Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/203

 Hence, by Rolle's Theorem (§105), $$F'(x)$$ must vanish for at least two values of $$x$$, one lying between $$x_0$$ and $$x_1$$, say $$x'$$, and the other lying between $$x_1$$ and $$x_2$$ say $$x''$$; that is,

$$F'(x') = 0,\ F'(x'') = 0.$$

Again, for the same reason, $$F(x)$$ must vanish for some value of $$x$$ between $$x'$$ and $$x$$, say $$x_3$$; hence

$$F''(x_3)\ =\ 0.$$

Therefore the elements $$\alpha', \beta', R'$$ of the circle passing through the points $$P_0, P_1, P_2$$ must satisfy the three equations

$$F(x_0) = 0,\ F'(x') = 0,\ F''(x_3) = 0.$$

Now let the points $$P_1$$ and $$P_2$$ approach $$P_0$$ as a limiting position; then $$x_1, x_2, x', x'', x_3$$ will all approach $$x_0$$ as a limit, and the elements $$\alpha, \beta, R$$ of the osculating circle are therefore determined by the three equations

$$F(x_0) = 0,\ F'(x_0) = 0,\ F''(x_0) = 0;$$

or, dropping the subscripts, which is the same thing,

Solving (B) and (C) for $$x - \alpha$$ and $$y - \beta$$, we get $$\left( \tfrac{d^2 y}{dx^2} \ne 0 \right)$$,

hence the coördinates of the center of curvature are