Page:Somerville Mechanism of the heavens.djvu/108

83. The circle ${\displaystyle {\text{A}}m{\text{B}}}$, fig. 22, which coincides with a curve or curved surface through an indefinitely small space on each side of ${\displaystyle m}$ the point of contact, is called the curve of equal curvature, or the oscillating circle of the curve ${\displaystyle {\text{MN}}}$, and ${\displaystyle om}$ is the radius of curvature.

In a plane curve the radius of curvature ${\displaystyle r}$, is expressed by

and in a curve of double curvature it is

${\displaystyle ds}$ being the constant element of the curve.

Let the angle ${\displaystyle com}$ be represented by ${\displaystyle \theta }$, then if ${\displaystyle Am}$ be the indefinitely small but constant element of the curve ${\displaystyle MN}$, the triangles ${\displaystyle com}$ and ${\displaystyle ADm}$ are similar; hence ${\displaystyle mA:mD::om:mc}$ or ${\displaystyle ds:dx::l:\sin \theta }$, and ${\displaystyle \sin \theta ={\frac {dx}{ds}}}$. In the same manner ${\displaystyle \cos \theta ={\frac {dy}{ds}}}$,

But ${\displaystyle d.\cos \theta =-d\theta \sin \theta }$, and ${\displaystyle d\theta =-{\frac {d.\cos \theta }{\sin \theta }}}$; also ${\displaystyle d.\sin \theta =d\theta \cos \theta }$, and ${\displaystyle d\theta ={\frac {d.\sin \theta }{\cos \theta }}}$; but these evidently become

Now if ${\displaystyle om}$ the radius of curvature be represented by ${\displaystyle r}$, then ${\displaystyle moA}$ being the indefinitely small increment ${\displaystyle d\theta }$ of the angle ${\displaystyle com}$, we have ${\displaystyle r:ds::l:d\theta }$; for the sine of the infinitely small angle is to be considered as coinciding with the arc: hence ${\displaystyle d\theta ={\frac {ds}{r}}}$, whence ${\displaystyle =-{\frac {ds.dy}{d2s}}={\frac {dsdx}{d^{2}y}}}$. But ${\displaystyle dx^{2}+dy^{2}=ds^{2}}$, and as ${\displaystyle ds}$ is constant ${\displaystyle dx.d^{2}x+dyd^{2}y=0}$. Whence ${\displaystyle {\frac {d^{2}s}{d^{2}y}}=-{\frac {dy}{dx}}}$, or ${\displaystyle \left({\frac {d^{2}x}{d^{2}y}}\right)^{2}={\frac {dy^{2}}{dx^{2}}}}$,