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:

(i)

The circle $\mathcal{C}$ has the same tangent at $P$ as the graph of $f$ ;

(ii)

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

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

   
   

2. The radius of curvature is given by

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

   
   

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:

$$\forall x \in (0,+\infty), f'(x)= 1+\ln x ,$$

$$\quad f''(x)=-\frac {1}{x^2}.$$

Thus, the curvature at $P$ is equal to

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

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.

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

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

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:

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

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

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

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

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

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

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

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:

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

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

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

Thus, the curvature at $A$ is

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