3.4 Tangent Planes and Local Flatness

The last section measured two slopes at a point of a surface. This section makes the central claim of differential calculus: near a point where $f$ is differentiable, those two slopes are all you need, because the curved surface is locally indistinguishable from a plane. The plane through $(a, b, f(a,b))$ whose $x$ -slope is $f_x(a,b)$ and whose $y$ -slope is $f_y(a,b)$ is the tangent plane, the graph of the linear approximation

$L(x, y) = f(a, b) + f_x(a, b)\,(x - a) + f_y(a, b)\,(y - b).$

Claims deserve tests, and the only fair test of "locally flat" is to zoom in and look. The figure is built for exactly that: the zoom is performed by editing the plotted window, not by trusting a picture.

A green-to-yellow saddle-shaped surface z = x squared minus y squared, pierced by a half-transparent red-and-blue flat plane. The two touch at a single dark marked point P at (1, 0.5, 0.75), where the plane is tangent to the saddle. — The saddle $z = x^2 - y^2$ pierced by its half-transparent tangent plane $z = 2x - y - 0.75$ , the two touching at the single marked dark point $P = (1, 0.5, 0.75)$ . The camera is aimed at $P$ , not the origin, so shrinking both objects' plotted domains zooms the scene in on the point of tangency.

Explore

Orbit around $P$ . The plane is tangent at $P$ , yet away from $P$ it slices straight through the saddle. Find a region where the plane runs above the surface and one where it runs below.
Orbit until you sight edge-on along the plane, so it reads as a line. Near $P$ , does the gap between surface and plane open quickly or slowly as you move away?
Predict before acting: if the plotted window shrinks to $x$ from $0.8$ to $1.2$ and $y$ from $0.3$ to $0.7$ on both objects, will you still be able to tell which patch is the curved one?
Retype the saddle's domain fields to those values, then the plane's. Each edit regenerates instantly. Zoom in on the two small patches and orbit edge-on again. Can you tell them apart?
Push it: set both domains to $x$ from $0.95$ to $1.05$ , $y$ from $0.45$ to $0.55$ . At this window the saddle is its tangent plane, to the eye.
Widen both domains back to $-2$ , $2$ and watch curvature return. Zooming never changed the function — only the window. Moving away from $P$ , does the gap open fastest in the $x$ -direction, the $y$ -direction, or diagonally?

What differentiability means

Differentiability at $P$ is precisely the success of this experiment: the error $f(x,y) - L(x,y)$ shrinks faster than your distance to $P$ , so every step of zoom makes the surface look flatter, without limit. No other plane through the point survives the test — tilt the plane by any amount and the edge-on gap reappears at some zoom level. That is why the tangent plane is worth naming: near $P$ , evaluating the easy linear $L$ is as good as evaluating $f$ , and the next chapters spend the profit.

Example 1 Building the plane, then cashing it in

Find the tangent plane to $f(x, y) = x^2 - y^2$ at $P = (1, 0.5)$ , use it to estimate $f(1.1, 0.6)$ and $f(1.2, 0.5)$ , and compare with the true values.

The partials are $f_x = 2x$ and $f_y = -2y$ , so at $(1, 0.5)$ : $f = 0.75$ , $f_x = 2$ , $f_y = -1$ , and

$L(x, y) = 0.75 + 2(x - 1) - (y - 0.5) = 2x - y - 0.75,$

exactly the plane in the scene. Now estimate. First, $L(1.1, 0.6) = 2.2 - 0.6 - 0.75 = 0.85$ , while the true value is $f(1.1, 0.6) = 1.21 - 0.36 = 0.85$ — exactly equal. That is no accident: the error is $f - L = (x-1)^2 - (y-0.5)^2$ , itself a tiny saddle centered at $P$ , and it vanishes along the two diagonals through $P$ ; the point $(1.1, 0.6)$ sits on one. Second, $L(1.2, 0.5) = 2.4 - 0.5 - 0.75 = 1.15$ against the true $f(1.2, 0.5) = 1.44 - 0.25 = 1.19$ : an error of $0.04 = (0.2)^2$ , the square of the step. Quadratic error is the signature of a correct tangent plane — halve your distance to $P$ and the error drops by four.

\left[ 2 \cdot x,\; -2 \cdot y \right]

Both slice slopes of the saddle at once ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The two partials every tangent plane to the saddle is built from, computed symbolically: $f_x = 2x$ and $f_y = -2y$ . Evaluated at $P = (1, 0.5)$ they give the plane's slopes $2$ and $-1$ — the coefficients of $L(x, y) = 2x - y - 0.75$ .

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.