Questions: Level Sets and Surfaces in 3D

Question 1

For the function f(x, y, z) = x² + y² + z², what geometric object is the level set f = 9?

• A. A point at the origin
• B. A sphere of radius 9 centered at the origin
• C. A sphere of radius 3 centered at the origin
• D. A circle of radius 3 in the xy-plane

Answer: (C)

The level set f = 9 is the set of all (x, y, z) satisfying x² + y² + z² = 9, which is a sphere of radius √9 = 3, not 9. This is a common mistake: the level value k is not the radius — the radius is √k. A level set lives in the *domain* space (here ℝ³), not the output space, and takes the shape determined by the equation f = k.

Question 2

At a point P on a level surface f(x, y, z) = k, in which direction does the gradient vector ∇f(P) point?

• A. Along the surface — tangent to the level surface at P
• B. Perpendicular to the surface — normal to the level surface at P
• C. Toward the nearest critical point of f
• D. In the direction of decreasing f

Answer: (B)

The gradient ∇f points in the direction of maximum increase of f. Moving along the level surface keeps f constant, so the directional derivative along any tangent direction is zero — meaning all tangent vectors are orthogonal to ∇f. Therefore ∇f is perpendicular (normal) to the level surface at P. This is the key geometric fact that makes the gradient essential: it always points away from (perpendicular to) the level surfaces of f, which is why the tangent plane to the level surface is exactly the plane perpendicular to ∇f(P).

Question 3

For a function f: ℝ² → ℝ, a level set f(x, y) = k is typically a single point in the plane.

• T. True
• F. False

Answer: False

A level set f(x, y) = k is typically a curve in the plane — the set of all (x, y) satisfying the equation. For example, f(x, y) = x² + y² = 4 gives a circle; f(x, y) = x + y = 1 gives a line. A single point would only arise in degenerate cases, such as f(x, y) = x² + y² = 0 (giving just the origin). The collection of level curves for all values of k forms the contour map used in topographic charts.

Question 4

The tangent plane to the level surface f(x, y, z) = k at a point P is the plane through P perpendicular to ∇f(P).

• T. True
• F. False

Answer: True

This follows directly from the gradient's perpendicularity to level surfaces. Any vector v tangent to the surface at P satisfies ∇f(P)·v = 0 (moving along the surface keeps f constant). The set of all such tangent vectors forms the tangent plane, and the plane is characterized by being perpendicular to ∇f(P). This gives a practical method for finding tangent planes to implicitly defined surfaces: compute the gradient and use it as the normal vector.

Question 5

Why is the implicit representation f(x, y, z) = k more flexible than the explicit form z = g(x, y) for describing surfaces? Give an example of a surface that illustrates this advantage.

Model answer: The explicit form z = g(x, y) requires solving for z as a single-valued function of x and y, which fails for surfaces where a vertical line hits more than one point. The implicit form f(x, y, z) = k imposes an equation without singling out any variable, allowing it to represent such surfaces. Example: the sphere x² + y² + z² = 1. For the upper hemisphere, z = √(1 − x² − y²), and for the lower hemisphere, z = −√(1 − x² − y²), requiring two explicit functions. The single equation x² + y² + z² = 1 captures the whole sphere.

The implicit representation is also more natural for level sets, since they are defined by fixing the output value of f — no algebraic manipulation needed. Other surfaces that can't be globally expressed as z = g(x, y): the torus, cylinders oriented along any axis, and any closed surface. This is why implicit forms appear throughout geometry, physics (equipotential surfaces, wavefronts), and machine learning (decision boundaries).