Question

+
find the radius of curvature of curve x^2+y^2=25
at point (4,3)​

Answer 1

Answer:

Given curve is circle with center atvirigin and point 4,3 satisfies the equation radius is 5

Answer 2

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

We know,

Radius of curvature for cartesian curve y = f(x) is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \: R = \dfrac{ {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ |y_2| }}}$

Now,

Given curve is

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 25$

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

$\rm :\longmapsto\:2x + 2yy_1 = 0$

$\rm :\longmapsto\:x + yy_1 = 0 - - - (1)$

$\bf\implies \: y_1= - \: \dfrac{x}{y}$

$\bf\implies \: y_1 \: at \: (4,3)= - \: \dfrac{4}{3} - - (2)$

From equation (1), we have

$\rm :\longmapsto\:x + yy_1 = 0$

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

$\rm :\longmapsto\:1 + y\dfrac{d}{dx} y_1 + y_1 \: \dfrac{d}{dx}y = 0$

$\rm :\longmapsto\:1 + yy_2 + y_1 \times y_1 = 0$

$\rm :\longmapsto\: yy_2 = - 1 - {y_1}^{2}$

$\bf\implies \:y_2 = \: - \: \dfrac{1 + {y_1}^{2} }{y}$

$\bf\implies \:y_2 \: at \: (4,3) = \: - \: \dfrac{1 + {\bigg( - \dfrac{4}{3} \bigg) }^{2} }{3} = - \dfrac{25}{27}$

Hence,

We have

$\rm :\longmapsto\:y_1 = \: - \: \dfrac{4}{3}$

and

$\rm :\longmapsto\:y_2 = \: - \: \dfrac{25}{27}$

Consider,

$\rm :\longmapsto\:1 + {y_1}^{2}$

$\rm \: = \: \: 1 + {\bigg( - \: \dfrac{4}{3} \bigg) }^{2}$

$\rm \: = \: \: 1 + \dfrac{16}{9}$

$\rm \: = \: \: \dfrac{9 + 16}{9}$

$\rm \: = \: \: \dfrac{25}{9}$

Now,

Consider,

$\rm :\longmapsto\: {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} }$

$\rm \: = \: \: \: {\bigg( \dfrac{25}{9} \bigg) }^{\dfrac{3}{2} }$

$\rm \: = \: \: \: {\bigg( \dfrac{5}{3} \bigg) }^{2 \times \dfrac{3}{2} }$

$\rm \: = \: \: \: {\bigg( \dfrac{5}{3} \bigg) }^{3}$

$\rm \: = \: \: \: \dfrac{125}{27}$

Hence,

$\bf :\longmapsto\: {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } = \dfrac{125}{27}$

So,

Radius of curvature is

$\rm :\longmapsto\:{ \: R = \dfrac{ {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ |y_2| }}$

$\rm \: = \: \: \dfrac{\dfrac{125}{27} }{ \: \: \: \bigg| - \dfrac{25}{27}\bigg| \: \: \: }$

$\rm \: = \: \: \dfrac{\dfrac{125}{27} }{ \: \: \: \bigg|\dfrac{25}{27}\bigg| \: \: \: }$

$\rm \: = \: \: \dfrac{125}{27} \times \dfrac{27}{25}$

$\rm \: = \: \: 5$

$\bf\implies \:Radius \: of \: curvature \: (R) = 5 \: units$

Additional Information :-

Radius of curvature for polar curve is given by

$\rm :\longmapsto\:{ \: R = \dfrac{ {\bigg( {r}^{2} + {r_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ | {r}^{2} + 2 {r_1}^{2} - r r_2| }}$

Radius of curvature for parametric curve is given by

$\rm :\longmapsto\:{ \: R = \dfrac{ {\bigg( {x_1}^{2} + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ |x_1y_2 - x_2y_1| }}$