Question

find the radius of Curvature the curve x⁴+y⁴=2 at the point (1, 1)​

Answer 1

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ radius of curvature (r) to the curve y = f(x) is,

$\boxed{\bf\: r = \dfrac{ {(1 + {y_1}^{2}) }^{ \frac{3}{2} } } { |y_2| }}$

where,

$\rm :\longmapsto\: y_1\: = \dfrac{dy}{dx}$

$\rm :\longmapsto\: y_2\: = \dfrac{ {d}^{2} y}{d {x}^{2} }$

Given that,

$\rm :\longmapsto\: {x}^{4} + {y}^{4} = 2$

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

$\rm :\longmapsto\: {4x}^{3} + {4y}^{3}y_1 = 0$

$\rm :\longmapsto\: {4y}^{3}y_1 = - {4x}^{3}$

$\rm :\implies\:y_1 = - \: \dfrac{ {x}^{3} }{ {y}^{3} } - - (1)$

$\rm :\implies\:(y_1)_ {(1,1)}= - \: \dfrac{ {1}^{3} }{ {1}^{3} }$

$\rm :\implies \: \boxed{\bf\: \:(y_1)_ {(1,1)}= - \:1}$

On differentiating equation (1) w. r. t. x, we get

$\rm :\longmapsto\:y_2 = - \dfrac{ {y}^{3}\dfrac{d}{dx} {x}^{3}- {x}^{3}\dfrac{d}{dx} {y}^{3}}{ {y}^{6} }$

$\rm :\longmapsto\:y_2 = - \: \dfrac{ {y}^{3} {(3x}^{2}) - {x}^{3}( {3y}^{2})y_1}{ {y}^{6} }$

$\rm :\longmapsto\:(y_2)_{(1,1)}= - \: \dfrac{ {1}^{3} {(3 \times (1)}^{2}) - {1}^{3}( {3 \times 1}^{2})( - 1)}{ {1}^{6} }$

$\rm :\longmapsto\:(y_2)_{(1,1)}= - (3 + 3)$

$\rm :\implies\:\boxed{\bf\: \:(y_2)_{(1,1)}= - \: 6}$

Now,

we know that,

Radius of curvature is given by

$\rm :\longmapsto\:{\bf\: r = \dfrac{ {(1 + {y_1}^{2}) }^{ \frac{3}{2} } } { |y_2| }}$

On substituting the values, we get

$\rm :\longmapsto\:{\bf\: r = \dfrac{ {(1 + {( - 1)}^{2}) }^{ \frac{3}{2} } } { | - 6| }}$

$\rm :\longmapsto\:{\bf\: r = \dfrac{ {(1 + 1) }^{ \frac{3}{2} } } {6}}$

$\rm :\longmapsto\:{\bf\: r = \dfrac{ {(2) }^{ \frac{3}{2} } } {6}}$

$\rm :\longmapsto\:{\bf\: r = \dfrac{ {( {( \sqrt{2}) }^{2} ) }^{ \frac{3}{2} } } {6}}$

$\rm :\longmapsto\:{\bf\: r = \dfrac{2 \sqrt{2}} {6}}$

$\rm :\longmapsto\:{\bf\: r = \dfrac{ \sqrt{2}} {3}}$

Additional Information :-

Radius of Curvature for parametric curves :-

$\boxed{\bf\: r = \dfrac{ {( {x_1}^{2} + {y_1}^{2} )}^{ \frac{3}{2} } }{x_1y_2 - y_1x_2} }$

Radius of Curvature for polar curves :-

$\boxed{\bf\: r = \dfrac{ {( {r}^{2} + {r_1}^{2} )}^{ \frac{3}{2} } }{ {r}^{2} + {2r_1}^{2} - r {r_2}^{2} } }$