# Strange use of differentials - is $d{\bf x} \cdot d{\bf x}$ a dot product?

If ${\bf x}(s)$ is a curve in $\mathbb{R}^3$ on a surface parameterized by its arc length $s$, and ${\bf N}$ is the surface normal at ${\bf x}$, consider the following equality (with "$\cdot$" being the dot product):

\begin{align*} \frac{d{\bf x}}{ds} \cdot \frac{d{\bf N}}{ds} = \frac{d{\bf x} \cdot d{\bf N}}{d{\bf x} \cdot d{\bf x}} \end{align*}

This comes from a book on differential geometry. Now, the LHS is a perfectly fine dot product between two vectors. But to me, the RHS does not make any sense at all.

I do understand that this result can easily be arrived at if you treat $d{\bf x}$ and $d{\bf N}$ as real vectors (e.g. the part where $d{\bf x} \cdot d{\bf x}=ds^2$). But this kind of manipulation seems rather sloppy...

I would really like to understand this from first principles. A derivative like $\frac{d{\bf x}}{ds}$ is defined in terms of limits,

\begin{align*} \frac{d{\bf x}}{ds}(s) = \lim_{\Delta s \rightarrow \infty} \frac{{\bf x}(s + \Delta s) - {\bf x}(s)}{\Delta s} \end{align*}

The extension from scalar $x$ to a vector-valued $\bf x$ is obvious. But how does this definition apply to the RHS of the equation given above? Does a dot product of differentials like $d{\bf x} \cdot d{\bf x}$ even mean anything? How do you set up the limits so that the expression can be evaluated (after all, the RHS should evaluate to a real number)?

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As in the definition of an ordinary derivative, you could consider the difference quotient $\frac{\Delta{\bf x} \cdot \Delta{\bf N}}{\Delta{\bf x} \cdot \Delta{\bf x}}$, where $\Delta$ denotes taking the difference between the value of that quantity at some point on the curve and its value at the given point, and then take the limit $\lvert\Delta{\bf x}\rvert \rightarrow 0$ -- if that exists, it would be a natural interpretation of that notation -- I would presume this is what the book means, since it would make sense of the equation you quote. –  joriki Feb 11 '11 at 21:58
What does your book do with the right hand side? Maybe it pays to look one step ahead. –  Christian Blatter Feb 12 '11 at 12:13