Radius of curvature

By analogy with the circle (see (12.1)), the radius of curvature of a curve at a point is defined as the reciprocal of the curvature of the curve at that point. Denoting the radius of curvature by $R$, we have^12.2

$$R = \frac{1}{K}.$$

Or, substituting the values of $x$ from (12.3) and (12.4),

$$R = \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} }$$ (12.5)

and^12.3

$$R = \frac{ \left[ \rho^2 + \left( \frac{d\rho}{d\theta} \right)^2... ...- \rho \frac{d^2 \rho}{d\theta^2} + 2 \left( \frac{d\rho}{d\theta} \right)^2 }.$$ (12.6)

Example 12.5.1   Find the radius of curvature at any point of the catenary $y = \frac{a}{2}(e^{\frac{x}{a}} + e^{-\frac{x}{a}})$.

Solution. $\frac{dy}{dx} = \frac{1}{2} (e^{\frac{x}{a}} - e^{-\frac{x}{a}})$; $\frac{d^2 y}{dx^2} = \frac{1}{2a} (e^{\frac{x}{a}} - e^{-\frac{x}{a}})$. Substituting in (12.5),

$$\begin{array}{ll} R &= \frac{\left[ 1 + \left( \frac{e^{\fr... ...{a}} - e^{-\frac{x}{a}})^2}{4} \\ = \frac{y^2}{a}. \end{array}$$

If the equation of the curve is given in parametric form, find the first and second derivatives of $y$ with respect to $x$ from (11.5) and (11.6), namely:

$$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} },$$

and

$$\frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 },$$

and then substitute^12.4 the results in (12.5).

Example 12.5.2   Find the radius of curvature of the cycloid $x = a(t -\sin\, t)$, $y = a(t -\cos\, t)$.

Solution. $\frac{dx}{dt} = a(1 - \cos\, t)$, $\frac{dy}{dt} = a \sin\, t$; $\frac{d^2 x}{dt^2} = a \sin\, t$, $\frac{d^2 y}{dt^2} = a \cos\, t$. Substituting the previous example and then in (12.5), we get

$\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}$, $\frac{d^2 y}{dx^2} = \frac{ a(1 - \cos t) a \cos t - a \sin t a \sin t }{ a^3 (1 - \cos t)^3 } = \frac{1}{a(1 - \cos t)^2}$, and $R = \frac{ \left[ 1 + \left( \frac{\sin t}{1 - \cos t} \right)^2 \right]^{\frac{3}{2}} }{ -\frac{1}{a(1 - \cos t)^2} } = -2a \sqrt{2 - 2 \cos t}$.