The sets $X_a$ and $Y_b$ of equations $x^2+y^3+z = a$ and $x+y+z = b$, respectively, are smooth surfaces in $R^3$.
What values of $a$ and $b$ guarantee that the intersection of $X_a$ and $Y_b$ is a smooth curve?
I'm trying to use implicit function theorem for this. So F(x, y, z) = ($x^2+y^3+z - a$, $x+y+z-b$) = 0. (I don't know how to make vectors in Latex so forgive me for poor formatting.)
I calculated the Jacobian for F, all 1's in the top row and other stuff in the second row; row-reduced it, but don't know what to do from here.
Thanks!
You know that $F^{-1}(0)$ is a smooth curve as long as $0$ is a regular value, that is, as long as for every point $(x,y,z)$ which maps to $0$, the Jacobian at $(x,y,z)$ is surjective (rank $2$). From your row reduction, you can see that points of the domain which do NOT have rank $2$ Jacobian are those for which $x = \frac{1}{2}$ and $y = \pm\frac{\sqrt{3}}{3}$. So as long as $a$ and $b$ are chosen such that no point of this form maps to $0$, you can say that $0$ is a regular value, and therefore $F^{-1}(0)$ is a manifold.