# Homework Help: Vector calculus, surfaces, and planes.

1. Sep 20, 2014

### PhysicsKid0123

I have attached an image... Okay, so I have been stuck on this problem for like 2 hours now and I have no idea how to find r(x). I know the trace is the intersection of the plane and the surface. My first attempt was to substitute the plane y+2x=0 equation for the surface equation by solving for y getting y=-2x. So I got 3x^2-(-2x)^2+2x+2(-2x) = z = f(x,y).
So I got an equation which is only a function of x e.g f(x) = -x^2-2x. However, how do I go from this to a vector equation r(x) so I can differentiate to get the tangent vector equation.

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2. Sep 20, 2014

### LCKurtz

You could use the vector parameterization of the surface $x = x, y = y, z = 3x^3 - y^2 +2x+2y$ so for the surface$$\vec R(x,y) = \langle x,y,3x^3 - y^2 +2x+2y\rangle$$Now put $y = -2x$ in that and you will have a vector equation for the required curve.

 Fixed typo in parameterization $x^2$ should have been $x^3$.

Last edited: Sep 20, 2014
3. Sep 20, 2014

### PhysicsKid0123

okay, I haven't solved it yet using what you've told me, but why would that work? How would I know if using x and not x^2 or 4x would be the correct parametrization? Just seems arbitrary how your define the position vector of the surface. mhmmm....

4. Sep 20, 2014

### LCKurtz

Parameterizations of surfaces need two variables. Since $z$ is given in terms of $x$ and $y$, it is most natural to use them as the parameters.