Skip to main content
QLDSpecialist MathematicsSyllabus dot point

Topic 2: Rates of change and differential equations

Construct and interpret slope (direction) fields of a first-order differential equation, sketch solution curves through given points, and relate the qualitative behaviour of solutions to the field

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on slope fields. Covers building a direction field from dy/dx, reading slopes at grid points, sketching solution curves through given initial points, and identifying equilibrium solutions, with a verified worked example and the curve-crossing trap.

Generated by Claude Opus 4.76 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

QCAA wants you to understand a first-order differential equation dydx=F(x,y)\frac{dy}{dx} = F(x, y) geometrically, before solving it. A slope field draws a short line segment at each grid point with gradient given by FF, and the solution curves follow these segments. The assessable skill is reading and constructing the field and sketching solution curves that stay tangent to it.

The answer

What a slope field shows

A first-order differential equation dydx=F(x,y)\frac{dy}{dx} = F(x, y) assigns a slope to every point (x,y)(x, y) in the plane. The slope field (direction field) is a grid of short segments, each drawn with the gradient F(x,y)F(x, y) at its location. A solution curve is any curve that is tangent to these segments everywhere it passes.

Constructing the field

At each grid point, evaluate F(x,y)F(x, y) to get the slope, then draw a short segment with that gradient. For example, if dydx=x\frac{dy}{dx} = x, the slope at (2,1)(2, 1) is 22, so the segment there rises steeply; along the yy-axis (x=0x = 0) all segments are horizontal.

Reading qualitative behaviour

The field reveals behaviour without an explicit formula. Where F(x,y)=0F(x, y) = 0, segments are horizontal, marking possible turning points or equilibria. Where FF is large, the field is steep. Regions where F>0F > 0 have solutions increasing left to right; where F<0F < 0, decreasing.

Equilibrium solutions

If dydx=F(y)\frac{dy}{dx} = F(y) depends only on yy, then any value y=cy = c with F(c)=0F(c) = 0 gives a constant (equilibrium) solution: a horizontal line that the field segments lie along. Nearby solution curves either approach (stable) or move away from (unstable) this equilibrium.

Sketching a solution through a point

Given an initial point, start there and draw a smooth curve that follows the direction of the nearby segments, always staying tangent to the field. The curve threads through the segments like a path following arrows. Different initial points give different members of the solution family.

Solutions do not cross

For a well-behaved equation, distinct solution curves never cross, because the slope at any point is uniquely determined by F(x,y)F(x, y). Two curves crossing would require two different tangents at one point, which is impossible.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 QCAA4 marksThe slope field for the differential equation dy/dx = -0.5(y - 4)/x, x not equal to 0, using -6 less than or equal to x less than or equal to 6 and -6 less than or equal to y less than or equal to 6 is shown. Point A is marked at (-4, -2). a) Determine the value of the slope at point A. b) Use the slope field to sketch the solution curve for the equation given that when x = -6, y = 3.5.
Show worked answer →

a) [2 marks] Substitute the coordinates of A = (-4, -2) into dy/dx = -0.5(y - 4)/x:
dy/dx = -0.5(-2 - 4)/(-4) = -0.5(-6)/(-4) = 3/(-4) = -3/4.
So the slope at A is -3/4.

b) [2 marks] Starting at (-6, 3.5), the curve sits in the second quadrant where y < 4 and x < 0, so dy/dx = -0.5(y - 4)/x is negative; the curve falls as it moves right. As x approaches 0 from the left, the 1/x factor makes the slope steep, so the curve is asymptotic to the y-axis in the third quadrant. A correct sketch follows the field's tangent segments through (-6, 3.5), has a negative slope in the second quadrant, and becomes asymptotic near x = 0; there is no solution curve crossing into x > 0 from this branch (the line x = 0 is excluded).