How to check when two manifolds $X_a$ and $Y$ intersect transversally The question is to find the values of $a$ for which $X_a = \\{(x,y,z)\in R^3 | x^2+y^2+z^2 = a\\}$ and $Y=\\{(x,y,z)\in R^3 | x+y^2+2z = 1\\}$ intersect transversally in $R^3$. If $Y$ had $x^2$ instead of $x$, this would have been much easier. To find $X_a \cap Y$, other than getting $x^2+z^2-x-2z = a-1$, I am not sure how to simply further to get a condition on $a$. Any hint is appreciated. Thanks. Following the hints, $T(X_a) = \\{(u,v,w)| xu+yv+zw=0\\}$ and $T(Y) = \\{(u,v,w)|xu+2yv+2zw=0\\}$. These two tangent spaces are the same when $x=0$ which means $y^2+z^2 = a$ and $y^2+2z=1$. Thus $(z-1)^2=a$ which means $a\geq 0$ which means $X_a$ and $Y$ intersect nontransversally when $a\geq 0$ and intersect transversally when $a<0$. Does this look right?

From calculus 3, the gradient of $f(x,y,z)=x^2+y^2+z^2$ evaluated at any $(x,y,z)$ satisfying $f(x,y,z)=a$ is a vector that is perpendicular to the surface $f(x,y,z)=a$. Similarly for $g(x,y,z)=x+y^2+2z$, its gradient, and the surface $g=1$.

Note that $X_a$ and $Y$ are surfaces and, generically, they will be transverse. So it will be easier to find where they are _not_ transverse. If this happens, the tangent space of their intersection will fail to span 3-space. So, their tangent spaces will be the same. this happens whenever the normals are parallel, hence Jason's comment.

This is a long-winded way of saying use the method of lagrange multipliers: $$\
abla f=\lambda\
abla g$$ And, as usual in this situation, careful with your case analysis!