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

Question

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

in progress 0
Adalyn 4 months 2021-07-18T23:12:16+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-18T23:13:33+00:00

    \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| }}

    0
    2021-07-18T23:13:40+00:00

    Answer:

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

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