I saw this in another question
$$\kappa(\theta) =\frac{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}{(x'(\theta)^2+y'(\theta)^2)^{3/2}}$$ Polar curvature
I know that the general form of parametric curve for polar as fallow
$$ <r\cos(t), r\sin(t)> $$
what is the curvature form of polar equation $$ r = \cos(5\theta) $$
Note: For a plane curve in a polar coordinates $\rho = g(\varphi)$, we can use this expression for curvature $K$: $$ K=\frac{\rho^2+2\rho'^2-\rho\rho''}{(\rho^2+\rho'^2)^{3/2}}. $$ We have $\rho=\cos(5\varphi)$, so $\rho'=-5\sin(5\varphi)$ and $\rho''=-25\cos(5\varphi)$, hence it will hold: $$ \begin{align} K = & \frac{\cos^2(5\varphi)+50\sin^2(5\varphi)+25\cos^2(5\varphi)}{(\cos^2(5\varphi)+25\sin^2(5\varphi))^{3/2}}=\frac{26\cos^2(5\varphi)+50\sin^2(5\varphi)}{(\cos^2(5\varphi)+25\sin^2(5\varphi))^{3/2}}\\ &\\ = &\frac{38\cos^2(5\varphi)-12\cos^2(5\varphi)+38\sin^2(5\varphi)+12\sin^2(5\varphi)}{(13\cos^2(5\varphi)-12\cos^2(5\varphi)+13\sin^2(5\varphi)+12\sin^2(5\varphi))^{3/2}}\\ &\\ = & \frac{38-12(\cos^2(5\varphi)-\sin^2(5\varphi))}{(13-12(\cos^2(5\varphi)-\sin^2(5\varphi)))^{3/2}}\\ &\\ = & \boxed{\frac{38-12\cos(10\varphi)}{(13-12\cos(10\varphi))^{3/2}}}. \end{align} $$