Answer to Question #95117 in Differential Geometry | Topology for Ashritha

Question #95117
Find the radius of curvature of follium decart x^3+y^3=3axy at the point (3a,3a)
1
Expert's answer
2019-09-24T11:35:24-0400

To find the radius of curvature of the curve given in the form "y = y({x_0})" at some point "({x_0},y({x_0})) = ({x_0},{y_0})" we can use the following formula


"R({x_0}) = \\frac{{{{[1 + {{(y'({x_0}))}^2}]}^{3\/2}}}}{{\\left| {y''({x_0})} \\right|}}"


To find the derivatives we can use the method of implicit differentiation. Let's write the equation of the curve


"{x^3} + {y^3} - 3axy = 0"


and differentiate both halves


"3{x^2} + 3{y^2}y' - 3ay - 3axy' = 0"

group the terms


"3({y^2} - ax)y' + 3{x^2} - 3ay = 0"


and expressing "y'"


"y' = \\frac{{3ay - 3{x^2}}}{{3({y^2} - ax)}} = \\frac{{ay - {x^2}}}{{{y^2} - ax}}"

Remembering this result return to the expression


"3{x^2} + 3{y^2}y' - 3ay - 3axy' = 0"

Differentiate it once again


"6x + 6y{(y')^2} + 3{y^2}y'' - 3ay' - 3ay' - 3axy'' = 0"

group the terms


"3({y^2} - ax)y'' + 6(x + y{(y')^2} - ay') = 0"

"y'' = 2\\frac{{ay' - y{{(y')}^2} - x}}{{{y^2} - ax}}"

Now we can do all calculations using "x = {x_0} = 3a" and "y = {y_0} = 3a" . Let's begin with the first derivative.


"y'({x_0}) = \\frac{{a{y_0} - x_0^2}}{{y_0^2 - a{x_0}}} = \\frac{{3{a^2} - 9{a^2}}}{{9{a^2} - 3{a^2}}} = - 1"

And with the second derivative, using the value of the first


"y''({x_0}) = 2\\frac{{ay'({x_0}) - {y_0}{{(y'({x_0}))}^2} - {x_0}}}{{y_0^2 - a{x_0}}} = 2\\frac{{ - a - 3a - 3a}}{{9{a^2} - 3{a^2}}} = - 2\\frac{{7a}}{{6{a^2}}} = - \\frac{7}{{3a}}"

At least we can use the formula for the radius of curvature


"R({x_0}) = \\frac{{{{[1 + {{(y'({x_0}))}^2}]}^{3\/2}}}}{{\\left| {y''({x_0})} \\right|}} = \\frac{{{{(1 + 1)}^{3\/2}}}}{{\\frac{7}{{3\\left| a \\right|}}}} = \\frac{{3\\left| a \\right|}}{7} \\cdot {2^{3\/2}} = 2\\sqrt 2 \\frac{{3\\left| a \\right|}}{7} = \\frac{{6\\sqrt 2 }}{7}\\left| a \\right|"

This is the answer



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