Vector fields — definition, visualization: Worked examples

This page is the "no gaps" drill for Vector fields — definition, visualization. We will walk through every kind of situation a vector field can throw at you: each quadrant, the origin (where things can blow up), constant fields, swirls, gradient fields, an inverse-square law, and a word problem. Each worked example is tagged with the cell of the scenario matrix it fills.

Before line one, four tiny reminders in plain words so no symbol appears unearned:

Why $a^{2} + b^{2}$ and not something else? Because the arrow's two legs $a$ (rightward) and $b$ (upward) are at right angles, so the straight-line length is the hypotenuse — the Pythagorean theorem is the only tool that turns two perpendicular legs into one length.

The scenario matrix

Every case class this topic can present, and which example covers it:

Cell	Scenario class	Covered by
A	Constant / degenerate field (arrow never changes)	Ex 1
B	Radial field across all four quadrants (signs of $x, y$ )	Ex 2
B $^{'}$	On-axis edge cases ( $x = 0$ or $y = 0$ , one component vanishes)	Ex 2b
C	Field at the origin (zero / degenerate input)	Ex 3
D	Rotational field + perpendicularity test	Ex 4
E	Gradient field from a scalar (uses partial derivatives)	Ex 5
F	Limiting behaviour of an inverse-square field ( $r \to 0$ , $r \to \infty$ )	Ex 6
G	Direction vs magnitude split (unit vector + length separately)	Ex 7
H	Real-world word problem (river current)	Ex 8
I	Exam-style twist (is a mystery field a gradient field?)	Ex 9

The goal: after Ex 9 you have seen every sign, every quadrant, the axes, the origin, both limits, and the two flavours of "special" field.

Ex 1 — Constant field (Cell A)

Worked example Statement

$F (x, y) = (3, 0)$ . Find the arrow at $(0, 0)$ , at $(5, - 2)$ , and at $(- 1, 7)$ . Describe the whole field.

Forecast: guess before reading — do the arrows change from point to point?

Read the components. $P (x, y) = 3$ , $Q (x, y) = 0$ . Why this step? The rule only depends on $x, y$ through $P$ and $Q$ . If neither $P$ nor $Q$ contains $x$ or $y$ , the output cannot change.
Evaluate at each point. $F (0, 0) = (3, 0)$ , $F (5, - 2) = (3, 0)$ , $F (- 1, 7) = (3, 0)$ . Why this step? Substituting different points changes nothing because $3$ and $0$ have no $x$ or $y$ in them.
Describe. Every arrow points straight right, length $3^{2} + 0^{2} = 3$ . A uniform "steady east wind." Why this step? Direction $= (3, 0)$ normalised $= (1, 0)$ ; length $3$ everywhere.

Verify: length $= 9 + 0 = 3$ ✅. This is the degenerate case — a single vector cloned everywhere (the "steel-man" trap from the parent note).

Ex 2 — Radial field, all four quadrants (Cell B)

Figure — Vector fields — definition, visualization

Worked example Statement

$F (x, y) = (x, y)$ . Find and describe the arrow in each quadrant: at $(2, 1)$ , $(- 2, 1)$ , $(- 2, - 1)$ , $(2, - 1)$ .

Forecast: which way does the arrow point relative to the origin at each of these?

Quadrant I — $(2, 1)$ : arrow $(2, 1)$ . Both parts positive → points up-and-right, away from origin. Why this step? Output equals the position vector, so it points from origin toward the point, i.e. outward.
Quadrant II — $(- 2, 1)$ : arrow $(- 2, 1)$ . Left and up → still outward (up-left). Why this step? $x < 0$ makes $P < 0$ (leftward), $y > 0$ makes $Q > 0$ (upward): the arrow again reaches away from the centre.
Quadrant III — $(- 2, - 1)$ : arrow $(- 2, - 1)$ . Down-left, outward.
Quadrant IV — $(2, - 1)$ : arrow $(2, - 1)$ . Down-right, outward. Why these steps? In every quadrant the signs of $(P, Q)$ copy the signs of $(x, y)$ , so the arrow always points radially outward — the "explosion / source" pattern. Look at the figure: all four arrows fan away from the centre dot.

Verify: length at $(2, 1)$ is $4 + 1 = 5 \approx 2.236$ . General length $= x^{2} + y^{2} = r$ : arrows get longer the farther out you go ✅.

