Math, asked by aruns23, 1 month ago

find the radius of curvature for the ellipse x²/a²+y²/b²=1​

Answers

Answered by mathdude500
12

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

Given equation of ellipse is

\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} }  + \dfrac{ {y}^{2} }{ {b}^{2} }  = 1

Now, the parametric coordinates of ellipse are

\rm :\longmapsto\:x = a \: cost

and

\rm :\longmapsto\:y = b \: sint

Now,

Consider,

\rm :\longmapsto\:x = a \: cost

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

\rm :\longmapsto\:x' =  - a \: sint

Again, on differentiating both sides w. r. t. t, we get

\rm :\longmapsto\:x'' =  - a \: cost

Now, Consider

\rm :\longmapsto\:y \:  =  \: b \: sint

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

\rm :\longmapsto\:y' \:  =  \: b \: cost

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

\rm :\longmapsto\:y'' \:  =  -  \: b \: sint

Now, we know,

Radius of curvature of curve in parametric coordinates is given by

\boxed{ \bf{ \:  \rho \:  =  \: \dfrac{ {\bigg( {(x')}^{2} +  {(y')}^{2}  \bigg) }^{\dfrac{3}{2} } }{ |x'y'' - y'x''| } }}

So, on substituting the values evaluated above, we get

\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {( - asint)}^{2}  +  {(bcost)}^{2} \bigg) }^{ \dfrac{3}{2} } }{ |ab {sin}^{2}t + ab {cos}^{2}t  | }

\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2}  {sin}^{2}t  +  {b}^{2}  {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab( {sin}^{2}t +{cos}^{2}t ) | }

\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2}  {sin}^{2}t  +  {b}^{2}  {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab( 1 ) | }

\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2}  {sin}^{2}t  +  {b}^{2}  {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ |ab| }

\rm :\longmapsto\: \rho = \dfrac{ {\bigg( {a}^{2}  {sin}^{2}t  +  {b}^{2}  {cos}^{2}t \bigg) }^{ \dfrac{3}{2} } }{ ab }

Additional Information :-

1. Radius of curvature for cartesian curve y = f(x) is

\boxed{ \bf{ \:  \rho = \dfrac{ {\bigg( 1 +  {(y')}^{2} \bigg) }^{\dfrac{3}{2} } }{ |y''| }  \: }}

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