7.2 Slope fields

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Slope fields

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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 lets you visualize how solution curves behave without solving the differential equation. You can think of it as a picture of the “direction” solution curves follow at many points - similar to how a family of antiderivatives shows all possible solutions to a derivative.

Sketching slope fields

A slope field is a diagram made of many short line segments. Each segment shows the slope a solution curve would have at a particular point on a grid.

Suppose we have the differential equation

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

This equation tells you: at any point $(x, y)$ , the slope of the solution curve is $x + y$ .

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 $(0, 0)$ , the slope is $\frac{d y}{d x} = 0 + 0 = 0$ . Draw a short horizontal segment.
At $(0, 1)$ , the slope is $\frac{d y}{d x} = 0 + 1 = 1$ . Draw a short segment with slope 1 (rising 1 for every 1 to the right).

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

Particular solutions and initial conditions

A slope field shows many possible solution curves. An initial condition picks out one specific curve, called a particular solution.

For example, the initial condition $y (0) = 1$ means the solution curve must pass through $(0, 1)$ .

To sketch the particular solution:

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. The curve should follow the “tilt” of the segments.
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

Sometimes you’ll be given a slope field and asked to identify the matching equation. The answer choices might be differential equations or specific solutions.

If the choices are differential equations, pick 2-3 points and compare slopes:

Choose points where the segments are clearly horizontal (slope $0$ ) or clearly steep.
Plug the point’s coordinates into each candidate $\frac{d y}{d x} = f (x, y)$ .
The correct equation should match the slope shown in the field at those points.

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 clear which curve fits, differentiate each candidate $y = f (x)$ to get $y^{'}$ , then compare those 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 $\frac{d y}{d x} = 0$ whenever $y = 0$ or $y = 2$ (for any value of $x$ ).

Also, the slopes don’t change as you move left or right (as $x$ changes). That suggests the differential equation does not depend on $x$ . This rules out choices $a) d y / d x = - y / x$ and $b) d y / d x = (x - 2)^{3}$ , since both include $x$ .

Between the remaining choices:

Choice $c)$ gives slope $0$ for both $y = 0$ and $y = 2$ because $y^{2} - 2 y = y (y - 2)$ .
Choice $d)$ gives slope $0$ only when $y = - 2$ .

So the correct choice is $c)$ .

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 $x = 1$ . That suggests the slope is undefined there, which often happens when the differential equation has a denominator of $(x - 1)$ (or something similar). On the AP exam, this area would generally be depicted with vertical line segments.

That rules out choice $d) y = (x - 1)^{2}$ because it’s a polynomial, so it’s smooth and defined at $x = 1$ .

The remaining choices all have a vertical asymptote at $x = 1$ . Differentiate each one to compare slopes:

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 check a couple of $x$ -values using the slope field.

Along $x = 0$ , the segments have positive slope. But for choice $c)$ , when $x = 0$ we get $d y / d x = \frac{1}{- 1} = - 1$ , which is negative. So eliminate $c)$ .
Along $x = 2$ , the segments also have positive slope. For choice $a)$ , when $x = 2$ we get $d y / d x = - \frac{2}{1 ^{3}} = - 2$ , which is negative. So eliminate $a)$ .

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

Slope fields

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

Sketching slope fields

A slope field is a diagram made of many short line segments. Each segment shows the slope a solution curve would have at a particular point on a grid.

Suppose we have the differential equation

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

This equation tells you: at any point $(x, y)$ , the slope of the solution curve is $x + y$ .

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 $(0, 0)$ , the slope is $\frac{d y}{d x} = 0 + 0 = 0$ . Draw a short horizontal segment.
At $(0, 1)$ , the slope is $\frac{d y}{d x} = 0 + 1 = 1$ . Draw a short segment with slope 1 (rising 1 for every 1 to the right).

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

Particular solutions and initial conditions

A slope field shows many possible solution curves. An initial condition picks out one specific curve, called a particular solution.

For example, the initial condition $y (0) = 1$ means the solution curve must pass through $(0, 1)$ .

To sketch the particular solution:

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. The curve should follow the “tilt” of the segments.
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

Sometimes you’ll be given a slope field and asked to identify the matching equation. The answer choices might be differential equations or specific solutions.

If the choices are differential equations, pick 2-3 points and compare slopes:

Choose points where the segments are clearly horizontal (slope $0$ ) or clearly steep.
Plug the point’s coordinates into each candidate $\frac{d y}{d x} = f (x, y)$ .
The correct equation should match the slope shown in the field at those points.

Match the differential equation to the slope field shown.

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

Solution

(spoiler)

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

Between the remaining choices:

Choice $c)$ gives slope $0$ for both $y = 0$ and $y = 2$ because $y^{2} - 2 y = y (y - 2)$ .
Choice $d)$ gives slope $0$ only when $y = - 2$ .

So the correct choice is $c)$ .

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

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

Solution

(spoiler)

That rules out choice $d) y = (x - 1)^{2}$ because it’s a polynomial, so it’s smooth and defined at $x = 1$ .

The remaining choices all have a vertical asymptote at $x = 1$ . Differentiate each one to compare slopes:

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 check a couple of $x$ -values using the slope field.

