Curvature of a plane curve.

Definition 6.13.1   Let $ f$ be a function and $ P$ be a point on the graph of $ f$ . Denote by $ \mathcal{C}$ the circle (if it exists) verifying the following properties:
The circle $ \mathcal{C}$ has the same tangent at $ P$ as the graph of $ f$ ;
It lies on the same side of the tangent as the graph does;
This circle is called the circle of curvature of the graph at $ P$ . Its radius is called the radius of curvature at $ P$ , and its center is called the center of curvature at $ P$ .

Proposition 6.13.2   Suppose that $ f$ is differentiable at least twice at $ x_0$ and denote by $ P$ the point whose coordinates are $ (x_0, f(x_0))$ . Then:
  1. The curvature $ \kappa$ at $ P$ is given by the formula

    $\displaystyle \kappa = \frac {\vert f''(x_0)\vert}{\left( 1 + (f'(x_0)^2) \right)^{3/2}};$    

  2. The radius of curvature is given by

    $\displaystyle \rho = \frac {1}{\kappa}.$    

Example 6.13.3   Let $ f(x)=x \; \ln x$ , for $ x>0$ . Let $ P$ be the point of intersection of the graph of $ f$ with the $ x-$ axis, i.e. $ x_0=1$ .

The function $ f$ is differentiable more tna twice over $ (0,+\infty)$ . We have:

$\displaystyle \forall x \in (0,+\infty),$ $\displaystyle f'(x)= 1+\ln x ,$    
$\displaystyle \quad$ $\displaystyle f''(x)=-\frac {1}{x^2}.$    

Thus, the curvature at $ P$ is equal to

$\displaystyle \kappa = \frac {\vert-1\vert}{(1+1^2)^{3/2}} = \frac {1}{2 \sqrt{2}}.$    

The radius of curvature is equal to $ 2\sqrt{2}$ . See Figure 16.

Figure 16: The circle of curvature at a point of a graph.
\mbox{\epsfig{file=Curvature1.eps, height=7cm}}\end{figure}

The tangent to the graph at $ P$ has equation $ y=x-1$ , and the normal $ \mathcal{L}$ at $ P$ has equation $ y=-x+1$ . The center of curvature is the point $ A$ on $ \mathcal{L}$ at a distance of $ 2\sqrt{2}$ from $ P$ and ``inside'' the curve; thus the coordinates of $ A$ are $ (-1,2)$ .

Proposition 6.13.4   Suppose that the curve is given by a parametric representation of the form

\begin{displaymath}\begin{cases}x= x(t) y=y(t) \end{cases}\end{displaymath}    

where the functions are differentiable at least twice at $ t=t_0$ . Denote by $ P$ the point whose coordinates are $ (x(t_0),y(t_0))$ .

The curvature at $ P$ is given by the formula:

$\displaystyle \kappa=\frac {x'(t_0)y''(t_0)-x''(t_0)y'(t_0)}{(x'(t_0)^2+y'(t_0)^2)^{3/2}}.$    

Physicists use to denote $ \dot{x}$ instead of $ x'$ and $ \ddot{x}$ instead of $ x''$ ; the above formula becomes:

$\displaystyle \kappa =\frac {\vert\dot{x}\ddot{y} - \ddot{x}\dot{y}\vert}{(\dot{x}^2+\dot{y}^2)^{3/2}}.$    

Example 6.13.5   The curve $ C$ is given by the parametric equations

\begin{displaymath}\begin{cases}x=3 \cos t  y= 2 \sin t \end{cases}.\end{displaymath}    

Find the curvature and the radius of curvature at the right end of the major axis.

The curve is an ellipse, whose cartesian equation is

$\displaystyle \frac {x^2}{9}+\frac{y^2}{4}=1.$    

The major axis is the $ x-$ axis and the point we have to consider is $ A(3,0)$ . This point corresponds to $ t=0$ .

We use the formula in Proposition 13.4. We have:

\begin{displaymath}\begin{cases}x=3 \cos t  y= 2 \sin t \end{cases} \Longright...
...begin{cases}x''(t)=-3 \cos t  y''(t)= - 2 \sin t \end{cases}.\end{displaymath}    

We compute these derivatives for $ t=0$ ; we have:

\begin{displaymath}\begin{cases}x'(0)= 0  y'(0)= 2 \end{cases} \qquad \text{and} \qquad \begin{cases}x''(0)=-3  y''(0)= 0 \end{cases}.\end{displaymath}    

Thus, the curvature at $ A$ is

$\displaystyle \kappa = \frac {\vert \cdot 0 - (-3) \cdot 2\vert}{(0^2+2^2)^{3/2}} =\frac {6}{4^{3/2}} = \frac 34.$ (6.2)

Noah Dana-Picard 2007-12-28