Slope (direction) fields (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Interpret and sketch slope (direction) fields for first-order differential equations and sketch solution curves”

7 min readUpdated 2 Jun 2026

Quick answer

A slope field draws a short segment at each point with gradient given by $\tfrac{dy}{dx} = f(x, y)$ . A solution curve is one that is tangent to the field everywhere, so you sketch solutions by following the segments. Points where $\tfrac{dy}{dx} = 0$ give horizontal segments and possible equilibrium lines.

Jump to a section

What this dot point is asking
Why slope fields are useful
What a slope field shows
Reading and drawing the field
Connecting the field to the solution
Sketching solution curves
Exam-style questions

What this dot point is asking

SCSA wants you to read a slope field, construct one from a differential equation, and sketch solution curves through given points by following the field.

Why slope fields are useful

Many differential equations cannot be solved in closed form, yet their solution curves still have a definite shape. The slope field is a way to see that shape directly from the equation, without integrating. It is also a powerful conceptual tool even when the equation is solvable: it shows at a glance how the whole family of solutions behaves, where curves rise or fall, where they flatten out, and which equilibrium levels attract or repel nearby solutions. For SCSA this means a slope field question tests understanding of the differential equation itself, not just the mechanics of solving it.

What a slope field shows

A first-order differential equation $\tfrac{dy}{dx} = f(x, y)$ assigns a gradient to every point of the plane. The slope field is a grid of short line segments, each drawn with the gradient that the equation prescribes at that point. The field is a picture of the entire family of solutions at once, before any are found explicitly.

Reading and drawing the field

To draw the field, evaluate $f(x, y)$ at sample points and draw a short segment of that gradient. Patterns help: if $f$ depends only on $x$ , all segments in a vertical line are parallel; if $f$ depends only on $y$ , all segments in a horizontal line are parallel.

Isoclines

An isocline is a curve $f(x, y) = c$ along which every segment has the same gradient $c$ . Sketching a few isoclines makes the whole field quick to draw.

The isocline method turns a slope field into a manageable sketch. Choose a few values of $c$ , draw the curve $f(x, y) = c$ , and along that curve draw all segments with the single gradient $c$ . For $\dfrac{dy}{dx} = x + y$ , the isoclines $x + y = c$ are parallel lines, and along each one the segments all share the gradient $c$ . Doing this for several values of $c$ fills the plane efficiently without evaluating $f$ at every grid point separately.

Connecting the field to the solution

A slope field and the analytic solution describe the same family from two viewpoints. The field is the qualitative picture; solving the equation gives the explicit curves. The two should always agree, which is a useful check: after solving, confirm that the solution curves through a few points have the gradient the field prescribes there. For $\dfrac{dy}{dx} = -\dfrac{x}{y}$ the field segments are perpendicular to the radial lines, and the solution $x^2 + y^2 = K$ is a family of concentric circles, exactly the curves that cut every radius at right angles.

Sketching solution curves

Solutions follow the field

A solution curve through a given point is the curve that stays tangent to the segments everywhere. It never crosses a segment; it flows along them. Different starting points give different members of the solution family.

Where $\tfrac{dy}{dx} = 0$ the segments are horizontal; a horizontal line on which this holds for all $x$ is an equilibrium (constant) solution that other curves approach or leave.

Worked example

Describe the slope field of $\dfrac{dy}{dx} = x$ and sketch a solution

Here the gradient depends only on $x$ . Along any vertical line $x = c$ every segment has the same gradient $c$ : steeply negative for large negative $x$ , flat at $x = 0$ , steeply positive for large positive $x$ .

A solution curve has gradient matching $x$ , so it is concave up, flat where it crosses $x = 0$ , and steepening on either side. Indeed solving gives $y = \tfrac{x^2}{2} + C$ , a family of upward parabolas. A curve through the origin follows the field to give $y = \tfrac{x^2}{2}$ .

Common traps

Crossing the segments: A solution curve must be tangent to the field; it should not cut across the segments.
Treating it as a single curve: A slope field shows a whole family; each starting point gives a different solution.
Ignoring equilibrium lines: Horizontal segments along a constant $y$ signal an equilibrium solution worth marking.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2023-style6 marks

Calculator-assumed. A slope field is given by

\dfrac{dy}{dx} = y - 1

. (a) Describe the segments along the lines

y = 1

y = 2

and

y = 0

. (b) Identify any equilibrium solution and state whether it is stable.

Show worked answer →

Reading a slope field that depends only on $y$ .

(a) Since the gradient depends only on $y$ , all segments on a given horizontal line are parallel. On $y = 1$ : gradient $1 - 1 = 0$ , horizontal segments. On $y = 2$ : gradient $2 - 1 = 1$ , segments sloping up at $45^\circ$ . On $y = 0$ : gradient $0 - 1 = -1$ , segments sloping down at $45^\circ$ .

(b) The equilibrium is where $\dfrac{dy}{dx} = 0$ , i.e. $y = 1$ . Just above ( $y > 1$ ) the gradient is positive so curves move away upward; just below ( $y < 1$ ) the gradient is negative so curves move away downward. Solutions move away from $y = 1$ , so it is an unstable equilibrium.

Markers reward the parallel-on-horizontals observation, the three gradients, the equilibrium $y = 1$ , and the unstable classification.

WACE 2021-style4 marks

Calculator-free. The slope field for

\dfrac{dy}{dx} = -\dfrac{x}{y}

has segments perpendicular to the lines through the origin. Sketch the solution curve through

(0, 4)

and identify the family of solution curves.

Show worked answer →

Recognising circular solution curves.

The gradient $-\dfrac{x}{y}$ is the negative reciprocal of $\dfrac{y}{x}$ , the gradient of the radial line from the origin, so each segment is perpendicular to the radius at that point. Curves perpendicular to all radii are circles centred at the origin.

Solving by separation confirms it: $y\,dy = -x\,dx$ , so $\dfrac{y^2}{2} = -\dfrac{x^2}{2} + C$ , that is $x^2 + y^2 = K$ . Through $(0, 4)$ : $K = 16$ , the circle $x^2 + y^2 = 16$ of radius $4$ .

Markers reward the perpendicular-to-radius observation, the family of concentric circles, and the specific circle of radius $4$ through $(0, 4)$ .

Practise this

Sources & how we know this

Mathematics Specialist ATAR course: syllabus and support materials — SCSA
Past ATAR course examinations — SCSA