Tangent planes and linear approximations to surfaces: Worked examples

Where the one formula comes from

Everything on this page rests on a single equation. Before we use it, let us build it, so nothing is a mystery later.

Step 1 — start from the 1-D idea. In one variable the tangent line is $y \approx f (a) + f^{'} (a) (x - a)$ : height, plus slope times step. We want the 2-D twin.

Step 2 — write the most general non-vertical plane through the base point. $z = f (a, b) + A (x - a) + B (y - b)$ Why this form? Setting $x = a, y = b$ forces $z = f (a, b)$ , so it passes through the base point automatically. The unknowns $A, B$ are the plane's slopes in the $x$ - and $y$ -directions — we now pick them to match the surface.

Step 3 — match the x-slice. Freeze $y = b$ . The plane collapses to $z = f (a, b) + A (x - a)$ , a line of slope $A$ . The surface's x-slice has slope $f_{x} (a, b)$ . For the plane to hug the surface here, $A = f_{x} (a, b)$ .

Step 4 — match the y-slice. Freeze $x = a$ the same way: $B = f_{y} (a, b)$ .

Figure — Tangent planes and linear approximations to surfaces

Now that we know why the formula holds, we spend the rest of the page exhausting the cases.

The scenario matrix

Every tangent-plane problem lands in one of these cells. Each example below is tagged with the cell it fills.

#	Cell (what makes it different)	Example
A	Both slopes positive — plain paraboloid	Ex 1
B	Slopes of mixed sign (one $+$ , one $-$ )	Ex 2
B $^{'}$	Both slopes negative — completing the four-quadrant sign story	Ex 2b
C	A zero slope — flat in one direction, tilted in the other	Ex 3
D	Degenerate surface: $f$ is already a plane (approximation is exact)	Ex 4
E	Breakdown case: partials exist but surface is NOT differentiable (no tangent plane)	Ex 5
F	Real-world word problem — differentials for measurement error	Ex 6
G	Exam twist: given an implicit level surface $F (x, y, z) = 0$ , use the normal vector	Ex 7
H	Limiting behaviour: error grows like distance $^{2}$ — quantifying trust	Ex 8

Example 1 — Cell A: both slopes positive

Worked example Paraboloid

$f (x, y) = x^{2} + y^{2}$ at $(1, 2)$ , estimate $f (1.05, 2.1)$ Forecast: Both partials $2 x, 2 y$ are positive at $(1, 2)$ , so stepping in $+ x$ and $+ y$ should both raise the estimate above $f (1, 2) = 5$ . Guess: a bit above $5$ .

Here the base point is $(a, b) = (1, 2)$ .

Step 1 — height. $f (1, 2) = 1 + 4 = 5$ . Why this step? The plane must touch the surface at the base point; that height is the constant term.

Step 2 — slopes. $f_{x} = 2 x \Rightarrow f_{x} (1, 2) = 2$ ; $f_{y} = 2 y \Rightarrow f_{y} (1, 2) = 4$ . Why this step? These are the two tilts the plane must match.

Step 3 — steps. $x - a = 1.05 - 1 = 0.05$ ; $y - b = 2.1 - 2 = 0.1$ . Why this step? The formula multiplies each slope by the offset from the base point, not by $x$ or $y$ themselves.

Step 4 — assemble. $L = 5 + 2 (0.05) + 4 (0.1) = 5 + 0.1 + 0.4 = 5.5$ Why this step? We are simply walking along the flat plane: start at the height, then add each slope times its matching step. That sum IS the plane's height above $(1.05, 2.1)$ — our estimate of the surface.

Verify: True value $f (1.05, 2.1) = 1.1025 + 4.41 = 5.5125$ . Error $= 0.0125$ , tiny — and above $5$ as forecast. ✓

Figure s02 shows the flat red patch touching the bowl at $(1, 2, 5)$ , tilting upward both ways.

Example 2 — Cell B: mixed-sign slopes

Worked example Saddle

$f (x, y) = x^{2} - y^{2}$ at $(2, 1)$ , estimate $f (2.1, 0.9)$ Forecast: $f_{x} = 2 x > 0$ but $f_{y} = - 2 y < 0$ . Stepping $+ x$ raises, stepping $- y$ also raises (negative slope times negative step). Guess: clearly above $f (2, 1) = 3$ .

