0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Commented Jul 9, 2022 at 20:59

2 Answers 2

2
$\begingroup$

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

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .