Question

6. Find the radius of curvature of if Y = Log sinx at a x = pi/2

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

For a curve y = f(x) the radius of curvature

$= \displaystyle \sf{ \frac{ { \bigg[ 1+ \bigg( { \frac{dy}{dx} \bigg)}^{2} \bigg] }^{ \frac{3}{2} } }{ \big| \: \frac{ {d}^{2}y }{d {x}^{2} } \big| } \: }$

TO DETERMINE

The radius of curvature of

$\displaystyle \: \sf{y = \log \sin x \: \: \: \: at \: \: \: x = \frac{\pi}{2} \: }$

CALCULATION

Given

$\displaystyle \: \sf{y = \log \sin x \: \: \: \: \: }$

Differentiating both sides with respect to x two times we get

$\displaystyle \: \sf{ \frac{dy}{dx} = \frac{ \cos x}{ \sin x} = \cot x \: \: \: }$

$\displaystyle \: \sf{ \frac{ {d}^{2}y }{d {x}^{2} } = - { \cosec}^{2} x \: }$

Now

$\displaystyle \: \sf{ At\: x = \frac{\pi}{2} \: \: \: we \: have \: \frac{dy}{dx} = \cot \frac{\pi}{2} \: = 0\: \: }$

$\displaystyle \: \sf{ At\: x = \frac{\pi}{2} \: \: we \: \: have \: \: \frac{ {d}^{2}y }{d {x}^{2} } = - { \cosec}^{2} \frac{\pi}{2} = - 1 \: }$

Hence the required radius of curvature

$= \displaystyle \sf{ \frac{ { \bigg[ 1+ \bigg( { \frac{dy}{dx} \bigg)}^{2} \bigg] }^{ \frac{3}{2} } }{ \big| \: \frac{ {d}^{2}y }{d {x}^{2} } \big| } \: }$

$= \displaystyle \sf{ \frac{ { \bigg[ 1+ 0 \bigg] }^{ \frac{3}{2} } }{ \big| \: - 1 \: \big| } \: }$

$= \displaystyle \sf{ \frac{1}{1} }$

$= \displaystyle \sf{ 1}$

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