# Show that the curvature of a parametric curve is invariant under rigid motions.

Show that the curvature of a parametric curve is invariant under rigid motions.

My attempt

Let $$A:\mathbb{R}^3 \rightarrow \mathbb{R}^3$$ such that $$A(\rho)=\rho+v$$.

Let $$\rho:\mathbb{R}^3 \rightarrow \mathbb{R}^3$$ such that $$\rho$$ is a linear and orthogonal map

Let $$M := A\circ \rho$$ a rigid motion.

Let $$\alpha :\mathbb{R}^3 \rightarrow \mathbb{R}^3$$ a parametric curve parametrized by arc lenght, ie $$|\alpha(t)|=1$$ for all $$t\in\mathbb{R}^3$$.

We need to show that $$|M''(\alpha(t))| = k(s)$$ where $$k(s) = |\alpha ''(s)|$$

Note that:

$$M(\alpha (t)) = \rho(\alpha(t)) + v \implies M'(\alpha (t)) = \rho '(\alpha(t))\alpha(t)$$

Then

$$M''(\alpha (t)) = (\rho '(\alpha(t))\alpha(t))' = \rho''(\alpha(t))\alpha'(t) + \rho'(\alpha(t))\alpha'(t)$$

Here i'm stuck. can someone help me?

• Using $\rho$ as both a function and a (variable) point in $\Bbb R^3$ is beyond confusing. Moreover, is $\alpha$ a function from $\Bbb R^3$ to $\Bbb R^3$? Why are you mixing variables $s$ and $t$? I suggest you do a thorough proofreading and rewriting. Commented Jul 9, 2022 at 20:59

Let $$\alpha : I \longrightarrow \mathbb{R}^3$$ be a regular curve parametrized by the arc length. Let $$M$$ be a rigid motion. Let $$\beta = M \circ \alpha$$. Then since $$M$$ is g rigid motion is of the form $$Mx = Ax + b$$. Then $$k_\beta(s) = \det(\beta'(s), \beta''(s)) = \det((M \circ \alpha)'(s), (M \circ \alpha)''(s)) = \det(A\alpha'(s), A\alpha''(s))=\det(A)\det(\alpha'(s),\alpha''(s))=\begin{cases} k_\alpha(s) \quad \text{ if } M \text{ is direct } \\ - k_\alpha(s) \quad \text{ if } M \text{ is inverse } \end{cases}$$ Since $$A$$ is an orthogonal matrix
Let $$f = T_q \circ L_A$$, $$A \in \mathit{O}(3)$$, $$q \in \mathbb{R}^3$$ be a rigid motion. Define $$\beta(s) = A\alpha(s) + q$$, then $$\beta''(s) = A\alpha''(s)$$ ($$A$$ is a constant) and so $$k_\beta(s) = \lvert \beta''(s) \rvert = \lvert A\alpha''(s) \rvert = \lvert \alpha''(s) \rvert = k_\alpha(s)$$. Notice here that the step $$\lvert A\alpha''(s) \rvert = \lvert \alpha''(s) \rvert$$ uses the fact that $$\lvert Ax \rvert = \lvert x \rvert$$ which holds as $$A$$ is orthogonal. Note that I am assuming $$f$$ is orientation preserving in this answer.