Tangent planes and linear approximations to surfaces: Question bank

Before you start, recall the two objects everything hangs on:

Recall The formula and the normal (peek only if needed)

Tangent plane at $(a, b)$ : $z = f (a, b) + f_{x} (a, b) (x - a) + f_{y} (a, b) (y - b)$ , with normal vector $n = ⟨ f_{x}, f_{y}, - 1 ⟩$ . The right-hand side is the linear approximation $L (x, y)$ .

True or false — justify

A surface with both partial derivatives existing at a point always has a tangent plane there.

False. Partials only measure slope along the

x

- and

y

-axes; a surface can be smooth along both axes yet crumpled diagonally, so you need full differentiability, not just existence of partials.

f

is differentiable at

(a, b)

then

f_{x}

and

f_{y}

exist there.

True. Differentiability is the stronger condition; it forces the linear approximation to match every direction, which in particular includes the two axis directions that define the partials.

f_{x}

and

f_{y}

are continuous near

(a, b)

, the tangent plane is guaranteed to exist.

True. Continuity of both partials near the point is a sufficient condition for differentiability, and differentiability is exactly what guarantees the plane hugs the surface.

The tangent plane and the surface touch only at the single point

(a, b, f (a, b))

False in general. They must touch and share slope there, but the plane can also cross or re-touch the surface elsewhere (e.g. a plane tangent at one point of a wavy surface may hit it again far away).

The linear approximation

L (x, y)

is a plane, so it is valid for all

(x, y)

False.

L

is defined everywhere as an equation, but it only approximates

f

well near

(a, b)

; the error grows roughly like the square of the distance, so far away it is worthless.

Every point of a smooth surface has exactly one tangent plane.

True for a graph $z = f (x, y)$ that is differentiable there — the two matched slopes

f_{x}, f_{y}

pin down a unique non-vertical plane.

If the tangent plane is horizontal at

(a, b)

, then both partials are zero there.

True. A horizontal plane has zero slope in every direction, so in particular

f_{x} (a, b) = f_{y} (a, b) = 0

— this is exactly a critical point.

The differential

d z

always equals the true change

Δ z

False.

d z = f_{x} d x + f_{y} d y

is the tangent-plane prediction; the true change

Δ z = f (x, y) - f (a, b)

equals

d z

only in the limit of infinitely small steps.

Two different points on a surface can share the same tangent plane.

True. For example a flat region, or a cylinder, or symmetric points of a bump can all have parallel or even identical tangent planes.

Spot the error

"Tangent plane:

z = f (a, b) + f_{x} (a, b) x + f_{y} (a, b) y

The slope terms must multiply the displacement from the base point, not the raw coordinates: it should be

(x - a)

and

(y - b)

. As written it doesn't even pass through

(a, b, f (a, b))

"The normal vector to

z = f (x, y)

⟨ f_{x}, f_{y}, 1 ⟩

The

z

-component is

- 1

, not

+ 1

: writing the surface as

F = f (x, y) - z = 0

gives

\nabla F = ⟨ f_{x}, f_{y}, - 1 ⟩

. The sign flip matters for orientation (it points generally downward in

z

"To find the tangent plane I set

A = f_{x} (x, y)

using the current point

(x, y)

The slopes are frozen at the base point:

A = f_{x} (a, b)

, a number, not a function of the moving

(x, y)

. A plane has constant slopes.

f

is continuous at

(a, b)

, so it's differentiable there and has a tangent plane."

Continuity is far too weak; a cone tip (

x^{2} + y^{2}

at the origin) is continuous but has no tangent plane. You need the partials to exist and the error

f - L

to vanish faster than distance.

"Since both partials of

f = \frac{x y}{x ^{2} + y ^{2}}

(extended by

0

) exist at the origin, its tangent plane is

z = 0

The partials exist and are zero, but

f

isn't even continuous at the origin (it takes value

\frac{1}{2}

along

y = x

), so it is not differentiable and has no tangent plane — the naive

z = 0

is a false conclusion.

