Question

radius of curvature of the curve x^2/3 + y^2/3 = a^2/3​

Answer 1

SOLUTION

TO DETERMINE

The radius of curvature at any point (x, y) of the curve

$\displaystyle \sf{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } }$

EVALUATION

Here the given equation of the curve is

$\displaystyle \sf{ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = {a}^{ \frac{2}{3} } }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{2}{3} {x}^{ - \frac{1}{3} } + \frac{2}{3} {y}^{ - \frac{1}{3} }y_1 = 0 }$

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

$\displaystyle \sf{ \implies y_1 = - \frac{ {y}^{ \frac{1}{3} } }{ {x}^{ \frac{1}{3} } } }$

Again Differentiating both sides of Equation (1) with respect to x we get

$\displaystyle \sf{ \implies - \frac{1}{3} {x}^{ - \frac{4}{3} } - \frac{1}{3} {y}^{ - \frac{ 4}{3} } {y_1}^{2} + {y}^{ - \frac{1}{3} }y_2 = 0 }$

$\displaystyle \sf{ \implies {y}^{ - \frac{1}{3} }y_2 = \frac{1}{3} {x}^{ - \frac{4}{3} } + \frac{1}{3} {y}^{ - \frac{ 4}{3} } {y_1}^{2} }$

$\displaystyle \sf{ \implies {y}^{ - \frac{1}{3} }y_2 = \frac{1}{3} {x}^{ - \frac{4}{3} } + \frac{1}{3} {y}^{ - \frac{ 2}{3} } {x}^{ - \frac{2}{3} } }$

$\displaystyle \sf{ \implies {y}^{ - \frac{1}{3} }y_2 = \frac{1}{3} {x}^{ - \frac{2}{3} } \bigg( {x}^{ - \frac{2}{3} } + {y}^{ - \frac{ 2}{3} } \bigg) }$

$\displaystyle \sf{ \implies {y}^{ - \frac{1}{3} }y_2 = \frac{1}{3} {x}^{ - \frac{2}{3} } \bigg( \frac{{x}^{ \frac{2}{3} } + {y}^{ \frac{ 2}{3} } }{{x}^{ \frac{2}{3} } {y}^{ \frac{ 2}{3} } } \bigg) }$

$\displaystyle \sf{ \implies {y}^{ - \frac{1}{3} }y_2 = \frac{1}{3} {x}^{ - \frac{4}{3} } {y}^{ - \frac{2}{3} } {a}^{ \frac{2}{3} } }$

$\displaystyle \sf{ \implies y_2 = \frac{1}{3} {x}^{ - \frac{4}{3} } {y}^{ - \frac{1}{3} } {a}^{ \frac{2}{3} } }$

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

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

$\displaystyle \sf{ = \frac{{ \bigg(1 + \frac{ {y}^{ \frac{2}{3} } }{ {x}^{ \frac{2}{3} } } \bigg)}^{ \frac{3}{2} } }{ \frac{1}{3} {x}^{ - \frac{4}{3} } {y}^{ - \frac{1}{3} } {a}^{ \frac{2}{3} } } }$

$\displaystyle \sf{ = \frac{{ \bigg( {x}^ \frac{2}{3} + {y}^{ \frac{2}{3} } \: \bigg)}^{ \frac{3}{2} } }{ \frac{1}{3} {x}^{ - \frac{1}{3} } {y}^{ - \frac{1}{3} } {a}^{ \frac{2}{3} } } }$

$\displaystyle \sf{ = \frac{{ \bigg( {a}^{ \frac{2}{3} } \: \bigg)}^{ \frac{3}{2} } }{ \frac{1}{3} {x}^{ - \frac{1}{3} } {y}^{ - \frac{1}{3} } {a}^{ \frac{2}{3} } } }$

$\displaystyle \sf{ = \frac{{ a \: } }{ \frac{1}{3} {x}^{ - \frac{1}{3} } {y}^{ - \frac{1}{3} } {a}^{ \frac{2}{3} } } }$

$\displaystyle \sf{ = 3 {a}^{ \frac{1}{3} } {x}^{ \frac{1}{3} } {y}^{ \frac{1}{3} } }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. find the radius of curvature for the curve xy3=a4 at (a,a)

https://brainly.in/question/22270223

2. Divergence of r / r3 is (a) zero at the origin (b) zero everywhere (c) zero everywhere except the origin (d) nonzero

https://brainly.in/question/22316220