Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/211

But	${\tfrac {d\beta }{d\alpha }}=\tan \tau$ = slope of tangent to the evolute at C, and
	${\tfrac {dy}{dx}}=\tan \tau$ = slope of tangent to the given curve at the corresponding point $P(x,y)$.

Substituting the last two results in (F), we get

$\tan \tau '=-{\tfrac {1}{\tan \tau }}.$

Since the slope of one tangent is the negative reciprocal of the slope of the other, they are perpendicular. But a line perpendicular to the tangent at P is a normal to the curve. Hence

A normal to the given curve is a tangent to its evolute.

Again, squaring equations (D) and (E) and adding, we get

(G)	$\left({\frac {d\alpha }{ds}}\right)^{2}+\left({\frac {d\beta }{ds}}\right)^{2}=\left({\frac {dR}{ds}}\right)^{2}.$

But if $s'=$ length of arc of the evolute, the left-hand member of (G) is precisely the square of ${\tfrac {ds'}{ds}}$ (from (34), §94, where $t=s,s=s',x=\alpha ,y=\beta$). Hence (G) asserts that

$\left({\frac {ds'}{ds}}\right)^{2}=\left({\frac {dR}{ds}}\right)^{2},{\text{ or }}{\frac {ds'}{ds}}=\pm {\frac {dR}{ds}}.$

That is, the radius of curvature of the given curve increases or decreases as fast as the arc of the evolute increases. In our figure this means that

$P_{1}C_{1}-PC={\text{arc}}\ CC_{1}.$

The length of an arc of the evolute is equal to the difference between the radii of curvature of the given curve which are tangent to this arc at its extremities.

Thus in Illustrative Example 4, §118, we observe that if we fold $Q^{v}P^{v}(=4a)$ over to the left on the evolute, $P^{v}$ will reach to O', and we have:

The length of one arc of the cycloid (as OO'Q^v) is eight times the length of the radius of the generating circle.

121. Involutes and their mechanical construction. Let a flexible ruler be bent in the form of the curve $C_{1}C_{9}$ the evolute of the curve. $P_{1}P_{9}$, and suppose a string of length $R_{9}$, with one end fastened at $C_{9}$, to be