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

2 thoughts on “+<br />find the radius of curvature of curve x^2+y^2=25<br />at point (4,3)​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   We know,

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

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

   Now,

   Given curve is

   [tex]\rm :\longmapsto\: {x}^{2} + {y}^{2} = 25 [/tex]

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

   [tex]\rm :\longmapsto\:2x + 2yy_1 = 0[/tex]

   [tex]\rm :\longmapsto\:x + yy_1 = 0 – – – (1)[/tex]

   [tex]\bf\implies \: y_1= – \: \dfrac{x}{y} [/tex]

   [tex]\bf\implies \: y_1 \: at \: (4,3)= – \: \dfrac{4}{3} – – (2)[/tex]

   From equation (1), we have

   [tex]\rm :\longmapsto\:x + yy_1 = 0[/tex]

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

   [tex]\rm :\longmapsto\:1 + y\dfrac{d}{dx} y_1 + y_1 \: \dfrac{d}{dx}y = 0[/tex]

   [tex]\rm :\longmapsto\:1 + yy_2 + y_1 \times y_1 = 0[/tex]

   [tex]\rm :\longmapsto\: yy_2 = – 1 – {y_1}^{2} [/tex]

   [tex]\bf\implies \:y_2 = \: – \: \dfrac{1 + {y_1}^{2} }{y} [/tex]

   [tex]\bf\implies \:y_2 \: at \: (4,3) = \: – \: \dfrac{1 + {\bigg( – \dfrac{4}{3} \bigg) }^{2} }{3} = – \dfrac{25}{27} [/tex]

   Hence,

   We have

   [tex]\rm :\longmapsto\:y_1 = \: – \: \dfrac{4}{3} [/tex]

   and

   [tex]\rm :\longmapsto\:y_2 = \: – \: \dfrac{25}{27} [/tex]

   Consider,

   [tex]\rm :\longmapsto\:1 + {y_1}^{2} [/tex]

   [tex] \rm \: = \: \: 1 + {\bigg( – \: \dfrac{4}{3} \bigg) }^{2} [/tex]

   [tex]\rm \: = \: \: 1 + \dfrac{16}{9} [/tex]

   [tex]\rm \: = \: \: \dfrac{9 + 16}{9} [/tex]

   [tex]\rm \: = \: \: \dfrac{25}{9} [/tex]

   Now,

   Consider,

   [tex]\rm :\longmapsto\: {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } [/tex]

   [tex]\rm \: = \: \: \: {\bigg( \dfrac{25}{9} \bigg) }^{\dfrac{3}{2} } [/tex]

   [tex]\rm \: = \: \: \: {\bigg( \dfrac{5}{3} \bigg) }^{2 \times \dfrac{3}{2} } [/tex]

   [tex]\rm \: = \: \: \: {\bigg( \dfrac{5}{3} \bigg) }^{3} [/tex]

   [tex]\rm \: = \: \: \: \dfrac{125}{27}[/tex]

   Hence,

   [tex]\bf :\longmapsto\: {\bigg(1 + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } = \dfrac{125}{27} [/tex]

   So,

   Radius of curvature is

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

   [tex]\rm \: = \: \: \dfrac{\dfrac{125}{27} }{ \: \: \: \bigg| – \dfrac{25}{27}\bigg| \: \: \: } [/tex]

   [tex]\rm \: = \: \: \dfrac{\dfrac{125}{27} }{ \: \: \: \bigg|\dfrac{25}{27}\bigg| \: \: \: } [/tex]

   [tex]\rm \: = \: \: \dfrac{125}{27} \times \dfrac{27}{25} [/tex]

   [tex]\rm \: = \: \: 5[/tex]

   [tex]\bf\implies \:Radius \: of \: curvature \: (R) = 5 \: units[/tex]

   Additional Information :-

   Radius of curvature for polar curve is given by

   [tex]\rm :\longmapsto\:{ \: R = \dfrac{ {\bigg( {r}^{2} + {r_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ | {r}^{2} + 2 {r_1}^{2} – r r_2| }}[/tex]

   Radius of curvature for parametric curve is given by

   [tex]\rm :\longmapsto\:{ \: R = \dfrac{ {\bigg( {x_1}^{2} + {y_1}^{2} \bigg) }^{\dfrac{3}{2} } }{ |x_1y_2 – x_2y_1| }}[/tex]

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2. Answer:

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

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