Question

Find the curvature p at the origin for the curve y-x = x^2 + 2xy + y^2.

Answer 1

$\textbf{Given:}$

$y-x=x^2+2xy+y^2$

$\textbf{To find:}$

$\text{Curvature of the given curve}$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{The curvature of the curve at the point (x,y) is given by}$

$\bf\,K=\dfrac{y''(x)}{[1+(y'(x))^2]^{\frac{3}{2}}}$

$y-x=x^2+2xy+y^2$

$\implies\,y-x=(x+y)^2$

$\textbf{Differentiate with respect to x}$

$\dfrac{dy}{dx}-1=2(x+y)(1+\dfrac{dy}{dx})$

$\dfrac{dy}{dx}-1=2(x+y)+2(x+y)\dfrac{dy}{dx}$

$\dfrac{dy}{dx}-2(x+y)\dfrac{dy}{dx}=2x+2y+1$

$\dfrac{dy}{dx}(1-2x-2y)=2x+2y+1$

$\dfrac{dy}{dx}=\dfrac{-(2x+2y+1)}{2x+2y-1}$

$\textbf{Differentiate again with respect to x}$

$\dfrac{d^2y}{dx^2}=\dfrac{-[(2x+2y-1)(2+2\frac{dy}{dx})-(2x+2y+1)(2+2\frac{dy}{dx})]}{(2x+2y-1)^2}$

$\dfrac{d^2y}{dx^2}=\dfrac{-(2+2\frac{dy}{dx})[2x+2y-1-2x-2y-1]}{(2x+2y-1)^2}$

$\dfrac{d^2y}{dx^2}=\dfrac{-(2+2\frac{dy}{dx})(-2)}{(2x+2y-1)^2}$

$\dfrac{d^2y}{dx^2}=\dfrac{4+4\frac{dy}{dx}}{(2x+2y-1)^2}$

$\text{At (0,0)}$

$\dfrac{dy}{dx}=\dfrac{-(2(0)+2(0)+1)}{2(0)+2(0)-1}=1$

$\dfrac{d^2y}{dx^2}=\dfrac{4+4(1)}{(2(0)+2(0)-1)^2}=8$

$\text{Now,}$

$\bf\,K=\dfrac{y''(x)}{[1+(y'(x))^2]^{\frac{3}{2}}}$

$K=\dfrac{8}{[1+(1)^2]^{\frac{3}{2}}}$

$K=\dfrac{8}{2^{\frac{3}{2}}}$

$K=\dfrac{8}{\sqrt{8}}$

$K=\sqrt{8}$

$\therefore\textbf{The curvature of the given curve is \bf\sqrt{8} }$