Hello good sirs and ladies.
I'm doing a course in Geometry and I have fallen over a question I'm having quite a hard time determining exactly what to do on, The question as I'm reading it is.
Determine for $0<a<3/2$ the points $p$ in $L$, for which the rank of the jacobimatrix $Df(p)$ is strictly less than 2.
Let $p$ be; $$p \in L= \left \{ (x,y,z) \mid f(x,y,z)=c \right \}$$ Where $f(x,y,z)=\left \langle x^2+y^2+z^2, (x-\frac{1}{2} )^2 +y^2 \right \rangle $ and $c=(1,a^2)$
I've determined the jacobian to be. $$Df(p)=\begin{bmatrix} 2x & 2y & 2z\\ 2x-1 & 2y & 0 \end{bmatrix}$$
Which as I've understood it means I'll have to find alle the x and y which fulfills the following $(x-\frac{1}{2} )^2 +y^2=a^2$, with $0<a<3/2$. But which also fulfills $2x-1\propto 2x$.
But this makes me very confused, because it makes me feel like I'm going to end up guessing. I proposed to myself that the procedure for determining $a$'s and $x$'s would be to keep y at $0$ and than to determine the $x$'s corresponding to $x^2-x=a^2-1/4$, which seems silly to me.
Please help, any advice is useful
If the rank of matrix is strictly less than 2, it means it's either 1 or 0.
If it's $0$, all element in your matrix are zero. However, it can't be as it implies $2x=2x-1=0$
If it's $1$, the rows a linearly dependent $(2x,2y,2z)=k(2x-1,2y,0)$. This can only be when $y=z=0$.
Can you do the rest?
P.S. Geometrically, this is very easy question. $f(x,y,z)$ defines two level-surfaces. Since sphere $x^2+y^2+z^2=1$ has no special points, the rank of Jacobi matrix cannot be 0. It will be 1 when two surfaces: sphere and cylinder have the same tangent plane or in other words touch. Or where the line of intersection has special points. One can easily see when it happens.