| Term / concept | Short meaning / formula |
|---|---|
| Slope (direction) at a point | Value of dy/dx = f(x,y) evaluated at that (x,y); drawn as a small line segment with that slope. |
| Slope (direction) field | Grid of short line segments at sample (x,y) showing direction dy/dx = f(x,y) — visualises solution behaviour. |
| Solution curve (integral curve) | Curve tangent at every point to the segment shown in the slope field; passes through chosen initial point. |
📌 What a slope field is (plain explanation)
- A slope field (direction field) is a graphical representation of the differential equation dy/dx = f(x,y): at many grid points (x,y) we draw a short line segment whose slope equals f(x,y).
- It does not give exact solutions but shows the family of possible solution curves and their qualitative behaviour (increasing/decreasing, tendencies, equilibrium lines).
- Integral curves (solutions) are curves that at every point are tangent to the short segments; different initial conditions pick different integral curves.
🧠 Examiner tip
- When asked to sketch a slope field, compute and label slopes at a few representative grid points (method marks).
- If asked for behaviour as x→∞, use slope signs to explain whether solutions increase or decrease and whether they approach equilibrium curves.
- If the question gives dy/dx = f(x,y) and an initial point, indicate the direction of movement from that point (up/down/left/right) using computed slope sign.
📌 How to calculate and draw a slope field (step-by-step)
- 1) Choose a grid: pick x-values and y-values (e.g., x = −2, −1, 0, 1, 2 and y = −2, −1, 0, 1, 2).
- 2) Evaluate f(x,y): for each grid point compute m = f(x,y) = dy/dx. Example: if dy/dx = x − y then at (1,0) m = 1 − 0 = 1.
- 3) Convert slope to a short segment: draw a small line centered at (x,y) with slope m. For steep slopes, make the segment short; for m = 0 draw horizontal small line.
- 4) Repeat: do for every grid point — the pattern of segments is the slope field.
- 5) Sketch solution curves: from a chosen initial point follow the local tangents to trace the integral curve that stays tangent to each segment.
🌍 Real-world connection
- Slope fields help visualize differential models in physics (e.g., velocity fields), ecology (population change rate depending on population and environment) and engineering (temperature gradients).
- They are used when closed-form solutions are difficult — slope fields show qualitative behaviour without solving analytically.
📌 Practical worked example (step-by-step with grid calculations)
Problem: Consider dy/dx = x − y. Draw a small slope field on the grid x = −1, 0, 1 and y = −1, 0, 1 and sketch the integral curve through (0,1).Step A — compute slopes at grid points (table):
| (x,y) | dy/dx = x − y |
|---|---|
| (−1,1) | −1 − 1 = −2 |
| (0,1) | 0 − 1 = −1 |
| (1,1) | 1 − 1 = 0 |
| (−1,0) | −1 − 0 = −1 |
| (0,0) | 0 − 0 = 0 |
| (1,0) | 1 − 0 = 1 |
| (−1,−1) | −1 − (−1) = 0 |
| (0,−1) | 0 − (−1) = 1 |
| (1,−1) | 1 − (−1) = 2 |
Interpretation: at (0,1) slope = −1 → small line segment of slope −1; at (1,1) slope = 0 → horizontal tiny segment; at (1,−1) slope = 2 → steep upwards segment.
Step B — draw segments: for each grid point draw a short segment centered at (x,y) with the computed slope (keep segments short to avoid overlap). The pattern shows tendency: around y ≈ x the slopes are small; above that slopes negative (downward); below that slopes positive (upward).
Step C — sketch integral curve through (0,1): start at (0,1) and follow local tangent direction given by segments (slope −1) — the curve will move down-right then approach a direction where slope tends to zero (near line y = x). This visual method shows the qualitative behaviour: solutions move toward the line y = x as x increases.
🔍 TOK perspective
- Slope fields show qualitative knowledge — discuss strengths and limits of qualitative vs analytic solutions.
- Consider how choice of sampling grid influences the representation (visual bias).
❤️ CAS / community
- Work with local environmental group to model pollutant dispersion rate dy/dt = f(x,y) and visualise direction fields for policy discussion.
📌 Short practice question (for self-test)
Q: For dy/dx = y(1 − x), compute slopes at (0,0), (1,1), (0,1) and explain whether solutions starting at (0,1) initially increase or decrease.Solution (brief):
• At (0,0): dy/dx = 0(1 − 0) = 0 → horizontal segment.
• At (1,1): dy/dx = 1(1 − 1) = 0 → horizontal segment (equilibrium line x = 1 gives slope 0 when y nonzero).
• At (0,1): dy/dx = 1(1 − 0) = 1 → upward segment (solution through (0,1) initially increases).
Explain: since slope positive at (0,1) the integral curve rises as x increases from 0.
• At (0,0): dy/dx = 0(1 − 0) = 0 → horizontal segment.
• At (1,1): dy/dx = 1(1 − 1) = 0 → horizontal segment (equilibrium line x = 1 gives slope 0 when y nonzero).
• At (0,1): dy/dx = 1(1 − 0) = 1 → upward segment (solution through (0,1) initially increases).
Explain: since slope positive at (0,1) the integral curve rises as x increases from 0.
📝 Paper tip
- When asked to sketch, include at least 6–9 labeled grid points with computed slopes — examiners expect explicit calculations at some points.
- If asked to describe behaviour, refer to equilibrium curves (where f(x,y)=0) and sign of f on either side.