# Find the radius of curvature of a particle

1. Aug 17, 2015

### PlickPlock

1. The problem statement, all variables and given/known data
A particle is moving on a path parameterized as such:
$$x(t)=a\sinωt \quad y(t)=b\cosωt$$
Find the radius of curvature ρ as a function of time. Give your answer in Cartesian coordinates.
2. Relevant equations

$$\frac{1} {Radius~of~curvature}=|\frac{de_t}{ds}|$$, where et is the unit tangent vector and s the path length.
3. The attempt at a solution
$$\frac{de_t}{ds}=\frac{de_t}{dt}\frac{dt}{ds}=\frac{1}{v}\frac{de_t}{dt}$$, where v is the speed.
$$\frac{de_t}{dt}=\frac{d}{dt}\frac{\vec v}{v}$$ so$$|\frac{de_t}{ds}|=\frac{1}{v^2}|{\vec a}|$$
Evaluating the derivative gives me $$\frac{\sqrt{a^2ω^4sin^2ωt+b^2ω^4cos^2ωt}}{a^2ω^2cos^2ωt+b^2ω^2sin^2ωt}$$ which is wrong. At which step did I make a mistake?

2. Aug 17, 2015

### SteamKing

Staff Emeritus

The osculating circle is given by these coordinates (parametrically):

https://en.wikipedia.org/wiki/Osculating_circle

3. Aug 17, 2015

### PlickPlock

Thanks; that formulation is much easier to work with.
As for the definition I used, it was illustrated using a circle. The full equation would be:
$$|\frac{de_Θ}{ds}|=\frac{1}{R}e_r$$
, where $$e_Θ \quad and \quad e_r$$ are the unit tangent vector and unit normal vector respectively.
What I gleaned from watching online tutorials is that the rate of change of the direction of motion w.r.t path length gives the curvature.