2
$\begingroup$

I am looking for explanations how we could classify the curve (spiral) defined by differential equation in polar coordinates: $r'^2+r^2=(kt)^2$, $r(t=0)=0$, k- Const.

For small angles the curve is similar to Galileo spiral which then transforms into Archimedean spiral: a) if $\phi \in (0,t)$, t is quite small, then $r(\phi) \approx k/2 *\phi^2$ b) if $\phi \in (c, + \infty)$, c is quite large positive number, then $r(\phi)=k\phi+o(\phi)$, as $\phi \to \infty$.

I thought that this is a 'mechanical' curve. Indeed, according to OEIS the junction point of the curve and the ray uniformly rotated in the origin coordinates moves uniformly accelerated. However I failed to find any references to be sure that is a mechanical curve. The graphic of the curve is also available to see.

Graphic of the curve

So what's the right way to classify the curve?

Any explanations are highly welcomed.

PS I think the curve is not an algebraic one as well.

$\endgroup$
5
  • $\begingroup$ Let $k=1$; Your differential equation can be replaced by the following arclength constraint $s(\theta_0):=\int\limits_0^{\theta_0}\sqrt{r^2+(dr/d\theta)^2}\,d\theta=\frac12\theta_0^2$. Isn't it ? $\endgroup$
    – Jean Marie
    Aug 18 '20 at 12:04
  • $\begingroup$ have you produced a graphical representation of this curve ? $\endgroup$
    – Jean Marie
    Aug 18 '20 at 12:07
  • $\begingroup$ @ Jean Marie Yes, there is a graphical representation of this curve (please see the link above) and due to the definition (uniform acceleration) $S(t)$ is about $t^2$, right $\endgroup$ Aug 18 '20 at 17:34
  • $\begingroup$ Thanks. The spiral of Galileo is also described and plotted here but is it exactly the same curve ? $\endgroup$
    – Jean Marie
    Aug 18 '20 at 17:34
  • $\begingroup$ For very small $t$ the spiral similar to Galileo one, approximately. Of course not the same. $\endgroup$ Aug 18 '20 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.