Center of curvature

In this section, we discuss how the center of curvature can be thought of geometricaly as the limiting position of the intersection of normals at neighboring points. Let the equation of a curve be

$\displaystyle y = f(x).
$ (14.10)

Figure 14.4: Geometric visualization of the center of curvature.
\includegraphics[height=4cm,width=6cm]{circle-of-curvature3.eps}
The equations of the normals to the curve at two neighboring points $ P_0$ and $ P_1$ are (using (6.2) [§6.3]),

$\displaystyle (x_0 - X) + (y_0 - Y) \frac{dy_0}{dx_0} = 0,
\ \ \ \
(x_1 - X) + (y_1 - Y) \frac{dy_1}{dx_1} = 0.
$

If the normals intersect at $ C'=(\alpha',\beta')$, the coordinates of this point must satisfy both equations, giving

\begin{displaymath}\begin{cases}(x_0 - \alpha') + (y_0 - \beta) \frac{dy_0}{dx_0...
... - \alpha') + (y_1 - \beta') \frac{dy_1}{dx_1} = 0. \end{cases}\end{displaymath} (14.11)

Now consider the function of x defined by

$\displaystyle \phi(x) = (x - \alpha') + (y - \beta') \frac{dy}{dx},
$

in which $ y = f(x)$ using (14.10). Then equations (14.11) show that

$\displaystyle \phi(x_0) = 0,\ \ \ \phi(x_1) = 0.
$

But then, by Rolle's Theorem (§13.1), $ \phi'(x)$ must vanish for some value of $ x$ between $ x_0$ and $ x_1$ say $ x'$. Therefore $ \alpha'$ and $ \beta'$ are determined by the two equations

$\displaystyle \phi(x_0) = 0,\ \ \ \phi'(x') = 0.
$

If now $ P_1$ approaches $ P_0$ as a limiting position, then $ x'$ approaches $ x_0$, giving

$\displaystyle \phi(x_0) = 0,\ \ \ \phi'(x_0) = 0,
$

and $ C'(\alpha',\beta')$ will approach as a limiting position the center of curvature $ C(\alpha,\beta)$ corresponding to $ P_0$ on the curve. For if we drop the subscripts and write the last two equations in the form

$\displaystyle (x - \alpha') + (y - \beta') \frac{dy}{dx} = 0,
\ \ \ \
1 + \left( \frac{dy}{dx} \right)^2 + (y - \beta') \frac{d^2 y}{dx^2} = 0,
$

it is evident that solving for $ \alpha'$ and $ \beta'$ will give the same results as solving (14.4) and ((14.5) for $ \alpha$ and $ \beta$. Hence we have the following result.

Theorem 14.3.1   The center of curvature $ C$ corresponding to a point $ P$ on a curve is the limiting position of the intersection of the normal to the curve at $ P$ with a neighboring normal.

david joyner 2008-11-22