Ex 2b — On the axes: the edge cases (Cell B $^{'}$ )

Worked example Statement

Still with $F (x, y) = (x, y)$ , look at the points that sit on the axes (not in any open quadrant): $(3, 0)$ , $(- 3, 0)$ on the $x$ -axis, and $(0, 4)$ , $(0, - 4)$ on the $y$ -axis. What happens to each component?

Forecast: on the $x$ -axis, which part of the arrow becomes zero?

On the $x$ -axis ( $y = 0$ ). $F (3, 0) = (3, 0)$ points purely right; $F (- 3, 0) = (- 3, 0)$ points purely left. Why this step? When $y = 0$ the second component $Q = y = 0$ vanishes, so the arrow has no vertical part — it lies flat along the axis. The sign of the surviving part $P = x$ flips as you cross the origin (right for $x > 0$ , left for $x < 0$ ).
On the $y$ -axis ( $x = 0$ ). $F (0, 4) = (0, 4)$ points purely up; $F (0, - 4) = (0, - 4)$ points purely down. Why this step? Now $P = x = 0$ vanishes, so the arrow has no horizontal part — it lies straight along the $y$ -axis, flipping sign at the origin.
Read the lesson. The axes are exactly where one component is zero — the boundary lines between quadrants. The arrow is still radial (outward), just pointing along the axis. No sign contradiction: crossing an axis is precisely where a component changes sign smoothly through $0$ . Why this step? This fills the gap between quadrants: every point either is in a quadrant (Ex 2), on an axis (here), or at the origin (Ex 3).

Verify: $F (3, 0) = (3, 0)$ , length $3$ ; $F (0, - 4) = (0, - 4)$ , length $4$ ; in each case one component is exactly $0$ ✅.

Ex 3 — At the origin: the degenerate input (Cell C)

Worked example Statement

For the radial field $F (x, y) = (x, y)$ , what is the arrow exactly at the origin $(0, 0)$ ? Is its direction defined?

Forecast: can you draw an arrow of length zero with a direction?

Evaluate. $F (0, 0) = (0, 0)$ — the zero vector. Why this step? Plug $x = 0, y = 0$ into the rule; both components vanish (this is the point where both axes' zero-cases from Ex 2b meet).
Length. $∣ F (0, 0) ∣ = 0^{2} + 0^{2} = 0$ . Why this step? A zero-length arrow is a dot, not a spear — it has no direction.
Direction is undefined here. The unit direction $\frac{F}{∣ F ∣} = \frac{( 0 , 0 )}{0}$ divides by zero. Why this step? We cannot normalise the zero vector — division by $0$ is forbidden. So the origin is a critical / equilibrium point: nothing pushes there.

Verify: length $0$ ✅. Physically: a particle placed exactly at the source feels no push — it sits still (unstable, but still). This is the general lesson: always check whether your field is zero (or undefined) at special points before claiming a direction.

Ex 4 — Rotation field + perpendicularity (Cell D)

Worked example Statement

$F (x, y) = (- y, x)$ . Find the arrows at $(1, 0)$ , $(0, 1)$ , $(- 1, 0)$ , $(0, - 1)$ and prove the field is a pure swirl.

Forecast: clockwise or counter-clockwise?

Evaluate. $F (1, 0) = (0, 1)$ up; $F (0, 1) = (- 1, 0)$ left; $F (- 1, 0) = (0, - 1)$ down; $F (0, - 1) = (1, 0)$ right. Why this step? Each output is the position vector turned $9 0^{\circ}$ counter-clockwise. Follow up → left → down → right: that traces a counter-clockwise circle (see figure).
Perpendicularity test — the dot product. Using the dot-product formula $u \cdot v = u_{1} v_{1} + u_{2} v_{2}$ from the definition callout above, take $u = F = (- y, x)$ and $v =$ the radius $(x, y)$ : $F \cdot (x, y) = (- y) (x) + (x) (y) = - y x + x y = 0.$ Why this tool and not another? The dot product answers exactly one geometric question: "are these two arrows at right angles?" — it equals zero precisely when they are perpendicular. Since $F \cdot (x, y) = 0$ everywhere, each arrow is at $9 0^{\circ}$ to the radius, i.e. tangent to the circle. No outward part means no spreading, pure rotation.
Length. $∣ F ∣ = (- y)^{2} + x^{2} = x^{2} + y^{2} = r$ . Why this step? Arrows lengthen with distance $r$ , just like the radial field, but they curl instead of fanning out.

