# Can the tangent line be defined independently of the derivative?

The graph of the function $$f:x \mapsto x^{1/3}$$ has a 'vertical tangent' at $$x=0$$:

Although this idea is certainly geometrically sound, from what I understand the tangent line is defined by the derivative, not vice versa. In other words, the tangent line to a function at the point $$(a,f(a))$$ is simply the line given by the equation $$y - f(a) = f'(a)(x-a) \, ,$$ where $$f'(a)$$ is of course defined as a limit. Since $$f'(0)$$ does not exist in this case, I'm unsure if we can truly say that the graph has a vertical tangent. The intuitive idea of a tangent 'just touching' the curve breaks down when we consider, for instance, the graph of a linear function, where the tangent touches the graph of the function itself at infinitely many points. Nevertheless, I have heard people say that the tangent line is fundamentally a geometric concept. Although the slope of the tangent line 'agrees' with the derivative if the derivative exists, there are instances where the tangent line is a meaningful concept even when the derivative doesn't exist. If this be the case, then what is the formal definition of a tangent?

• Vertical lines having undefined slope is a consequence the coordinate-biased convention that "slope" is "change-in-$y$ over change-in-$x$" (aka, "rise over run"). Nevertheless, vertical lines (in particular, vertical tangent lines) exist despite their algebraically problematic slopes. A way around the problem is to separate the "rise" and "run" components into a vector: $(\text{run},\text{rise})$. With this, either component can vanish yet still provide a meaningful description of the line's direction. Conveniently, a curve parameterized by $(x(t),y(t))$ has tangent vector $(x'(t),y'(t))$. – Blue Feb 17 at 10:38
• @Blue: Thank you, it does help to think of this phenomenon simply as a by-product of our coordinate system. I'm not that familiar with curve parameterisation. Is it possible if you give me an example of how $f: x \mapsto x^{1/3}$ can be parameterised? – Joe Feb 17 at 10:48
• See Wikipedia's "Parametric equations" entry; also, this ancient answer of mine. We parameterize an $xy$ relation by assigning $x$ and $y$ to be formulas in a third variable (say, $t$) that fit. Eg, circle $x^2+y^2=1$ can be parameterized by $(x,y)=(\cos t,\sin t)$, since $(\cos t)^2+(\sin t)^2=1$. Your relation $y=x^{1/3}$ can be parameterized by $(t^3,t)$ (which meets the origin when $t=0$); the tangent's direction vector is $(x',y')=(3t^2,1)$, which is non-problematically $(0,1)$ when $t=0$. – Blue Feb 17 at 11:03
• @Blue: It amazes me that you always have an existing answer that is pertinent to my question :) If you wish, you can turn your comments into an answer. – Joe Feb 17 at 12:08

You won't find a definition of the tangent line that is completely independent of the general concept of derivatives, since they are so intimately connected. But from the point of view of differential geometry, you can most certainly have vertical tangent lines. To this end, we have to slightly shift our point of view: we're not looking for tangent lines to the graph of a function, but to some smooth set of points, which happens to be easily described as the graph of a function. More rigorously, we're interpreting the graph as a smooth submanifold of $$\mathbb R^2$$. In this case, the tangent line at $$p$$ can be parameterized as $$t_p(\tau)=p+\tau v$$, where $$v$$ is a nonzero vector tangent to the manifold at $$p$$, in this case $$(0,1)$$, for instance. Of course, to find such a vector, we'll end up using derivatives anyway: we'd parametrize the manifold, say by the smooth curve $$\gamma:\tau\mapsto(\tau^3,\tau)$$, and then take its derivative at $$\tau=\gamma^{-1}(p)$$ (since $$\tau=\gamma^{-1}(p)$$ yields $$p$$ when inserted into $$\gamma$$). This derivative is a vector tangent to the parameterized curve/manifold at $$p=0$$.