Tangent Plane to an Implicitly Defined Surface: 04-BS-1, May 2015, Q5

Question 5: Tangent Plane to an Implicitly Defined Surface (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. Implicit surface $G(x,y,z)=xy^2z^3-y-2=0$; point $(3,4,\tfrac12)$.

Find. The equation of the tangent plane at that point.

Approach. Confirm the point lies on the surface, compute the gradient $\nabla G=(G_x,G_y,G_z)$ (which is normal to the level surface $G=0$), and write the tangent plane through the point with that normal.

Confirm the point is on the surface. $xy^2z^3=3(4)^2\big(\tfrac12\big)^3=3(16)\big(\tfrac18\big)=6$, and $2+y=2+4=6$ — equal, so the point satisfies $G=0$.
Partial derivatives of $G$. $$G_x=y^2z^3,\qquad G_y=2xyz^3-1,\qquad G_z=3xy^2z^2.$$
Evaluate the gradient at $(3,4,\tfrac12)$. With $z^3=\tfrac18,\ z^2=\tfrac14$: $$G_x=16\cdot\tfrac18=2,\qquad G_y=2(3)(4)\big(\tfrac18\big)-1=3-1=2,\qquad G_z=3(3)(16)\big(\tfrac14\big)=36.$$
Assemble the tangent plane. $$G_x(x-3)+G_y(y-4)+G_z\big(z-\tfrac12\big)=0\ \Rightarrow\ 2(x-3)+2(y-4)+36\big(z-\tfrac12\big)=0.$$ Expanding: $2x+2y+36z-6-8-18=0\Rightarrow2x+2y+36z=32$. Dividing by $2$: $$\boxed{x+y+18z=16}.$$

Quantity	Result
$\nabla G$ at $(3,4,1/2)$	$(2,\,2,\,36)$
Tangent plane	$x+y+18z=16$

Concept & Theory

Topic. Tangent plane to a level surface $G(x,y,z)=0$ via the gradient.

Key relations. For a surface defined implicitly by $G(x,y,z)=0$, $\nabla G$ at any point on the surface is normal to it, so the tangent plane there is $\nabla G(x_0,y_0,z_0)\cdot\big(x-x_0,\,y-y_0,\,z-z_0\big)=0$.

Why this works. $\nabla G$ points in the direction of steepest increase of $G$, which is always orthogonal to the level sets of $G$ (curves/surfaces of constant $G$); since the surface here is the level set $G=0$, the gradient at any point on it is exactly the surface normal — no need to solve for $z$ explicitly as $z=f(x,y)$ first.

Common pitfall. Missing the product-rule cross term in $G_y=2xyz^3-1$ (both the explicit $-y$ and the $y^2$ inside the product contribute) — differentiating $xy^2z^3$ carefully term-by-term for each variable avoids this.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.6 Directional Derivatives and the Gradient Vector — tangent planes to level surfaces).

Question 5 of 8: Tangent Plane to an Implicitly Defined Surface

Question 5: Tangent Plane to an Implicitly Defined Surface (20 marks)