AP Calculus AB and BC

Slope Fields and Euler's Method Explained

Q: Do I have to solve the differential equation to sketch its slope field?

No, and that is the whole point. You only evaluate \frac{dy}{dx} = f(x, y) at each grid point and draw a segment with that slope. Slope fields work even when the equation cannot be solved by hand.

A slope field draws a tiny segment at every grid point whose slope equals dy/dx there, so you see the shape of solution curves without solving the equation. Euler's method (BC only, Topic 7.5) approximates one value by stepping along that field: new y = old y + (step size)(slope).

What a slope field shows

You are looking at a slope field problem when a differential equation is written in the form $\frac{dy}{dx} = f(x, y)$ and the task is to sketch it, match it to a graph, or describe its solution curves, not to find a formula. A slope field (FUN-7.C.1) is a grid of short line segments. At each point the segment is drawn with the slope that the differential equation assigns to that point. Because $\frac{dy}{dx}$ is the slope of the solution curve passing through a point, the field is a picture of every solution at once.

\frac{dy}{dx} = f(x, y)

This is the method-selection idea behind Topics 7.3 and 7.4: a slope field gives qualitative information (where solutions rise, fall, flatten, or steepen) with no integration required. You never solve the equation to draw the field. You only evaluate $f(x, y)$ at grid points. That makes slope fields the right tool when the equation is hard or impossible to solve by hand but you still need to understand how its solutions behave. Unit 7 is worth 5 to 10 percent of both the AB and BC exams, and this qualitative reasoning shows up in both multiple-choice and free-response.

Sketching a slope field (Topic 7.3)

The procedure is pure evaluation. There is no antiderivative anywhere in it.

Read the slope rule $f(x, y)$ straight off the right side of the differential equation.
Take each grid point $(x, y)$ and substitute to get the numerical slope there.
Draw a short segment at that point with that slope: flat when the slope is $0$ , tilted up to the right when positive, down when negative, and steeper as the magnitude grows.
Repeat across the whole grid. Segments that share a slope reveal the shape of the solutions.

A fast shortcut is the isocline: set $f(x, y)$ equal to a constant to find every point sharing one slope. Setting $f(x, y) = 0$ locates all the horizontal segments, where solution curves have horizontal tangents: the candidate locations for a peak or a valley (you must still check that $\frac{dy}{dx}$ changes sign, or the sign of $\frac{d^2y}{dx^2}$ , to confirm an extremum). For $\frac{dy}{dx} = x - y$ , the zero-slope isocline is the line $y = x$ , so every solution levels off the moment it crosses that line. Example 1 builds this field point by point.

Reasoning with slope fields on the exam

Topic 7.4 turns the picture into reasoning, and three question types come up again and again. Matching an equation to a field: test one or two easy points instead of the whole grid. If $f$ depends only on $x$ , every segment in the same vertical column is identical; if $f$ depends only on $y$ , every segment in the same horizontal row is identical. That single observation eliminates most wrong answers instantly. Sketching a particular solution: start at the given initial-condition point and draw a curve that stays tangent to the nearby segments, following them like a current (FUN-7.C.3). Your curve hugs the segments. It never crosses through one.

As a quick matching example, suppose one candidate is $\frac{dy}{dx} = x$ and another is $\frac{dy}{dx} = y$ . The first depends only on $x$ , so its segments are flat all along the $y$ -axis and grow steeper to the left and right; the second depends only on $y$ , so its segments are flat all along the $x$ -axis and grow steeper up and down. Reading which axis carries the horizontal segments settles the match in seconds.

Confirming concavity is the third type and the one students lose points on. Differentiate $\frac{dy}{dx}$ again with respect to $x$ , treating $y$ as a function of $x$ , to get $\frac{d^2y}{dx^2}$ . Its sign tells you where the solution is concave up or down. This links Unit 7 back to the derivative reasoning of Unit 5 and is a standard free-response step. Example 2 works it in full.

