If $y^2+x^2=R^2$ and $k=1 / R$, then $k$ is equal to

$\frac{\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{2\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{\left|y^{\prime \prime}\right|}{2 \sqrt{\left(1+y^{\prime 2}\right)^3}}$.

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

y^2+x^2=R^2

and

k=1 / R

, then

k

is equal to

View answer and explanation

Correct option: B

\frac{\left|y^{\prime \prime}\right|}{\sqrt{\left(1+y^{\prime 2}\right)^3}}

Option B (

\frac{{\left|y^{\prime \prime}\right|}}{{\sqrt{{\left(1+y^{\prime 2}\right)^3}}}}

) is correct because it represents the curvature (and hence the reciprocal of the radius) of the given circle.

Practice All Chapter MCQs →

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}