In differential geometry , the radius of curvature , r , is the reciprocal of curvature . For a curve , it is equal to the radius of the circular arc that best approximates the curve at that point. For surfaces , the radius of curvature is the radius of a circle that best fits a normal section or combination .

## Definition

In the case of a space curve , the radius of curvature is the length of the curvature vector .

In the case of a plane curve , then r is the absolute value of

{\displaystyle R\equiv \left|{\frac {ds}{d\varphi }}\right|={\frac {1}{\kappa }},}

where s is the arc length from a fixed point on the curve, is the tangent angle and is the curvature .

## Formula

### 2D. In

If the curve is given as y ( x ) in Cartesian coordinates , then the radius of curvature is (assuming that the curve is differentiable on the order of 2):

{\displaystyle R=\left|{\frac {\left(1+y'^{\,2}\right)^{\frac {3}{2}}}{y''}}\right|,\qquad {{where}}\quad y'={\frac {dy}{dx}},\quad y''={\frac {d^{2}y}{dx^{2}}},}

and | Jade | Indicates the absolute value of z .

If the curve is given parametrically by the x ( t ) and y ( t ) functions , then the radius of curvature is

{\displaystyle R=\left|{\frac {ds}{d\varphi }}\right|=\left|{\frac {\left({{\dot {x}}^{2}+{\dot {y}}^{2}}\right)^{\frac {3}{2}}}{{\dot {x}}{\ddot {y}}-{\dot {y}}{\ddot {x}}}}\right|,\qquad {{where}}\quad {\dot {x}}={\frac {dx}{dt}},\quad {\ddot {x}}={\frac {d^{2}x}{dt^{2}}},\quad {\dot {y}}={\frac {dy}{dt}},\quad {\ddot {y}}={\frac {d^{2}y}{dt^{2}}}.}

Presumably, this result can be interpreted as

{\displaystyle R={\frac {\left|\mathbf {v} \right|^{3}}{\left|\mathbf {v} \times \mathbf {\dot {v}} \right|}},\qquad {{where}}\quad \left|\mathbf {v} \right|={\big |}({\dot {x}},{\dot {y}}){\big |}=R{\frac {d\varphi }{dt}}.}

### in n dimensions

If  : → n is a parametrized curve in n then the radius of curvature at each point of the curve,  : → , is given by

{\displaystyle \rho ={\frac {\left|{\boldsymbol {\gamma }}'\right|^{3}}{\sqrt {\left|{\boldsymbol {\gamma }}'\right|^{2}\,\left|{\boldsymbol {\gamma }}''\right|^{2}-\left({\boldsymbol {\gamma }}'\cdot {\boldsymbol {\gamma }}''\right)^{2}}}}}.

As a special case, if f ( t ) is a function from to , then the radius of curvature of its graph , ( t ) = ( t , f ( t ) ) , is

{\displaystyle \rho (t)={\frac {\left|1+f'^{\,2}(t)\right|^{\frac {3}{2}}}{\left|f''(t)\right|}}.}

Let be as above, and t ok . We want to find the radius of a parametrized circle which corresponds to in its zeroth , first, and second derivatives t . Obviously the radius will not depend on position ( t ) , only on velocity ‘( t ) and acceleration “( t ) . There are only three independent scalars that can be obtained from two vectors v and w , namely v v , v w _, and w w . Thus the radius of curvature must be a function of three scalars .

‘ ( t ) | 2 , | ” ( t ) | 2 and ‘ ( t ) ” ( t ) .

The general equation for a parametrized circle in n is

{\displaystyle \mathbf {g} (u)=\mathbf {a} \cos h(u)+\mathbf {b} \sin h(u)+\mathbf {c} }

where c n is the center of the circle (irrelevant since it disappears in derivatives), , b are perpendicular vectors of length ( that is a a = b b = 2and a b = 0 ) , and h  : → is an arbitrary function that is twice differentiable at t .

The corresponding derivative of g is

{\displaystyle {\begin{aligned}|\mathbf {g} '|^{2}&=\rho ^{2}(h')^{2}\\\mathbf {g} '\cdot \mathbf {g} ''&=\rho ^{2}h'h''\\|\mathbf {g} ''|^{2}&=\rho ^{2}\left((h')^{4}+(h'')^{2}\right)\end{aligned}}}

If we now equate these derivatives of g then for the corresponding derivatives of g at t we get

{\displaystyle {\begin{aligned}|{\boldsymbol {\gamma }}'(t)|^{2}&=\rho ^{2}h'^{\,2}(t)\\{\boldsymbol {\gamma }}'(t)\cdot {\boldsymbol {\gamma }}''(t)&=\rho ^{2}h'(t)h''(t)\\|{\boldsymbol {\gamma }}''(t)|^{2}&=\rho ^{2}\left(h'^{\,4}(t)+h''^{\,2}(t)\right)\end{aligned}}}

The three unknowns (in these three equations , h  ( t ) and h “( t ) ) can be solved for , giving the formula for the radius of curvature:

{\displaystyle \rho (t)={\frac {\left|{\boldsymbol {\gamma }}'(t)\right|^{3}}{\sqrt {\left|{\boldsymbol {\gamma }}'(t)\right|^{2}\,\left|{\boldsymbol {\gamma }}''(t)\right|^{2}-{\big (}{\boldsymbol {\gamma }}'(t)\cdot {\boldsymbol {\gamma }}''(t){\big )}^{2}}}}}

or, omitting the parameter t for readability,

{\displaystyle \rho ={\frac {\left|{\boldsymbol {\gamma }}'\right|^{3}}{\sqrt {\left|{\boldsymbol {\gamma }}'\right|^{2}\;\left|{\boldsymbol {\gamma }}''\right|^{2}-\left({\boldsymbol {\gamma }}'\cdot {\boldsymbol {\gamma }}''\right)^{2}}}}.}

## Example

### semicircle and circle

In the plane one upper half of the radius of the semi-circle for a

{\displaystyle y={\sqrt {a^{2}-x^{2}}},\quad y'={\frac {-x}{\sqrt {a^{2}-x^{2}}}},\quad y''={\frac {-a^{2}}{\left(a^{2}-x^{2}\right)^{\frac {3}{2}}}},\quad R=|-a|=a.}

For a semicircle of radius a in the lower half plane

{\displaystyle y=-{\sqrt {a^{2}-x^{2}}},\quad R=|a|=a.}

One of the radius of the circle is equal to a radius of curvature a .

## multiple points

In an ellipse with major axis 2 a and minor axis 2 b , the radius of curvature of any point in the vertices of the major axis is the smallest, R = B2 / a , and any point on the minor axis has the greatest radius of curvature, R = a2 / b.

## Application

• See the Cesaro equation for use in differential geometry .
• for the radius of curvature of the Earth (a flat ellipsoid its approximation); See also: arc measurement
• The radius of curvature is also used in the three part equation for the bending of the beam .