Euler's method: stepping through the field (BC only)

BC only

Euler's method is Topic 7.5 and appears on the BC exam only (FUN-7.C.4). AB students can stop after the slope field sections above.

A slope field shows the shape of solutions; Euler's method puts a number on one of them. It approximates the value of a particular solution at a target $x$ by taking short straight steps along the slope field. Start at the initial condition, follow the current slope for a small step of width $h$ , land on a new point, read the new slope there, and repeat until you reach the target.

y_{n+1} = y_n + h \cdot f(x_n, y_n), \qquad x_{n+1} = x_n + h

Organize the work in a table. Each row records the current point, the slope $f(x_n, y_n)$ there, the step change $\Delta y = h \cdot f(x_n, y_n)$ , and the resulting next value of $y$ . The table below runs Euler's method for $\frac{dy}{dx} = x + y$ with $y(0) = 1$ and $h = 0.5$ , which is worked step by step in Example 3.

Step $n$	$x_n$	$y_n$	Slope $x_n + y_n$	$\Delta y = h \cdot \text{slope}$	$y_{n+1}$
0	0	1	1	0.5	1.5
1	0.5	1.5	2	1	2.5

A smaller step size $h$ means more rows and less error. Because every step follows a straight tangent line, the estimate drifts to one side of the true curve in a predictable way: when the solution is concave up, the tangent steps sit below it and Euler underestimates; when it is concave down, the steps sit above and Euler overestimates. Decide which case you are in from the sign of $\frac{d^2y}{dx^2}$ , never by guessing.

Common mistakes

Solving the differential equation when the problem only asks for a sketch. A slope field needs evaluation, not integration.
Drawing curved marks. Every field segment is straight; only the solution curve you sketch through them is allowed to bend.
In Euler's method, forgetting to multiply the slope by $h$ . The step change is always $\Delta y = h \cdot f(x_n, y_n)$ , not just $f(x_n, y_n)$ .
Reusing the starting slope for every step. Recompute $f(x_n, y_n)$ at each new point before taking the next step.
Guessing whether the estimate is an overestimate or underestimate. Read it off the concavity of the solution instead.

Worked examples

Worked example

Sketching the slope field for dy/dx = x - y

Compute the slopes of $\frac{dy}{dx} = x - y$ at the nine grid points with $x = -1, 0, 1$ and $y = -1, 0, 1$ , then describe the resulting slope field.

The slope rule is $f(x, y) = x - y$ . At each grid point, subtract the $y$ -coordinate from the $x$ -coordinate to get the slope of the segment drawn there.
Bottom row, $y = -1$ : at $(-1, -1)$ the slope is $-1 - (-1) = 0$ ; at $(0, -1)$ it is $0 - (-1) = 1$ ; at $(1, -1)$ it is $1 - (-1) = 2$ .
Middle row, $y = 0$ : at $(-1, 0)$ the slope is $-1 - 0 = -1$ ; at $(0, 0)$ it is $0$ ; at $(1, 0)$ it is $1$ .
Top row, $y = 1$ : at $(-1, 1)$ the slope is $-1 - 1 = -2$ ; at $(0, 1)$ it is $0 - 1 = -1$ ; at $(1, 1)$ it is $1 - 1 = 0$ .
Draw a short segment at each point with that slope: flat where the slope is $0$ , tilted up to the right where it is positive, down where it is negative, and steeper as the magnitude grows toward $\pm 2$ .
Notice the flat segments at $(-1, -1)$ , $(0, 0)$ , and $(1, 1)$ . They all lie on the line $y = x$ , which is the zero-slope isocline found by setting $x - y = 0$ . Along that line every solution curve momentarily levels off.

The slope at each point equals $x - y$ , ranging from $-2$ at the top left to $2$ at the bottom right, with horizontal segments all along the line $y = x$ .

Worked example

Concavity of a solution from the equation

For the same equation $\frac{dy}{dx} = x - y$ , the particular solution passes through $(0, 2)$ . Is that solution increasing or decreasing there, is it concave up or concave down, and where are its tangent lines horizontal?

