3D plots — surface, wireframe: Worked examples

Everything below builds only on the parent note and these prerequisites: the parent topic, NumPy vectorization, Matplotlib figure and axes objects, Colormaps in Matplotlib, and Contour plots.

The scenario matrix

Before any code, here is the full space of cases a 3D-surface problem can throw at you. Each later example is tagged with the cell it covers.

#	Case class	What's special	Example
A	Positive bowl (normal case)	$z \geq 0$ everywhere, smooth	Ex 1
B	Negative / saddle heights	$z$ takes both signs — sign matters	Ex 2
C	Zero / degenerate input — flat function	$z =$ constant, surface is a plane	Ex 3
D	Single-point / empty grid	`len(x)=1` or `len=0` — edge inputs	Ex 4
E	Shape / transpose bug	`Z.shape` vs `(len(y),len(x))`	Ex 5
F	Limiting behaviour — ripple far out	$r \to \infty$ , crest radii	Ex 6
G	Real-world word problem	terrain / height reading	Ex 7
H	Exam twist — wireframe stride count	how many lines actually drawn	Ex 8

We hit every cell A–H below.

Example 1 — Cell A: the positive bowl

Worked example The normal case

Build x = y = np.linspace(-2, 2, 5), X, Y = np.meshgrid(x, y), Z = X**2 + Y**2. What is Z.shape, the minimum height, and its location?

Forecast: guess the shape and where the bowl bottoms out before reading on.

x has len 5, y has len 5. meshgrid returns shape (len(y), len(x)) = (5, 5). Why this step? The parent formula says shape is $(n, m) = ($ len(y), len(x) $)$ . We must know this to sanity-check Z.
Z = X**2 + Y**2 is vectorized — it runs on the whole $5 \times 5$ array elementwise, so Z.shape == (5,5) too. Why this step? plot_surface demands X,Y,Z all the same 2D shape.
Each term is a square, so $z \geq 0$ . The smallest possible is $0 + 0 = 0$ , reached only where $x = 0$ and $y = 0$ . Why this step? The plot is literally $f$ drawn; the bowl's floor sits at the minimum of $f$ .
Look at the figure: a symmetric bowl, deepest at the centre.

Verify: $f (0, 0) = 0^{2} + 0^{2} = 0$ . Any other point, e.g. $f (2, 0) = 4 > 0$ . So the min is $0$ at the origin. ✓

Example 2 — Cell B: negative and saddle heights

Worked example Signs both ways — the saddle

Z = X**2 - Y**2 on the same grid. Find the height at $(2, 0)$ , at $(0, 2)$ , and at $(0, 0)$ . Is there a single lowest point?

Forecast: guess the sign of $z$ along the two axes before computing.

Along the $x$ -axis ( $y = 0$ ): $z = x^{2} \geq 0$ . At $(2, 0)$ : $z = 4$ . Why this step? Fixing $y = 0$ collapses $f$ to a 1D parabola opening up — a valley wall.
Along the $y$ -axis ( $x = 0$ ): $z = - y^{2} \leq 0$ . At $(0, 2)$ : $z = - 4$ . Why this step? Fixing $x = 0$ gives $- y^{2}$ , a parabola opening down — a ridge falling away.
At the origin: $z = 0 - 0 = 0$ . This point is a saddle: up in one direction, down in the perpendicular one. Why this step? Because signs disagree along the two axes, no single minimum exists — the surface has both positive and negative regions.
Look at the figure — the classic Pringle-chip shape.

[!mistake] "A saddle has a lowest point at the middle." No — the centre is neither a max nor a min. Go along $+ x$ and it rises; along $+ y$ it falls. That's why $z$ spans both signs and a diverging colormap (like coolwarm) reads best here.

Verify: $f (2, 0) = 4$ , $f (0, 2) = - 4$ , $f (0, 0) = 0$ ; signs opposite ⇒ saddle. ✓

Example 3 — Cell C: degenerate flat function

Worked example A constant-height plane

Z = np.full(X.shape, 3.0) (every height is 3). What does the surface look like, and does plot_surface still work?

