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  4. If x2+y2=R2(R>0) then k=(y prime prime/√(1+y prime 2)3) where k in terms of R alone is equal to

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Q. 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

Continuity and Differentiability

Solution:

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