Curve Perpendicular to Surface at Point

[jegues]

Homework Statement

It is known that the curve C,

[tex]x = 4t -2t^{2} -1, y = 2t^{3} - 1, z=3t^{2} -5[/tex]

and the surface,

[tex]x^{2}y + xz + 1 = 0[/tex]

intersect at the point (1,1,-2). You need NOT show this. Prove that the curve is perpendicular to the surface at the point.

Homework Equations

The Attempt at a Solution

I need to get my ideas straightened out before I can start attacking this one.

We want to show that the curve C is perpendicular to the surface at the point (1,1,-2).

If we stop and think about what we mean by perpendicular to the surface at a point, it means perpendicular to the tangent plane of that surface, at that point.

If we take the Gradient of the surface evaluated at the point we will get a vector that is perpendicular to the tangent plane of the surface at that point.

Now I think what I want to show is that the curve is pointing in the same(or opposite) direction of the Gradient of the surface at that point, thus the curve will also be perpendicular.

I could take the cross product between the Gradient of the surface and the curve, and if is this equals 0 then I know that they are parallel. But then again, I don't have a vector representation of the curve.

I think the part I'm confused about is how to represent my curve C in the form of a vector.

One can note that when t = 1, we obtain the point (1,1,-2) on the curve.

Do I have my ideas straight? Can anyone help me finish this one off?

Thanks again!

 

[micromass]

But the curve is given in vector notation:

[tex] (4t-2t^2-1,2t^3-1,3t^2-5)[/tex]

For the rest, your reasoning seems good...