Slope Fields

A slope field (or direction field) is a picture of a differential equation dy/dx = f(x, y). At many points (x, y), you draw a short line segment whose slope is f(x, y). Solution curves follow the segments, so a slope field shows what the solutions look like without solving the equation.

Before this lesson: Verifying Solutions to Differential Equations and Equation of a Tangent Line

Seeing a differential equation

The equation dy/dx = x + y doesn’t tell you y directly. It tells you the slope of the solution curve at any point. At (1, 0), the slope is 1 + 0 = 1. At (−1, −1), it is −2.

Draw a tiny tangent segment with that slope at each point of a grid, and a pattern appears, like iron filings around a magnet. Any solution curve must run along the segments everywhere it goes.

How to draw a slope field

1. Pick a grid of points, such as all (x, y) with x and y in −2, −1, 0, 1, 2.
2. At each point, find the slope m = f(x, y).
3. Draw a short segment through the point with slope m: flat for m = 0, rising for m positive, falling for m negative, steeper for bigger |m|.
4. To sketch a solution, start at the initial point and follow the segments.

A horizontal line y = c where dy/dx = 0 for every x is an equilibrium solution: a solution that never changes.

Slope field examples

Example 1: Finding slopes at points

For dy/dx = x + y, find the slope at (0, 0), (1, 0), (−1, 1), (1, 1) and (−1, −1).

(0, 0): \; 0 + 0 = 0

(1, 0): \; 1 + 0 = 1

(-1, 1): \; -1 + 1 = 0

(1, 1): \; 1 + 1 = 2

(-1, -1): \; -1 + (-1) = -2

Answer: the slopes are 0, 1, 0, 2 and −2. The slope is 0 wherever y = −x.

Example 2: Slopes that depend only on y

Describe the slope field of dy/dx = y, and connect it to the solutions y = Ceˣ.

y = 0: \; \dfrac{dy}{dx} = 0

y = 1: \; \dfrac{dy}{dx} = 1, \quad y = 2: \; \dfrac{dy}{dx} = 2

y = -1: \; \dfrac{dy}{dx} = -1

\dfrac{d}{dx}\left(Ce^{x}\right) = Ce^{x} = y

Answer: the slopes are the same along each horizontal line. Curves above the axis rise faster and faster; curves below fall faster and faster, just like y = Ceˣ.

Example 3: A field of circles

Sketch the slope field of dy/dx = −x/y at a few points and show that the solutions are circles.

(1, 1): \; -\dfrac{1}{1} = -1, \quad (1, -1): \; -\dfrac{1}{-1} = 1

(0, 2): \; -\dfrac{0}{2} = 0

(2, 0): \; -\dfrac{2}{0} \text{ is undefined}

x^2 + y^2 = r^2 \Rightarrow 2x + 2y\dfrac{dy}{dx} = 0

\dfrac{dy}{dx} = -\dfrac{x}{y}

Answer: the segments circle around the origin, and the solution curves are circles x² + y² = r² (each half above or below the x-axis).

Example 4: Matching an equation to a field

A slope field has flat segments all along y = 2, rising segments below y = 2 and falling segments above it. The slopes are the same along each horizontal line. Which equation is it: dy/dx = y − x, dy/dx = x² or dy/dx = 2 − y?

\text{Same along horizontal lines} \Rightarrow f \text{ depends on } y \text{ only}

2 - y = 0 \text{ at } y = 2

y = 0: \; 2 - 0 = 2 \gt 0

y = 3: \; 2 - 3 = -1 \lt 0

Answer: dy/dx = 2 − y. Every solution moves toward the equilibrium y = 2.

Example 5: Reading a population model

A fish population P (in thousands) follows dP/dt = 0.5P(1 − P/100), where t is in years. Use slopes to describe what happens to a population that starts at 20, and to one that starts at 120.

P = 0 \text{ or } P = 100: \; \dfrac{dP}{dt} = 0

P = 20: \; 0.5(20)(1 - 0.2) = 8

P = 50: \; 0.5(50)(1 - 0.5) = 12.5

P = 120: \; 0.5(120)(1 - 1.2) = -12

Answer: starting at 20, the population rises toward 100 thousand. Starting at 120, it falls toward 100 thousand. P = 100 is the carrying capacity.

