Direction Fields (Slope Fields)
At every point $(x, y)$ in the plane, the differential equation $\dfrac{dy}{dx} = f(x, y)$ prescribes a slope. Draw a tiny line segment with that slope at each grid point and the picture reveals every possible solution at once — without solving the equation. This visual tool is called a direction field (or slope field). In this lesson you will learn to read, sketch, and use direction fields to trace solution curves.
Consider the differential equation $\dfrac{dy}{dx} = x$. Without solving it — describe what you expect the solution curves to look like and how the slopes change as $x$ increases.
A direction field turns a differential equation into a visual. At each grid point $(x_0, y_0)$, evaluate $\dfrac{dy}{dx} = f(x_0, y_0)$ and draw a short line segment with that slope. The collection of all segments reveals the flow of solutions.
To sketch by hand:
- Choose a grid of $(x, y)$ points.
- At each point, compute the slope $m = f(x, y)$.
- Draw a short segment of slope $m$ centred at the point.
- To sketch a solution: start at an initial condition $(x_0, y_0)$ and follow the flow.
Key facts
- A direction field plots the slope $f(x,y)$ at each grid point as a short segment
- Isoclines are curves where $f(x,y) = c$ (constant slope)
- Solution curves follow the flow of the field and never cross
Concepts
- Why direction fields represent the entire family of solutions simultaneously
- How to identify equilibrium solutions from horizontal segments
- The connection between initial conditions and particular solution curves
Skills
- Evaluate and plot slope segments for a given $\dfrac{dy}{dx} = f(x,y)$
- Sketch solution curves through a specified initial condition
- Use isoclines to simplify the sketching process
Consider $\dfrac{dy}{dx} = x$. The slope at $(x,y)$ depends only on $x$, not on $y$. This means all points in a vertical column share the same slope — the isoclines are vertical lines $x = c$.
Slopes become steeper and positive to the right, and steeper and negative to the left. The horizontal isocline is $x = 0$ (the $y$-axis). To sketch a solution, place your pen at the initial condition and move in the direction the segments point — the curve should be tangent to every segment it passes through.
Answer to the hook: The solutions to $\dfrac{dy}{dx} = x$ are $y = \dfrac{x^2}{2} + C$ — upward-opening parabolas. Every parabola in this family is a solution; the direction field reveals this family without any algebra.
Consider $\dfrac{dy}{dx} = x$. The slope at $(x,y)$ depends only on $x$ , not on $y$. This means all points in a vertical column share the same slope — the isoclines are vertical lines $x = c$.
Pause — copy the three features to read from a direction field: where $\frac{dy}{dx}=0$ (horizontal tangents), where $\frac{dy}{dx}>0$ (increasing), and where $\frac{dy}{dx}<0$ (decreasing) into your book.
Quick check: For the direction field of $\dfrac{dy}{dx} = y$, what are the isoclines?
We just saw that in a direction field each slope segment at $(x,y)$ shows the value of $\frac{dy}{dx}$ there, and the pattern of segments reveals where solutions increase, decrease, or have horizontal tangents. That raises a question: given an initial condition $y(x_0)=y_0$, how do you trace a solution curve through that point using the field? This card answers it → start at $(x_0,y_0)$ and follow the local slope segments in both directions, staying tangent to the field at every point.
Given a direction field and an initial condition $y(x_0) = y_0$:
- Locate the point $(x_0, y_0)$ on the direction field.
- Move to the right by following the slope segments (increasing $x$).
- Move to the left as well to trace the complete curve.
- The curve must be smooth and tangent to every segment it crosses.
Because the initial condition is unique, each starting point gives a distinct solution curve. Two solution curves cannot intersect (uniqueness theorem).
Given a direction field and an initial condition $y(x_0) = y_0$:
Pause — copy the curve-tracing rule: place your pencil at the initial condition $(x_0,y_0)$ and follow the slope segments, adjusting direction continuously to stay tangent to the field into your book.
Did you get this? True or false: two distinct solution curves in a direction field can intersect at a point.