Forecast: guess the shape drawn.

np.full((5,5), 3.0) makes a $5 \times 5$ array of all 3's — same shape as X,Y. Why this step? The shape rule still applies even when the function is constant.
Every $(x, y)$ maps to $z = 3$ , so the "surface" is a flat horizontal plane floating at height 3. Why this step? Degenerate input (a constant function) is still valid — the drape just has no bumps.
cmap will paint the whole sheet a single color, because there's no height variation to map. Why this step? A colormap encodes differences in $z$ ; with $z$ constant there's nothing to differentiate.

Verify: Z.min() == Z.max() == 3.0 and Z.shape == (5,5). A valid, if boring, surface. ✓

Example 4 — Cell D: single-point and empty grids

Worked example Edge inputs

(a) x = np.linspace(0, 0, 1), y = np.linspace(-1, 1, 3). What is X.shape? (b) x = np.array([]), y = np.array([1,2]). What is X.shape?

Forecast: predict both shapes.

(a) len(x)=1, len(y)=3. Shape is (len(y), len(x)) = (3, 1). Why this step? The rule $(n, m)$ never breaks; a single-column grid is legal but too thin to draw a real surface.
(a) With only one x-column you cannot form a patch (a patch needs $\geq 2$ rows and $\geq 2$ columns). plot_surface runs without error but draws essentially a line, not a sheet. Why this step? A surface facet is a little quadrilateral spanning two rows × two columns; with one column there's nothing to fill.
(b) len(x)=0, len(y)=2. Shape is (2, 0) — an empty grid with zero columns. Nothing gets plotted. Why this step? Empty input is degenerate: Z is empty too, so the drape covers no area.

Verify: (a) shape (3,1); (b) shape (2,0). Both follow the $(l e n (y), l e n (x))$ rule. ✓

Example 5 — Cell E: the transpose bug

Worked example Shape mismatch caught by

.shape x = np.linspace(0,1,4), y = np.linspace(0,1,3). A student builds Z by looping as Z[j][i] (swapped) and gets shape (4,3). Why does the plot come out garbled, and what shape should Z have?

Forecast: guess the correct Z.shape.

meshgrid(x,y) gives X.shape == Y.shape == (len(y), len(x)) == (3, 4). Why this step? This is the target every Z must match.
The student's Z is (4,3) — the transpose. plot_surface(X, Y, Z) needs all three equal; (3,4) ≠ (4,3) (they're only equal by luck when the grid is square). Why this step? Mismatched shapes either raise an error or silently misalign heights to the wrong floor cells → garbled surface.
Fix: prefer vectorized Z = f(X, Y) (auto-correct shape), or if you must loop, index Z[i, j] with i over rows (y) and j over columns (x). Why this step? Letting broadcasting handle it removes the whole class of transpose bugs (see meshgrid and broadcasting).

Verify: X.shape == (3,4) but the buggy Z.shape == (4,3); (3,4) != (4,3) ⇒ mismatch detected. ✓

Example 6 — Cell F: limiting behaviour, the ripple

Worked example Where are the crests of

Z = sin(sqrt(X**2+Y**2))? Let $r = x^{2} + y^{2}$ be the distance from the origin. At which radii $r$ is the surface at a crest ( $z = 1$ ), and where is the first trough ( $z = - 1$ )?

Forecast: guess the radius of the first ring crest.

$z = sin (r)$ . A sine crest is where its argument equals $\frac{π}{2}$ plus full turns: $r = \frac{π}{2} + 2 π k$ . Why this step? $sin$ peaks at $π /2$ ; because $r$ only depends on distance, every point at that radius has the same height → concentric rings.
First crest: $k = 0 \Rightarrow r = π /2 \approx 1.5708$ . Why this step? This is the innermost bright ring; it's the Forecast-then-Verify target from the parent note.
First trough ( $z = - 1$ ): $sin r = - 1$ at $r = \frac{3 π}{2} \approx 4.712$ . Why this step? The wave keeps oscillating outward; troughs and crests alternate every $π$ in $r$ .
Limiting behaviour: as $r \to \infty$ the surface keeps rippling forever between $- 1$ and $+ 1$ — it never settles. Look at the concentric rings in the figure.