Verify: dot product $- y x + x y = 0$ ✅. At $(1, 0)$ length $= 0 + 1 = 1$ ✅. At $(3, 4)$ : $F = (- 4, 3)$ , dot with $(3, 4)$ is $(- 4) (3) + (3) (4) = - 12 + 12 = 0$ ✅, length $16 + 9 = 5 = r$ ✅.

Ex 5 — Gradient field from a scalar (Cell E)

Worked example Statement

Let $f (x, y) = x^{2} + 3 y^{2}$ . Build the gradient field $F = \nabla f$ and evaluate at $(1, 1)$ and $(- 2, 0)$ .

Forecast: which direction should the arrows point — uphill or downhill on $f$ ?

What is $\nabla$ ? The gradient $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ collects the two partial derivatives. Why this tool? A partial derivative $\partial f / \partial x$ answers "how fast does $f$ rise if I nudge only $x$ ?" Stacking both nudge-rates into an arrow gives the single direction of steepest increase — see Gradient and Directional Derivatives.
Differentiate. $\frac{\partial}{\partial x} (x^{2} + 3 y^{2}) = 2 x$ (treat $y$ as a frozen constant, so $3 y^{2}$ has zero $x$ -rate). $\frac{\partial}{\partial y} (x^{2} + 3 y^{2}) = 6 y$ . Why this step? "Partial" means freeze the other variable; then it's ordinary one-variable differentiation.
Assemble the field. $F (x, y) = (2 x, 6 y)$ .
Evaluate. $F (1, 1) = (2, 6)$ ; $F (- 2, 0) = (- 4, 0)$ . Why this step? Direct substitution. At $(- 2, 0)$ the arrow points left because $2 x = - 4 < 0$ ; the field points away from the minimum at the origin — uphill, as promised.

Verify: $F (1, 1) = (2 \cdot 1, 6 \cdot 1) = (2, 6)$ ✅. Length there $= 4 + 36 = 40 \approx 6.325$ . This $F$ is a conservative (gradient) field because it came from a scalar $f$ — bookmark this for Ex 9. See Conservative Fields and Potential Functions.

Ex 6 — Inverse-square field: both limits (Cell F)

Worked example Statement

For the 2D inverse-square-style field $F (r) = \frac{r}{∣ r ∣ ^{3}}$ with $r = (x, y)$ and $∣ r ∣ = r = x^{2} + y^{2}$ (the distance from the origin defined above), find $∣ F ∣$ and describe what happens as $r \to 0$ and as $r \to \infty$ .

Forecast: does the push get stronger or weaker near the centre?

Magnitude. $∣ F ∣ = \frac{∣ r ∣}{∣ r ∣ ^{3}} = \frac{1}{∣ r ∣ ^{2}} = \frac{1}{r ^{2}}$ . Why this step? $r /∣ r ∣$ is the outward unit arrow (length exactly $1$ ); dividing by $∣ r ∣^{2}$ shrinks the length to $1/ r^{2}$ . So strength falls off as one-over-distance-squared.
Limit $r \to \infty$ . $\frac{1}{r ^{2}} \to 0$ . Why this tool — a limit? A limit answers "what value does this approach as $r$ heads somewhere I can't plug in directly?" Far away, $1/ r^{2}$ approaches $0$ : the push fades to nothing (see the shrinking outer arrows in the figure).
Limit $r \to 0^{+}$ . $\frac{1}{r ^{2}} \to + \infty$ . Why this step? Close to the centre, dividing $1$ by a tiny squared number gives a huge value — the field blows up at the origin. So $r = 0$ is excluded from the domain (undefined there), unlike the radial field of Ex 3 which was merely zero.

Verify: at $r = 2$ , $∣ F ∣ = 1/4 = 0.25$ ✅; at $r = 1$ , $∣ F ∣ = 1$ ✅; at $r = 1/2$ , $∣ F ∣ = 1/ (1/4) = 4$ ✅ — strengthening toward the centre. See Flux and the Divergence Theorem for why this field's flux is special.

Ex 7 — Splitting direction from magnitude (Cell G)

Worked example Statement

For $F (x, y) = (x, y)$ at the point $(3, 4)$ , write the arrow as (unit direction) $\times$ (magnitude).

