Question

Find the radius of curvature of follium decart x^3+y^3=3axy at the point (3a,3a)

Answer 1

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