# Formulas of Curvature and Radius of Curvature

The commonly used results and formulas of curvature and radius of curvature are as shown below:

1. Curvature ${\rm K}$ and radius of curvature $\rho$ for a Cartesian curve is ${\rm K} = \frac{{\left| {\frac{{{d^2}y}}{{d{x^2}}}} \right|}}{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{3/2}}}}$ and $\rho = \frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{3/2}}}}{{\left| {\frac{{{d^2}y}}{{d{x^2}}}} \right|}} = \frac{1}{{\rm K}}$

2. If the equation of the curve is given by the implicit relation $f\left( {x,y} \right) = 0$, then ${\rm K} = \frac{{\left| { – {{\left( {{f_y}} \right)}^2}{f_{xx}} + 2{f_x}{f_y}{f_{xy}} – {{\left( {{f_x}} \right)}^2}{f_{yy}}} \right|}}{{{{\left[ {{{\left( {{f_x}} \right)}^2} + {{\left( {{f_y}} \right)}^2}} \right]}^{3/2}}}}$ and $\rho = \frac{1}{{\rm K}}$

3. If the curve is defined by parametric equations $x = f\left( t \right)$ and $y = f\left( t \right)$ then ${\rm K} = \frac{{\left| {f’\left( t \right)g”\left( t \right) – g’\left( t \right)f”\left( t \right)} \right|}}{{{{\left[ {{{f’}^2}\left( t \right) + {{g’}^2}\left( t \right)} \right]}^{3/2}}}}$ and so  $\rho = \frac{1}{{\rm K}} = \frac{{{{\left[ {{{f’}^2}\left( t \right) + {{g’}^2}\left( t \right)} \right]}^{3/2}}}}{{\left| {f’\left( t \right)g”\left( t \right) – g’\left( t \right)f”\left( t \right)} \right|}}$

4. For the curve $r = f\left( \theta \right)$ i.e., the curve in polar coordinates ${\rm K} = \frac{{\left| {{r^2} + 2{{\left( {\frac{{dr}}{{d\theta }}} \right)}^2} – r\frac{{{d^2}r}}{{d{\theta ^2}}}} \right|}}{{{{\left[ {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \right]}^{3/2}}}}$ and thus $\rho = \frac{1}{{\rm K}}\frac{{{{\left[ {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \right]}^{3/2}}}}{{\left| {{r^2} + 2{{\left( {\frac{{dr}}{{d\theta }}} \right)}^2} – r\frac{{{d^2}r}}{{d{\theta ^2}}}} \right|}}$

5. For the pedal curve $r = f\left( p \right)$ then, $\rho = r\frac{{dr}}{{dp}}$

6. If $\left( {\alpha ,\beta } \right)$ are the coordinates of the center of curvature of the curve $y = f\left( x \right)$ at $\left( {{x_1},{y_1}} \right)$ then $\alpha = {x_1} – \frac{{\frac{{dy}}{{dx}}\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$ and $\beta = {y_1} + \frac{{1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$