Slope fields | Differential equations | Achievable AP Calculus AB

What you’ll learn:

What slope fields represent
How to sketch a slope field from a differential equation
How to interpret and match slope fields to differential equations and solutions

A slope field shows how solution curves behave without having to solve the differential equation. Essentially, these look like the family of antiderivatives for a given derivative.

Sketching slope fields

A slope field is a diagram that shows many short line segments, each representing the slope of a solution curve at a grid of points.

Suppose we have a differential equation

$\frac{d y}{d x} = x + y$

This tells us that the slope of the solution curve at any point $(x, y)$ is equal to $x + y$ .

To construct the slope field:

Pick a set of points $(x, y)$ (the graph provided will usually be small, roughly 5 units or fewer in every direction from the origin).
At each point, calculate the slope using the given $\frac{d y}{d x} = f (x, y)$ .
Draw a small line segment at that point with the calculated slope.

For example, at point $(0, 0),$ the slope $\frac{d y}{d x} = 0$ . So at that point on the grid, you would draw a horizontal line segment (with a slope of $0$ ).

Picking another point such as $(0, 1)$ , the slope $\frac{d y}{d x} = 1$ . So at that point, you’d draw a slanted line segment with a slope of $1$ .

Here’s the slope field for the given differential equation - notice the horizontal line segment at $(0, 0)$ and the slanted one at $(0, 1)$ :

Particular solutions and initial conditions

Once a slope field is drawn, you can sketch a particular solution that satisfies a specific initial condition, such as $y (0) = 1$ . To do this:

Locate the point that corresponds to the initial condition e.g. $y (0) = 1$ means the point $(0, 1)$ .
Sketch a curve that is tangent to the slope segments as you move through the slope field. It should “flow” with the direction of the field.
This curve represents the unique solution to the differential equation that passes through the given point and helps you visualize how the solution behaves.

Using the slope field for $\frac{d y}{d x} = x + y$ , shown below is the solution curve that passes through $(0, 1)$ :

Practice sketching the slope fields for the following differential equations and check it on this slope field plotter in Desmos by changing the differential equation $F (x, y)$ .

$d y / d x = 2 x$

$d y / d x = 1$

$d y / d x = 1 - y$

$d y / d x = x / y$

$d y / d x = y / x$

You may notice that for $y^{'} = 2 x$ , the slope field resembles the family of parabolas $y = x^{2} + C$ , and for the 2nd problem $y^{'} = 1$ , the solution curves appear as straight lines of the form $y = x + C$ .

This illustrates the purpose of slope fields - they visually represent the family of antiderivatives and show how solution curves behave, even for functions that cannot be written explicitly.

Reasoning with slope fields

You may be shown a slope field and asked to identify the matching equation. The choices could be differential equations or specific solutions.

If the choices are differential equations, select 2-3 points (especially where the line segments are horizontal or vertical line segments) and plug the coordinates into the choices to see which one gives the correct slope.

If the answer choices are specific solutions of the form $y = f (x)$ , the slope field will resemble the family of curves $f (x) + C$ . But if it’s not immediately obvious which is the correct solution curve, differentiating each function can help you match the slopes to the slope field.

Match the differential equation to the slope field shown.

Matching a differential equation to a slope field

$a) d y / d x = - y / x$
$b) d y / d x = (x - 2)^{3}$
$c) d y / d x = y^{2} - 2 y$
$d) d y / d x = y + 2$

Solution

(spoiler)

The key features of this slope field are the horizontal line segments along the lines $y = 0$ and $y = 2$ , indicating that $\frac{d y}{d x} = 0$ as long as $y = 0$ or $y = 2$ for any value of $x$ .

Additionally, the slopes don’t change moving horizontally (as $x$ changes), which suggests that the differential equation does not depend on $x$ . This rules out choices $a) d y / d x = - y / x$ and b) $d y / d x = (x - 2)^{3}$ , both of which have $x$ .

Between the remaining answer choices, only choice $c)$ gives slope $0$ for both $y = 0$ or $y = 2$ (factor into $d y / d x = y (y - 2)$ ). Choice $d)$ gives slope $0$ only when $y = - 2$ instead.

The slope field from a certain differential equation is shown. Which of the following could be a specific solution to the differential equation?

Reasoning with slope fields

$a) y = \frac{1}{( x - 1 ) ^{2}}$
$b) y = - \frac{1}{x - 1}$
$c) y = ln ∣ x - 1∣$
$d) y = (x - 1)^{2}$

Solution

(spoiler)

The key feature of this slope field is the empty region along $x = 1$ , suggesting the slope is undefined there, possibly due to an asymptote. On the AP exam, this area would generally be depicted with vertical line segments.

As such, choice $d) y = (x - 1)^{2}$ can be eliminated because this solution curve is a smooth and continuous polynomial and cannot explain the undefined slope behavior at $x = 1$ .

Since the remaining answer choices all have the vertical asymptote $x = 1$ , we can differentiate each to see if the corresponding $\frac{d y}{d x}$ matches the slope field pattern.

The corresponding differential equations are:

$a) \frac{d y}{d x} = - \frac{2}{( x - 1 ) ^{3}}$

$b) \frac{d y}{d x} = \frac{1}{( x - 1 ) ^{2}}$

$c) \frac{d y}{d x} = \frac{1}{x - 1}$

Now pick any point on the slope field to check. For example, the line segments along the line $x = 0$ have positive slope. This means choice $c) \frac{d y}{d x} = \frac{1}{x - 1}$ can be eliminated because when $x = 0$ , $d y / d x = - 1$ .

