# Thread: How to know if implicit definition of a variable possible?

1. ## How to know if implicit definition of a variable possible?

The point P = (-2,1,1) satisfies z^3 + x y^2 z + 1 = 0
Can we implicitly define z in terms of x and y using this equation in a region about P?
Find dz/dx, dz/dy, d2z/dx2 if it makes sense to do so. (note: these are partial derivatives)

I am not sure how to know if we can implicitly define, but is it correct to find partial differentials as follows?
z^3 + x y^2 z + 1 = 0
3z^2 dz/dx + y^2 (x dz/dx + z) = 0
3z^2 dz/dx + xy^2 dz/dx + zy^2 = 0-------------(1)
dz/dx = -zy^2/(3z^2 + xy^2)
------------------------
z^3 + x y^2 z + 1 = 0
3z^2 dz/dy + x(2yz + y^2 dz/dy) = 0
dz/dy = -2xyz/(3z^2 + xy^2)
-----------------------

I can get d2z/dx2 by differentiating (1) and replacing dz/dx by its expression.

Is this method correct? Please let me know how to decide if we can implicitly define z in terms of x and y in a region around P.

2. ## Re: How to know if implicit definition of a variable possible?

Is this correct?
Let F(x,y,z) = z^3 + xy^2 z + 1
F(x,y,z) is continuous at P.
Partial derivative dF/dz = 3z^2 + xy^2
At P: dF/dz = 3(1)^2 + (-2)1^2 = 1 (not= 0)
Therefore the implicit definition is possible.

dz/dx = -(dF/dx)/(dF/dz) = -(y^2 z)/(3z^2 + xy^2)
dz/dy = -(dF/dy)/(dF/dz) = -2xyz/(3z^2 + xy^2)

Is this correct? But how to find the d2z/dx2 ?

Note: I have used 'd' here but it is actually 'del' (partial derivative)

3. ## Re: How to know if implicit definition of a variable possible?

I think that's correct, you need $\displaystyle \frac{\partial{F}}{\partial{z}}\ne{0}$ for the function to be defined. The situation is more complicated if you're trying to define a function of more than one variable.

To get $\displaystyle \frac{\partial^2z}{\partial{x}^2}$, I think you can do an implicit differentiation on your equation (1), or you could also differentiate $\displaystyle \frac{\partial{z}}{\partial{x}}$ directly using the quotient rule. It's pretty much the same calculation.

- Hollywood