By a change of orientation, the tangent vector of a curve changes its direction

Question

I am studying Differential Geometry - Do Carmo, and it claims that: If $\alpha: I \to \mathbb{R}^3$ is a curve then by a change of orientation, the tangent vector changes its direction; that is, if $\beta(-s) = \alpha(s),$ then $$\dfrac{d\beta}{d(-s)}(-s) = -\dfrac{d\alpha}{ds}(s)$$ I dont get it, what is the definition of $\dfrac{d\beta}{d(-s)}(-s)$? how does the above equality holds?

Answer 1

Let the original variable be called $s$, i.e. $\alpha$ is a function of $s$. Now define a new variable, $t$, by $$t:=-s, \quad \hbox{i.e.} \quad s=-t.$$ Define the function $\beta(t) := \alpha(s(t))$. Now we can evaluate the derivative of this function with respect to its variable using the chain rule, $$\frac{d\beta}{d t} = \frac{d\alpha}{d s}\frac{ds}{dt} = -\frac{d\alpha}{d s}.$$

In differential geometry, curves are maps from some subset of $\mathbb{R}$ to the given manifold (in this case, $\mathbb{R}^3$). In the current example $\alpha$ and $\beta$ are two different curves. The curve $\alpha(s)$ is parameterized by $s$, and the curve $\beta(t)$ is parameterized by $t$. At a given point where $s=-t$ we have $\beta(t) = \alpha(s)$, but the tangent vectors are opposite: $\dot\alpha = -\dot\beta$.

(Think of tangent vectors as velocity vectors. As one travels along $\alpha(s)$ the velocity is in a given direction, but as one travels along $\beta(t)$, with $t=-s$, the velocity is in the opposite direction.)

Answer 2

I believe this is simply saying that the derivative of B(-s) with respect to (-s) can be restated as derivative of a(s) wrt (-s). using the chain rule, we get da/ds*ds/d(-s) but the second term just becomes -1 and we have -da/ds as a function of s. Sorry for the crude notation, I hope that this helps!