Radius of Curvature in Whewell Form

Theorem

Let $C$ be a curve defined by a real function which is twice differentiable.

The radius of curvature $\kappa$ of $C$ at a point $P$ can be expressed in the form of a Whewell equation as:

$\rho = \size {\dfrac {\d s} {\d \psi} }$

where:

$s$ is the arc length of $C$

$\psi$ is the turning angle of $C$

$\size {\, \cdot \,}$ denotes the absolute value function.

Proof

By definition, the radius of curvature $\rho$ is given by:

$\rho = \dfrac 1 {\size \kappa}$

where $\kappa$ is the curvature, given in Whewell form as:

$\kappa = \dfrac {\d \psi} {\d s}$

Hence the result.

$\blacksquare$

Sources

• 1969: J.C. Anderson, D.M. Hum, B.G. Neal and J.H. Whitelaw: Data and Formulae for Engineering Students (2nd ed.) ... (previous) ... (next): $4.$ Mathematics: $4.4$ Differential calculus: $\text {(i)}$ Radius of curvature