"The linear approximation of

x^{2} + y^{2}

at the origin is fine because the function is nice everywhere else."

At the origin this surface is a cone point — the partials don't exist there, so no tangent plane and no linear approximation exists at that point, regardless of behaviour elsewhere.

Why questions

Why must the tangent plane contain both the

x

-cross-section tangent line and the

y

-cross-section tangent line?

Because two distinct lines through a common point uniquely determine a plane; matching both partial-slope lines is exactly what forces the plane to share the surface's slope in every direction (by linearity of directional slopes).

Why does linear approximation error grow like the square of the distance, not linearly?

Differentiability kills the first-order (linear) part of the error by construction, so the leading leftover term is second-order — governed by curvature — hence it scales like distance squared.

Why do we write the surface as the level set

F (x, y, z) = f (x, y) - z = 0

to get the normal?

On a level surface the gradient

\nabla F

is always perpendicular to it, so

\nabla F = ⟨ f_{x}, f_{y}, - 1 ⟩

hands us the normal directly, and the tangent plane is everything perpendicular to that normal.

Why is the tangent-plane idea just the 2D twin of the single-variable tangent line?

Both replace a curved object by its best flat local match: [[Tangent line and linear approximation (single variable)|

y \approx f (a) + f^{'} (a) (x - a)

]] uses one slope, the plane uses two slopes

f_{x}, f_{y}

— same "value plus slope times step" recipe, one extra direction.

Why does the multivariable chain rule depend on the tangent-plane/linear-approximation idea?

The chain rule computes how

f

changes as its inputs move; near a point that change is captured by the linear approximation

d z = f_{x} d x + f_{y} d y

, which is precisely the tangent-plane's predicted change.

Why isn't "slope in the

x

-direction and

y

-direction" enough to guarantee smoothness?

Those are just two of infinitely many directions; a function can behave linearly along the axes but blow up along a diagonal, so probing only two directions misses that failure.

Edge cases

A surface has a sharp cone tip at

(a, b)

(like

x^{2} + y^{2}

at the origin). Does a tangent plane exist?

No. At a cone tip the slope depends on which direction you approach, so no single plane matches all directions — the partials fail to exist there.

At a saddle point both partials are zero. Is the tangent plane horizontal, and does the surface stay below it nearby?

The plane is horizontal (

f_{x} = f_{y} = 0

), but the surface rises on one side and falls on the other, so it does not stay on one side — horizontal tangent plane does not mean local max or min.

Consider a vertical wall like

x = c

. Can it be written as a tangent plane

z = f (a, b) + \dots

No. The form

z = f (a, b) + A (x - a) + B (y - b)

only produces non-vertical planes (finite slopes); vertical tangent planes escape this framework entirely.

f_{x} (a, b) = f_{y} (a, b) = 0

, is the linear approximation useless?

Not useless — it correctly predicts

L (x, y) = f (a, b)

, i.e. "flat to first order." It just means the interesting behaviour is second-order (curvature), so first-order gives a constant estimate.

For a genuinely flat surface (a plane itself), how good is the tangent-plane approximation?

Exact everywhere. A plane is its own linear approximation, the error

E

is identically zero, and

d z = Δ z

for any step, near or far.

What happens to the linear-approximation error at the base point

(a, b)

itself?

It is exactly zero:

L (a, b) = f (a, b)

by construction, since all displacement terms

(x - a), (y - b)

vanish there.

Connections

Partial derivatives — the traps about "slope in only two directions."
Differentiability of multivariable functions — the heart of every "partials aren't enough" item.
The gradient vector — the normal-vector sign trap.
Directional derivatives — why two axis slopes force all directions.
Tangent line and linear approximation (single variable) — the 1D mirror.
The chain rule (multivariable) — why the linear-approximation engine matters downstream.