Implicit Mapping Theorem for a Surface

[Parmenides]

The question, from Edward's Advanced Calculus (which is becoming a rather frustrating book), asks if the following surface can be represented as a function of $z(x,y)$ near the point $(0,2,1)$:

$$xy - ylog(z) + sin(xz) = 0$$
Naturally, this should invite me to use the Implicit Mapping Theorem, but I've seen this theorem presented mostly for systems of equations, rather than a single case. But anyhow, let the surface be called $G({\bf{x}},{\bf{y}})$, where ${\bf{x}} = (x,y)$ and ${\bf{y}} = z$. Then,
$$\frac{\partial{G}}{\partial{z}} = xcos(xz) - \frac{y}{z}$$
So at $(0,2,1)$, $G'((0,2),(1)) = -2$. Thus, since the invertible "matrix" is non-singular, there exists a neighborhood near $(0,2,1)$ for which the surface can be represented as a function of $z$.

I suspect this is correct, but being confused by some notation, I just wanted to make sure. Thanks.

 

[mighty2000]

Dear Edward,

Thank you for your question. You are correct in using the Implicit Mapping Theorem to approach this problem. The theorem can indeed be applied to a single equation, as long as it meets the criteria of being continuously differentiable and having a non-singular matrix.

In this case, your calculations and reasoning are correct. The partial derivative of G with respect to z is indeed -2 at the point (0,2,1), and since it is non-singular, there exists a neighborhood near (0,2,1) where the surface can be represented as a function of z.

I hope this helps clarify any confusion you had. Keep up the good work and don't let the frustration with the book discourage you. Sometimes, approaching a problem from a different angle can help make it more understandable. Best of luck with your studies.