Tangent planes and linear approximations to surfaces: Visual walkthrough

Before anything: a tiny dictionary, so nothing is a mystery.

Step 1 — Zoom in until the hill looks flat

WHAT. Take any smooth hill and zoom your camera into one spot. As you zoom, the curviness melts away and the patch under you looks like a flat, tilted board.

WHY. This flat board is the whole idea of a tangent plane. If we can find the equation of this board, we can use it as a cheap stand-in for the real hill near that spot. Finding the board is the entire goal of the page — everything below is just pinning down its exact tilt.

PICTURE. On the left the surface bulges. On the right (zoomed in), the same patch is nearly a plane. The red dot is our base camp $(a, b, f (a, b))$ .

Figure — Tangent planes and linear approximations to surfaces

Step 2 — Slice the hill two ways to read its two slopes

WHAT. Cut the hill with two vertical walls through base camp:

Wall A: freeze $y = b$ , let $x$ move. The hill's edge along this wall is a curve in the $x$ -direction.
Wall B: freeze $x = a$ , let $y$ move. That edge is a curve in the $y$ -direction.

WHY. A tilted board has exactly two independent tilts: left–right and front–back. Each slice reveals one of them. Slicing turns a scary 3D question ("how does the hill tilt?") into two friendly 1D questions ("what's the slope of this curve?"), which we already know how to answer with an ordinary derivative.

PICTURE. The blue curve lives in Wall A (moving in $x$ ); the green curve lives in Wall B (moving in $y$ ). Both pass through the red base-camp point.

Step 3 — Draw the tangent line inside each slice

WHAT. In each slice, replace the slice-curve by its tangent line at base camp. Blue slice → blue tangent line; green slice → green tangent line.

WHY. This is the 1D linear approximation we trust: $predicted height y \approx height at base f (a) + slope f^{'} (a) step taken (x - a) .$ Each symbol: $f (a)$ is where you start, $f^{'} (a)$ is how fast height changes per unit step, and $(x - a)$ is how far you stepped from base. Multiply slope by step, add to the start — you get the predicted height. We're just going to do this twice and glue the results.

PICTURE. Two straight tangent lines, one per slice, kissing each curve exactly at the red dot.

Step 4 — Write the most general non-vertical plane

WHAT. Every non-vertical plane through the point $(a, b, z_{0})$ can be written $z = z_{0} + A (x - a) + B (y - b) .$

WHY. We don't yet know the tilts, so we leave them as unknown letters $A$ and $B$ and solve for them using the two slices. Let's read every symbol:

$z_{0}$ ::: the plane's height directly above base camp. When $x = a$ and $y = b$ , the last two terms vanish and $z = z_{0}$ .
$A$ ::: the plane's tilt in the $x$ -direction (its own $x$ -slope).
$(x - a)$ ::: how far you've stepped in $x$ away from base camp.
$B$ ::: the plane's tilt in the $y$ -direction.
$(y - b)$ ::: how far you've stepped in $y$ away from base camp.

PICTURE. A flat tilted sheet floating above the floor, its height-above-base $z_{0}$ marked, its two step-arrows $(x - a)$ and $(y - b)$ drawn on the floor.

Step 5 — Force the plane to touch the hill (match the height)

WHAT. Demand that the plane and the hill agree at base camp: set $z_{0} = f (a, b)$ .

WHY. A tangent plane must touch the surface, not float above or below it. At the exact spot $(a, b)$ the plane's height is $z_{0}$ and the hill's height is $f (a, b)$ ; touching means they're equal. So the unknown $z_{0}$ is now known.

Now the plane reads $z = f (a, b) + A (x - a) + B (y - b) .$

PICTURE. The floating sheet slides down until its base-camp point sits exactly on the red surface point — contact made.

Step 6 — Match the $x$ -tilt: solve for $A$

WHAT. Walk along Wall A only (freeze $y = b$ ). On the plane this collapses to $z = f (a, b) + A (x - a),$ a straight line of slope $A$ . For the plane to hug the hill, this must equal the blue slice's tangent, whose slope is $f_{x} (a, b)$ . Matching slopes gives $A = f_{x} (a, b)$ .

WHY. In Wall A the $(y - b)$ term is zero (since $y = b$ ), so $B$ disappears and only $A$ survives. That isolates $A$ perfectly — walls do the algebra for us. The plane's $x$ -slope must equal the hill's $x$ -slope or the sheet would peel away from the blue curve.

