Year 12s: what will be the first way you research your future options? >>

Parametrization of a 2D plane in three dimensions

Ok, I don't really understand how I can't do this, but it's late and I'm tired so I would appreciate some help.

We've got a unit cube, centred on the origin, in the first octant. How can I parametrize each face using two variables?

My first guess was that for the face in the xz plane, we could have

r(x,y) = (x,y,x)

But this turns out to be wrong.

Help, please?

Reply 1

generalebriety

If it's in the xz plane, you want y = 0.

Reply 2

Sun Ra and his Arkestra

Yeah, I considered that, too, but that's even worse cause then there is no way to get a non-zero value of the vector product of the partial derivative with respect to x and the partial derivative with respect to y. The partial derivative with respect to y is just the zero vector. Do not want. Where am I going wrong?

The worst bit is I know exactly what the vector surface element of this plane is, I just have to play the game for this problem sheet

Reply 3

generalebriety

I don't know what exactly you're looking for; if the y-coordinate of this plane isn't changing, why are you trying to take derivatives with respect to it?

Reply 4

Sun Ra and his Arkestra

I've got to find the vector surface element. Which, I'm pretty damn sure is dS = (0,-1,0)^Tdxdt, am I right? I've got to do it the long winded way though finding the vector product of the partial derivatives with respect to x and y. This involves parametrizing the surface. At the same time, though, you're right, y should not be involved. But, it has to be? This makes no sense. I have done much harder questions than this.

Reply 5

generalebriety

You're meant to be differentiating with respect to x and z, as those are your parameters. You'll end up crossing (1,0,0) with (0,0,1), which is (0,-1,0) as you predicted.