In differential geometry, the **radius of curvature**, R, is the reciprocal of the curvature. For a curve, it equals the radius of the circular arc which 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 combinations thereof.^{[1]}^{[2]}^{[3]}

## 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^{[3]}

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

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

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

- $R=\left|\frac{{\left(1+{y}^{\prime \phantom{\rule{thinmathspace}{0ex}}2}\right)}^{\frac{3}{2}}}{{y}^{\u2033}}\right|,\phantom{\rule{2em}{0ex}}{\textstyle \text{where}}\phantom{\rule{1em}{0ex}}{y}^{\prime}=\frac{dy}{dx},\phantom{\rule{1em}{0ex}}{y}^{\u2033}$
In the case of a plane curve, then R is the absolute value of^{[3]}

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

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

If the curve is given in Cartesian coordinates as <

In the case of a plane curve, then R is the absolute value of^{[3]}

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

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

- $R=|\frac{{\left(1+{y}^{\prime \phantom{\rule{thinmathspace}{0ex}}2}\right)}^{}}{}$
*y*(*x*), then the radius of curvature is (assuming the curve is differentiable up to order 2):
and |*z*| denotes the absolute value of z.

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

- $R=\left|\frac{ds}{d\phi}\right|=|\frac{{\left({\dot{x}}^{2}+{\dot{y}}^{2}\right)}^{\frac{3}{2}}}{\dot{x}\ddot{y}-\stackrel{If\; the\; curve\; is\; givenparametricallyby\; functions}{}}$
*x*(*t*) and *y*(*t*), then the radius of curvature is
Heuristically, this result can be interpreted as^{[2]}

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

## Formula

If **γ** : ℝ → ℝ^{n} is a parametrized curve in ℝ^{n} then the radius of curvature at each point of the curve, *ρ* : ℝ → ℝ, is given by^{[3]}

- $\rho ={\frac {\left|{\boldsymbol {\gamma }}'\right|^{3}}{\sqrt {\left|{\boldsymbol$
If **γ** : ℝ → ℝ^{n} is a parametrized curve in ℝ^{n} then the radius of curvature at each point of the curve, *ρ* : ℝ → ℝ, is given by^{[3]}

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

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

*f*(*t*) is a function from ℝ to ℝ, then the radius of curvature of its graph, **γ**(*t*) = (*t*, *f*(*t*)), is
- $\rho (t)={\frac {\left|1+f'^{\,2}(t)\right|^{\frac {3}{2}}}{\left|f''(t)\right|}}.$

### Derivation

Let **γ** be as above, and fix t. We want to find the radius ρ of a parametrized circle which matches γ in its zeroth, first, and second derivatives at t. Clearly the radius will not depend on the position **γ**(*t*), only on the velocity **γ**′(*t*) and acceleration **γ**″(*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 the three scalars |**γ** be as above, and fix t. We want to find the radius ρ of a parametrized circle which matches γ in its zeroth, first, and second derivatives at t. Clearly the radius will not depend on the position **γ**(*t*), only on the velocity **γ**′(*t*) and acceleration **γ**″(*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 the three scalars |**γ**′(*t*)|^{2}, |**γ**″(*t*)|^{2} and **γ**′(*t*) · **γ**″(*t*).^{[3]}

The general equation for a parametrized circle in ℝ^{n} is

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

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

The relevant derivatives of **g** work out to be

- ℝ
^{n} is
where **c** ∈ ℝ^{n} is the center of the circle (irrelevant since it disappears in the derivatives), **a**,**b** ∈ ℝ^{n} are perpendicular vectors of length ρ (that is, **a** · **a** = **b** · **b** = *ρ*^{2}^{} and **a** · **b** = 0), and *h* : ℝ → ℝ is an arbitrary function which is twice differentiable at t.

The relevant derivatives of **g** work out to be

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