Question

Find the radius of curvature of the curve xy= 12 at (3, 4)​

Answer 1

$\textbf{Given:}$

$\textsf{Curve is xy=12}$

$\textbf{To find:}$

$\textsf{Radius of curvature of the given curve at (3,4)}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{xy=12}$

$\implies\mathsf{y=\dfrac{12}{x}=12\,x^{-1}}$

$\mathsf{\dfrac{dy}{dx}=-12x^{-2}}$

$\mathsf{\dfrac{d^2y}{dx^2}=24x^{-3}}$

$\mathsf{At\;(3,4),}$

$\mathsf{\dfrac{dy}{dx}=-12(3)^{-2}=\dfrac{-12}{9}=\dfrac{-4}{3}}$

$\mathsf{\dfrac{d^2y}{dx^2}=24(3)^{-3}=\dfrac{24}{27}=\dfrac{8}{9}}$

$\underline{\textsf{Radius of curvature}}$

$\boxed{\mathsf{R=\dfrac{\left(1+\left(\dfrac{dy}{dx}\right)^2\right)^\frac{3}{2}}{\left|\dfrac{d^2y}{dx^2}\right|}}}$

$\implies\mathsf{R=\dfrac{\left(1+\left(\dfrac{-4}{3}\right)^2\right)^\frac{3}{2}}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\left(1+\dfrac{16}{9}\right)^\frac{3}{2}}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\left(\dfrac{9+16}{9}\right)^\frac{3}{2}}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\left(\dfrac{25}{9}\right)^\frac{3}{2}}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\left(\dfrac{5^2}{3^2}\right)^\frac{3}{2}}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\left(\dfrac{5}{3}\right)^3}{\left|\dfrac{8}{9}\right|}}$

$\implies\mathsf{R=\dfrac{\dfrac{125}{27}}{\dfrac{8}{9}}}$

$\implies\mathsf{R=\dfrac{125}{27}{\times}\dfrac{9}{8}}$

$\implies\mathsf{R=\dfrac{125}{3}{\times}\dfrac{1}{8}}$

$\implies\boxed{\mathsf{R=\dfrac{125}{24}}}$

$\textbf{Answer:}$

$\mathsf{Radius\;of\;curvature\;of\;the\;given\;curve\;at\;(3,4)\;is\;\dfrac{125}{24}}}$