Slope fields
Master slope fields for Year 12 Specialist Mathematics in Queensland (QCAA, Unit 4). A slope field is a grid of short segments that pictures a first-order differential equation — each segment has the gradient the equation gives at that point, so you can see how solutions behave without solving anything.
You will learn to read the slope at any grid point, find equilibrium solutions where the field is horizontal, match a differential equation to its field, and sketch a solution curve through a given point — the qualitative view that underpins the logistic equation and Newton’s law of cooling later in the course.
Theory
A slope field (direction field) pictures a first-order differential equation without solving it. In Year 12 Specialist Mathematics (QCAA, Queensland, Unit 4) it is a grid of short segments, each with the gradient the equation \(\dfrac{dy}{dx}=f(x,y)\) gives at that point. This page shows how to draw a field, read its equilibria and behaviour, match a field to its equation, and sketch a solution curve through a point.
A first-order differential equation \(\dfrac{dy}{dx}=f(x,y)\) states the gradient of a solution curve at every point of the plane. A slope field (also called a direction field or gradient field) makes this visible: at a grid of points \((x,y)\) you draw a short segment whose gradient is the value of \(f(x,y)\) there.
The segment at \((x_0,y_0)\) is horizontal where \(f=0\), steep where \(|f|\) is large, and rises or falls with the sign of \(f\). Reading the field means evaluating \(f\) at points, not solving anything.
A solution curve (integral curve) is a curve that is tangent to the field everywhere — it always follows the segments. Exactly one solution curve passes through each point, so a solution curve never crosses another.
An equilibrium (constant) solution is a horizontal line \(y=c\) where \(f=0\) all along it; the field is a row of horizontal segments there. Nearby solutions may bend towards it (stable) or away from it (unstable), which is how the field reveals long-term behaviour.
A first-order differential equation gives the gradient at each point:
The slope of the segment at a grid point \((x_0,y_0)\) is just the value there:
Equilibrium (constant) solutions are the lines where the field is horizontal:
How to draw and read a slope field
- Evaluate \(f(x,y)\) at each grid point to get the slope there.
- Draw a short segment of that gradient centred on the point (steeper where \(|f|\) is larger, horizontal where \(f=0\)).
- Find the equilibria: solve \(f=0\) for the horizontal lines, and check the sign of \(f\) on each side to see if solutions bend towards or away.
- Sketch a solution curve from a given point by moving so the curve stays tangent to the segments — following the field without crossing an equilibrium line.
Substitute each point into \(f(x,y)=x+y\):
| \(\text{at }(0,2):\ \dfrac{dy}{dx}\) | \(=\) | \(0+2\) |
| \(=\) | \(2\) | |
| \(\text{at }(-2,1):\ \dfrac{dy}{dx}\) | \(=\) | \(-2+1\) |
| \(=\) | \(-1\) |
The segment slopes are \(2\) at \((0,2)\) and \(-1\) at \((-2,1)\).
Horizontal segments have zero slope; set \(2x-y=0\):
| \(2x-y\) | \(=\) | \(0\) |
| \(y\) | \(=\) | \(2x\) |
Now the slope at the point \((0,3)\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(2(0)-3\) |
| \(=\) | \(-3\) |
The segments are horizontal along \(y=2x\); at \((0,3)\) the slope is \(-3\).
Constant solutions have \(\dfrac{dy}{dx}=0\); solve \(y(3-y)=0\):
| \(y(3-y)\) | \(=\) | \(0\) |
| \(y\) | \(=\) | \(0 \ \text{ or }\ y=3\) |
Check the sign of the slope at \(y=1\) (between the equilibria):
| \(\dfrac{dy}{dx}\) | \(=\) | \(1(3-1)\) |
| \(=\) | \(2 > 0\) |
The equilibria are \(y=0\) and \(y=3\). From \(y=1\) the slope is positive, so the solution rises and approaches \(y=3\).
The field depends on \(x\) only; antidifferentiate, then use \((0,1)\):
| \(y\) | \(=\) | \(\dfrac{x^2}{2}+c\) |
| \(1\) | \(=\) | \(\dfrac{0^2}{2}+c\) |
| \(c\) | \(=\) | \(1\) |
| \(y\) | \(=\) | \(\dfrac{x^2}{2}+1\) |
The solution curve is \(y=\dfrac{x^2}{2}+1\); it follows the field with its minimum at \((0,1)\).
Common pitfalls
Frequently asked questions
What is a slope field in Specialist Maths?
It is a grid of short segments, one at each point, whose gradient is the value of \(\dfrac{dy}{dx}=f(x,y)\) there. It pictures the differential equation without solving it.
How do you draw a slope field?
Evaluate \(f(x,y)\) at each grid point and draw a short segment of that slope: horizontal where \(f=0\), steeper where \(|f|\) is large, rising or falling with the sign of \(f\).
What is an equilibrium solution?
A constant solution \(y=c\) where \(f=0\) all along the line, so the field is a row of horizontal segments. Checking the sign of \(f\) on each side shows whether nearby solutions bend towards it or away.
How do you sketch a solution curve from a slope field?
Start at the given point and move so the curve stays tangent to the segments, following the field. The curve never crosses the segments and never crosses an equilibrium line.
How do you match a differential equation to its slope field?
Find where the field is horizontal (solve \(f=0\)) and test the slope at one or two grid points. The equation whose zero-slope locus and sample slopes match the picture is the right one.