Answer to Question #286743 in Differential Geometry | Topology for Riya

Question

Find the curvature, the radius and the center of curvature at a point.

r=1+ cos theta ,theta=π/2

Answer 1

$r=f(\theta)=1+cos\theta \Rightarrow f(\frac{\pi}{2})=1\\ f'(\theta)=-sin\theta\Rightarrow f'(\frac{\pi}{2})=-1\\ f''(\theta)=-cos\theta \Rightarrow f''(\frac{\pi}{2})=0\\ \therefore \text{Radius of curvature}\\= \dfrac{(f(\frac{\pi}{2})^2+f'(\frac{\pi}{2})^2)^\frac{3}{2}} {f(\frac{\pi}{2})^2+2f'(\frac{\pi}{2})^2-f(\frac{\pi}{2})f''(\frac{\pi}{2})}\\ =\dfrac{2^\frac{3}{2}}{3}=\dfrac{2\sqrt 2}{3}\\ \text{Now, curvature $=$}\dfrac{1}{\text{Radius of Curvature}}=\dfrac{3}{2\sqrt 2}$

$\text{And, Center of curvature = }((y_0+R),(x_0+f'(x_0))R)\\ \texttt{[Where $x_0=\frac{\pi}{2}, y_0=f(\frac{\pi}{2})=1$]}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= (1+ \frac{2\sqrt 2}{3}, (\frac{\pi}{2}-1)\frac{2\sqrt 2}{3})\\$