PICTURE. The plane, sliced by Wall A, shows one straight line lying on top of the blue tangent line — same steepness $A = f_{x}$ .

Step 7 — Match the $y$ -tilt: solve for $B$

WHAT. Mirror Step 6 with Wall B (freeze $x = a$ ). The plane collapses to $z = f (a, b) + B (y - b)$ , slope $B$ , which must equal the green slice's slope $f_{y} (a, b)$ . So $B = f_{y} (a, b)$ .

WHY. Same trick, other direction: freezing $x = a$ kills the $A$ term, isolating $B$ . The plane's front–back tilt must equal the hill's front–back tilt.

Substituting both solved letters:

PICTURE. Both walls at once: the plane lies on the blue tangent (slope $f_{x}$ ) and the green tangent (slope $f_{y}$ ) — fully pinned down.

The slopes $f_{x}, f_{y}$ come from Partial derivatives; the vector $⟨ f_{x}, f_{y}, - 1 ⟩$ that stands perpendicular to this plane is the The gradient vector view.

Step 8 — The degenerate & edge cases (never get surprised)

WHAT. Three "what if" situations the formula must survive:

Flat spot (both tilts zero). If $f_{x} (a, b) = f_{y} (a, b) = 0$ — a summit, a valley floor, or a saddle — the formula becomes $z = f (a, b)$ : a perfectly horizontal plane. Correct: on a hilltop the flat sticker lies level.
Vertical tangent (formula refuses to exist). For a cone $f = x^{2} + y^{2}$ at the tip $(0, 0)$ , the slopes blow up (the point is sharp, not smooth). Zooming in never flattens a needle-tip, so no tangent plane exists — and our "non-vertical plane" form $z = z_{0} + A (x - a) + B (y - b)$ honestly can't represent a vertical wall. The method warns you by failing.
Smooth along walls but wild diagonally. Partials can exist yet the surface still won't flatten in every direction (e.g. $f = \frac{x y}{x ^{2} + y ^{2}}$ near the origin). Our slices only checked two directions; a rogue diagonal can betray us. That's why the plane genuinely fits only when $f$ is differentiable, not merely sliceable.

WHY show these. Because a reader who only saw the happy case would trust the formula everywhere. Each edge case marks a boundary of validity — see Differentiability of multivariable functions for the precise error-shrinks-faster-than-distance test, and Directional derivatives for slopes in the diagonal directions the walls missed.

PICTURE. Three mini-panels: a level cap on a dome, a spike with no plane, and a twisty saddle-ish surface where axis-slices lie but the diagonal doesn't.

The one-picture summary

Everything at once: the hill, base camp, the two slices with their tilts $f_{x}$ and $f_{y}$ , and the tangent plane resting on both tangent lines. The plane's height above any nearby floor point is the linear approximation $L (x, y)$ .

Recall Feynman retelling — the whole walk in plain words

I stood on a chocolate hill and zoomed my eyes in until the ground looked like a flat tilted board (Step 1). To learn how the board should tilt, I sliced the hill with two glass walls: one front-to-back, one left-to-right (Step 2). Each slice showed a curve, and I drew the straight tangent line touching each curve at my feet (Step 3). Two crossing lines fix one flat sheet. So I wrote the most general flat sheet, leaving its two tilts as blanks $A$ and $B$ (Step 4). I slid the sheet down until it touched the hill under my feet — that set its height (Step 5). Then I demanded the sheet lean exactly like the left-right slice: that filled in $A = f_{x}$ (Step 6). Same for front-back: $B = f_{y}$ (Step 7). Gluing it together: height, plus each tilt times each step — the tangent plane. Finally I checked the weird spots: on a summit the sheet is level, on a needle-tip there's no sheet at all, and on a sneaky twisty surface the two slices can lie to me — which is exactly why "differentiable" (not just "has slopes") is the password (Step 8).

Connections

Parent topic — Tangent planes
Partial derivatives — where the tilts $f_{x}, f_{y}$ come from.
The gradient vector — the normal $⟨ f_{x}, f_{y}, - 1 ⟩$ perpendicular to our plane.
Differentiability of multivariable functions — the password behind Step 8.
Tangent line and linear approximation (single variable) — the 1D line we glued twice.
The chain rule (multivariable) — runs on this linear-approximation engine.
Directional derivatives — slopes in the diagonal directions our two walls missed.