Confusion about a tangent line approaching an asymptote

Question

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 $t \to b$ ?

Answer 1

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 $x+y+a=0$:

Answer 2

My answer

While the limit $\lim_{t \to 1} T_t'(h) = \lim_{t \to 1} \beta'(t)$ does not exist, the following limit of unit vectors does: $$ \lim_{t \to 1} \dfrac{\beta'(t)}{\lvert \beta'(t) \rvert } = \dfrac{h}{\sqrt{2} \lvert h \rvert} \cdot (-1,1) = \dfrac{ \mathrm{sgn}(h) \cdot (-1,1)}{\sqrt{2}} = \pm \dfrac{(-1,1)}{\sqrt{2}} $$ Thus $$\lim_{t \to 1} \dfrac{\beta'(t) \cdot \ell'(k)}{\lvert \beta'(t) \rvert \lvert \ell'(k) \rvert} = \mathrm{sgn}(h) \dfrac{(-1,1) \cdot (-1,1)}{\sqrt{2}\sqrt{2}} = \mathrm{sgn}(h) = \pm 1 = \lim_{t \to 1} \cos{\angle{\left( T_t'(h), \ell'(k) \right)}} $$ This states that $T_t[\mathbb{R}]$ is parallel to $L = \ell[\mathbb{R}]$ in the limit as $t \to 1$. It is trivial to show that $T_t[\mathbb{R}]$ and $L$ intersect at at least one point for all $t \in (-\infty,1)$. If two parallel lines intersect, then they are the same line. We can thus say that $T_t[\mathbb{R}]$ approaches $L$ as $t \to 1$, and we're done.

Answer 3

I expect that this isn’t a technique used in do Carmo, but I’d switch to homogeneous coordinates, which allows one to represent both points and lines in the plane as elements of $\mathbb R^3$ (equivalence classes of them, strictly speaking).

Let $\hat\beta(t)$ be homogeneous coordinates of the point $\beta(t)$, i.e., $\beta(t)$ with a third coordinate of $1$ appended to it. The tangent line at $t$ is then $$\hat\beta\times\hat\beta' = {3a\over(t+1)^2(t^2-t+1)^2} \left(t(t^3-2), 1-2t^3, 3at^2\right).$$ Since this is a homogeneous coordinate vector, we can drop the nonzero factor on the left without changing the line represented by the vector. The limit as $t\to-1$ is then $(3,3,3a)$, which, after removing the common factor of $3$, corresponds to the implicit Cartesian equation $x+y+a=0$.

Answer 4

Observe,

$$x(t)+y(t)+a=0\Leftrightarrow\frac{y(t)+a}{x(t)}=-1.$$

We have

$$\beta’(s)=(\frac{-3a(1+2s^3)}{(1-s^3)^2},\frac{3as(s^3+2)}{(1-s^3)^2}).$$

Now,

$$\lim_{s\rightarrow 1}\frac{y(t)+a}{x(t)}=[\frac{3as(s^3+2)}{(1-s^3)^2}+a]\cdot[\frac{(1-s^3)^2}{-3a(1+2s^3)}]=\frac{-3}{3}=-1.$$

This implies the desired result,

$$\beta’(s)\rightarrow x+y+a=0\text{ as }s\rightarrow1.$$