Slope fields | Year 12 | Ascent Library

01 / LEARN

The idea, step by step

Each short segment is local information

For $y^{'} = f (x, y)$ , calculate $f (x, y)$ at a grid point and draw a short line segment of that gradient. The segment shows the direction in which a solution through the point would move as $x$ increases. It is not a complete solution curve.

For $y^{'} = x - y$ , all points on $y = x$ have horizontal segments. Above that line $x - y < 0$ , so slopes are negative; below it slopes are positive. Drawing a few carefully chosen slopes is more useful than guessing a smooth curve.

Follow the directions, not the grid lines

A solution curve should be locally tangent to the field segments. Begin at the stated point and follow the slope directions smoothly. If the equation is autonomous, $y^{'} = f (y)$ , every point at the same height has the same slope.

An equilibrium is a constant solution. It occurs where $f (y) = 0$ for an autonomous equation. Check the signs above and below to decide whether nearby solutions move toward or away from the equilibrium. Do not call a line of zero slopes an equilibrium unless the entire constant function satisfies the equation.

A common trap

Every point with zero slope is a constant solution.

Instead: A constant solution must have the same value of $y$ for every $x$ and satisfy the equation everywhere.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

For $y^{'} = x - y$ , find the slopes at $(0, 1)$ , $(1, 1)$ and $(2, 1)$ .

Substitute the coordinates into $x - y$ .
The values are $0 - 1 = - 1$ , $1 - 1 = 0$ and $2 - 1 = 1$ .
Draw descending, horizontal and ascending short segments at those points.

- 1, 0, 1

Example 2

Worked example

For $y^{'} = 2 - y$ , describe the equilibrium and nearby solution behaviour.

Set $2 - y = 0$ to obtain $y = 2$ .
Below 2 the derivative is positive, and above 2 it is negative.
Solutions on both sides move toward $y = 2$ as $x$ increases.

y = 2

is a stable equilibrium.

Check your understanding

For $y^{'} = y - 3$ , is $y = 3$ stable?

Show answer

No; it is unstable.

Above 3 the derivative is positive and below 3 it is negative, so solutions move away.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find slopes for $y^{'} = x y$ at $(1, 2)$ , $(2, - 1)$ and $(0, 3)$ .
2 marks
Where are the slopes zero for $y^{'} = x + y$ ?
2 marks
For $y^{'} = y (4 - y)$ , identify constant equilibria.
2 marks
For $y^{'} = y (4 - y)$ , state slope signs in $y < 0$ , $0 < y < 4$ and $y > 4$ .
2 marks
For $y^{'} = x^{2}$ , what do segments at $(2, 1)$ and $(2, 5)$ have in common?
2 marks
For $y^{'} = - 2 y$ , describe whether a positive solution increases or decreases and whether it crosses the equilibrium.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find slopes for $y^{'} = x y$ at $(1, 2)$ , $(2, - 1)$ and $(0, 3)$ .
2 marks
Multiply the coordinates in each ordered pair.
Use each result as the local line gradient.
$2, - 2, 0$
Where are the slopes zero for $y^{'} = x + y$ ?
2 marks
Set $x + y = 0$ .
Solve for the line $y = - x$ ; this is not a constant equilibrium.
On $y = - x$ .
For $y^{'} = y (4 - y)$ , identify constant equilibria.
2 marks
Constant solutions need derivative zero.
$y (4 - y) = 0$ gives the two values.
$y = 0$ and $y = 4$ .
For $y^{'} = y (4 - y)$ , state slope signs in $y < 0$ , $0 < y < 4$ and $y > 4$ .
2 marks
Check the signs of both factors in each interval.
The sign pattern makes $y = 4$ stable for nearby positive values.
Negative, positive, negative.
For $y^{'} = x^{2}$ , what do segments at $(2, 1)$ and $(2, 5)$ have in common?
2 marks
The equation depends only on $x$ .
At $x = 2$ , $x^{2} = 4$ regardless of $y$ .
Both have gradient 4.
For $y^{'} = - 2 y$ , describe whether a positive solution increases or decreases and whether it crosses the equilibrium.
2 marks
For $y > 0$ , $- 2 y < 0$ .
$y = C e^{- 2 x}$ with $C > 0$ stays positive for every finite $x$ .
It decreases toward zero; the nonzero exponential solution does not cross zero.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

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