Question

Find a point in the first quadrant on the curve y = x + 1/x that is closest to the origin.

Answer 1

Consider the square distance between a point on the curve and the origin:$$r^2=x^2+(x+1/x)^2\\r^2=2x^2+1/x^2+2$$We wish to minimize this, so differentiate:$$2r\frac{dr}{dx}=4x-2/x^3$$At a local minimum we expect \(dr/dx=0\) hence:$$0=4x-2/x^3\\2/x^3=4x\\2=4x^4\\\frac12=x^4\\x=\frac1{\sqrt[4]2}$$Notice we neglect the negative root since we only take into consideration the first quadrant. To determine whether this is indeed a local minimum, consider the second derivative:$$2r\frac{d^2r}{dx^2}+2\frac{dr}{dx}^2=4x-2/x^3\\2r\frac{d^2r}{dx^2}=\frac4{\sqrt[4]2}-\frac2{\sqrt[4]8}>0$$Considering we want \(r>0\) this tells us we have positive curvature i.e. \(d^2r/dx^2\) at this stationary point and therefore we do indeed have a local minimum.

Answer 2

that earlier line should read:$$2r\frac{d^2r}{dx^2}+2\left(\frac{dr}{dx}\right)^2=4x-2/x^3\\2r\frac{d^2r}{dx^2}=\frac4{\sqrt[4]2}-\frac2{\sqrt[4]8}>0$$notice we use the fact \(dr/dx=0\) to reduce the equation