Question

Prove that the magnitude of the acceleration of a particle moving along a curve, C, with a constant speed is proportional to the curvature of C.

Short answer

It is proved thatthe magnitude of the acceleration of a particle moving along a curve,C, with a constant speed is proportional to the curvature ofC.

Step-by-step solution

Step 1. Given Information.

The formula for the Curvature of a Space Curve is$\kappa =\frac{||r\text{'}\left(t\right)\times r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3}.$

Step 2. Prove.

To prove that the magnitude of the acceleration of a particle moving along a curve, C, with a constant speed is proportional to the curvature of C,we will use the formula of curvature κ of C at a point on the curve.

So,

$$\begin{gathered} \kappa =\frac{||r\text{'}\left(t\right)\times r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3} \\ \kappa =\frac{||r\text{'}\left(t\right)||||r\text{'}\text{'}\left(t\right)||\sin \theta }{{||r\text{'}\left(t\right)||}^3} \\ \kappa =\frac{||r\text{'}\left(t\right)||||r\text{'}\text{'}\left(t\right)||\sin {\displaystyle \frac{\mathrm{\pi }}{2}}}{{||r\text{'}\left(t\right)||}^3}\left[\mathrm{speed}\mathrm{is}\mathrm{constant}\right] \\ \kappa =\frac{||r\text{'}\left(t\right)||||r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3}\left[\sin \frac{\mathrm{\pi }}{2}=1\right] \\ \kappa =\frac{||r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^2} \end{gathered}$$

Now, the denominator of kis the square of the speed, which we presumed as constant.

Thus,

$\kappa \propto ||r\text{'}\text{'}\left(t\right)||.$

Hence proved.