4
$\begingroup$

So I'm a set of practice problems regarding this but I don't quite understand how to think about this...

Example of a problem:

$x^3 (y^3 +z^3 )=0$ and $(x-y)^3 -z^2 -7=0$

A point lies on the surfaces $(1,-1,1)$

Show that in a neighborhood of this point, the curve of intersection of the surfaces can be described by a pair of equations $y=f(x)$, $z=g(x)$.

Now we just covered the inverse function theorem and implicit function theorem for multi-variable vector-valued functions.

I believe somehow I need to use the implicit function theorem..... but I don't know how.... for instance:

let $F(x,y,z)= (x^3 (y^3 +z^3) , (x-y)^3 -z^2 -7)$

Using the implicit function theorem:

$$\frac{\partial (f_1,f_2)}{\partial y\partial z}= \begin{bmatrix} ∂f_1/∂y & ∂f_1/∂z \\ ∂f_2/∂y & ∂f_2/∂z \end{bmatrix} $$ Now: $\det [∂(f_1,f_2)/∂y∂z] = -6(x^3 y^2 z)+(9z^2 x^3 (x-y)^2 )$. Evaluated at the point $(1,-1,1)$ we find determinant not equal to zero.

Thus we may apply the Implicit Function theorem.

And now I'm confused. I feel that I should somehow set $z,y$ equal to something but its not clear what I should set equal to what... My book's proof of the theorem (which oddly enough contains no examples of application in this context) using cramer's rule [linear algebra tool which I understand] but I don't see how this would help.

So I know I should be apply use to the implicit function theorem to find a $y=f(x)$ and $z=g(x)$ but I really don't see how.

$\endgroup$

1 Answer 1

5
$\begingroup$

Define $f: \mathbb{R}^2 \times \mathbb{R} \to \mathbb{R}^2$ by $f(w,x) = (x^3 (w_1^3 +w_2^3 ), (x-w_1)^3 -w_2^2 -7)^T$.

Note that ${ \partial f((-1,1)^T, 1) \over \partial w} = \begin{bmatrix} 3 & 3 \\ -24 & -2 \end{bmatrix}$ is invertible, and $f((-1,1)^T, 1) = (0,0)^T$, hence there is a differentiable function $\omega: U \to V$ where $U$ a neighbourhood of $1$ and $V \subset \mathbb{R}^2$ is a neighbourhood of $(-1,1)^T$ such that $f(\omega(x),x) = 0$ for all $x \in U$.

The curve $x \mapsto \omega(x)$ is the curve you are looking for.

$\endgroup$
1
  • 2
    $\begingroup$ I'm trying to understand what you wrote thus far. The fact that it is invertible is okay, this can be seen in two ways: 1)the Determinate at that point consists of "reals" thus it must be invertible 2) Since det is non-zero we know by the Implicit function theorem there exists a function g such that g(f(X)=X. And it makes sense(and i already understood this part) that the inverse function will be C_1 . I'm not sure how this helps me find X -> w(x). $\endgroup$
    – CTfunction
    Commented Mar 13, 2014 at 6:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .