Parameterize a closed surface.

Question

We are told to find a tangent plane of the surface

$$x^2 +2y^2+3z^2=36$$ at the point $(1,2,3)$.

Is it possible to parameterize this surface in 2 variables, perhaps with a spherical or cylindrical coordinate system?

I attempted to solve it by creating

$$F(x,y,z)=x^2 +2y^2+3z^2-36$$

and noting that the gradient of a function is normal to its surface, and all you need to define a tangent plane is a normal vector and a positional vector and you can denote the plane as $$r\cdot n=a\cdot n$$ Is this approach correct? Is there a better way to do this?

Go on! The quicker the better, and you are almost done. No need to parametrize, so your title is misleading. — Anne Bauval, Jan 07 '23 at 22:19

score 1 · Answer 1 · answered Jan 07 '23 at 22:24

1

Since $\nabla F(1,2,3)=(2,8,18)$, the plane that you're after is $$ 2(x-1)+8(y-2)+18(z-3)=0; $$ in other words, it's the plane $x+4y+9z=36$.

answered Jan 07 '23 at 22:24

Another User

5,048

score 1 · Answer 2 · answered Jan 07 '23 at 22:26

1

$\nabla(x^2+2y^2+3z^2) = (2x,4y,6x)$ at $(1,2,3)$ is $(2,8,18)$ and is the normal vector to your tangent plane.

Can you parametierize this plane now?

answered Jan 07 '23 at 22:26

user317176

11,017

but how would you parameterize this surface for future reference? – zak zaki Jan 07 '23 at 23:55
The surface of the ellipsoid? Quasi-spherical coordinates. $x = 6\cos\theta\sin \phi, y = 3\sqrt 2\sin\theta\sin\phi, z = 2\sqrt 3 \cos\phi$ – user317176 Jan 08 '23 at 00:53

Parameterize a closed surface.

2 Answers2