Slope Fields (AI HL)
A slope field lets you picture the whole family of solutions to a differential equation before you've solved anything. At every grid point, a short line segment shows the gradient a solution curve would have if it passed through that point - so reading the diagram is really just repeated substitution into \(f(x,y)\). This page covers how to interpret a slope field and the mistakes that lose marks. It's part of the broader Differential Equations topic.
20 questions on this sub-topic.
Reading a slope field
Covered under IB syllabus reference AHL5.15: slope fields and their diagrams. Students will be required to use and interpret slope fields. There's no booklet formula to learn here - the skill is substitution and reading a diagram.
Gradient at a point
\(\dfrac{dy}{dx} = f(x,y)\)
Substitute the coordinates of any point into \(f(x,y)\) - the result is the gradient of the tangent segment drawn there, whether or not you can solve the equation itself.
Sketching a solution curve
Start at the given initial point and draw a smooth curve that stays tangent to the local segments at every point you pass through.
The curve should flow with the segments' direction, not cross them at an angle.
Need the full syllabus wording and the wider calculus formula table? See Differential Equations.
Worked example
The slope field for \(\dfrac{dy}{dx} = x - y\) is shown.
State the value of \(\dfrac{dy}{dx}\) at the point \((2, -1)\).
Worked solution
\(\dfrac{dy}{dx} = 2-(-1)\) M1
\(= 3.\) A1
A quantity satisfies \(\dfrac{dy}{dx} = (y-2)(y-6).\)
(a)(i) Find the equilibrium solution with \(y<4.\)
(a)(ii) Find the equilibrium solution with \(y>4.\)
(b) Determine, with a reason, whether the equilibrium at \(y=2\) is stable or unstable.
(c) State the long-term behaviour of a solution starting at \(y = 4.\)
Worked solution
(a)(i) \((y-2)(y-6)=0\) M1
\(\Rightarrow y=2\) A1
(a)(ii) or \(y=6.\) A1
(b) For \(y\) slightly below \(2,\) both factors are negative so \(\dfrac{dy}{dx}>0\) (increasing toward 2); R1
for \(y\) slightly above \(2\) (but below 6), the first factor is positive and the second negative so \(\dfrac{dy}{dx}<0\) (decreasing toward 2). Solutions approach \(y=2\) from both sides, so it is stable. R1
(c) Since \(4\) lies strictly between the two equilibria, \(\dfrac{dy}{dx}<0\) there, M1
so the solution decreases toward \(y=2.\) A1
Common mistakes
- Substituting \(x\) and \(y\) the wrong way round. A point is written \((x,y)\) - swapping the coordinates before substituting into \(f(x,y)\) gives a gradient for the wrong point entirely.
- Treating the slope field as the solution curve itself. The diagram only shows local gradients; the actual solution through a given point is the smooth curve you sketch by following those segments, and different starting points give different curves.
- Ignoring the sign of the gradient when sketching. A positive gradient segment slopes up to the right and a negative one slopes down to the right - matching the direction, not just the steepness, is what makes a sketch match the field.
Ready to practise properly?
20 slope-field questions, marked instantly like the real exam.
Quick answers
What does a slope field show?
A slope field draws a short line segment at each grid point with gradient \(f(x,y)\), for a differential equation \(\dfrac{dy}{dx} = f(x,y)\). Following the segments from a starting point traces out an approximate solution curve without ever solving the equation algebraically.
How do I find the gradient at a point on a slope field?
Substitute that point's \(x\) and \(y\) coordinates into \(f(x,y)\) from the differential equation \(\dfrac{dy}{dx} = f(x,y)\). The result is the gradient of the tangent segment drawn at that point.
Where do I do this on my GDC?
See the Using your GDC section on the Differential Equations page for calculator-specific steps.