Worked examples · 3 in a row, reveal as you go
For $\dfrac{dy}{dx} = x - y$, compute the slope at the points $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$.
Find the equilibrium solutions of $\dfrac{dy}{dx} = y(2 - y)$ and describe the behaviour of solution curves near each one.
A direction field shows that all slope segments are horizontal along $y = x^2$. Which differential equation matches this field?
Fill the gap: For $\dfrac{dy}{dx} = x - y$, the zero-slope isocline is the line $y =$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $\dfrac{dy}{dx} = y(2-y)$, the equilibrium solution $y = 2$ is stable because nearby solutions are attracted to it.
Activities · practice with the ideas
For $\dfrac{dy}{dx} = x + y$, compute the slope at $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$. Which isocline passes through all points with slope 2?
Find the equilibrium solutions of $\dfrac{dy}{dx} = (y-1)(y+3)$ and classify each as stable or unstable using a sign analysis.
A direction field has all slope segments horizontal along the curve $y = 2x$. Suggest a differential equation that could produce this field.
For $\dfrac{dy}{dx} = -y$, describe what the solution curves look like and state the family of solutions without solving the ODE formally.
Explain why the uniqueness theorem guarantees that solution curves in a direction field cannot cross, and give a geometric reason based on the definition of the field.
Odd one out: Three of these statements about direction fields are correct. Which one is NOT?
Earlier you predicted what the solution curves to $\dfrac{dy}{dx} = x$ would look like.
The actual answer is $y = \dfrac{x^2}{2} + C$ — a family of upward-opening parabolas. The key insight: slopes are zero at $x=0$, negative for $x<0$, and positive for $x>0$, exactly matching the shape of a parabola. Did your intuition lead you here?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the slope assigned by the direction field of $\dfrac{dy}{dx} = x - y$ at the point $(3, 1)$. (1 mark)
Q2. Find the equilibrium solutions of $\dfrac{dy}{dx} = y^2 - 4$ and state the zero-slope isocline. (2 marks)
Q3. For $\dfrac{dy}{dx} = y - x$, find the equation of the zero-slope isocline, describe the sign of $\dfrac{dy}{dx}$ above and below it, and sketch the behaviour of a solution starting at $(0, 2)$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $(0,0)$: 0; $(1,0)$: 1; $(0,1)$: 1; $(1,1)$: 2. Isocline for slope 2: $x+y=2$, so $y=2-x$.
2. Equilibria: $y=1$ and $y=-3$. For $y>1$: $f>0$ (curves rise, $y=1$ is unstable from above). For $-3<y<1$: $f<0$ (curves fall toward $y=-3$). For $y<-3$: $f>0$ (curves rise toward $y=-3$). So $y=-3$ is stable, $y=1$ is unstable.
3. $f(x,y)=0$ on $y=2x$. Candidate: $\dfrac{dy}{dx}=y-2x$.
4. Slopes are $-y$: positive where $y<0$, negative where $y>0$ — curves decay toward zero. Family: $y=Ce^{-x}$.
5. At a crossing point, the two curves would each require the slope prescribed by the field, yet they would arrive at the same point from different directions — meaning two different slopes at one point, contradicting the fact that $f$ assigns a unique slope at each point.
Q1 (1 mark): $f(3,1)=3-1=2$ [1].
Q2 (2 marks): $y^2-4=0 \Rightarrow y=\pm 2$ [1]. Zero-slope isocline: $y=2$ and $y=-2$ (two horizontal lines) [1].
Q3 (3 marks): Isocline $y-x=0 \Rightarrow y=x$ [1]. Above $y=x$: $y-x>0$, so $\dfrac{dy}{dx}>0$ (slopes positive, curves rise). Below $y=x$: $\dfrac{dy}{dx}<0$ (curves fall) [1]. Starting at $(0,2)$ (above the isocline): the curve initially rises away from $y=x$, consistent with $\dfrac{dy}{dx}>0$ there [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering direction field questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.