Base point $(a, b) = (2, 1)$ .

Step 1 — height. $f (2, 1) = 4 - 1 = 3$ . Why this step? Constant term = surface height at the base point.

Step 2 — slopes. $f_{x} = 2 x \Rightarrow 4$ ; $f_{y} = - 2 y \Rightarrow - 2$ . Why this step? This is a saddle: it curves up in $x$ , down in $y$ , so the two slopes carry opposite signs. Getting a sign wrong here is the classic error.

Step 3 — steps. $x - 2 = 0.1$ ; $y - 1 = - 0.1$ . Why this step? Same offset rule as before — but note the $y$ -step is negative because we moved from $y = 1$ down to $0.9$ . The sign of the step matters as much as the sign of the slope.

Step 4 — assemble. $L = 3 + 4 (0.1) + (- 2) (- 0.1) = 3 + 0.4 + 0.2 = 3.6$ Why this step? Height plus slope-times-step, twice. The negative slope times the negative step gives a positive contribution — this is exactly why the estimate lands above $3$ , matching the forecast.

Verify: True $f (2.1, 0.9) = 4.41 - 0.81 = 3.6$ . Error $= 0$ to two decimals — above $3$ as forecast. ✓

Figure s03 shows the saddle and its tilted tangent plane — the red plane cuts through the surface, not just resting on top.

Example 2b — Cell B $^{'}$ : both slopes negative

Worked example

$f (x, y) = 10 - x^{2} - y^{2}$ (downward bowl) at $(1, 1)$ , estimate $f (1.1, 1.2)$ Forecast: $f_{x} = - 2 x < 0$ and $f_{y} = - 2 y < 0$ — both slopes negative. Stepping $+ x$ and $+ y$ (both positive steps) should lower the estimate below $f (1, 1) = 8$ . This completes the four-quadrant sign story: we have now seen $(+, +)$ , $(+, -)$ , and $(-, -)$ .

Base point $(a, b) = (1, 1)$ .

Step 1 — height. $f (1, 1) = 10 - 1 - 1 = 8$ . Why this step? The plane touches the dome at the base point; this height is the constant term.

Step 2 — slopes. $f_{x} = - 2 x \Rightarrow - 2$ ; $f_{y} = - 2 y \Rightarrow - 2$ . Why this step? This is an upside-down paraboloid, so it slopes downward in both directions — both partials are negative.

Step 3 — steps. $x - 1 = 0.1$ ; $y - 1 = 0.2$ . Why this step? Both steps are positive (we moved to larger $x$ and $y$ ); the negative slopes will pull the estimate down.

Step 4 — assemble. $L = 8 + (- 2) (0.1) + (- 2) (0.2) = 8 - 0.2 - 0.4 = 7.4$ Why this step? Height plus slope-times-step, twice. Two negative contributions drag the estimate below $8$ , exactly as the downward dome demands.

Verify: True $f (1.1, 1.2) = 10 - 1.21 - 1.44 = 7.35$ . Error $= 0.05$ , small — and below $8$ as forecast. ✓

Example 3 — Cell C: a zero slope

Worked example

$f (x, y) = sin x + y^{2}$ at $(\frac{π}{2}, 1)$ , estimate $f (\frac{π}{2} + 0.1, 1.1)$ Forecast: $f_{x} = cos x$ , and $cos \frac{π}{2} = 0$ . At the top of the sine wave the surface is momentarily flat in the $x$ -direction, so the $x$ -step contributes (almost) nothing. Guess: change comes mostly from $y$ .

Base point $(a, b) = (\frac{π}{2}, 1)$ .

Step 1 — height. $f (\frac{π}{2}, 1) = sin \frac{π}{2} + 1 = 1 + 1 = 2$ . Why this step? The constant term is the surface height at the base point — the level from which both slope contributions are measured.

Step 2 — slopes. $f_{x} = cos x \Rightarrow cos \frac{π}{2} = 0$ ; $f_{y} = 2 y \Rightarrow 2$ . Why this step? A zero partial is legal and important: it means the tangent plane is level along that axis. Zero is a slope, not a missing one.

Step 3 — steps. $x - a = 0.1$ ; $y - b = 0.1$ . Why this step? Offsets from the base point, one per input. Even though the $x$ -slope is $0$ , we still record its step honestly — the formula will multiply it by zero for us.

