Sketching Slope Fields — AP® Calculus AB Notes

A differential equation such as $\frac{dy}{dx} = x + y$ does something remarkable: it tells you the slope of a solution curve at every point of the plane, before you know a single solution. Pick the point $(1, 2)$ and the equation says any solution passing there has slope $1 + 2 = 3$ . A slope field turns that information into a picture.

What a slope field is

At each point of a grid, draw a short segment centered on the point, tilted to the slope the differential equation gives there. That is all. The collection of segments is the slope field (also called a direction field). It is a map of directions: a solution curve passing through a point must run along the little segment there.

Short segments at the 25 integer points with x and y from −2 to 2. Each segment has slope x + y: slope 0 along the line y = −x, steep positive slopes up to 4 at (2, 2), and steep negative slopes down to −4 at (−2, −2). The nine points with x and y from −1 to 1 are labeled with their slopes. — **Figure 1.** The slope field of $\frac{dy}{dx} = x + y$ at the 25 integer points from $-2$ to 2. The crimson numbers are the slopes at the nine boxed points. LevlPrep original diagram.

How to sketch one

List the points you are asked about. A point is $(x, y)$ : the first number is $x$ .
Substitute each point into the right side to get a number, the slope there.
Draw a short segment through the point with that slope. Positive slopes rise to the right, negative slopes fall, slope 0 is horizontal. If the slope is undefined (a division by 0), draw a vertical segment.
Make the size of the slope visible: a slope of 3 looks steeper than a slope of 1, and a slope of $\frac{1}{2}$ looks gentle. Exact angles are not needed; the sign and the rough steepness are.

Worked example. Sketch the slope field of $\frac{dy}{dx} = x + y$ at the nine points with $x$ and $y$ in $\{-1, 0, 1\}$ .

Compute $x + y$ at each point and organize the results in a grid, with $y$ going up the page:

	$x = -1$	$x = 0$	$x = 1$
$y = 1$	0	1	2
$y = 0$	$-1$	0	1
$y = -1$	$-2$	$-1$	0

Now draw: flat segments on the diagonal from $(-1, 1)$ to $(1, -1)$ , gentle rises of slope 1 at $(0, 1)$ and $(1, 0)$ , a steeper rise of slope 2 at $(1, 1)$ , and the mirror-image falls below the diagonal. The figure above shows the result.

Shortcuts that save time

Before computing 9 or 25 slopes, look at the equation's form. Three patterns cover most exam questions.

If $\frac{dy}{dx}$ depends on	then	Example
$x$ alone	segments in each vertical column are parallel	$\frac{dy}{dx} = \cos x$
$y$ alone	segments in each horizontal row are parallel	$\frac{dy}{dx} = y^2 - 1$
both	look for lines where the slope is 0 or constant	$\frac{dy}{dx} = 2x - y$

Find where the slope is 0 first. For $\frac{dy}{dx} = y^2 - 1$ that is the rows $y = 1$ and $y = -1$ : every segment there is flat. For $\frac{dy}{dx} = 2x - y$ it is the slanted line $y = 2x$ . For $\frac{dy}{dx} = x^2(y - 2)$ it is both the row $y = 2$ and the column $x = 0$ , because a product is 0 when either factor is.

Then look for undefined slopes. In $\frac{dy}{dx} = \frac{x}{y}$ , every point on the $x$ -axis (except the origin) has $y = 0$ in the denominator, so the segments there are vertical.

Matching an equation to a field

To decide which equation produced a given field, test a few easy points. Points on the axes make the arithmetic quick: at $(0, 1)$ and $(1, 0)$ most expressions are one short computation. Check the sign first, then the steepness. Also check where the field is flat: if the flat segments lie on $y = 2x$ , the right side must be 0 whenever $y = 2x$ .

Common slips

Swapping coordinates: at $(0, 2)$ the value of $x$ is 0, not 2.
Drawing a segment that is not centered on its point, or that runs into the next one: keep segments short.
Getting the sign right but ignoring steepness, so slopes of 0.5 and 4 look the same.
Forgetting that a factor of 0 makes the whole slope 0.