Forecast: what's the length, and what's the pure "which-way" arrow?

Magnitude. $∣ F (3, 4) ∣ = 3^{2} + 4^{2} = 25 = 5$ . Why this step? Pythagoras on legs $3$ and $4$ — the classic $3 - 4 - 5$ triangle.
Unit direction. $\frac{F}{∣ F ∣} = \frac{( 3 , 4 )}{5} = (\frac{3}{5}, \frac{4}{5}) = (0.6, 0.8)$ . Why this step? Dividing an arrow by its own length gives a length- $1$ arrow pointing the same way — this is pure direction, stripped of magnitude.
Reassemble. $(0.6, 0.8) \times 5 = (3, 4) = F$ ✅. Why this step? Confirms the split: direction times length rebuilds the original arrow. This is exactly how field-plotters draw equal-length arrows (use step 2) but colour by step 1.

Verify: $0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1$ (genuinely a unit vector) ✅; $5 \times 0.6 = 3$ , $5 \times 0.8 = 4$ ✅.

Ex 8 — Word problem: river current (Cell H)

Worked example Statement

A river's surface current is modelled by $v (x, y) = (0, 1 - x^{2})$ metres per second, where $x$ is distance from the centre-line ( $x = 0$ mid-river) and $∣ x ∣ \leq 1$ are the banks. Find the current at the centre, at $x = 0.5$ , and at the banks. Where is it fastest?

Forecast: where does water move quickest — middle or edge?

Read the model. $P = 0$ (no sideways drift), $Q = 1 - x^{2}$ (downstream speed depends only on how far from centre). Why this step? $P = 0$ means the water only flows straight down-river; the sideways part is zero everywhere.
Centre $x = 0$ . $v = (0, 1 - 0) = (0, 1)$ → $1 m/s$ downstream.
Half-way $x = 0.5$ . $v = (0, 1 - 0.25) = (0, 0.75)$ → $0.75 m/s$ .
Banks $x = \pm 1$ . $v = (0, 1 - 1) = (0, 0)$ → zero current (water sticks at the edge). Why these steps? Substituting the given $x$ values into $1 - x^{2}$ . The pattern: speed is largest at the centre and drops to zero at the banks.
Fastest point. $1 - x^{2}$ is biggest when $x^{2}$ is smallest, i.e. $x = 0$ . Why this step? Subtracting the smallest possible non-negative number ( $x^{2} = 0$ ) leaves the largest result.

Verify: $1 - 0^{2} = 1$ , $1 - 0. 5^{2} = 0.75$ , $1 - 1^{2} = 0$ ✅. Units: $Q$ is in m/s, matching a velocity field — this is a genuine Parametric Curves and Velocity style flow. Direction $(0, 1)$ has length $1$ , so at the centre magnitude equals $1$ m/s exactly ✅.

Ex 9 — Exam twist: is this field a gradient field? (Cell I)

Worked example Statement

A field $F (x, y) = (2 x, 6 y)$ is dropped in front of you (from Ex 5, but pretend you don't know its origin). Is it a gradient field, i.e. is there a scalar $f$ with $\nabla f = F$ ? If so, find $f$ explicitly.

Forecast: guess yes or no, then check.