Step 4 — assemble. $L = 2 + 0 (0.1) + 2 (0.1) = 2 + 0 + 0.2 = 2.2$ Why this step? Height plus slope-times-step, twice. The zero $x$ -slope kills the $x$ -contribution, so the whole change comes from the $y$ -term — the flat-in- $x$ picture made numerical.

Verify: True $f = sin (1.6708) + 1.21 = 0.99500 + 1.21 = 2.20500$ . The tiny $0.005$ gap is exactly the curvature the flat $x$ -slope ignored. Forecast confirmed: nearly all the change came from $y$ . ✓

Example 4 — Cell D: degenerate surface (already a plane)

Worked example

$f (x, y) = 3 x - 2 y + 7$ at $(4, 5)$ , estimate $f (4.3, 4.6)$ Forecast: The surface is a flat plane. The tangent plane of a plane is the plane itself — so the "approximation" should be exact, error zero, no matter how far we step.

Base point $(a, b) = (4, 5)$ .

Step 1 — height. $f (4, 5) = 12 - 10 + 7 = 9$ . Why this step? Same as always — the constant term is the surface value at the base point; nothing about a flat surface changes this rule.

Step 2 — slopes. $f_{x} = 3$ , $f_{y} = - 2$ — constant everywhere. Why this step? For a linear function the partials don't depend on the point; the plane never curves.

Step 3 — steps. $x - 4 = 0.3$ ; $y - 5 = - 0.4$ . Why this step? Offsets from the base point as usual; here the $y$ -step is negative because we dropped from $y = 5$ to $4.6$ .

Step 4 — assemble. $L = 9 + 3 (0.3) + (- 2) (- 0.4) = 9 + 0.9 + 0.8 = 10.7$ Why this step? Height plus slope-times-step, twice. Because the true surface has no curvature, this "walk along the plane" reproduces the surface exactly, not approximately.

Verify: True $f (4.3, 4.6) = 12.9 - 9.2 + 7 = 10.7$ . Error exactly $0$ — the degenerate case where linear approximation is not an approximation at all. ✓

Example 5 — Cell E: breakdown (no tangent plane)

Worked example Does

$f (x, y) = \frac{x y}{x ^{2} + y ^{2}}$ (with $f (0, 0) = 0$ ) have a tangent plane at the origin? Forecast: The parent's warning says "partials existing is not enough." Let's watch it fail.

Base point $(a, b) = (0, 0)$ .

Step 1 — compute the partials at the origin. Along the $x$ -axis ( $y = 0$ ): $f (x, 0) = 0$ for all $x$ , so $f_{x} (0, 0) = 0$ . Along the $y$ -axis ( $x = 0$ ): $f (0, y) = 0$ , so $f_{y} (0, 0) = 0$ . Why this step? Both partials exist and equal $0$ .

Step 2 — write down the candidate plane $L$ . Feeding $f (0, 0) = 0$ , $f_{x} (0, 0) = 0$ , $f_{y} (0, 0) = 0$ into our derived formula: $L (x, y) = 0 + 0 \cdot (x - 0) + 0 \cdot (y - 0) = 0.$ Why this step? This is the only plane the formula can produce — the flat sheet $z = 0$ . The differentiability test now asks: does this candidate actually hug the surface?

Step 3 — test the diagonal $y = x$ . $f (x, x) = \frac{x \cdot x}{x ^{2} + x ^{2}} = \frac{x ^{2}}{2 x ^{2}} = \frac{1}{2}$ Why this step? Along $y = x$ the function is stuck at $\frac{1}{2}$ , no matter how close to the origin — but the candidate plane $L$ says $z = 0$ there. They disagree by $\frac{1}{2}$ arbitrarily close to the base point.

Step 4 — check the differentiability limit. The error over distance is $\frac{f - L}{x ^{2} + y ^{2}} = \frac{1/2 - 0}{2 ∣ x ∣} \to \infty$ as $x \to 0$ along $y = x$ . Why this step? Differentiability demands this ratio $\to 0$ . Here it blows up. So $f$ is not differentiable at $(0, 0)$ .

