Signed curvature and turning angle

Question

I am trying to find the signed curvature of a function, I have so far that

$$g'(t)=(\cos(\cosh(t)), \sin(\cosh(t)))$$

I know that $g'$ is unit speed so i don't have to parametrize by arc length, and I know that the direction of $g'$ is measured by the angle $\phi$ such that $g'(t)=(\cos(\phi t), \sin(\phi t))$ but I'm unsure how to apply this here.

the implicit $g$ is an arc-parameterization because $||g'||=1$, so together with $g''$ you should be doing fine — janmarqz, Jan 19 '14 at 17:04

janmarqz · Answer 1 · 2014-01-20T18:06:42.437

0

To find $x(t),y(t)$ one is compelled to solve this integrals: $$x(t)=\int^t_0\cos(\cosh(u))du\quad \mbox{and}\quad y(t)=\int^t_0\sin(\cosh(u))du,$$ but you don't need to.

Now, if we choose $g(t)=(x(t),y(t))$ we can compute the signed curvature with $$k(t)=\frac{x'(t) y''(t) - y'(t) x''(t)}{(x'(t)^2 + y'(t)^2)^{3/2}}.\qquad (*)$$

Since we already have $g'(t)$ it is remaining to compute $g''(t)$.

This is $g''(t)=(-\sin(\cosh(t))\sinh(t),\cos(\cosh(t))\sinh(t))$

Substituting into $(*)$ then we are going to have $$k(t)=\sinh(t),$$ as the signed curvature for such a $g(t)$.

edited Jan 20 '14 at 18:06

answered Jan 20 '14 at 01:21

janmarqz

10,538

if we take advantage that $g$ is arc-parameterized then the unsigned curvature is just $|g''|$ which happens to be $|\sinh(t)|$ – janmarqz Jan 20 '14 at 01:23
for the turning angle please tell me the definition – janmarqz Jan 20 '14 at 01:28

Signed curvature and turning angle

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