Question

Find the radius of curvature at any point on the curve xy = c².

Answer 1

Given curve is xy = c²

= y = \(\cfrac{c^2}x\)

\(\therefore\) y' = \(\cfrac{-c^2}{x^2}\)

y'' = \(\cfrac{2c^2}{x^2}\)

\(\therefore\) Radius of curvature is

ρ  = \(\cfrac{(1+(y)^2)^{3/2}}{y''}\) = \(\cfrac{(\cfrac{1+c^4}{x^4})^{3/2}}{\cfrac{2c^2}{x^3}}\)

= \(\cfrac{(\cfrac{x^4+c^4}{x^6})^{3/2}}{\cfrac{2c^2}{x^3}}\) = \(\cfrac{(x^4+c^4)^{3/2}}{2c^2x^3}\)

= \(\cfrac{(x^4+x^2y^2)^{3/2}}{2x^4y}\) (\(\because\) c² = xy)

= \(\cfrac{x^3(x^2 +y^2)^{3/2}}{2x^4y}\)

= \(\cfrac{(x^2 +y^2)^{3/2}}{2xy}\)