the correct quantity to investigate. you may find that the limit doesn't exist because \begin{align} \beta'(t) &= 3a \frac{2 t^{3}+1}{\left(1-t^{3}\right)^{2}}\binom{-1}{t} + 3a\frac{t}{1-t^{3}}\binom{-1}{1} \\ &= \underbrace{\frac{3a}{(1-t^3)^2}}_{=O(\frac1{(t-1)^2})}\ \underbrace{\binom{-2t^3 -1-t(1-t^3)}{2t^4+t+t(1-t^3)}}_{=\binom{-3}{3} + O(t-1)} \\ &= O\left(\frac1{(t-1)^2}\right) , \quad t\to 1 \end{align} explodes in norm as $t\to 1$ (for $a\neq 0$). This isn't really indicative of the geometry, as the size of this tangent vector is really decided by the choice of parameterisation (e.g. an arc-length parameterisation would have $|\beta'|=1$). So one should instead renormalise $\beta'$ to be e.g. a unit tangent vector, i.e. consider instead
$$ \tau := \frac{\beta'}{|\beta'|}$$
Also, the dot product $a\cdot b$ is maximised when $a,b$ are parallel, which to me is a little awkward to use. I prefer to use the fact that a (nonzero) vector is parallel to a line iff the dot product with a normal vector of the line is zero(appropriately centered). In this case, one normal would be $\binom{1}{1}$, and I leave it to you to verify that from the above calculations,
$$\tau \cdot \binom{1}{1} \to 0.$$
Desmos graph of $\beta$,$\beta'$ and the line $x+y+a=0$: