5.15 Slope fields

Learning objectives

• Calculate the gradient $\frac{\mathrm{d}y}{\mathrm{d}x}$ at a point from a differential equation, and explain how a slope field is built from these gradients.

• Use technology to draw the slope field of a differential equation.

• Sketch the solution curve through a given point on a slope field, following the segments smoothly and showing asymptotic behaviour.

• Explain why solution curves of a differential equation do not cross.

• Find the equation of an isocline, and in particular of the curve on which $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, and sketch it on a slope field.

• Determine where solution curves have local maxima and minima, and which is which, from the curve on which $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.

• Interpret patterns in a slope field, including fields of autonomous equations, equilibrium solutions and long-term behaviour, and match differential equations to their fields.

Introduction

Separation of variables, taught in subtopic 5.14, Setting up differential equations, solves only equations whose right-hand side splits into an $x$-part times a $y$-part. An equation as simple as $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$ cannot be separated, and many models from science cannot be solved exactly at all. Yet every first-order differential equation gives one piece of information directly: the gradient of the solution curve at any point. Drawing that gradient as a short line segment at many points produces a picture of all the solutions at once.

This picture, a slope field, shows the shape of the solutions without any formula: where solutions rise and fall, where they turn, what value they settle towards and which starting points lead to which long-term behaviour. Weather maps that show the wind as short arrows across a region use the same idea: a light object carried by the wind follows a path that runs along the arrows.

Slope fields: a differential equation as a picture

Gradients from the equation

A first-order differential equation written as $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y)$ gives the gradient of any solution curve that passes through the point $(x, y)$. For $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$, a solution through $(2, -1)$ has gradient $2 - (-1) = 3$ there, and a solution through $(1, 1)$ has gradient $0$: it is horizontal at that point. No solving is needed, only substitution.

Repeating this at every point of a grid gives a table of gradients. Each gradient is then drawn as a short segment through its grid point, a small piece of the tangent to whichever solution passes there. The table shows the gradients of $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$ at the integer points from $-2$ to $2$.

The zeros lie on a diagonal, where $y = x$; the gradients grow more positive towards the lower right and more negative towards the upper left. The same pattern appears in the finished field, drawn with a segment every half unit.

The Khan Academy video on creating a slope field builds one point by point in exactly this way, from a table of gradients to the completed picture.

YouTube

Generating a slope field with technology

Drawing hundreds of segments by hand is slow, so slope fields are generated with software. In GeoGebra the command SlopeField(x - y) draws the field of $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$ in the current window; the expression inside the brackets is the right-hand side of the equation, with $x$ and $y$ as the variables. Some GDCs have a differential-equation graphing mode that draws fields as well. Whatever the tool, the window should cover the region of interest, and a coarser grid (fewer segments) is often easier to read than a dense one.

Solution curves

Following the field

A solution of the differential equation is a function whose graph has, at every point, exactly the gradient the equation prescribes there. On a slope field this means the graph runs along the segments: it is tangent to the field at every point it passes through. Starting at a chosen point and following the directions of nearby segments traces out that solution, much as a leaf floating on a stream follows the current.

Each starting point gives a different particular solution, and together the solution curves form the general solution, the whole family. For $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$, the exact general solution is $y = x - 1 + C\mathrm{e}^{-x}$ (it can be checked by substitution, as in subtopic 5.14). Three of its curves are drawn on the field: through $(0, 2)$ with $C = 3$, through $(0, -2)$ with $C = -1$, and the straight line $y = x - 1$ with $C = 0$.

The curve through $(0, 2)$ starts above the dashed line $y = x$, where the segments slope downward, so it falls. It meets the line $y = x$ at about $(1.10, 1.10)$, where it is momentarily horizontal, and then rises below the line. The curve through $(0, -2)$ rises all the way. As $x$ increases, the term $C\mathrm{e}^{-x}$ shrinks to zero, so every solution approaches $y = x - 1$: this line is an asymptote of every other solution, which the field shows without any algebra.

The Khan Academy video on visualising solutions draws solution curves on a field in the same way, starting from different points and following the segments.

YouTube

Solution curves do not cross

Two solution curves of $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y)$ cannot cross. At a crossing point both would pass through the same $(x, y)$, so both would have the same gradient $f(x, y)$, and from there the equation fixes the gradient at every next point in the same way for both. The two curves would therefore be the same curve. Solutions may come very close (as all solutions of $x - y$ approach $y = x - 1$) but they never meet or cross. This holds wherever $f$ is a smooth function; at points where the equation is undefined, such as where a denominator is zero, the rule can fail.

How to sketch a solution curve

A good sketch on a given slope field shows the right shape, passes through the right point and has the right long-term behaviour. These steps produce one.

1. Mark the given point clearly.

2. From the point, move a little to the right in the direction of the nearest segments, then adjust to the new segments; repeat. The curve should run between segments parallel to them, never across them.

3. Do the same to the left of the point, so the curve covers the whole region shown.

4. Wherever the curve crosses the curve on which $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, make it horizontal: that is a turning point, a local maximum or minimum of the solution.

5. Show asymptotic behaviour: a curve approaching a line should get closer and closer without meeting it.

6. Draw one smooth curve with no corners and no flat sections that the field does not show.

Isoclines and the curve where dy/dx = 0

Isoclines

Points where the field has the same gradient lie on curves called isoclines (from the Greek for "equal slope"). For $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y)$, the isocline for gradient $c$ is the curve $f(x, y) = c$. Along it every segment is parallel, which makes isoclines a quick way to draw a field by hand: draw the curve, then add parallel segments along it.

For $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y$, the isocline for gradient $c$ is $x - y = c$, or $y = x - c$: a family of parallel straight lines. The isocline for $c = 0$ is $y = x$, and the isocline for $c = 1$ is $y = x - 1$. That second line has gradient 1 itself, equal to the gradient of every segment on it, so the line runs along its own segments and is a solution. This is why $y = x - 1$ appeared as a straight-line solution in the field of $x - y$.

The nullcline holds every turning point

The most useful isocline is the one with gradient zero. At a local maximum or local minimum of a solution, the gradient is 0 (subtopic 5.6, Stationary points), so every turning point of every solution lies on the curve $f(x, y) = 0$. Finding this curve locates all the maxima and minima before a single solution is drawn.

The nullcline also splits the plane into regions where solutions rise ($f > 0$) and where they fall ($f < 0$). For $x - y$, points above $y = x$ have $x - y < 0$, so solutions fall there; points below have $x - y > 0$, so solutions rise. A solution moving from the falling region into the rising region turns at a minimum, and one moving from the rising region into the falling region turns at a maximum. The curve through $(0, 2)$ in the field of $x - y$ passes from above $y = x$ to below it, so its turning point at $(\ln 3, \ln 3) \approx (1.10, 1.10)$ is a minimum.

The slope field of $\frac{\mathrm{d}y}{\mathrm{d}x} = x^2 - y$ has a nullcline that is a parabola rather than a line.

Deciding which turning points are maxima

For a nullcline that is a line or curve with a gradient of its own, the direction in which solutions cross it decides the type of turning point. A solution crosses the nullcline horizontally, so if the nullcline slopes down from left to right, every solution passes from below it to above it as $x$ increases; if the nullcline slopes up, every solution passes from above it to below it. Combined with the sign of $f$ on each side, this identifies maxima and minima without drawing any solution.

Patterns in slope fields

Equations that depend on x only

When the right-hand side contains only $x$, as in $\frac{\mathrm{d}y}{\mathrm{d}x} = \cos x$ or $\frac{\mathrm{d}y}{\mathrm{d}x} = 2x$, the gradient is the same at every point of a vertical line. The field repeats exactly as it moves up or down, so every solution curve is a vertical translation of every other. This matches the general solution $y = \int f(x)\,\mathrm{d}x + c$ of subtopic 5.14, where changing $c$ shifts the curve vertically.

Autonomous equations

When the right-hand side contains only the unknown, as in $\frac{\mathrm{d}P}{\mathrm{d}t} = 0.5P\left(1 - \frac{P}{10}\right)$, the rate of change does not depend on the time. The gradient is then the same along every horizontal line, so each row of the field is identical and solutions are horizontal translations of one another: starting the same population later gives the same curve shifted to the right.

The values of $y$ where $f(y) = 0$ give horizontal segments all the way across the field, so the horizontal lines at those heights are themselves solutions, the equilibrium solutions. The field of the logistic equation $\frac{\mathrm{d}P}{\mathrm{d}t} = 0.5P\left(1 - \frac{P}{10}\right)$ has two: $P = 0$ and $P = 10$.

Between 0 and 10 the segments slope upward, steepest at $P = 5$ where $\frac{\mathrm{d}P}{\mathrm{d}t} = 0.5 \times 5 \times 0.5 = 1.25$, so populations grow with an S-shaped curve towards 10, the carrying capacity. Above 10 the segments slope downward, so a population that starts above the carrying capacity falls towards it. Solutions on both sides approach $P = 10$, so it is a stable equilibrium; solutions near $P = 0$ move away from it, so $P = 0$ is unstable. The same reasoning applied to the cooling field $\frac{\mathrm{d}T}{\mathrm{d}t} = -0.1(T - 20)$ shows a single stable equilibrium at room temperature, $T = 20$.

Stability can be read from the signs of the rate on either side of each equilibrium, without drawing the field: if the rate is positive just below and negative just above, nearby solutions move towards the equilibrium from both sides.

Reading a slope field

Most features of a slope field can be predicted from its equation, and most features of an equation can be recognised in its field. The table collects the patterns that identify a field.

Slope fields in the world

Wind maps such as earth.nullschool.net show the wind across the globe as short moving lines on a grid, drawn from the Global Forecast System model of the United States National Oceanic and Atmospheric Administration (NOAA). Each line shows the wind's direction at that place, so the map is a direction field in two dimensions, and a balloon drifting with the wind follows a path tangent to the lines, as a solution curve follows its field. The analogy holds only while the wind pattern stays the same; real winds change from hour to hour, so a real balloon follows a changing field.

Exam focus

Slope fields are examined on HL Paper 1 in short questions of 2 to 7 marks, with the field printed on the paper. The usual parts are: calculate $\frac{\mathrm{d}y}{\mathrm{d}x}$ at a given point (substitution); write down or find the equation of the curve on which the maxima and minima of the solutions lie (set the right-hand side to zero and rearrange); sketch that curve on the field; sketch the solution curve through one or two given points. Markschemes award marks for the correct gradient, the curve through the given point, turning points on the nullcline and clear asymptotic behaviour.

The examiners' reports describe these questions as often left blank, and the commonest error as drawing a solution through the point where $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ instead of the curve of all such points. Sketches also lose marks for flat sections, missing turning points and curves that touch, cross or stop at an asymptote.

The slope field of $\frac{\mathrm{d}y}{\mathrm{d}x} = \sin x - \frac{y}{2}$, drawn for $-1 \le x \le 7$ and $-2.5 \le y \le 3$, is used in the following question.

Part (e) needs no solving of the differential equation: once the $x$-coordinate of a turning point is known, the nullcline gives its $y$-coordinate. Questions of this kind reward understanding of the link between $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ and turning points more than calculation.

Practice questions

The slope field of $\frac{\mathrm{d}y}{\mathrm{d}x} = y - x$ for $-3 \le x \le 3$ and $-3 \le y \le 3$ is used in questions 1 to 4.

1. Calculate the gradient of the solution curve through $(1, 3)$ and through $(2, -2)$. [2]

2. Write down the equation of the curve on which every segment is horizontal. [1]

3. Show that the line $y = x + 1$ is a solution curve of the differential equation. [2]

4. (a) Sketch the solution curve through $(0, 0)$. (b) Explain why it has a local maximum at the origin. (c) Describe the long-term behaviour of the solutions through $(0, 2)$ and $(0, 0)$ as $x$ increases. [6]

5. Find the equation of the isocline on which the gradient is 2 for $\frac{\mathrm{d}y}{\mathrm{d}x} = xy$, and describe the slope field on the coordinate axes. [3]

6. For $\frac{\mathrm{d}y}{\mathrm{d}x} = x - y^2$, write down the equation of the curve on which all turning points of solutions lie, and describe its shape. [2]

7. For the autonomous equation $\frac{\mathrm{d}y}{\mathrm{d}t} = y(3 - y)(y + 2)$, find the equilibrium solutions and determine whether each is stable or unstable. [5]

8. Explain why every solution curve of $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 4$ is a vertical translation of the curve through the origin, and write down the equation of that curve. [3]

9. State, with a reason, which of $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x}$ and $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y}{x}$ has a slope field in which the segments all point along lines through the origin. [2]

Summary

• A slope field draws a short segment with gradient $f(x, y)$ at each grid point for $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y)$; the gradient at a point comes from substituting into the equation.

• Software draws fields, for example GeoGebra's SlopeField(f(x, y)); checking two or three gradients confirms the field is right.

• A solution curve is tangent to the segments everywhere; sketch it smoothly through the given point, in both directions, showing turning points and asymptotes.

• Solution curves never cross, because at a shared point they would have the same gradient and so be the same curve.

• An isocline $f(x, y) = c$ joins points with equal gradient; an isocline that is a line with gradient $c$ is itself a solution.

• The nullcline $f(x, y) = 0$ holds every local maximum and minimum of every solution; it is not itself a solution in general.

• A solution passing from a region where $f > 0$ to one where $f < 0$ has a maximum on the nullcline; from $f < 0$ to $f > 0$, a minimum.

• Fields of $\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$ repeat vertically (solutions are vertical translations); fields of autonomous equations $\frac{\mathrm{d}y}{\mathrm{d}x} = f(y)$ repeat horizontally, with equilibrium solutions where $f(y) = 0$, stable when nearby solutions approach them.

Self-check

• To do:

  I can calculate $\frac{\mathrm{d}y}{\mathrm{d}x}$ at any point by substituting into the differential equation.

• To do:

  I can explain how a slope field is built from a grid of gradients.

• To do:

  I can generate a slope field with technology and check it against the equation.

• To do:

  I can sketch a smooth solution curve through a given point in both directions.

• To do:

  I can explain why two solution curves cannot cross.

• To do:

  I can find the equation of an isocline for a given gradient.

• To do:

  I can find and sketch the curve on which $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, and avoid drawing a solution through a single point instead.

• To do:

  I can decide whether a turning point on the nullcline is a maximum or a minimum from the signs of $\frac{\mathrm{d}y}{\mathrm{d}x}$ on each side.

• To do:

  I can use the nullcline to find a coordinate of a turning point.

• To do:

  I can recognise an autonomous equation from its field and find its equilibrium solutions.

• To do:

  I can decide whether an equilibrium is stable or unstable from the field or the sign of the rate.

• To do:

  I can show approach to an asymptote clearly without touching or crossing it.

• To do:

  I can match a differential equation to its slope field from its features.