# Why normal vector's formula for implicit function is different than for explicit function?

Find tangent plane to surface: $$z = x^2 + y^2 - 3$$ at point $$P(2, 1, 2)$$

The normal vector to the plane I am looking for is defined as:

$$\vec{n} = [\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, -1] \Big|_{P(2,1,2)}$$

I think I understand everything. Problem is when I am dealing with an implicit form of function.

Find tangent plane to $$4x^2 + 2y^2 + z^2 - 12 = 0$$ at point $$P(1, -\sqrt{2}, -2)$$

So the normal vector is defined as:

$$\vec{n} = [\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}] \Big|_{P(1, -\sqrt{2}, -2)}$$

I am still getting good results. But I simply cannot understand what is going on.

The second, implicit function, could be rewritten in explicit form as something like: $$z = \pm \sqrt{-4x^2 - 2y^2 + 12} = f(x, y)$$

So... it is really confusing to me why in the first normal vector's equation there's $$-1$$ and in the second there's $$\frac{\partial f}{\partial z}$$ instead. After all I am dealing with function of two variables in both cases, right?

And, after I convert the function from implicit to explicit form (and even before conversion), shouldn't I use $$\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z}$$ notation instead of $$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$$?

To show what I am struggling with to understand: let's say I have function $$y = x^2 = f(x)$$. Clearly I can find $$y'$$, but I cannot find $$x'$$, right?

Can someone help me understand this, how is it possible that the first normal vector equation can be that much different than the implicit one?

Are those two functions: implicit and explicit, even equivalent or not?

• If you have $z=F(x,y)$ you can think of it as a level set (with level $0$) for $F(x,y)-z=G(x,y,z)$. Note that $\frac {\partial G}{\partial z}=-1$. – lulu Jun 3 '19 at 23:02

No. In the second case you're dealing with a function of three variables. You can see that because you have all three variables on one side of the equation $$4x^2 + 2y^2 + z^2 - 12 = 0$$ and a constant on the other. Also, as you point out, the normal vector includes $$\frac{\partial f}{\partial z}$$ as a component, so $$f$$ must be a function of $$z$$ (in addition to $$x$$ and $$y$$).
If you want to solve the second question using the first method, you can write $$z = - \sqrt{-4x^2 - 2y^2 + 12}$$ (Not $$\pm$$, you have to choose the negative branch to pass through $$(1, -\sqrt{2}, -2)$$. Then you get \begin{align*} \frac{\partial z}{\partial x} &= \frac{1}{2}(-4x^2-2y^2+12)^{-1/2}(-8x) = -\frac{4x}{\sqrt{-4x^2 - 2y^2 + 12}} \\ \frac{\partial z}{\partial y} &= \frac{1}{2}(-4x^2-2y^2+12)^{-1/2}(-4x) = -\frac{2y}{\sqrt{-4x^2 - 2y^2 + 12}} \\ \end{align*} At $$x=1$$ and $$y=-\sqrt{2}$$ this gives a normal vector of $$\vec n_1 = \left[-\frac{4\cdot 1}{-2},-\frac{2\cdot -\sqrt{2}}{-2},-1\right] = \left[2,-\sqrt{2},-1\right]$$ This seems different from the normal vector given by the second method: $$\vec n_2 = \left.\left[8 x,4y,2z\right]\right|_{(1,-\sqrt{2},-2)} = \left[8,-4\sqrt{2},-4\right]$$ But as you can see $$\vec n_2 = 4 \vec n_1$$. “Normal” is more of a line than a vector; any two normal vectors will be linearly dependent.
Note that:$$z=x^2+y^2-3\iff\overbrace{x^2+y^2-z}^{\phantom{f(x,y,z)}=f(x,y,z)}=3.$$So, if you apply both methods, the you get the same results. That $$-1$$ comes from $$\frac{\partial f}{\partial z}$$.