# Implicit Function Theorem Application

1. Aug 16, 2012

### Fractal20

1. The problem statement, all variables and given/known data
Let f(x,y) = xe$^{x^2 + y^2}$ + y^3 + cos x

a) Show there exists a differentiable function $\phi$(y) such that x = $\phi$(y) solves f(x,y) = 1 in some neighborhood of (0,0) in R^2. (Meaning (x,y) satisfies f(x,y) = 1 for (x,y) near (0,0) if and only if x = $\phi$(y)) Prove your answer

b) Find d$\phi$/dy as a function of x and y

2. Relevant equations

3. The attempt at a solution
I can only presume this is an implicit function theorem related question. I have been having some trouble with this theorem. The book I have been going through (Calculus, Boyce and Diprima) states the theorem very particularly. They only discuss y as = a function of x and don't talk about the corresponding case of x = a function of y. Moreover, they require that the original function is 0 at the point in question (the wikipedia page on implicit function theorem also seems to stipulate the function = 0 in there initial statement as well...?). And they provide no proof so it is hard for me to get any real grasp of it to apply it in a differing situation.

Anywho, I have tried looking around for more info and came across this old thread, https://www.physicsforums.com/showthread.php?t=29475, where mathwonk was pretty helpful about this theorem. From reading this it seemed that all I needed to check is that the x partial of f is non zero. I got the x partial as = e$^{x^2 + y^2}$ + xe $^{x^2 + y^2}$(2x). this is nonzero at (0,0) so that checks out. So then the implicit function theorem says that since the function is not horizontal there: fx $\neq$ 0, then we can solve for x in terms of y. This resulting function $\phi$(y) solves it.

Is this in any way adequate? I'm really just groping in the dark...

as for b) I want to just differentiate f with respect to y and remembering that x is an implicit function of y. Then solve for d$\phi$/dy. However, the implicit functions section of my Boyce and Diprima Calc book always explicitly has the function in question = 0. This then facilitates the solving for d phi /dy. Is the point here that we are only interested around f(x,y) = 1 and thus the partial derivative here and thus the 1 turns into a 0?

Basically I have no intuitive sense of this. My Calc book didn't back anything up. I tried to look up proofs elsewhere, but it all appears to require a much higher mathematics. Thanks a bunch!

2. Aug 16, 2012

### voko

x vs y is a non-issue. They are just labels, so whatever is true for one is true for the other - just swap them in the equations.

F(x, y) = 1 can be trivially converted to G(x, y) = 0 - can you see how?

3. Aug 16, 2012

### Fractal20

I had thought about that but for some reason was still put off. So g(x,y) = f(x,y) - 1. I guess that certainly makes sense for part b) and taking the derivatives since g' = f'. And then for a) it just means that if g(x,y) = 0 then f(x,y) must equal 1? So in trying to use the implicit function theorem, do I need to convert to such a g(x,y) or is it sufficient to stay with f(x,y). And is what I wrote in anyway a "proof"? Thanks Voko!

4. Aug 16, 2012

### voko

You need to form a function g(x, y) that matches the conditions of the theorem as you have it given to you, check all of its conditions, and that should do it.

Unless, of course, you are required to have your result without using the theorem (this I can't tell, ask your instructor if unsure). In the latter case you would pretty much have to repeat the argument in the proof of the theorem.