Question

What is the radius of curvature of the curve x4+y4=2 at (1,1)

Answer 1

SOLUTION

TO DETERMINE

The radius of curvature of the curve

$\sf{ {x}^{4} + {y}^{4} = 2 }$

at (1,1)

EVALUATION

Here the given equation of the curve is

$\sf{ {x}^{4} + {y}^{4} = 2 }$

Differentiating both sides with respect to x we get

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

$\sf{ \implies \: {x}^{3} + {y}^{3} \: y_1 = 0 } \: \: \: \: \: \: - - (1)$

Putting x = 1 & y = 1 we get

$\sf{ \implies \: {(1)}^{3} + {(1)}^{3} \: y_1 = 0 }$

$\sf{ \implies \: \: y_1 = - 1 }$

Again Differentiating equation (1) with respect to x we get

$\sf{ \implies \: 3{x}^{2} + {y}^{3} \: y_2 +3 {y}^{2} \: {y_1}^{2} = 0 } \:$

$\sf{Putting \: \: x = 1 \: , \: y = 1, y_1 = - 1 \: we \: \: get }$

$\sf{ \implies \: 3 + \: y_2 + 3 = 0 } \:$

$\sf{ \implies \: \: y_2 = - 6 } \:$

Hence the the required radius of curvature at any point (x, y) is

$\displaystyle \sf{ R = \frac{{(1 + {y_1}^{2})}^{ \frac{3}{2} } }{y_2} }$

$\displaystyle \sf{ = \frac{{(1 + {y_1}^{2})}^{ \frac{3}{2} } }{y_2} }$

$\displaystyle \sf{ = \frac{{(1 + 1)}^{ \frac{3}{2} } }{ - 6} }$

$\displaystyle \sf{ = \frac{{(2)}^{ \frac{3}{2} } }{ - 6} }$

$\displaystyle \sf{ = - \frac{2 \sqrt{2} }{6} }$

$\displaystyle \sf{ = - \frac{ \sqrt{2} }{3} }$

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