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In case of Folium of Descartes $x^3+y^3=3axy$ , prove that the radius of curvature at the point $(3a/2,3a/2)$ is numerically equal to $3\sqrt2 a/16$.

The solution given in the book is as follows:

If $f(x,y)=x^3+y^3=3axy=0$ we may easily obtain $f_x(3a/2,3a/2)=(9/4)a^2=f_y(3a/2,3a/2);f_{xx}(3a/2,3a/2)=9a=f_{yy}(3a/2,3a/2);f_{xy}(3a/2,3a/2)=-3a$. Now, we know that radius of curvature at a point on a curve (for $f(x,y)=0$) $\rho=\frac{(f_x^2+f_y^2)^{\frac{3}{2}}}{f_{xx}f_y^2-2f_{yx}f_xf_y+f_{yy}f_x^2}$ and hence substituting the values we get $\rho=3\sqrt2 a/16$.

However, I am not quite getting it. I have a really fundamental doubt . What do they mean by $f_x, f_{xx},f_{yy},f_{xy}$, etc ? I have no idea about it.

Arthur
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  • $f_x$ indicates the partial derivative $\frac{\partial f}{\partial x}$. Similarly $f_{xy}$ indicates the mixed partial derivative $\frac{\partial^2 f}{\partial x \partial y}$, and so on for the others. – Lemmon Oct 27 '22 at 08:54
  • @Lemmon Well , $\frac{\delta f}{\delta x}$ means differentiating x but keeping all other variables as constant and same goes for $f_y$ then , right? Does $f_{xy}$ means first differentiating with respect to $x$ keeping all others as constant and differentiating again the resulting function with respect to $y$ keeping all other variables as constant , right?... – Arthur Oct 27 '22 at 09:25
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    Exactly. For a partial derivative you derive with respect to one variable while holding all others constant. And $f_{xy}$ is just $(f_x)_y$, correct. Further, if the second partial derivatives are continuous, then the order doesn't matter, by Clairaut's Theorem. To get the curly d in MathJax you can use \partial. – Lemmon Oct 27 '22 at 09:48

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