Answer to Question #124395 in Differential Geometry | Topology for nisha

Question

(a) Find radius of curvature of curve:

x²+ xy + y²= 4 at point (–2, 0)

Answer 1

To find Radius of curvature

formula: R= $\frac{(1+y'(x))^{2/3}}{\left | y''(x)) \right |}$

Solution :

y'(x)= $\frac{-(2x+y)}{x+2y}$

At (-2,0), y'(x)= -(-4+0)/(-2)=-2

Y"(x)= [(x+2y)(-2-y')+(2x+y)(1+2y')]/(x+2y)²= (3xy'-3y)/(x+2y)²

At (-2,0), y''(x)= -6)(-2)/4 = 3

Hence substituting in formula , R= $\frac{(1-2)^{2/3}}{\left | 3 \right |}$ =-1/3