Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/205

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Therefore, by (40), §102, the curvature $K = 0$; and from (42), §103, and (50), §117, we see that in general $\alpha, \beta, R$ increase without limit as the second derivative approaches zero. That is, if we suppose P with its tangent to move along the curve to P', at the point of inflection Q the curvature is zero, the rotation of the tangent is momentarily arrested, and as the direction of rotation changes, the center of curvature moves out indefinitely and the radius of curvature becomes infinite.

Illustrative Example 1. Find the coördinates of the center of curvature of the parabola $y^2 = 4 px$ corresponding (a) to any point on the curve; (b) to the vertex.

Solution. $\frac{dy}{dx} = \frac{2p}{y}; \frac{d^2 y}{dx^2} = -\frac{4p^2}{y^3}.$

(a) Substituting in (E),§116,

$\alpha = x + \frac{y^2 + 4 p^2}{y^2} \cdot \frac{2p}{y} \cdot \frac{y^3}{4 p^2} = 3x + 2p.$

$\beta = y - \frac{y^2 + 4 p^2}{y^2} \cdot \frac{y^3}{4 p^2} = -\frac{y^3}{4 p ^2}.$

Therefore $\left( 3x + 2p, -\frac{y^3}{4 p^2} \right)$ is the center of curvature corresponding to any point on the curve.

(b) $(2 p, 0)$ is the center of curvature corresponding to the vertex $(0, 0)$.

118. Center of curvature the limiting position of the intersection of normals at neighboring points. Let the equation of a curve be

(A)	$y\ =\ f(x).$

The equations of the normals to the curve at two neighboring points $P_0$ and $P_1$ are

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

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

(B)	$\begin{cases}\left(x_0 - \alpha'\right) + \left(y_0 - \beta\right) \frac{dy_0}{dx_0} = 0, \\ \left(x_1 - \alpha'\right) + \left(y_1 - \beta'\right) \frac{dy_1}{dx_1} = 0.\end{cases}$