2

I am having trouble solving Exercise 2.2.4 of Andrew Pressley's Elementary Differential Geometry

[2.2.4] Let $k$ be the signed curvature of a plane curve $C$ expressed in terms of its arc-length. Show that, if $C_a$ is the image of C under the dilation $v\mapsto av$, then the signed curvature of $C_a$ in terms of it's arc length $s$ is $\frac1a k(\frac{s}a)$

Please help me with the proof of this result.

Try

The new curve after applying the dilation is given by

$Y_a (t) = aY(t) \rightarrow Y_a'(t) =aY'(t)$

Let $s_a$ be arclength of $Y_a\rightarrow s_a(t) = \int_{t_0}^t |Y_a'(u)|du = a\int_{t_0}^t|Y'(u)|du = as(t)$ where $s(t)$ is arclength of $Y$

Thus we get, $s_a = as \;\;\;\ldots(1)$

Suppose, $Y_a'(s_a) = (\cos \phi_a(s_a), \sin\phi_a(s_a))$ where $\phi_a(s_a)$ is the angle measured from $(1,0) $ to $Y_a'(s_a)$

and suppose

$Y'(s) = (\cos \phi(s), \sin\phi(s))$ where $\phi(s)$ is the angle measured from $(1,0) $ to $Y'(s)$

Dilation will preserve the angle, will only change the length, so we must have

$$\phi_a = \phi\rightarrow \frac{d\phi_a}{ds} = \frac{d\phi}{ds} \rightarrow \frac{d\phi_a}{ds_a}\times \frac{ds_a}{ds} = k(s) \rightarrow k_a(s_a)\times a = k(s_a / a) \rightarrow k_a(s_a) = \frac1a k_a(s_a/a)$$

chesslad
  • 2,533
  • Please do not post unsearchable images of text and equations. Instead, typeset using MathJax. – David G. Stork Nov 11 '20 at 21:27
  • @DavidG.Stork I have written down the question, kindly remove the downvote if possible. I am looking for sincere help – chesslad Nov 11 '20 at 21:31
  • Please make an effort. Show us what you've tried. If you know definitions (which you MUST know), this problem is straightforward. But you must show us what you know and what notation you use, and do not just expect us to do your homework for you. – Ted Shifrin Nov 11 '20 at 22:29
  • @TedShifrin please see my attempt. I didn't want to show because it is quite embarrassing – chesslad Nov 11 '20 at 22:45
  • No, the dilation is giving $Y_a(s) = aY(s)$. Now let $s_a$ be the arclength of $Y_a$. How are the two arclengths related? Now fix what you have done. – Ted Shifrin Nov 11 '20 at 22:54
  • @TedShifrin thanks for the correction. can you check it again. – chesslad Nov 11 '20 at 23:23
  • So $\phi_a(s_a) = \phi(s)$. So you have to do the last bit of stuff more carefully, but, yes, you're very close now. – Ted Shifrin Nov 11 '20 at 23:53
  • @TedShifrin I had that doubt, now I think I'll be able to show it correctly. Thanks very much and your book on differential geometry is great – chesslad Nov 12 '20 at 00:12

0 Answers0