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 itdescribe 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 confidencethat 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.
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