Multivariate calculus, find the normal vector

[jaguar7]

Homework Statement

Consider the surface x^2 / 4 + y^2 + z^2 = 3

(a) what type of quadric surface is this? (a spheroid...)

(b) find the outward unit normal vector to this surface at the point (2,1,1).

Homework Equations

How do we find a normal vector? Does one just take grad(f(2,1,1)) and then turn that into a unit vector to the the "unit normal vector"?

edit: can we define the function implicitly by subtracting three from both sides, then, as wikipedia says, we can just take the gradient at (2,1,1) to get the normal?

The Attempt at a Solution

 

[micromass]

You'll need to find the partial derivatives to a=(2,1,1) first. If you found them, then

$$(1,0,\frac{\partial f}{\partial x}(a))~\text{and}~(0,1,\frac{\partial f}{\partial y}(a))$$

these vectors span the plane tangent on a. All you have to do then, is to find a vector perpendicular to the plane (thus perpendicular to the two vectors above).

 

[jaguar7]

how do we take the partial derivatives when the function is not defined explicitly?

 

[micromass]

You can easily make it explicit, by

$$z=\sqrt{3-\frac{x^2}{4}-y^2}$$

I know there is a negative part to, but we ignore it because were interested in the point (2,1,1).
So your function is

$$f:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\rightarrow \sqrt{3-\frac{x^2}{4}-y^2}$$.

 

[micromass]

Or you could also find the partial derivatives by implicit differentiation. (i.e. apply the implicit function theorem). Perhaps this is easier...

 

[jaguar7]

how would one take the partial derivatives of that, well, i guess we would take set y=constant, then 1/2 (3 - x^2/4 - y^2) ^(-1/2) * (x/2) for dz/dx and 1/2 (3 - x^2/4 - y^2) ^(-1/2) * (2y)for dz/dy, right?

then we have the gradient...

**scratch that... (if we did it that way, I would not know how to find the normal and found an easier way...)

it is true that the gradient of Ax^2 + By^2 + Cz^2 + D = 0 is (2Ax, 2By, 2Cz).

so therefor we can define the function in the problem implicitly by subtracting 3 from both sides, and then take the gradient to be (x/2, 2y, 2z), and simply plug our point (2,2,1) into our gradient formula to get our normal vector.