Questions: Parametric Surfaces and Tangent Vectors

Question 1

For the cylinder r(u, v) = ⟨cos u, sin u, v⟩, the tangent vectors are r_u = ⟨−sin u, cos u, 0⟩ and r_v = ⟨0, 0, 1⟩, giving r_u × r_v = ⟨cos u, sin u, 0⟩ with magnitude 1. What does this magnitude of 1 represent geometrically?

• A. The curvature of the cylinder surface at that point is 1
• B. The local area scaling factor — a small rectangle of area du·dv in the parameter domain maps to a surface patch with the same area 1·du·dv
• C. The angle between r_u and r_v is 1 radian
• D. The speed at which the parametrization traces the surface as u and v increase simultaneously

Answer: (B)

The surface area element is dS = |r_u × r_v| du dv. When |r_u × r_v| = 1, a unit square in parameter space maps to a unit-area patch on the surface — the standard cylindrical parametrization is an isometry (area-preserving). For more complex surfaces such as a sphere, the magnitude varies with position: near the poles of a spherical parametrization |r_φ × r_θ| = sin φ, which shrinks to zero at the poles, correctly reflecting that the parameter grid compresses to a point there.

Question 2

Why are r_u and r_v called 'tangent vectors' to the surface at a point r(u₀, v₀)?

• A. They point in the direction of steepest ascent along the surface at that point
• B. They are velocity vectors along the coordinate curves on the surface: holding v = v₀ and varying u traces a curve whose tangent vector is r_u = ∂r/∂u
• C. They are always perpendicular to each other, forming a natural orthonormal frame
• D. They equal the gradient of the height function at each point when the surface is expressed as z = g(x, y)

Answer: (B)

Holding v = v₀ fixed in r(u, v) traces a curve on the surface as u varies; the velocity of this curve is dr/du = r_u. Similarly, r_v is the velocity along the u = u₀ coordinate curves. These two vectors lie in the tangent plane and serve as its natural basis vectors — they are tangent to the two families of coordinate curves on the surface, just as the columns of a Jacobian span the tangent space for a coordinate change. They are generally not perpendicular unless the parametrization is orthogonal.

Question 3

The cross product n = r_u × r_v is always perpendicular to the tangent plane of the surface at the corresponding point.

• T. True
• F. False

Answer: True

The cross product of two vectors is perpendicular to both by definition. Since r_u and r_v lie in the tangent plane, their cross product is automatically perpendicular to the tangent plane — it is the surface normal. This is not a formula to memorize but a consequence of the cross product's geometric meaning. The normal's direction encodes orientation (which side of the surface faces outward), and its magnitude encodes local area scaling, making it the key quantity for all surface integral calculations.

Question 4

For a surface parametrized as r(u, v), the surface area element is simply dS = du dv — the same as an area element in the flat parameter plane.

• T. True
• F. False

Answer: False

The surface area element is dS = |r_u × r_v| du dv. The magnitude |r_u × r_v| accounts for how the parametrization stretches or compresses area as it maps from the flat parameter domain into 3D space. For a flat surface r(u, v) = ⟨u, v, 0⟩, we get |r_u × r_v| = 1, recovering dS = du dv as a special case. But for a curved surface this factor varies with position, and omitting it gives an incorrect answer for surface area — just as omitting the Jacobian gives incorrect answers in a change of variables for double integrals.

Question 5

Explain why |r_u × r_v| — rather than, say, |r_u|·|r_v| — is the correct surface area scaling factor when setting up surface integrals.

Model answer: A small parameter rectangle [u, u+du] × [v, v+dv] maps to a parallelogram on the surface with edge vectors r_u·du and r_v·dv. The area of a parallelogram with edges **a** and **b** is |**a** × **b**|, which equals |**a**||**b**| sin θ where θ is the angle between them. Using |r_u|·|r_v| would assume θ = 90°, which is only correct when r_u and r_v are perpendicular (an orthogonal parametrization). The cross product magnitude correctly accounts for the angle between the tangent vectors and gives the true area of the surface patch regardless of parametrization.

This is the direct 3D generalization of the 2D Jacobian story. In 2D, a change of variables (u,v) → (x,y) scales area by |det J| = area of the parallelogram formed by the Jacobian columns. Here, the Jacobian is a 3×2 matrix with columns r_u and r_v, and the generalized 'determinant' is |r_u × r_v|, the magnitude of the cross product. It equals √(det(JᵀJ)), the Gram determinant, and captures the area distortion of the map from the parameter plane into 3D space.