Is there a more geometrical definition of the tangent line of a curve? based on the intuitive idea that a tangent line only touches at one point

Question

I asked this question to my calculus teacher and it was a frustrating experience, basically he would say, over and over again, that a tangent line at a point is a line that goes through that point and has a slope equal to the instantaneous slope of the curve at that point. I understand that, I really do, but that is not what I was asking. This all started because, geometrically, a tangent touches the circle once vs a secant touches the circle twice. This of course is no longer the case with any random curve, so I was wondering what is the difference geometrically between a secant line and a tangent line, given that they both can touch the curve multiple times. After thinking about it for a bit, I came up with one possible geometric definition for tangent of a curve, and I would like for you guys to tell me if this is ok and equivalent to the usual definition with derivatives.

So let $f(x)$ be a curve and $t(x)$ be a line. We say $t(x)$ is a tangent to $f(x)$ at point $x_0$ iff:

1. $f(x_0) = t(x_0)$
2. $(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < \mid x - x_0 \mid < \delta \implies t(x) > f(x))$ or $(\exists \delta > 0)(\forall x \in \mathbb{R})(0 < \mid x - x_0 \mid < \delta \implies t(x) < f(x))$

In other words, a line is tangent to a curve at a point if it goes through that point and also if you look close enough (for some arbitrarily small $\delta$), all of the points of the line, except for the tangency point, are either on one side of the curve or the other (hence why I wrote two statements one with $t(x) > f(x)$ and another one with $t(x) < f(x)$).

This definition doesn't work for lines, since the tangent of a line is the line itself and this definition wouldn't allow that, but at least for the case of curves, what do you guys think? Can I modify it so that it works for lines too? I am thinking of changing the $<$ and $>$ for $\leq$ and $\geq$ maybe that would make it work for lines. Regardless of whether that definition works or not, it is an example of what I mean when I say "a more geometrical definition of tangent line".

Answer 1

If you consider the curve $y=x^3$ then by your definition there would be no tangent line at the point $(0,0)$ on the curve. With the standard definition there is a tangent through this point, namely $y=0$.

Answer 2

Your definition does not work for any inflection point; that is, points where the second derivative is zero. Consider $\sin(x)$ at the origin.

Answer 3

With your definition, a linear function would not be differentiable at any point, which is pretty odd to me. Indeed, one motivation to consider the tangent line at a point is to approximate a curve by a line near that point.

Answer 4

See here and here. Try implementing derivatives into your definition.

This is also a duplicate, see here.

Answer 5

The notion of a tangent line, representing a line that intersects the graph of a function at a single point, is typically applied locally within the function's domain. However, in a global context, this tangent line may intersect the curve elsewhere, which introduces complexity when dealing with arbitrary curves. To ensure the tangent line touches the function's graph at precisely one point, constraints on the functions and sets under consideration are necessary. Which I guess is the problem you are encountering.

$\textit{Convexity}$ is the key concept. A set $C$ is $\textit{convex}$ if $tx_1 + (1-t)x_2 \in C$ whenever $x_1,x_2 \in C$ and $t \in (0,1)$, or in words: any line segment connecting two points within the set lies entirely within it (Image taken from Wikipedia).

A function $f$ is $\textit{convex}$ if its domain, $\text{dom}(f)$, is a convex set and if for all $x_1,x_2 \in \text{dom}(f)$, and $t \in (0,1)$, it holds that \begin{equation}\label{convexity} f(tx_1 +(1-t)x_2) \leq t f(x_1) +(1-t)f(x_2). \end{equation} or in words: for all points within its domain, the function value lies below the line connecting any two points. This condition holds true for convex functions, while $\textit{strictly convex}$ functions enforce strict inequality in this relationship. A line is a convex function but not a strictly convex function. So the idea of tangent lines touching the graph of a function at a single point works for strictly convex functions.

In higher dimensions, the concept of a tangent line is generalized by the concept of $\textit{supporting hyper-planes}$ to sets $C$. If you want to learn more about the idea of tangency and convexity I recommend reading $\textit{Convex Analysis}$ by R. Tyrrell Rockafellar, this is an excellent book.

Answer 6

Here's one way you can visualize it. If you have a function, for example...

$f(x)=\frac{1}{20}(x+4)(x-5)^2+4$

and you want to find the tangent line at the point $(4,f(4))$, then imagine that the portion of the graph to the right of $x=4$ was erased, and try to draw a straight line that continues the slope of $f(x)$:

$\Rightarrow$$\Rightarrow$$\Rightarrow$$\Rightarrow$

Now extend the straight line and $f(x)$ to their full domains:

As you can see, even though the tangent line intersects $f(x)$ at more than one point, it is still tangent to the graph at $x=4$ because it follows the instantaneous slope of $f(x)$ at that point. But on a general scale, you could consider the green line as both secant and tangent to $f(x)$.

For reference, here is the Desmos graph I made: https://www.desmos.com/calculator/lmgxmx6u63

Answer 7

The most geometric definition of the tangent line to a curve is the one originally used by Leibniz (and probably before him). Namely, take two infinitely close (but distinct) points on the curve, and draw the (unique) line through those two distinct points. The resulting line is indistinguishable from the tangent line, in the sense that if one writes its equation as $\bar ax+\bar by=\bar c$ with $\bar a^2+\bar b^2=1$ (which can always be done via normalisation), the tangent line will have equation $ax+by=c$ where $a$ is the standard part of $\bar a$, and similarly for $b, c$. This approach needs neither derivatives nor slopes.