Question

If $x^2+y^2=R^2$ $(R>0)$ then $k=\frac{y^{\prime \prime }}{\sqrt{(1+y^{\prime 2}{)}^3}}$ where $k$ in terms of $R$ alone is equal to

• A. $-\frac{1}{R^2}$
• B. $-\frac{1}{R}$
• C. $\frac{2}{R}$
• D. $-\frac{2}{R^2}$

Answer 1

The correct option is B $-\frac{1}{R}$

$x^2+y^2=R^2$
$x=-y{y}^{\prime }=>y^{\prime }=\frac{-x}{y}$
$y^{\prime \prime }=\frac{-y+x{y}^{\prime }}{y^2}=\frac{-y^2-x^2}{y^3}=-\frac{R^2}{y^3}$
$k=\frac{y^{\prime \prime }}{\sqrt{(1+y^{\prime 2}{)}^3}}=\frac{-\frac{R^2}{y^3}}{\sqrt{(1+\frac{x^2}{y^2}{)}^3}}$
$=\frac{-\frac{R^2}{y^3}}{\sqrt{(\frac{R^2}{y^2}{)}^3}}$
$=-\frac{R^2}{R^3}=\frac{-1}{R}$