Finding planar curve $\beta(s)$ given curvature $k(s) = \frac{1}{s}$, $\beta(1) = (0, 0)$, $\beta'(1) = (1, 0)$

Question

Find the curve $\beta(s) = (x(s),y(s))$ with $|\beta'(s)|= 1$ for all $s >0$ such that $\beta(1) = (0,0)$, $\beta'(1) = (1,0)$ , and curvature $k(s) =\frac{1}{s}$ for all $s >0$.

I am guessing some sort of integral is required since we are given 'initial' conditions, but I do not see how to begin.

Answer 1

Once you have a continuous function $k:(a,b)\rightarrow\mathbb{R}$ there exists always a planar curve $\beta:(a,b)\rightarrow\mathbb{R}^2$ parametrized by arc-lenght and such that $k_\beta(s)=k(s)$, where $k_\beta(s)$ is the curvature of such $\beta$.

In your case since $|\beta'(s)|=1$, then $k_\beta(s)=\theta'(s)$ where $\beta'(s)=T(s)=(cos\theta(s); sin\theta(s))$. In particular \begin{equation*}\theta(s)=\int k_\beta(s)ds \end{equation*}is determined up to a constant (such constant will depend on the initial condition). In your case you have to integrate the function $\frac{1}{s}$ and once you have found $\theta(s)$ put it in the definition of $T(s)$ then integrate both coordinates to find $x(s), y(s)$ which are the components of the planar curve.

It's a useful exercise to do if your are approaching the first time to these topics.

EDIT Once you mandaged to solve the integral defining $\theta(s)$ you'll have an equation of the form $\theta(s)=f(s)+c$, where $f(s)$ is the function integrated and $c\in\mathbb{R}$. Put this (explicit) expression in $T(s)=\beta'(s)$ and use the hypothesis $\beta'(1)=(1,0)$, you'll find the constant $c$. The same will happen when you integrate $T(s)$ to find $x(s),y(s)$, you'll get another constant $\bar{c}$ and use the hypothesis $\beta(1)=(0,0)$

Answer 2

Let's write everything down. $$\left|\beta^{\prime}(s)\right|=1 \implies \left(x'(s)\right)^2+\left(y'(s)\right)^2=1$$ An arc length paramtrization of $\beta$ is going to be given by $$\left(\int_1^s \cos(\theta(t))dt,\int_1^s \sin(\theta(t))dt \right) $$ for some $\theta$. In the above equation we used the first initial condition. The curvature is given by $$\beta''(s)=\kappa(s) e_2$$ where $e_2$ is orthonormal to $\beta'$ so we have that $$|\beta''(s)|^2=\frac{1}{s^2}$$ Thus we get $$\sin^2(\theta(s))\theta'(s)^2+\cos^2(\theta(s))\theta'(s)^2=\frac{1}{s^2}$$ This implies $\theta'(s)=\frac{1}{s}$. Thus we get $\theta(s)=\ln(s)$ by the initial condition for $\beta'$.

Answer 3

EDIT1:

Curve behavior

The curve is natural or intrinsic. A rigid bent curved arc of this shape or can be moved anywhere in the plane merely by varying IC.

$$ \phi'=\frac{1}{s},\;\phi= \log\frac{s}{a}\;; \dfrac{s}{a}= e ^{\phi}$$

where $a$ is an arbitrary constant and $\phi$ is curve rotation with respect to x-axis.

$$ x=\int\cos \phi \;ds=\int\cos\log\frac{s}{a} ; y=\int\sin \phi \; ds=\int\sin \log\frac{s}{a} ds;$$

Curve's analytical parametrization

$$\boxed{x= \frac{s}{2} [{\sin (\log (s))+\cos (\log (s));\;y=\sin (\log (s))-\cos (\log (s))}] } $$

Calculated by numerical integration. BC $ (x=1,y=0) $

At start infinite curvature or Cusp occurs. Numerically, $s=0 $ cusp is avoided, a small value given for x to avoid a singularity.

As $s\to \infty $ the curve becomes flatter.

I have not found a reference to this curve in 2dCurves, Lockwood, McTutor or Mathworld