Page:Somerville Mechanism of the heavens.djvu/109

Chap II.] and adding one to each side of the last equation, it becomes

$\frac{dx^2+dy^2}{dx^2}=\frac{ds^2}{dx^2}=\frac{(d^2x)^2+(d^2y)^2}{(d^2y)^2}$

$\frac{dx}{d^2y}=\frac{ds}{\sqrt{(d^2x)^2+(d^2y)^2}}$|undefined

But it has been shown that $$r =\frac{(d^2y)^2}$$; hence in a plane curve the radius of curvature is

$r=\frac{ds^2}{\sqrt{(d^2x)^2+(d^2y)^2}}$|undefined.



We may imagine $$MN$$ to be the projection of a curve of double curvature on the plane $$x o y$$ then

$$r=\frac{ds^2}{\sqrt{(d^2x)^2+(d^2y)^2}}$$ will be the projection of the radius of curvature on $$xoy$$ and it is evident that a similar expression will be found for the projection of the radius of curvature on each of the other co-ordinate planes. In fact $$\frac{1}{2}\sqrt{(d^2x)^2+(d^2y)^2}$$ is the sagitta of curvature $$n\mathrm{E}$$; for

$(nm)^2=2r.n\mathrm{E}$

$r= \frac {(nm)^2}{2n\mathrm{E}}=\frac{ds^2}{2n\mathrm{E}}=\frac{ds^2}{\sqrt{(d^2x)^2+(d^2y)^2}}$|undefined

for the arc being indefinitely small, the tangent may be considered as coinciding with it. Thus the three projections of the sagitta of curvature of the surface, or curve of double curvature, are

$\frac{1}{2}\sqrt{(d^2x)^2+(d^2y)^2}; \frac{1}{2}\sqrt{(d^2x)^2+(d^2z)^2}; \frac{1}{2}\sqrt{(d^2y)^2+(d^2z)^2};$;

hence the sum of their squares is

$\frac{1}{2}\sqrt{(d^2x)^2+(d^2y)^2+(d^2z)^2}$;

and the radius of curvature of a surface, or curve of double curvature, is

$r=\frac{ds^2}{\sqrt{(d^2x)^2+(d^2y)^2)+(d^2z)^2}}$|undefined