Question

If $x^2+y^2=R^2(R>0)$ then $k=\frac{y^{\prime \prime }}{\sqrt{{\left(1+y^{\prime 2}\right)}^3}}$ 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

Answer: (B)

$2x+2y{y}^{\prime }=0$
$x+y{y}^{\prime }=0\Rightarrow y^{\prime }=-\frac{x}{y}\dots .(1)$
1$+y^{\prime \prime }+{\left(y^{\prime }\right)}^2=0$
$y^{\prime \prime }=-\frac{1+{\left(y^{\prime }\right)}^2}{y}$
$\text{ now }k=\frac{y^{\prime \prime }}{{\left(1+{\left(y^{\prime }\right)}^2\right)}^{3/2}}=-\frac{1+{\left(y^{\prime }\right)}^2}{y{\left(1+{\left(y^{\prime }\right)}^2\right)}^{3/2}}=-\frac{1}{y\sqrt{1+{\left(y^{\prime }\right)}^2}}=-\frac{1}{y\sqrt{1+\frac{x^2}{y^2}}}=-\frac{1}{\sqrt{y^2+x^2}}=-\frac{1}{R}$