Differential Equations

Second Order Equations

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Curvature of Plane Curves

  • Let a plane curve \(C\) be defined parametrically by the radius vector \(\mathbf{r}\left( t \right).\) While a point \(M\) moves along the curve \(C\), the direction of the tangent changes (Figure \(1\)).

    Definition of Curvature of a Plane Curve
    Figure 1.

    The curvature of the curve can be defined as the ratio of the rotation angle of the tangent \(\Delta \varphi \) to the traversed arc length \(\Delta s = M{M_1}.\) This ratio \(\large\frac{{\Delta \varphi }}{{\Delta s}}\normalsize\) is called the average curvature of the curve. When the point \({M_1}\) approaches the point \(M,\) we obtain the curvature of the curve at the point \(M:\)

    \[k = \lim\limits_{\Delta s \to 0} \frac{{\Delta \varphi }}{{\Delta s}} = \frac{{d\varphi }}{{ds}}.\]

    It is clear that the curvature \(k\) in the general case can be either positive or negative, depending on the direction of rotation of the tangent.

    If a curve is defined by the radius vector \(\mathbf{r}\left( t \right),\) its curvature is given by

    \[k = \frac{{\mathbf{r’} \times \mathbf{r^{\prime\prime}}}}{{{{\left| {\mathbf{r’}} \right|}^3}}},\]

    where \(\mathbf{r}’,\) \(\mathbf{r}^{\prime\prime}\) are the first and second derivatives of the radius vector.

    In this formula, the numerator contains the vector product of the vectors \(\mathbf{r}’\) and \(\mathbf{r}^{\prime\prime}.\)

    If the coordinates of a curve are specified parametrically as \(x\left( t \right)\) and \(y\left( t \right)\), the formula for calculating the curvature takes the form

    \[k = \frac{{x’y^{\prime\prime} – y’x^{\prime\prime}}}{{{{\left[ {{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]

    If a plane curve is given by an explicit function \(y = f\left( x \right),\) the curvature is calculated by the formula

    \[k = \frac{{y^{\prime\prime}}}{{{{\left[ {1 + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]

    In the case when a curve is given in polar coordinates in the form \(\rho = \rho \left( \varphi \right),\) its curvature \(k\) is defined by the expression

    \[{k\left( \varphi \right) }={ \frac{{{\rho ^2} + 2{{\left( {\rho’} \right)}^2} – \rho \rho^{\prime\prime}}}{{{{\left[ {{\rho ^2} + {{\left( {\rho’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.}\]

    The curvature of the curve is often understood as the absolute value of curvature, without taking into account the direction of rotation of the tangent. In this case, the above formulas remain valid, but the absolute value appears in the numerator. For example, the formula for the curvature when the coordinates \(x\left( t \right)\) and \(y\left( t \right)\) of a curve are given parametrically will look as follows:

    \[k = \frac{{\left| {x’y^{\prime\prime} – y’x^{\prime\prime}} \right|}}{{{{\left[ {{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]

    The reciprocal of the curvature, is called the radius of curvature:

    \[R = \frac{1}{{\left| k \right|}}.\]

    The circle with this radius and the center, located on the inner normal line, will most closely approximate the plane curve at the given point (Figure \(2\)).

    Osculating circle
    Figure 2.

    Such a circle is called the osculating circle.

  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Determine the radius of curvature of the straight line.

    Example 2

    Determine the equation of the railway transition curve.

    Example 3

    Find the curve whose radius of curvature is constant.

    Example 1.

    Determine the radius of curvature of the straight line.


    Let the line be given by the explicit equation \(y = ax + b,\) where \(a, b\) are some coefficients. We calculate the curvature \(k\) and the radius of curvature \(R\) of this straight line.

    The absolute value of the curvature is given by

    \[k = \frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y’} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}.\]

    In our case

    \[{y’ = \left( {ax + b} \right) = a,\;\;}\kern-0.3pt{y^{\prime\prime} = a’ = 0.}\]

    This immediately implies that the curvature of the straight line is equal to zero, and the radius of curvature is respectively equal to infinity.

    Page 1
    Problem 1
    Page 2
    Problems 2-3