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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?

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    $\begingroup$ 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 $\frac {\partial G}{\partial z}=-1$. $\endgroup$ – lulu Jun 3 '19 at 23:02
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After all I am dealing with function of two variables in both cases, right?

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.

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  • $\begingroup$ Thanks for effort. $\endgroup$ – weno Jun 3 '19 at 23:22
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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 $\frac{\partial f}{\partial z}$.

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