# Determining parametrization of curve from its acceleration

I am doing a project in which I have an object experiencing acceleration in a direction changing with time. I know the along-track and transverse acceleration components $a_{||}(t)$ and $a_{\perp}(t)$ in the comoving frame of the object, but not in an inertial reference frame.

As time passes $a_{\perp}(t)$ increases while $a_{||}(t)$ decreases, so I expect the motion of the object to be a spiral in an inertial reference frame, and it is a parametrization of this spiral I am looking for.

I have attempted to describe the curvature by $\kappa(t) = \frac{v(t)^{2}}{a_{\perp}(t)}$, where $v(t) = \sqrt{a_{\perp}(t)^{2} + a_{||}(t)^{2}} \ t$, and as expected I find the curvature to be increasing in time.

Now how do I find a parametrization for the resulting motion in an inertial frame? I have looked into frenet-serret basis but the only thing resembling a solution I have not been able to find a solution. The closes I have found is this parametrization for an arc-length parametrized curvature:

$\eta(t) = \int_0^t\cos\left( \int_0^t \kappa(s)ds\right)ds$

But I get curve that spirals outwards in time from origo, not inwards.

• What do mean by acceleration in the reference frame of the object? In the reference frame of the object, the object is at rest and its velocity, not to mention acceleration, should be zero! Commented May 2, 2017 at 18:30
• I have two problems with your question. Number 1, the curvature should have a dimension of inverse length, but it appears to be time. Number 2, given the curvature, the curve is found in the complex plane from $$z(s)=\int e^{i\int \kappa(s)ds}ds$$ where $s$ is the arc length. You seem to have only the real part of that (and I don't know what $\eta$ means). Commented May 2, 2017 at 18:40
• @velutluna I changed reference frame to comoving frame in the question. Commented May 2, 2017 at 18:59
• But how is an acceleration vector a combination of coframe elements? Commented May 2, 2017 at 20:12

I'm going to try out an idea here where we start with

$$z(s)=\int e^{i\int \kappa(s)ds}ds$$

and $|\ddot z(s)|=\kappa(s)$. Now, allow that $v=\frac{ds}{dt}$ is the velocity parallel to the curve, so that

$$z(t)=\int e^{i\int \kappa(t)vdt}vdt\\ \dot z=v(t)e^{i\int \kappa(t)vdt}\\ \ddot z=\left[\frac{dv}{dt}+iv^2(t)\kappa(t) \right]e^{i\int \kappa(t)vdt}$$ The terms in the bracket for $\ddot z$ represent the parallel acceleration and, perpendicular to it, the centrifugal acceleration. Notice that everything is dimensionally correct. Therefore, I think we can say

$$|\ddot z|=\sqrt{a_{||}(t)^{2} + a_{\perp}(t)^{2}}=\sqrt{\left(\frac{dv}{dt}\right)^2+[v^2(t)\kappa(t)]^2}$$

and solve for $\kappa$ as follows

$$\kappa(t)=\frac{a_{\perp}}{v^2}$$

since $\frac{dv}{dt}=a_{||}$. This is the equation we should have known in the first place, since $a_{\perp}=v^2/\rho=\kappa v^2$

We can then return to the equation for $z(t)$ to determine the curve (spiral) itself.

• I think you are right. Using $z(t)$ produces some funny looking results though. I shall have to look into my acceleration terms again. Commented May 3, 2017 at 8:25