Along $x = 0$ , the segments have positive slope. But for choice $c)$ , when $x = 0$ we get $d y / d x = \frac{1}{- 1} = - 1$ , which is negative. So eliminate $c)$ .
Along $x = 2$ , the segments also have positive slope. For choice $a)$ , when $x = 2$ we get $d y / d x = - \frac{2}{1 ^{3}} = - 2$ , which is negative. So eliminate $a)$ .

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

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Slope fields

7 min read

Font

Discuss

Feedback

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

Sketching slope fields

A slope field is a diagram made of many short line segments. Each segment shows the slope a solution curve would have at a particular point on a grid.

Suppose we have the differential equation

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

This equation tells you: at any point $(x, y)$ , the slope of the solution curve is $x + y$ .

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 $(0, 0)$ , the slope is $\frac{d y}{d x} = 0 + 0 = 0$ . Draw a short horizontal segment.
At $(0, 1)$ , the slope is $\frac{d y}{d x} = 0 + 1 = 1$ . Draw a short segment with slope 1 (rising 1 for every 1 to the right).

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

Particular solutions and initial conditions

A slope field shows many possible solution curves. An initial condition picks out one specific curve, called a particular solution.

For example, the initial condition $y (0) = 1$ means the solution curve must pass through $(0, 1)$ .

To sketch the particular solution:

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. The curve should follow the “tilt” of the segments.
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

Sometimes you’ll be given a slope field and asked to identify the matching equation. The answer choices might be differential equations or specific solutions.

If the choices are differential equations, pick 2-3 points and compare slopes:

Choose points where the segments are clearly horizontal (slope $0$ ) or clearly steep.
Plug the point’s coordinates into each candidate $\frac{d y}{d x} = f (x, y)$ .
The correct equation should match the slope shown in the field at those points.

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 $\frac{d y}{d x} = 0$ whenever $y = 0$ or $y = 2$ (for any value of $x$ ).

Between the remaining choices:

Choice $c)$ gives slope $0$ for both $y = 0$ and $y = 2$ because $y^{2} - 2 y = y (y - 2)$ .
Choice $d)$ gives slope $0$ only when $y = - 2$ .

So the correct choice is $c)$ .

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)

That rules out choice $d) y = (x - 1)^{2}$ because it’s a polynomial, so it’s smooth and defined at $x = 1$ .

The remaining choices all have a vertical asymptote at $x = 1$ . Differentiate each one to compare slopes:

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 check a couple of $x$ -values using the slope field.

Along $x = 0$ , the segments have positive slope. But for choice $c)$ , when $x = 0$ we get $d y / d x = \frac{1}{- 1} = - 1$ , which is negative. So eliminate $c)$ .
Along $x = 2$ , the segments also have positive slope. For choice $a)$ , when $x = 2$ we get $d y / d x = - \frac{2}{1 ^{3}} = - 2$ , which is negative. So eliminate $a)$ .

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

Slope fields

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

Sketching slope fields

A slope field is a diagram made of many short line segments. Each segment shows the slope a solution curve would have at a particular point on a grid.

Suppose we have the differential equation

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

This equation tells you: at any point $(x, y)$ , the slope of the solution curve is $x + y$ .

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 $(0, 0)$ , the slope is $\frac{d y}{d x} = 0 + 0 = 0$ . Draw a short horizontal segment.
At $(0, 1)$ , the slope is $\frac{d y}{d x} = 0 + 1 = 1$ . Draw a short segment with slope 1 (rising 1 for every 1 to the right).

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

Particular solutions and initial conditions

A slope field shows many possible solution curves. An initial condition picks out one specific curve, called a particular solution.

For example, the initial condition $y (0) = 1$ means the solution curve must pass through $(0, 1)$ .

To sketch the particular solution:

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. The curve should follow the “tilt” of the segments.
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

Sometimes you’ll be given a slope field and asked to identify the matching equation. The answer choices might be differential equations or specific solutions.

If the choices are differential equations, pick 2-3 points and compare slopes:

Choose points where the segments are clearly horizontal (slope $0$ ) or clearly steep.
Plug the point’s coordinates into each candidate $\frac{d y}{d x} = f (x, y)$ .
The correct equation should match the slope shown in the field at those points.

Match the differential equation to the slope field shown.

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

Solution

(spoiler)

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

Between the remaining choices:

Choice $c)$ gives slope $0$ for both $y = 0$ and $y = 2$ because $y^{2} - 2 y = y (y - 2)$ .
Choice $d)$ gives slope $0$ only when $y = - 2$ .

So the correct choice is $c)$ .

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

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

Solution

(spoiler)

That rules out choice $d) y = (x - 1)^{2}$ because it’s a polynomial, so it’s smooth and defined at $x = 1$ .

The remaining choices all have a vertical asymptote at $x = 1$ . Differentiate each one to compare slopes:

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 check a couple of $x$ -values using the slope field.

Along $x = 0$ , the segments have positive slope. But for choice $c)$ , when $x = 0$ we get $d y / d x = \frac{1}{- 1} = - 1$ , which is negative. So eliminate $c)$ .
Along $x = 2$ , the segments also have positive slope. For choice $a)$ , when $x = 2$ we get $d y / d x = - \frac{2}{1 ^{3}} = - 2$ , which is negative. So eliminate $a)$ .

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