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Will there be an isometry in $\mathbb{R}^2$ taking the curve $\alpha(t)=(\cos(t)+1, \sin(t)+2)$, where $t\in [0,\pi]$, to the curve $\beta (t)=(t,\sin(t))$, where $t\in[0,c]$ and $c$ is a constant.

I know isometries preserves distances and curvatures. So when $f:\mathbb{R}^n\to \mathbb{R}^n$ then $$f(\vec{p})=A(\vec{p})+\vec{a},$$ where $\vec{a}\in \mathbb{R}^N$ and $A$ is an orthogonal transformation. Thus an isometry is a composition of a rotation and a translation.

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    $\begingroup$ I do not understand the question. $\endgroup$ – copper.hat Sep 16 '13 at 5:54
  • $\begingroup$ @copper.hat I will rephrase it. Editing now. $\endgroup$ – Lays Sep 16 '13 at 5:56
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    $\begingroup$ Isometries also preserve invariants, e.g. curvatures. $\endgroup$ – Yuri Vyatkin Sep 16 '13 at 6:02
  • $\begingroup$ @copper.hat Is that more clear? $\endgroup$ – Lays Sep 16 '13 at 6:03
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    $\begingroup$ Draw a picture. $\endgroup$ – copper.hat Sep 16 '13 at 6:07
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Here is a variation on @Yuri Vyatkin's answer: the initial point of the curve $\beta$ is an inflection point and therefore its curvature is 0, whereas all points of $\alpha$ have curvature 1. Hence the curves are not isometric.

The curvature $\kappa$ can be computed as the inverse of the radius $r$ of the osculating circle at the point. To find the osculating circle, consider three infinitely close points and use the unique circle through these three points. In the case of the sine curve the three points can be taken to be the origin, an infinitely close point $(t,\sin t)$, and the "symmetric" point $(-t, -\sin t)$. The three points lie on a line. It follows that the radius of curvature is infinite and therefore the curvature is $0$.

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  • $\begingroup$ Thanks user! I see that for $\alpha$ the curvature $\kappa = \frac{dT}{\alpha'(t)}$ is equal to $1$ like you said. For $\beta$ I having trouble getting the curvature. I got that $T(t) = \frac{\langle 1, cost \rangle}{\sqrt{1+cos^2t}}$ then if I divide $T'(t)$ by $\beta$ how can I get $0$? $\endgroup$ – Lays Sep 16 '13 at 18:46
  • $\begingroup$ @Lays, see my edit to the answer. $\endgroup$ – Mikhail Katz Sep 22 '13 at 16:38

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