Questions: Tangent Planes to Surfaces

Question 1

For the surface z = f(x,y), which vector is normal to the tangent plane at the point (x₀, y₀, z₀)?

• A. ⟨f_x, f_y⟩ — the 2D gradient vector evaluated at (x₀, y₀)
• B. ⟨f_x, f_y, −1⟩ — the gradient components plus −1 as the z-component
• C. ⟨f_x, f_y, 1⟩ — the gradient components plus +1 as the z-component
• D. ⟨−f_x, −f_y, 0⟩ — the negative gradient in the xy-plane

Answer: (B)

The tangent plane equation f_x(x−x₀) + f_y(y−y₀) − (z−z₀) = 0 is in the form n·⟨x−x₀, y−y₀, z−z₀⟩ = 0, so the normal vector is the coefficient vector n = ⟨f_x, f_y, −1⟩. The −1 appears because z enters with a coefficient of −1 when the equation is rearranged. Option A is the most tempting wrong answer: the 2D gradient ∇f = ⟨f_x, f_y⟩ lives in the xy-plane and lies IN the tangent plane (it is not the normal to it). Confusing the 2D gradient with the 3D normal is the central misconception of this topic.

Question 2

For an implicitly defined surface F(x, y, z) = c, which vector is normal to the surface at a point?

• A. The 2D gradient ⟨F_x, F_y⟩ evaluated on the surface
• B. The 3D gradient ∇F = ⟨F_x, F_y, F_z⟩
• C. The unit tangent vector to any curve lying on the surface
• D. The Hessian matrix of F at the point

Answer: (B)

Because F is constant (= c) along the surface, any tangent direction v to the surface satisfies ∇F · v = 0. This means ∇F is perpendicular to every tangent vector, making it the normal. This unifies the explicit and implicit cases: for z = f(x,y), define F = f(x,y)−z, so ∇F = ⟨f_x, f_y, −1⟩ — recovering the explicit formula. The Hessian is a matrix of second derivatives and does not in general point normal to the surface.

Question 3

The 2D gradient vector ∇f = ⟨f_x, f_y⟩ is the normal vector to the tangent plane of the surface z = f(x,y).

• T. True
• F. False

Answer: False

The 2D gradient lives in the xy-plane and actually lies IN the tangent plane (projected down). The normal to the tangent plane in 3D is ⟨f_x, f_y, −1⟩ — which includes the −1 z-component arising from the z term in the plane equation. Confusing the 2D gradient with the 3D normal is the most common error in this topic. The gradient ∇f encodes the slope in each horizontal direction; the normal must also account for the z-direction.

Question 4

The tangent plane to z = f(x,y) at a point (x₀, y₀, z₀) contains the tangent line to the cross-sectional curve obtained by fixing y = y₀.

• T. True
• F. False

Answer: True

Setting y = y₀ in the tangent plane equation z − z₀ = f_x(x−x₀) + f_y(y−y₀) gives z − z₀ = f_x(x−x₀), which is exactly the tangent line in the xz-plane at y = y₀. Similarly, fixing x = x₀ gives the tangent line in the yz-plane. The tangent plane is precisely the unique plane that contains both of these tangent lines simultaneously — it is the natural 3D generalization of the 1D tangent line.

Question 5

Explain why the normal vector to the tangent plane of z = f(x,y) has −1 as its z-component, rather than +1 or some other value.

Model answer: Rearranging the tangent plane equation: f_x(x−x₀) + f_y(y−y₀) − (z−z₀) = 0. Written as a dot product n·⟨x−x₀, y−y₀, z−z₀⟩ = 0, the normal vector is the coefficient vector n = ⟨f_x, f_y, −1⟩. The −1 appears because z appears on the right side of z − z₀ = f_x(x−x₀) + f_y(y−y₀), so when moved to the left it becomes −(z−z₀), giving a coefficient of −1 for z.

This can also be understood from the implicit formulation: define F(x,y,z) = f(x,y) − z. Then the surface z = f(x,y) is the level set F = 0, and ∇F = ⟨f_x, f_y, −1⟩. The −1 comes from ∂F/∂z = −1. The sign matters: it points 'upward' out of the surface in the sense that increasing z means increasing z−f(x,y), so the outward normal has a negative z-component when the surface is a graph over the xy-plane.