Using the implicit function theorem in a system of equations.

Question

I want to see for which points the system of equations $$\begin{cases}x^2+y^2+z^2=16\\2x-y+z=4\end{cases}$$ define $(y,z)$ as an implicit function of $x$, and then calculate $(y'(x),z'(x))$.

I know I have to use the implicit function theorem. What I've done:

Computing the determinant of the partial derivatives I have $$\begin{vmatrix} 2y & 2z \\ -1 & 1 \notag \end{vmatrix} =2y+2z\neq0$$ if $(y,z)\neq0$ or $z\neq-y$

Therefore I will be able to define $(y,z)$ as an implicit function of $x$ as long as $(y,z)\neq0$ or $z\neq-y$. Is this correct? How do I calculate $(y'(x),z'(x))$?

Answer 1

Yes, your argument is fine. More precisely, define $F:\mathbb R^3\to \mathbb R^2$ by

$F(x,y,z)=(x^2+y^2+z^2-16,2x-y+z-4)$.

Then, check that

$\begin{pmatrix} (f_1)_y & (f_1)_z \\ (f_2)_y& (f_2)_z \end{pmatrix}=\begin{pmatrix} 2y & 2z \\ -1& 1 \end{pmatrix}$ and argue as you did, to show that as long as $y+z\neq 0$,

one has

$G(x):=F(x,y(x),z(x))=0$ in a neighborhood of $(x_0,y_0,z_0)$ such that $F(x_0,y_0,z_0)=0.$

Then,

$x^2+y^2(x)+z^2(x)-16=0$ and $2x-y(x)+z(x)-4=0$. To finish, differentiate these and solve simultaneously.

Answer 2

Defining $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z) = 2x-y+z$, you can rewrite your equations as $f(x,y,z)=16$ and $g(x,y,z)=4$. Now to define $y,z$ as implicit function of $x$, you'll get $f(x,y(x),z(x))=16$ and $g(x,y(x),z(x))=4$. Then opening those you get $x^2+y(x)^2+z(x)^2=16$ and $2x-y(x)+z(x)=4$. Then you get $y(x)=2x+z(x)-4$ and $x^2+(2x+z(x)-4)^2+z(x)^2=16$. From this you can calculate $z(x)$, and then further $y(x)$.

Answer 3

Your reasoning is correct. To obtain the derivatives, write

$$\begin{cases}2x+2yy'+2zz'=0\\2-y'+z'=0\end{cases}$$ and solve for $y',z'$.