Verify: $sin (π /2) = 1$ (crest), $sin (3 π /2) = - 1$ (trough), and $∣ sin r ∣ \leq 1$ always. ✓

Example 7 — Cell G: real-world word problem

Worked example A hill's height map

A hill's altitude in metres is $z = 100 - 4 x^{2} - y^{2}$ where $x, y$ are km east/north of a marker. What is the summit altitude and its location? At $x = 3, y = 0$ km, are you above or below sea level ( $z = 0$ )?

Forecast: guess the summit height before computing.

The two subtracted squares are largest-subtracted when $x = y = 0$ ; there they subtract nothing, so $z$ is maximised there. Why this step? Both $- 4 x^{2}$ and $- y^{2}$ only lower the height, so removing them (setting them to 0) gives the peak.
Summit: $z = 100 - 0 - 0 = 100$ m at the marker $(0, 0)$ . Why this step? Plug the maximising point into $f$ — the plot is just $f$ .
At $(3, 0)$ : $z = 100 - 4 (9) - 0 = 100 - 36 = 64$ m — still above sea level. Why this step? We evaluate $f$ at the queried coordinate and compare to 0.
Units check: $x^{2}$ is km², coefficient $4$ carries m/km² so $4 x^{2}$ is metres — consistent with $z$ in metres. ✓

Verify: summit $= 100$ m at origin; $f (3, 0) = 64 > 0$ (above sea level). ✓

Example 8 — Cell H: exam twist — counting wireframe lines

Worked example How many mesh lines does

rstride actually draw? A grid from linspace(-3,3,61) in both axes gives a $61 \times 61$ grid. You call plot_wireframe(X,Y,Z, rstride=10, cstride=10). How many horizontal (row) lines are drawn?

Forecast: guess the count before dividing.

There are 61 rows (indices 0…60). Stride 10 keeps rows $0, 10, 20, 30, 40, 50, 60$ . Why this step? rstride=k draws every $k$ -th row line, always including the first and last.
Count: $0, 10, 20, 30, 40, 50, 60$ → that's $⌊ 60/10 ⌋ + 1 = 6 + 1 = 7$ lines. Why this step? The number of kept indices from $0$ to $N - 1$ with step $k$ is $⌊(N - 1) / k ⌋ + 1$ .
So 7 row-lines (and by symmetry 7 column-lines) form the cage — clean and see-through, exactly the point of thinning a dense grid. Why this step? Confirms the parent's claim that stride trades detail for readability.

[!mnemonic] Lines drawn $= ⌊ \frac{N - 1}{stride} ⌋ + 1$ . "Divide the gaps, add one for the fencepost."

Verify: $⌊ 60/10 ⌋ + 1 = 7$ . ✓

Active Recall

Recall Min of

Z = X**2 + Y**2? $0$ at the origin — sum of two squares can't go below zero.

Recall Is

X**2 - Y**2 a bowl or a saddle? A saddle: up along $x$ , down along $y$ ; $z$ takes both signs.

Recall

meshgrid shape for len(x)=1, len(y)=3? (3, 1) — too thin to draw a real surface patch.

Recall First crest radius of

sin(sqrt(X**2+Y**2))? $r = π /2 \approx 1.57$ , where $sin = 1$ .

Recall Lines drawn by

rstride=10 on a 61-row grid? $⌊ 60/10 ⌋ + 1 = 7$ .

Prediction line — cover the answer:

Summit of

z = 100 - 4 x^{2} - y^{2}

and its height?

(0, 0)

, height

100

Connections

3D plots — surface, wireframe (parent)
meshgrid and broadcasting
NumPy vectorization
Matplotlib figure and axes objects
Colormaps in Matplotlib
Contour plots