What is the intuitive reason why derivatives can be tangent as well as normal?

Question

Background: Let $f:x \mapsto x^2$. I first learned to think of the derivative of a function as giving the slope of line tangent to the graph of the function at the point.

For example, since $f':x \mapsto2x$, the tangent to the graph of $f$ at $x=1$ has slope $f'(1)=2(1)=2$.

However, consider now the function $F:(x,y)\mapsto y-x^2$. The graph of $f$ is just the level set $F^{-1}\{0\}$. To find the normal to the graph at $x=1$, we take the line spanned by the vector $$\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}\right)=(-f',1).$$

The discrepancies in possible interpretations of differentiation don't stop with what was mentioned above. For example, consider now the graph of $f$, or equivalently $F^{-1}\{0\}$, to be the trace of the parametrized curve $\phi: t \mapsto (t,t^2)$. Then the tangent vectors are described by $$(\phi_1',\phi_2')=\left(\frac{d\phi_1}{dt},\frac{d\phi_2}{dt} \right) .$$ And yet simultaneously the following vector is normal to the trace of $\phi$: $$(\phi_1'',\phi_2'')=\left(\frac{d^2\phi_1}{d^2 t}, \frac{d^2 \phi_2}{d^2t} \right) .$$

Thus, despite the fact that tangents and normals are essentially opposite concepts, it seems like we can think of derivatives as characterizing either type of local behavior near a point.

Question: What is the "correct" intuition to think of differentiation? As producing tangents, as one might learn in high school, or as producing normals, as one might learn in college?

If this is an issue of duality, which type of duality? Does the duality require a choice of inner product (i.e. Riemannian metric), or is it independent of any such choice?

Further Discussion:
If we consider instead of just smooth curves arbitrary smooth manifolds, the ambiguity of "proper interpretation" remains. When applying the implicit function theorem, the partial derivatives span a vector space that is normal to the graph of the hypersurface, rather than tangent.

However, the total derivative, which defines the partial derivatives, also defines the directional derivatives, which are in one-to-one correspondence with the derivations of smooth functions which are the elements of the tangent space of the hypersurface.

I feel like this might be an issue of some type of duality which I was unaware of, with the kernel of the total derivative (considered as a linear transformation) corresponding to the tangent $(n-m)$-space and the image of the total derivative corresponding to the normal $m$-space.

The confusion might also be a result of failing to recognize implied identifications of covectors and vectors (like the identification of the outer product of two vectors in $\mathbb{R}^3$, which is a bivector, with a vector via Hodge duality and calling the resulting vector the cross product).

In either case, I have no idea how I could use either explanation to understand the discrepancies between derivatives as normals and tangents in the one-dimensional case (for smooth curves).

Is this related at all to curvature, e.g. the distinctions between Gaussian and principal curvatures, or between the curvature tensor and sectional curvatures? How much of this is linear algebra without an inner product, i.e. the distinction between vectors and covectors, between a vector space and its dual, and how much does it rely upon musical isomorphisms and Hodge duality and other canonical identifications between the tangent and cotangent bundle which implicitly assume the presence of a background Riemannian metric (inner product)?

Answer 1

For $f(x)$, you're basing your interpretation on the graph $y=f(x)$ and the slope of its tangent line at a point. This picture is drawn in a two-dimensional $xy$ coordinate system.

In order to find a corresponding interpretation for $F(x,y)$, you must think about the graph $z=F(x,y)$ and the slope(s) of its tangent plane at a point, so you must have a three-dimensional $xyz$ coordinate system in mind.

When you start talking about level curves for $F(x,y)$, which are curves in the $xy$ plane, it's not surprising that the interpretations don't seem to agree!

Answer 2

Both interpretations are correct. For simplicity and clarity, let's think of hypersurfaces in $\Bbb R^n$.

If you are working with an implicit description $F(x_1, \dots, x_n) = 0$ then $\left( \frac {\partial F} {\partial x_1}, \dots, \frac {\partial F} {\partial x_n} \right)$ is a vector orthogonal to the surface. At the same time, the kernel of $\Bbb d F$ gives the tangent space. For the sphere $x^2 + y^2 + z^2 = 1$, notice that $(2x, 2y, 2z)$ is a normal vector.

If, you are working with a parametric description (I include here implicit descriptions as a particular case) $(u_1, \dots, u_n) \mapsto f(u_1, \dots, u_n) \in \Bbb R^n$, then the vectors $\frac {\partial f} {\partial u_i}$ are tangent to your surface (they form a basis at each point). This is what you do with the function $x \mapsto x^2$: rewrite it as $x \mapsto (x, x^2) \in \Bbb R^2$ and notice that this is a parametrization of the graph; the derivative $(1, 2x)$ is a vector tangent to the curve.

Your question is slightly mistaken in that you force an alternative ("either tangent or normal") where, in fact, the concept of differentiability underlies both.