Finding the direction vector for the tangent of two Surfaces

Question

Question Well we have to find the direction vector of the tangent to two surfaces. I was thinking about finding the 2 planes. Then the parameter representation P(t) = B + Vt of the intersection and then v is the direction vector. But I got really weird fractions. So I was wondering if I determined the gradient for both these surfaces and then crossproduct the vectors we get to finally get a direction vector of the tangent But I am unsure if this will be right! I hope I have proved enough information.

Answer 1

Yes, you are right. One approach is to cross product normal vectors of the two tangent planes. The other way is intersecting them and find the driving vector of the obtained line. Both are straight forward. Let $F(x,y,z) = 4x^2 + 3y + z^3 - 11$ and $G(x,y,z) = x^3 + 2y^2 -z - 8$. Thus $\nabla F$ and $\nabla G$ are perpendicular to level curves $F(x,y,z) = 0$ and $G(x,y,z) = 0$ or equivalently $S_1$ and $S_2$. So we have: $$\nabla F \Big\vert_{(1,2,1)} = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z})\Big\vert_{(1,2,1)} = (8x, 3, 3z^2) \Big\vert_{(1,2,1)} = (8, 3, 3)$$ $$\nabla G \Big\vert_{(1,2,1)} = (3x^2, 4y, -1) \Big\vert_{(1,2,1)} = (3, 8, -1)$$ And the cross product $(8, 3, 3) \times (3, 8, -1) = (-27, 17, 55)$. I think you are writing $z$ as a function of $x,y$ and facing complex fractions when computing the partial derivatives especially for $S_1$ which contains third root. (z = $\sqrt[3]{11-4x^2-3y}$). Actually it is not necessary to calculate that since you can see the surface as the level curves of three variable function.