Question

kercises 43. 1. Find the radius of curvature for the curves
$y = log x\div x \: at \: x = 1$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:y = \dfrac{logx}{x}$

We know, Radius of Curvature for cartesian curve is given by

$\red{\rm :\longmapsto\:\boxed{\tt{ \rho \: = \: \dfrac{ {\bigg[1 + {y_{1}}^{2} \bigg]}^{\dfrac{3}{2} } }{ |y_{2}| } }}}$

Now,

$\rm :\longmapsto\:y = \dfrac{logx}{x}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:y_{1} = \dfrac{x\dfrac{d}{dx}logx - logx\dfrac{d}{dx}x }{ {x}^{2} }$

$\rm :\longmapsto\:y_{1} = \dfrac{x \times \dfrac{1}{x} - logx \times 1}{ {x}^{2} }$

$\rm :\longmapsto\:y_{1} = \dfrac{1 - logx}{ {x}^{2} }$

So,

$\rm :\longmapsto\:\boxed{\tt{ y_{1} \{ at \: x = 1 \} = \dfrac{1 - log1}{ {1}^{2} } = 1 - 0 = 1}}$

Now,

$\rm :\longmapsto\:y_{1} = \dfrac{1 - logx}{ {x}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:y_{2} = \dfrac{ {x}^{2} \dfrac{d}{dx}(1 - logx) - (1 - logx)\dfrac{d}{dx} {x}^{2} }{ {x}^{4} }$

$\rm :\longmapsto\:y_{2} = \dfrac{ {x}^{2} \times \dfrac{1}{x} - (1 - logx)2x }{ {x}^{4} }$

$\rm :\longmapsto\:y_{2} = \dfrac{ x - (1 - logx)2x }{ {x}^{4} }$

$\rm :\longmapsto\:y_{2} = \dfrac{ 1 - (1 - logx)2}{ {x}^{3} }$

$\rm :\longmapsto\:y_{2} = \dfrac{ 1 - 2 + 2logx}{ {x}^{3} }$

$\rm :\longmapsto\:y_{2} = \dfrac{ - 1 + 2logx}{ {x}^{3} }$

$\rm :\longmapsto\:\boxed{\tt{ y_{2} \: at \{x = 1 \}= \dfrac{ - 1 + 2log1}{ {1}^{3} } = - 1 + 0 = - 1 }}$

Now, On substituting the values, in the formula of radius of curvature, we get

$\rm :\longmapsto\:\rho \: = \: \dfrac{ {\bigg[1 + {y_{1}}^{2} \bigg]}^{\dfrac{3}{2} } }{ |y_{2}| }$

$\rm :\longmapsto\:\rho \: = \: \dfrac{ {\bigg[1 + {1}^{2} \bigg]}^{\dfrac{3}{2} } }{ | - 1| }$

$\rm :\longmapsto\:\rho \: = \: \dfrac{ {\bigg[2\bigg]}^{\dfrac{3}{2} } }{ 1 }$

$\bf\implies \: \rho \: = \: 2 \sqrt{2}$

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Formula Used :-

$\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}$

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{\tt{ \dfrac{d}{dx}k = 0 \: }}$

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } }}$

$\boxed{\tt{ log1 = 0}}$