Here x = a(θ – sinθ), y = a(1 – cosθ) .....(1)
∴ x' = a(1 – cosθ), y' = a sinθ
x" = a sinθ, y" = a cosθ ......(2)
Thus,
\(\rho = \frac{(x'^2 + y'^2)^{3/2}}{x'y'' - y'x''}\)
\(= \frac{[\{a(1 - \cos \theta)\}^2 + (a \sin \theta)^2]^{3/2}}{a(1 - \cos \theta)a\cos\theta - a\sin \theta . a.\sin \theta}\)
\(= \frac{[2a^2(1 - \cos \theta)]^{3/2}}{a^2\cos \theta - a^2(\cos^2\theta + \sin^2\theta)}\)
\(= \frac{(2a^2)^{3/2} (1 - \cos \theta)^{3/2}}{- a^2 (1 - \cos \theta)}\)
\(= -2\sqrt2a.(1 - \cos \theta)^{1/2}\)
\(\because (1 - \cos \theta ) = 2\sin^2 \frac 82\)
\(= -4 a \sin \frac \theta 2\) ......(3)
Now the length of the normal intercepted between the curve and the x-axis,
\(p = y \sqrt{1 + {y_1}^2}\)
\(= y\sqrt {1 + \left( \cfrac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\right)^2}\) as \(y_1 = \frac{dy}{dx}\)
\(= a(1 - \cos \theta) \sqrt{1 + \left(\frac{a \sin \theta}{a(1 - \cos \theta)}\right)^2}\)
\(= a .2\sin^2 \frac \theta 2 \sqrt{1 + \cot ^2 \frac \theta 2}\)
\(= 2 a \sin^2 \frac\theta2 \sqrt{cosec^2 \frac \theta 2}\)
\(p = 2a \sin \frac \theta 2\) ......(4)
From equation (3) and (4),
The radius of curvature ρ at any general point is twice the length of the normal (p).