Verify: $f$ is not even continuous at the origin (value $0$ on the axes, $\frac{1}{2}$ on the diagonal), so it cannot be differentiable, so no tangent plane exists — even though both partials are $0$ . ✓

Figure s04 shows the two paths meeting at the origin at different heights — the surface cannot be flattened here.

Example 6 — Cell F: real-world word problem

Worked example A closed metal box measures length

$x = 50$ cm, width $y = 30$ cm, height $h = 20$ cm, each measurable to within $0.2$ cm. Estimate the maximum error in the computed volume $V = x y h$ . Forecast: Three sources of error, all adding up. Since each side is large, even a $0.2$ cm slip multiplies into a chunky volume error. Guess: a few hundred cm³.

Step 1 — partials (the sensitivities). $V_{x} = y h = 30 \cdot 20 = 600$ , $V_{y} = x h = 50 \cdot 20 = 1000$ , $V_{h} = x y = 50 \cdot 30 = 1500$ . Why this step? Each partial says "how many cm³ does $V$ change per cm of error in that dimension." Using the gradient here is exactly the tangent-plane idea in 3 inputs.

Step 2 — the differential. $d V = V_{x} d x + V_{y} d y + V_{h} d h$ Why this step? $d V$ is the flat-plane prediction of the change $Δ V$ ; valid because $0.2$ cm is small versus each side.

Step 3 — worst case (all errors same sign). Take $d x = d y = d h = 0.2$ : $d V = 600 (0.2) + 1000 (0.2) + 1500 (0.2) = 120 + 200 + 300 = 620 cm^{3}$ Why this step? The differential is a sum of slope-times-step terms — the same tangent-plane recipe. It is largest when every step pushes the same way (all $+ 0.2$ ), so choosing all errors positive gives the worst-case (maximum) volume error we were asked for. Mixed signs would partly cancel and shrink it.

Verify (units + sanity): $V = 50 \cdot 30 \cdot 20 = 30000$ cm³, and $620/30000 \approx 2.07%$ . Each partial has units cm² (an area), times cm gives cm³ — correct. The percentage matches the rough rule $\frac{d V}{V} = \frac{d x}{x} + \frac{d y}{y} + \frac{d h}{h} = 0.004 + 0.0067 + 0.01 = 0.0207$ . ✓

Example 7 — Cell G: exam twist, implicit level surface

Sometimes $z$ cannot be cleanly isolated. Then we use the gradient and a normal vector. First the general tool, then the example.

Worked example Tangent plane to the ellipsoid

$x^{2} + 2 y^{2} + 3 z^{2} = 20$ at $(3, - 1, 3)$ Forecast: We can't easily solve for $z = f (x, y)$ cleanly. Instead treat it as $F = 0$ and use the normal $n = \nabla F$ , then apply $n \cdot (X - X_{0}) = 0$ .

Step 1 — write it as $F = 0$ . $F (x, y, z) = x^{2} + 2 y^{2} + 3 z^{2} - 20$ . Why this step? Moving everything to one side makes it a level surface, so $\nabla F$ gives the normal without isolating $z$ .

Step 2 — gradient (the normal). $\nabla F = ⟨ 2 x, 4 y, 6 z ⟩ (3, - 1, 3) n = ⟨ 6, - 4, 63 ⟩ .$ Why this step? This vector is perpendicular to the surface at that point — the $n$ our construction needs.

Step 3 — check the point is actually on the surface. $9 + 2 (1) + 3 (3) = 9 + 2 + 9 = 20$ ✓. Why this step? A tangent plane is meaningless if the point isn't on the surface — a favourite exam trap.

Step 4 — apply $n \cdot (X - X_{0}) = 0$ and simplify. $6 (x - 3) - 4 (y - (- 1)) + 63 (z - 3) = 0$ i.e. $6 (x - 3) - 4 (y + 1) + 63 (z - 3) = 0$ . Expand each piece: $6 x - 18 - 4 y - 4 + 63 z - 18 = 0$ , so $6 x - 4 y + 63 z = 40 .$ Why this step? This is the dot-product-equals-zero plane written out — "perpendicular to the normal, through the base point." Nothing was assumed; it is the general construction from the callout above.

