Tangent Line to the Intersection of Two Surfaces: 04-BS-1, May 2016, Q5

Question 5: Tangent Line to the Intersection of Two Surfaces 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two implicit surfaces $F_1=3x^2+2y^2-2z-1=0$ (a paraboloid) and $F_2=x^2+y^2+z^2-4y-2z+2=0$ (a sphere), meeting at the point $P=(1,1,2)$.

The two surfaces meet along a curve through $P$; the tangent line there is perpendicular to both surface normals, i.e. along $\nabla F_1\times\nabla F_2$.

Find. A parametric equation for the tangent line to the curve of intersection at $P$.

Approach. The intersection curve lies in both surfaces, so its tangent direction at $P$ is perpendicular to both gradients there — i.e. parallel to $\nabla F_1\times\nabla F_2$ evaluated at $P$.

Confirm $P$ lies on both surfaces. $F_1(1,1,2)=3+2-4-1=0$ ✓. $F_2(1,1,2)=1+1+4-4-4+2=0$ ✓.
Gradients at $P$. $$\nabla F_1=(6x,4y,-2)\Big|_P=(6,4,-2),\qquad \nabla F_2=(2x,2y-4,2z-2)\Big|_P=(2,-2,2).$$
Cross product for the tangent direction. $$\nabla F_1\times\nabla F_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\6&4&-2\\2&-2&2\end{vmatrix}=\big(4(2)-(-2)(-2)\big)\mathbf i-\big(6(2)-(-2)(2)\big)\mathbf j+\big(6(-2)-4(2)\big)\mathbf k=(4,-16,-20).$$ This simplifies (dividing by 4) to the direction $(1,-4,-5)$.
Write the tangent line. Through $P=(1,1,2)$ with direction $(1,-4,-5)$: $$\boxed{(x,y,z)=(1,1,2)+t(1,-4,-5),\quad t\in\mathbb R}$$ equivalently $x=1+t,\ y=1-4t,\ z=2-5t$.

Quantity	Result
$\nabla F_1(P)$	$(6,4,-2)$
$\nabla F_2(P)$	$(2,-2,2)$
Tangent direction	$(1,-4,-5)$
Tangent line	$(x,y,z)=(1,1,2)+t(1,-4,-5)$

Concept & Theory

Topic. Tangent line to the curve of intersection of two implicit surfaces.

Key relations. For a level surface $F=0$, $\nabla F$ at a point is normal to the surface there. A curve lying in both $F_1=0$ and $F_2=0$ has, at each point, tangent vector perpendicular to both $\nabla F_1$ and $\nabla F_2$ — hence parallel to $\nabla F_1\times\nabla F_2$.

Why this works. The intersection curve is tangent to both surfaces simultaneously, so at $P$ its direction vector must lie in the tangent plane of $F_1$ (perpendicular to $\nabla F_1$) and also in the tangent plane of $F_2$ (perpendicular to $\nabla F_2$). The cross product $\nabla F_1\times\nabla F_2$ is, up to scale, the unique vector satisfying both perpendicularity conditions at once, provided the two gradients are not parallel — i.e. the surfaces cross transversally at $P$ rather than being tangent to each other there, which holds here since $(6,4,-2)$ is clearly not a scalar multiple of $(2,-2,2)$.

Common pitfall. Forgetting to verify $P$ actually lies on both surfaces before differentiating (a point that satisfies neither equation makes the whole gradient calculation meaningless); sign errors expanding the $3\times3$ determinant, especially the middle ($\mathbf j$) term, which carries an extra minus sign from the cofactor expansion; forgetting the constant $-2$ (from the $-2z$ term) inside $F_1$ when computing $\nabla F_1$.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.6 Directional Derivatives and the Gradient Vector).

Question 5 of 8: Tangent Line to the Intersection of Two Surfaces

Question 5: Tangent Line to the Intersection of Two Surfaces 20 marks