Picture: a field of tiny tangent arrows: Imagine the (x, y)-plane covered in a grid of dots, and at every dot a short line segment drawn like a tiny ruler.
For a differential equation dy/dx = f(x, y) you cannot solve at every point, but you can find the gradient of a solution curve passing through any point: just substitute that point into f.
That gradient is the slope of the little segment drawn at the dot. The whole picture of segments is the slope field (also called the direction field) — a map of which way solution curves head everywhere.
IB-style question — reading a slope field for a savings model
The balance y (in thousands of euros) of an investment after x years is modelled by dy/dx = x + 0.5y.
(a) Find the gradient of the slope field at the point (2, 4).
(b) Describe, in words, the segment drawn at (2, 4).
Step by step
- (a) The gradient at a point is just f(x, y) evaluated there, so substitute x = 2 and y = 4.
- Work it out.
- (b) A gradient of 4 is steep and positive, so the little segment at (2, 4) slopes steeply uphill from left to right.
Final answer
(a) The gradient is 4. (b) The segment at (2, 4) is steep and rising — a solution curve through (2, 4) is increasing fast there, i.e. the balance is growing by about €4000 per year at that instant.
Build a small table of gradients: A neat exam habit: pick a few grid points and tabulate f(x, y). For dy/dx = x + 0.5y:
• (0, 0) → 0 (flat segment) • (1, 2) → 1 + 1 = 2 (gentle uphill) • (2, 4) → 2 + 2 = 4 (steep uphill) • (−1, −2) → −1 − 1 = −2 (downhill)
Reading the table left-to-right tells you the segments tilt upward more steeply as x and y grow, and tilt downward in the third quadrant — exactly the shape of the field.
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Isoclines: lines of equal tilt; f = 0: the flat line: An isocline is the set of points where the segments all have the same gradient — the curve f(x, y) = c. Along one isocline every little segment is parallel.
The most useful isocline is f(x, y) = 0: there the segments are flat (horizontal). A solution curve is momentarily level when it crosses this curve, so f = 0 is where solution curves have their maximum or minimum points (their stationary points).
To sketch a solution curve through a given starting point, place your pencil there and glide so the curve is always tangent to the segments it passes — like a leaf carried by the field.
IB-style question — cooling coffee slope field
A cup of coffee cools according to dy/dx = −0.2(y − 20), where y is the temperature in °C and x is the time in minutes (room temperature is 20°C).
(a) Find the equation of the isocline where the segments are flat, and explain its meaning.
(b) State the sign of the gradient at the point (0, 70), and what it tells you about the cooling.
Step by step
- (a) Flat segments occur where the gradient is zero, so set f(x, y) = 0.
- This is the horizontal line y = 20: along it every segment is flat, so a solution curve that reaches 20°C stops changing — it is the equilibrium temperature.
- (b) Substitute (0, 70) into f to get the gradient there.
- The gradient is negative, so the segment slopes downhill — the coffee is cooling at that instant.
Final answer
(a) The flat-segment isocline is y = 20°C: solution curves level off there because it is room temperature (the equilibrium). (b) At (0, 70) the gradient is −10, so the segment slopes steeply down — the coffee is cooling by about 10°C per minute at the start. As y approaches 20 the segments flatten, so the curve flattens out toward 20°C.
How a sketched solution curve behaves: Once you know the flat locus and a few signs, the solution curve's story tells itself:
• Where f > 0 the curve is increasing (segments tilt up). • Where f < 0 the curve is decreasing (segments tilt down). • Where f = 0 the curve has a turning point (a max if it changes + to −, a min if − to +).
The curve never crosses a segment — it always runs along the local direction, so its shape is forced by the field once you fix a starting point.