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

1.	What is the radius of curvature of the curve x4+y4=2 at (1,1)
Answer» 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 $\displaystyle \sf{ R = \frac{{(1 + {y_1}^{2})}^{ \frac{3}{2} } }{y_2} }$ $\displaystyle \sf{ = \frac{{(1 + {y_1}^{2})}^{ \frac{3}{2} } }{y_2} }$ $\displaystyle \sf{ = \frac{{(1 + 1)}^{ \frac{3}{2} } }{ - 6} }$ $\displaystyle \sf{ = \frac{{(2)}^{ \frac{3}{2} } }{ - 6} }$ $\displaystyle \sf{ = - \frac{2 \sqrt{2} }{6} }$ $\displaystyle \sf{ = - \frac{ \sqrt{2} }{3} }$ ━━━━━━━━━━━━━━━━ Learn more from Brainly :- 1. FIND the radius of curvature for the curve xy3=a4 at (a,a) brainly.in/question/22270223 2. radius of curvature of the curve x^2/3 + y^2/3 = a^2/3 brainly.in/question/13357557

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