Next, the line segments along the line $x = 2$ also have positive slope. However, plugging $x = 2$ into choice $a) \frac{d y}{d x} = - \frac{2}{( x - 1 ) ^{3}}$ gives a slope of $- 2$ .

This leaves choice $b) y = - \frac{1}{x - 1}$ as the solution curve that matches the given slope field.

Key points

Slope fields show solution behavior for differential equations
To draw a slope field from a differential equation, calculate the slope at each point $(x, y)$ and draw a line segment at that point with the specified slope.
Use patterns (horizontal or vertical lines) to recognize equation types.

What you’ll learn:

What slope fields represent
How to sketch a slope field from a differential equation
How to interpret and match slope fields to differential equations and solutions

A slope field shows how solution curves behave without having to solve the differential equation. Essentially, these look like the family of antiderivatives for a given derivative.

Sketching slope fields

A slope field is a diagram that shows many short line segments, each representing the slope of a solution curve at a grid of points.

Suppose we have a differential equation

$\frac{d y}{d x} = x + y$

This tells us that the slope of the solution curve at any point $(x, y)$ is equal to $x + y$ .

To construct the slope field:

Pick a set of points $(x, y)$ (the graph provided will usually be small, roughly 5 units or fewer in every direction from the origin).
At each point, calculate the slope using the given $\frac{d y}{d x} = f (x, y)$ .
Draw a small line segment at that point with the calculated slope.

For example, at point $(0, 0),$ the slope $\frac{d y}{d x} = 0$ . So at that point on the grid, you would draw a horizontal line segment (with a slope of $0$ ).

Picking another point such as $(0, 1)$ , the slope $\frac{d y}{d x} = 1$ . So at that point, you’d draw a slanted line segment with a slope of $1$ .

Here’s the slope field for the given differential equation - notice the horizontal line segment at $(0, 0)$ and the slanted one at $(0, 1)$ :

Particular solutions and initial conditions

Once a slope field is drawn, you can sketch a particular solution that satisfies a specific initial condition, such as $y (0) = 1$ . To do this:

Locate the point that corresponds to the initial condition e.g. $y (0) = 1$ means the point $(0, 1)$ .
Sketch a curve that is tangent to the slope segments as you move through the slope field. It should “flow” with the direction of the field.
This curve represents the unique solution to the differential equation that passes through the given point and helps you visualize how the solution behaves.

Using the slope field for $\frac{d y}{d x} = x + y$ , shown below is the solution curve that passes through $(0, 1)$ :

Practice sketching the slope fields for the following differential equations and check it on this slope field plotter in Desmos by changing the differential equation $F (x, y)$ .

$d y / d x = 2 x$

$d y / d x = 1$

$d y / d x = 1 - y$

$d y / d x = x / y$

$d y / d x = y / x$

This illustrates the purpose of slope fields - they visually represent the family of antiderivatives and show how solution curves behave, even for functions that cannot be written explicitly.

Reasoning with slope fields

You may be shown a slope field and asked to identify the matching equation. The choices could be differential equations or specific solutions.

Match the differential equation to the slope field shown.

Matching a differential equation to a slope field

$a) d y / d x = - y / x$
$b) d y / d x = (x - 2)^{3}$
$c) d y / d x = y^{2} - 2 y$
$d) d y / d x = y + 2$

Solution

(spoiler)

The key features of this slope field are the horizontal line segments along the lines $y = 0$ and $y = 2$ , indicating that $\frac{d y}{d x} = 0$ as long as $y = 0$ or $y = 2$ for any value of $x$ .

Between the remaining answer choices, only choice $c)$ gives slope $0$ for both $y = 0$ or $y = 2$ (factor into $d y / d x = y (y - 2)$ ). Choice $d)$ gives slope $0$ only when $y = - 2$ instead.

The slope field from a certain differential equation is shown. Which of the following could be a specific solution to the differential equation?

Reasoning with slope fields

$a) y = \frac{1}{( x - 1 ) ^{2}}$
$b) y = - \frac{1}{x - 1}$
$c) y = ln ∣ x - 1∣$
$d) y = (x - 1)^{2}$

Solution

(spoiler)

As such, choice $d) y = (x - 1)^{2}$ can be eliminated because this solution curve is a smooth and continuous polynomial and cannot explain the undefined slope behavior at $x = 1$ .

Since the remaining answer choices all have the vertical asymptote $x = 1$ , we can differentiate each to see if the corresponding $\frac{d y}{d x}$ matches the slope field pattern.

The corresponding differential equations are:

$a) \frac{d y}{d x} = - \frac{2}{( x - 1 ) ^{3}}$

$b) \frac{d y}{d x} = \frac{1}{( x - 1 ) ^{2}}$

$c) \frac{d y}{d x} = \frac{1}{x - 1}$

Next, the line segments along the line $x = 2$ also have positive slope. However, plugging $x = 2$ into choice $a) \frac{d y}{d x} = - \frac{2}{( x - 1 ) ^{3}}$ gives a slope of $- 2$ .

This leaves choice $b) y = - \frac{1}{x - 1}$ as the solution curve that matches the given slope field.

Key points

Slope fields show solution behavior for differential equations
To draw a slope field from a differential equation, calculate the slope at each point $(x, y)$ and draw a line segment at that point with the specified slope.
Use patterns (horizontal or vertical lines) to recognize equation types.