How did $\sqrt{\frac{\mathrm d\mathbf r}{\mathrm d s}\cdot \frac{\mathrm d\mathbf r}{\mathrm ds}}$ come in the variation of the action $\mathrm A$?

Question

I was reading about variational calculus and Euler-Lagrange equation of motion.

There the variation of action $\mathrm A$ is defined as:

$$\delta \mathrm A = \frac{\mathrm d}{\mathrm d\epsilon}\bigg|_{\varepsilon~=~0}\int_A^B n(\mathbf r(s,\varepsilon) )~\mathrm ds\tag I$$ where $s$ is the arc-length parameter and $\mathrm ds^2 = \mathrm d\mathbf r\cdot \mathrm d\mathbf r\,.$

After defining the variational derivative $\delta \mathbf r(s) = \frac{\mathrm d}{\mathrm d\epsilon}\bigg|_{\epsilon~=~0} \mathbf r(s,\varepsilon)$, the author
wrote $\mathrm{(I)}$ as: $$\delta \mathrm A = \delta \int_A^B n(\mathbf r(s) )\underbrace{\sqrt{\frac{\mathrm d\mathbf r}{\mathrm d s}\cdot \frac{\mathrm d\mathbf r}{\mathrm ds}}}_{?}~\mathrm ds\tag{II}$$

What I'm not getting is how the author wrote the term $\sqrt{\frac{\mathrm d\mathbf r}{\mathrm d s}\cdot \frac{\mathrm d\mathbf r}{\mathrm ds}};$ $||\mathrm{\dot r} || = 1,$ so isn't it trivial to write $\sqrt{\frac{\mathrm d\mathbf r}{\mathrm d s}\cdot \frac{\mathrm d\mathbf r}{\mathrm ds}}$ for it is equal to 1, isn't it? Or am I mistaking?

Could anyone tell me how he got the term?

Answer 1

If $\mathbf r$ is not parameterized with respect to arclength (which it usually isn't), then it is necessary to use the more general formula for a line integral (of a scalar field). A good explanation of how the formula is derived is given in the relevant Wiki article.