Increasing versus decreasing is set by $\frac{dy}{dx}$ . At $(0, 2)$ : $\frac{dy}{dx} = 0 - 2 = -2$ . The slope is negative, so the solution is decreasing at that point.
Concavity comes from $\frac{d^2y}{dx^2}$ . Differentiate both sides of $\frac{dy}{dx} = x - y$ with respect to $x$ , treating $y$ as a function of $x$ , so that $y$ differentiates to $\frac{dy}{dx}$ : this gives $\frac{d^2y}{dx^2} = 1 - \frac{dy}{dx}$ .
Substitute the original equation $\frac{dy}{dx} = x - y$ to write the second derivative in terms of $x$ and $y$ : $\frac{d^2y}{dx^2} = 1 - (x - y) = 1 - x + y$ .
Evaluate at $(0, 2)$ : $\frac{d^2y}{dx^2} = 1 - 0 + 2 = 3$ . This is positive, so the solution is concave up at $(0, 2)$ : it is decreasing while bending back upward.
Horizontal tangents occur where $\frac{dy}{dx} = 0$ , that is where $x - y = 0$ , or $y = x$ . The solution through $(0, 2)$ has a horizontal tangent at the moment it reaches the line $y = x$ , exactly the isocline from Example 1.

At $(0, 2)$ the solution is decreasing (slope $-2$ ) and concave up (second derivative $3$ ); its tangent line is horizontal wherever the solution crosses the line $y = x$ .

Worked example

Euler's method table (BC only)

Use Euler's method with step size $h = 0.5$ to approximate $y(1)$ for $\frac{dy}{dx} = x + y$ with the initial condition $y(0) = 1$ .

Set up the recursion $y_{n+1} = y_n + h \cdot f(x_n, y_n)$ with $f(x, y) = x + y$ and $h = 0.5$ . Reaching $x = 1$ from $x = 0$ in steps of $0.5$ takes two steps.
Step 1 from $(0, 1)$ . The slope is $f(0, 1) = 0 + 1 = 1$ . The change is $\Delta y = h \cdot \text{slope} = 0.5 \cdot 1 = 0.5$ . So $y_1 = 1 + 0.5 = 1.5$ at $x_1 = 0.5$ .
Step 2 from $(0.5, 1.5)$ . Recompute the slope at the new point: $f(0.5, 1.5) = 0.5 + 1.5 = 2$ . The change is $\Delta y = 0.5 \cdot 2 = 1$ . So $y_2 = 1.5 + 1 = 2.5$ at $x_2 = 1$ .
The two steps give $y(1) \approx 2.5$ , matching the table in the Euler's method section. As a check, the exact solution is $y = 2e^x - x - 1$ , so $y(1) = 2e - 2 \approx 3.44$ . Euler undershoots because this solution is concave up: $\frac{d^2y}{dx^2} = \frac{dy}{dx} + 1 = x + y + 1 > 0$ along the path, so the straight tangent steps ride below the true curve.

$y(1) \approx 2.5$ . Because the solution is concave up, this is an underestimate of the true value $2e - 2 \approx 3.44$ .

Frequently asked questions

Is Euler's method on the AP Calculus AB exam?

No. Euler's method is Topic 7.5 and is tested on the BC exam only. Slope fields (Topics 7.3 and 7.4) are on both AB and BC.

Does Euler's method overestimate or underestimate?

It depends on concavity. When the solution curve is concave up, the tangent steps stay below it and Euler underestimates; when it is concave down, Euler overestimates. Check the sign of $\frac{d^2y}{dx^2}$ to decide rather than guessing.

Do I have to solve the differential equation to sketch its slope field?

No, and that is the whole point. You only evaluate $\frac{dy}{dx} = f(x, y)$ at each grid point and draw a segment with that slope. Slope fields work even when the equation cannot be solved by hand.

What is the difference between a slope field and Euler's method?

A slope field is a qualitative picture of every solution at once; Euler's method is a numerical procedure that estimates one solution's value at a specific $x$ . The field shows shape, Euler produces a number.