The consistency test. For $F = (P, Q)$ to be a gradient, the cross partials must agree: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ . Why this tool? If $P = \partial f / \partial x$ and $Q = \partial f / \partial y$ , then both $\partial P / \partial y$ and $\partial Q / \partial x$ equal the same mixed second derivative $\partial^{2} f / \partial x \partial y$ — so they must match. This is the fast screening test (see Conservative Fields and Potential Functions).
Compute the cross partials. $\frac{\partial}{\partial y} (2 x) = 0$ and $\frac{\partial}{\partial x} (6 y) = 0$ . They match ( $0 = 0$ ) → candidate passes, so we now go hunting for $f$ . Why this step? $2 x$ has no $y$ , so its $y$ -rate is $0$ ; $6 y$ has no $x$ , so its $x$ -rate is $0$ .
Integrate $P$ with respect to $x$ to start building $f$ . We want $\partial f / \partial x = P = 2 x$ , so undo that derivative: $f (x, y) = \int 2 x d x = x^{2} + g (y) .$ Why this step? Anti-differentiating $2 x$ in $x$ gives $x^{2}$ . But the "constant of integration" is only constant with respect to $x$ — it may still depend on $y$ . So we write an unknown ==function of $y$ alone==, $g (y)$ , not just a number. This is the key subtlety: the constant of integration is a whole function $g (y)$ , because $\partial / \partial x$ annihilates any pure- $y$ term.
Fix $g (y)$ using the second component $Q$ . We also need $\partial f / \partial y = Q = 6 y$ . Differentiate our $f$ in $y$ : $\frac{\partial}{\partial y} (x^{2} + g (y)) = 0 + g^{'} (y) = g^{'} (y) .$ Set this equal to $6 y$ : $g^{'} (y) = 6 y$ . Why this step? The $x^{2}$ term contributes nothing to the $y$ -derivative, so whatever is left must supply all of $Q$ . That pins down $g$ .
Integrate to recover $g (y)$ , with its own constant. $g (y) = \int 6 y d y = 3 y^{2} + C$ , where $C$ is a genuine numerical constant (no variable can hide now — $g$ depends only on $y$ , and we have used up the $y$ -condition). Why this step? This second integration introduces the true constant of integration $C$ . It is a plain number because a potential is only ever determined up to an additive constant — shifting $f$ by any constant leaves $\nabla f$ unchanged.
Answer. $f (x, y) = x^{2} + 3 y^{2} + C$ . Choosing $C = 0$ gives the clean potential $f (x, y) = x^{2} + 3 y^{2}$ — exactly Ex 5's original scalar. So $F$ is conservative. ✅

Verify: $\nabla (x^{2} + 3 y^{2} + C) = (2 x, 6 y) = F$ for any constant $C$ ✅. Cross-partial check $\partial_{y} (2 x) = 0 = \partial_{x} (6 y)$ ✅.

Recall Quick self-test (cloze)

The direction of the zero vector is undefined because you cannot divide by zero. The dot product $u \cdot v = u_{1} v_{1} + u_{2} v_{2}$ being zero proves two arrows are perpendicular. On the $x$ -axis the radial field $(x, y)$ has its second component == $y = 0$ ==, so the arrow lies flat along the axis. As $r \to 0$ , the inverse-square magnitude $1/ r^{2} \to$ infinity (field blows up). A gradient candidate $(P, Q)$ passes the screen when $\partial P / \partial y =$ == $\partial Q / \partial x$ ==. When integrating $P$ in $x$ to recover a potential, the "constant" is really a ==function of $y$ == alone.

Which quadrant fails for a naive radial description?

None — the radial field

(x, y)

points outward in all four quadrants because signs of

(P, Q)

copy signs of

(x, y)

What happens on the axes for the radial field?

One component vanishes (

Q = 0

on the

x

-axis,

P = 0

on the

y

-axis), so the arrow points straight along that axis, still outward.

What makes the origin degenerate for the radial field?

F (0, 0) = (0, 0)

: zero length, so direction is undefined (can't normalise).

What limit distinguishes inverse-square from radial at the centre?

Inverse-square

\to \infty

(undefined at origin); radial

\to 0

(defined, just zero).

When recovering a potential, why is the constant of integration a function of

y

Because

\partial / \partial x

kills any term depending only on

y

, so anti-differentiating in

x

cannot see it.

Connections

Vector fields — definition, visualization — the parent topic this drill supports.
Gradient and Directional Derivatives — Ex 5 builds a gradient field.
Divergence and Curl — radial vs rotational contrast (Ex 2 vs Ex 4).
Conservative Fields and Potential Functions — Ex 9's recovery of $f$ .
Flux and the Divergence Theorem — inverse-square field of Ex 6.
Parametric Curves and Velocity — the river velocity field of Ex 8.

Worked examples — Vector fields — definition, visualization

The scenario matrix

Ex 1 — Constant field (Cell A)

Ex 2 — Radial field, all four quadrants (Cell B)

Ex 2b — On the axes: the edge cases (Cell B $^{'}$ )

Ex 3 — At the origin: the degenerate input (Cell C)

Ex 4 — Rotation field + perpendicularity (Cell D)

Ex 5 — Gradient field from a scalar (Cell E)

Ex 6 — Inverse-square field: both limits (Cell F)

Ex 7 — Splitting direction from magnitude (Cell G)

Ex 8 — Word problem: river current (Cell H)

Ex 9 — Exam twist: is this field a gradient field? (Cell I)

Connections

Concept Map