Answer to Question #271919 in Differential Geometry | Topology for Angel Nodado

Question

Find the curvature, the radius of curvature and the center of curvature having a parametric equations at the given point:

r = 4cos 2theta , theta = 1/12 * pi

Answer 1

r = 4cos2$\theta$

r₁= $\frac{dr}{d\theta}=-8sin2\theta$

r₂= $\frac{d²r}{d\theta^{2}}=-16cos2\theta$

Curvature at (r,$\theta)$ = $\frac{r²+2r_{1}^{2}-rr_{2}}{(r²+r_{1}²)^{3/2}}$

r($\theta=\frac{π}{12})=4cos(\frac{π}{6})=2\sqrt{3}$

r₁($\theta=\frac{π}{12})=$ $-8sin(\frac{π}{6})=-4$

r₂($\theta=\frac{π}{12})=$ $-16cos(\frac{π}{6})=-8\sqrt{3}$

rr₂ at $(\theta = \frac{π}{12})=$ $-2\sqrt{3}.8\sqrt{3}=-48$

Therefore curvature at $(\theta = \frac{π}{12})=$

$\frac{(2\sqrt{3})²+2(-4)²+48}{((2\sqrt{3})²+(-4)²)^{3/2}}$ = $\frac{92}{(28)^{3/2}}$ = $\frac{23}{2(7)^{3/2}}$

and radius of curvature = $\frac{2(7)^{3/2}}{23}$