Common mistakes

Related methods

Euler’s method

Following the segments with small straight steps is a numerical way to approximate a solution. That is Euler’s method, the next lesson.

Isoclines

An isocline is a curve where every segment has the same slope. For dy/dx = x + y, the isocline with slope k is the line x + y = k. Drawing a few isoclines makes a slope field quick to sketch.

Solving exactly

Some equations can be solved by separating the variables, coming up in this unit. The slope field is still a good way to check that a formula looks right.

Practice problems

1. For dy/dx = x − y, find the slope at (2, 1), (1, 2) and (0, 0).
2. For dy/dx = x², find the slopes at x = −1, 0 and 2. Why are the slopes the same along each vertical line?
3. Find the equilibrium solutions of dy/dx = y² − 1, and the slopes along y = 0 and y = 2.
4. Where are the segments horizontal in the slope field of dy/dx = x + y?
5. Show that the curve y = x² + 2 follows the slope field of dy/dx = 2x at x = 1.
6. One field has slopes that are the same along every vertical line. Is it dy/dx = cos x or dy/dx = y?
7. For dy/dx = −y/2, find the slope at (0, 4). Describe the solution through (0, 4).
8. For dy/dx = xy, where are the segments flat? What is the sign of the slope in each quadrant?
9. For dy/dx = x + y, use the slope at (0, 1) to write the tangent line there and estimate y(0.1).
10. A pie cools by dT/dt = −0.2(T − 70), with T in °F. Find the slopes at T = 170, 70 and 50. What happens over time?

(2, 1): \; 2 - 1 = 1; \; (1, 2): \; 1 - 2 = -1; \; (0, 0): \; 0

(-1)^2 = 1, \; 0^2 = 0, \; 2^2 = 4

y^2 - 1 = 0 \Rightarrow y = 1 \text{ or } y = -1

y = 0: \; 0 - 1 = -1; \; y = 2: \; 4 - 1 = 3

x + y = 0 \Rightarrow y = -x

\dfrac{dy}{dx} = 2(1) = 2

\dfrac{d}{dx}(x^2 + 2) = 2x = 2 \text{ at } x = 1

\dfrac{dy}{dx} = \cos x \text{ depends only on } x

-\dfrac{4}{2} = -2

\dfrac{dy}{dx} \lt 0 \text{ for } y \gt 0, \text{ and } \to 0 \text{ as } y \to 0

xy = 0 \Rightarrow x = 0 \text{ or } y = 0

\text{Q1: } (+)(+) \gt 0, \; \text{Q2: } (-)(+) \lt 0

\text{Q3: } (-)(-) \gt 0, \; \text{Q4: } (+)(-) \lt 0

m = 0 + 1 = 1

y - 1 = 1(x - 0) \Rightarrow y = x + 1

y(0.1) \approx 0.1 + 1 = 1.1

T = 170: \; -0.2(100) = -20; \; T = 70: \; 0; \; T = 50: \; -0.2(-20) = 4

Slope fields FAQ

What is a slope field?

It is a grid of short line segments that shows the slope dy/dx = f(x, y) at each point. Solution curves of the differential equation follow the segments.

How do you draw a slope field?

Choose grid points, compute f(x, y) at each one, and draw a short segment with that slope through the point.

How do you sketch a solution on a slope field?

Start at the initial point and draw a smooth curve that stays tangent to the nearby segments, like a leaf floating along a current.

What is an equilibrium solution?

It is a constant solution, y = c, where dy/dx = 0 for every x. On a slope field it is a row of flat segments.

Summary

• A slope field draws a short segment with slope f(x, y) at each grid point.
• Solution curves are tangent to the segments everywhere.
• If f depends only on x, slopes match along vertical lines; only on y, along horizontal lines.
• Rows of flat segments mark equilibrium solutions.
• Slope fields let you predict long-term behavior without solving.

Previous lesson: Verifying Solutions to Differential Equations

Next lesson: Euler’s Method

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