Verify: Plug the point into $6 x - 4 y + 63 z$ : $18 + 4 + 63 \cdot 3 = 18 + 4 + 18 = 40$ . ✓ And the coefficients $⟨ 6, - 4, 63 ⟩$ match $\nabla F$ , so the plane is correctly oriented. ✓

Example 8 — Cell H: how far can you trust it? (error $\sim$ distance²)

Worked example For

$f (x, y) = x^{2} + y^{2}$ at $(1, 2)$ (from Ex 1), how does the error behave as we step out along $y$ by amounts $0.1, 0.2, 0.4$ ? Forecast: The parent said "error grows like the square of the distance." Doubling the step should roughly quadruple the error. Watch.

Step 1 — the exact error formula. With $L = 2 x + 4 y - 5$ (from Ex 1) and $f = x^{2} + y^{2}$ , freeze $x = 1$ : $E = f - L = (1 + y^{2}) - (2 + 4 y - 5) = y^{2} - 4 y + 4 = (y - 2)^{2}$ Why this step? Writing the error in closed form lets us see the distance $^{2}$ law directly instead of trusting it: the error is literally the square of the $y$ -step $s = y - 2$ .

Step 2 — tabulate. Let $s = y - 2$ be the step:

$s = 0.1$ : $E = (0.1)^{2} = 0.01$
$s = 0.2$ : $E = (0.2)^{2} = 0.04$
$s = 0.4$ : $E = (0.4)^{2} = 0.16$ Why this step? Each doubling of $s$ multiplies $E$ by $4$ — the promised quadratic growth, exactly.

Verify: $0.04/0.01 = 4$ and $0.16/0.04 = 4$ . The error is literally $s^{2}$ ; the linear approximation is trustworthy only for small $s$ . ✓

Figure s05 plots the true parabola against its tangent line and shades the growing gap.

Recall Quick self-test across the matrix

Which cell has zero error at any distance? ::: Cell D — the surface is already a plane (zero curvature). Both partials are $0$ but no tangent plane exists — which cell and why? ::: Cell E — partials probe only the axes; along the diagonal $f$ jumps, so the differentiability limit blows up. A zero partial means the surface is... ::: level (flat) along that one axis at the point — Cell C. Both slopes negative with positive steps sends the estimate... ::: below the base height (downward dome) — Cell B $^{'}$ . For a level surface $F = 0$ , the tangent plane is perpendicular to... ::: the gradient $\nabla F$ — Cell G. Doubling your distance from the base point multiplies the error by roughly... ::: four (error $\sim$ distance $^{2}$ ) — Cell H.

Connections

Parent topic — the derivation these examples exercise.
Partial derivatives — every example starts by computing $f_{x}, f_{y}$ .
The gradient vector — Ex 7 uses $\nabla F$ as the normal.
Differentiability of multivariable functions — Ex 5 is the breakdown case.
Tangent line and linear approximation (single variable) — the 1-D idea Ex 8's error law mirrors.
The chain rule (multivariable) — used to differentiate $\cdot$ and $sin$ in the surfaces.
Directional derivatives — Ex 7's "perpendicular to normal" is a dot-product condition.

#flashcards/maths

Where does

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

come from?

Match the general plane

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

to the surface's two axis slices, forcing

A = f_{x} (a, b)

B = f_{y} (a, b)

Tangent plane estimate of

x^{2} + y^{2}

(1, 2)

for

(1.05, 2.1)

L = 5.5

(true

5.5125

For the saddle

x^{2} - y^{2}

(2, 1)

, what are the two partial slopes?

f_{x} = 4

f_{y} = - 2

Estimate of

10 - x^{2} - y^{2}

(1, 1)

for

(1.1, 1.2)

L = 7.4

(true

7.35

) — both slopes negative.

Why does

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

have no tangent plane at the origin?

Both partials are

0

but along

y = x

the value is

\frac{1}{2} \neq = 0

; the error/distance ratio

\to \infty

, so

f

is not differentiable.

Volume error of

V = x y h

(

50, 30, 20

) with

\pm 0.2

each?

d V = 620

cm³ (about

2.07%

General implicit-surface tangent plane at

X_{0}

n \cdot (X - X_{0}) = 0

with

n = \nabla F

; here

6 x - 4 y + 63 z = 40

How does the error of

x^{2} + y^{2}

grow along

x = 1

as a function of step

s = y - 2

E = s^{2}

— exactly quadratic in the step.