A curve along which the slope is constant, such as dy by dx equals zero giving horizontal segments.

How do you match a field to an equation?

Check a few sample points; the segment slopes should equal dy by dx there.

A curve along which the slope is constant, such as dy by dx equals zero giving horizontal segments.

How do you match a field to an equation?

Check a few sample points; the segment slopes should equal dy by dx there.

Year 12 Maths Extension 1 (2027) Further applications of calculus

Slope (direction) fields

Q: What is a slope field?

A grid of short segments whose slopes equal dy by dx from the equation at each point, showing how solutions flow.

Q: How do you find the slope at a point?

Substitute both coordinates into dy by dx.

20 practice questions 2 video lessons Theory + worked examples

Understand slope fields for NSW Year 12 Mathematics Extension 1. A slope field draws the gradient given by a differential equation at many points, revealing the shape of its solutions even when the equation is hard to solve.

You will learn to read and construct direction fields, match a field to its differential equation, and sketch the solution curve through a given initial condition — a visual way to understand differential equations in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 2 video(s)

ODE | Slope fields Watch
Slope Fields | Calculus Watch

Theory

A slope field draws a short segment at each point with gradient $\frac{d y}{d x}$ from the differential equation, so solution curves follow the segments. This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-05.

A slope field (direction field) draws a short segment at each point with gradient $\frac{d y}{d x}$ taken from the differential equation. Solution curves follow these segments — especially useful when the DE cannot be solved exactly.

At each point $(x, y)$ , the segment's slope is the value of $\frac{d y}{d x}$ there. A solution curve threads through the field, staying tangent to every segment it meets.

An isocline is a curve where the slope is constant — for example, $\frac{d y}{d x} = 0$ gives horizontal segments.

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. A syllabus task: choose the slope field that represents $\frac{d y}{d x} = x - y$ .

\frac{d y}{d x} = x - y

: horizontal along

y = x

; a solution threads through.

\frac{d y}{d x} = - \frac{x}{y}

: the solution curves are circles

x^{2} + y^{2} = C

slope at (x, y) = {\frac{d y}{d x} |}_{(x, y)} .

slope at (x,y) equals dy/dx evaluated at that point

Horizontal segments occur where $\frac{d y}{d x} = 0$ ; vertical segments where $\frac{d y}{d x}$ is undefined. Segments of equal slope lie on an isocline.

Read both coordinates. The slope generally depends on $x$ and $y$ , so substitute the full point — a solution curve is tangent to the field, never crossing a segment.

How to use a slope field

Evaluate the slope $\frac{d y}{d x}$ at the point, substituting both coordinates.
Find special features: set $\frac{d y}{d x} = 0$ for horizontal segments (an isocline).
Sketch a solution by following the segments, keeping tangent to them.
Match a field to a DE by checking a few sample slopes.

Example 1 — Slopes at points

For

\frac{d y}{d x} = 2 x - y

, find the slope at

(1, 3)

(0, - 1)

and

(2, 2)

Solution

At $(1, 3)$ : $2 - 3 = - 1$ . At $(0, - 1)$ : $0 + 1 = 1$ . At $(2, 2)$ : $4 - 2 = 2$ .

slopes -1, 1, 2

Slopes $- 1$ , $1$ , $2$ .

Example 2 — Slope on a curve

The curve

\frac{d y}{d x} = \frac{x}{y}

passes through

(3, 2)

. Find the slope there.

Solution

\frac{d y}{d x}

=

\frac{3}{2}

slope = 3/2

The slope is $\frac{3}{2}$ .

Example 3 — Pick the solution curve

\frac{d y}{d x} = - \frac{x}{y}

has solution curves

x^{2} + y^{2} = C

. Find the one through

(6, 8)

Solution

C

=

6^{2} + 8^{2} = 100

x^2 + y^2 = 100

$x^{2} + y^{2} = 100$ .

Example 4 — Horizontal segments

For

\frac{d y}{d x} = x - y

, find where the field segments are horizontal.

Solution

Set $\frac{d y}{d x} = 0$ .

x - y

=

0 \Rightarrow y = x

horizontal along the line y = x

Along the line $y = x$ , the segments are horizontal.

Common pitfalls

Using only $x$ . The slope usually depends on both coordinates — substitute

x

and

y

Crossing segments. Solution curves run tangent to the segments; they never cut across them.

Missing isoclines. Segments of equal slope lie on an isocline;

\frac{d y}{d x} = 0

gives the horizontal ones.

Undefined slopes. Where

\frac{d y}{d x}

is undefined (division by zero), segments are vertical.

Frequently asked questions

What is a slope field?

A grid of short segments whose slopes equal $\frac{d y}{d x}$ from the DE at each point, showing how solutions flow.

How do you find the slope at a point?

Substitute both coordinates into $\frac{d y}{d x}$ .

What is an isocline?

A curve along which the slope is constant, such as $\frac{d y}{d x} = 0$ giving horizontal segments.

How do solution curves relate to the field?

They stay tangent to the segments and never cross them.

How do you match a field to a DE?

Check a few sample points — the segment slopes should equal $\frac{d y}{d x}$ there.

← Previous subtopic

Differential equations – introduction & order

Next subtopic →

Solve dy/dx = f(x)

More in Further applications of calculus

Slope (direction) fields

📖 Theory

How to use a slope field

Common pitfalls

Frequently asked questions

Theory