Radius of Curvature of Curve $r^2=a^2\cos 2x$

Question

What is the radius of the polar curve $$r^2=a^2 \cos2\theta$$ at any point (r, $\theta$) using the expression of the radius of curvature of polar curves i.e. $$R= \frac{(r^2+r_1^2)^{3/2} } {r^2+2r_1^2-rr_2} $$

Answer 1

I'll try to do some part of the algebra. Deriving $r=a\sqrt{\cos 2\theta}$ using the chain rule gives $r'=-a\sqrt{\sec 2\theta}\sin 2\theta$. To find the second derivative we use both the product rule and chain rule. Bring both terms on the same denominator and you should get $$r''=-a \frac{3+\cos 4\theta}{2(\cos 2\theta)^{3/2}}.$$ Here we have used a double angle formula for $\cos$.

Then it's a simplification using trigoniometry. For instance, the fact $r^2+(r')^2$ should become $a^2 \sec 2\theta$. You can find this by puttin the terms on the same denominator and then using the main formula of trigoniometry: $\cos^2 \theta + \sin^2\theta = 1$.

The answer should be $R= \frac{a}{3}\sqrt{\sec 2\theta}$.