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?
Best Answer
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 $p=0$.