Tangent Plane and Tangent Line for Intersecting Surfaces: 04-BS-1, December 2016, Q5

Question 5: Tangent Plane and Tangent Line for Intersecting Surfaces (a) 10, (b) 10 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 $f=0$ and $g=9$, both passing through $P=(3,-1,1)$ (check: $f(P)=9+1+1-2-9=0$ ✓, $g(P)=9+1-1=9$ ✓).

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

Find. (a) The tangent plane to $g=9$ at $P$. (b) The tangent line to the intersection curve $f=0\cap g=9$ at $P$.

Approach. (a) $\nabla g$ at $P$ is normal to the level surface $g=9$; write the plane through $P$ with that normal. (b) The intersection curve's tangent direction is perpendicular to both $\nabla f$ and $\nabla g$ at $P$, i.e. parallel to $\nabla f\times\nabla g$.

(a) Gradient of $g$ at $P$. $\nabla g=(3,2y,-2z)\big|_P=(3,-2,-2)$.
(a) Tangent plane. Through $P=(3,-1,1)$ with normal $(3,-2,-2)$: $$3(x-3)-2(y+1)-2(z-1)=0\ \Rightarrow\ \boxed{3x-2y-2z=9}$$ (Check: $3(3)-2(-1)-2(1)=9+2-2=9$ ✓.)
(b) Gradient of $f$ at $P$. $\nabla f=(2x-3,2y+2,2z)\big|_P=(3,0,2)$.
(b) Cross product for the tangent direction. $$\nabla f\times\nabla g=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&0&2\\3&-2&-2\end{vmatrix}=\big(0(-2)-2(-2)\big)\mathbf i-\big(3(-2)-2(3)\big)\mathbf j+\big(3(-2)-0(3)\big)\mathbf k=(4,12,-6).$$ This simplifies (dividing by 2) to the direction $(2,6,-3)$.
(b) Write the tangent line. Through $P=(3,-1,1)$ with direction $(2,6,-3)$: $$\boxed{(x,y,z)=(3,-1,1)+t(2,6,-3),\quad t\in\mathbb R}$$

Quantity	Result
(a) $\nabla g(P)$	$(3,-2,-2)$
(a) Tangent plane	$3x-2y-2z=9$
(b) $\nabla f(P)$	$(3,0,2)$
(b) Tangent direction	$(2,6,-3)$
(b) Tangent line	$(x,y,z)=(3,-1,1)+t(2,6,-3)$

Concept & Theory

Topic. Tangent planes to level surfaces and tangent lines to curves of intersection, via the gradient.

Key relations. For a level surface $F(x,y,z)=k$, $\nabla F$ at a point is normal to the surface there, giving the tangent plane $\nabla F(P)\cdot(\mathbf r-P)=0$. A curve lying in both $F_1=0$ and $F_2=0$ has tangent direction $\nabla F_1\times\nabla F_2$ at each point, since that direction is simultaneously perpendicular to both normals.

Why this works. Part (b) reuses the point and one of the two gradients from part (a) — $g=9$ is one of the two surfaces whose intersection curve is being examined, so $\nabla g(P)$ computed in (a) is exactly one of the two vectors crossed in (b). Recognizing this reuse saves a repeated computation.

Common pitfall. Recomputing $\nabla g$ from scratch in part (b) instead of reusing part (a)'s result; sign errors in the $3\times3$ determinant's middle ($\mathbf j$) cofactor, which carries an extra minus sign; forgetting to verify $P$ lies on both surfaces before differentiating.

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

Question 5 of 8: Tangent Plane and Tangent Line for Intersecting Surfaces

Question 5: Tangent Plane and Tangent Line for Intersecting Surfaces (a) 10, (b) 10 marks