If ${x^2} + {y^2} = {R^2}(R > 0),$ then $K = \frac{{y''}}{{\sqrt {1 + {{(y')}^2}{)^3}} }}.$ Find $K$ in terms of $R.$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

{x^2} + {y^2} = {R^2}(R > 0),

then

K = \frac{{y''}}{{\sqrt {1 + {{(y')}^2}{)^3}} }}.

Find

K

in terms of

R.

View answer and explanation

Correct option: B

- \frac{1}{R}

Option B (

- \frac{1}{R}

) is correct because it is the curvature K of the circle expressed in terms of its radius R.

Practice All Chapter MCQs →

f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

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