The slope, everywhere at once AB & BC

A differential equation of the form $\frac{dy}{dx} = f(x, y)$ hands you a slope at every point. Pick a point, substitute both coordinates, and draw a short segment through the point with that slope. Do it on a lattice and you have a slope field.

Take $\frac{dy}{dx} = x + y$. At $(1, 2)$ the slope is $3$, so the segment there is steep and rising. At $(1, -1)$ the slope is $0$, so the segment is horizontal. At $(-2, 0)$ it is $-2$, falling. Nothing more sophisticated than substitution is involved, which is why this is the one topic in the unit where the arithmetic can be done in your head.

Two things the segments are not. They are not pieces of a single solution curve: each one belongs to whichever solution happens to pass through that point, and infinitely many solutions run through the field. And they have no length in any meaningful sense. A segment records a direction, so drawing it longer says nothing extra.

The payoff is that the field shows the whole family of solutions before any of them is found. A solution curve is any curve that stays tangent to the segments it passes, and once the field is drawn you can trace one with a pencil.

Doing every lattice point is slow and unnecessary. Three questions get most of a field on the page.

1. Where is the slope zero? Solve $f(x, y) = 0$. For $\frac{dy}{dx} = x + y$ that is the line $y = -x$, a diagonal of horizontal segments. For $\frac{dy}{dx} = y^{2} - 4$ it is the pair of horizontal lines $y = 2$ and $y = -2$. That curve or line is the spine of the picture.
2. Where is the slope undefined? A quotient such as $\frac{dy}{dx} = \frac{x}{y}$ has no slope where $y = 0$, and the segments there are drawn vertical. Solutions cannot cross that line horizontally, which shapes everything nearby.
3. Which side is positive? Pick one point in each region the zero set carves out and check the sign. Above $y = -x$ the sum $x + y$ is positive, so every segment there rises.

Then fill in a handful of representative points. Steepness grows as you move away from the zero set in most equations, so the field fans out.

One structural shortcut is worth more than any amount of point-plotting, and it is the subject of the next section.

If $\frac{dy}{dx}$ depends on $x$ alone, then moving straight up does not change the slope, so every segment in a column is parallel. If it depends on $y$ alone, moving sideways does not change the slope, so every segment in a row is parallel. If it depends on both, neither is true.

That single observation eliminates most candidates in a matching problem instantly, and it needs no substitution at all. A field whose columns are uniform cannot come from $\frac{dy}{dx} = x + y$, and a field whose rows are uniform cannot come from $\frac{dy}{dx} = x$.

When the structure does not settle it, test points. Choose points that separate the candidates rather than convenient ones, and stop as soon as a candidate fails. Testing $(1, 0)$ distinguishes $\frac{dy}{dx} = x - y$ from $\frac{dy}{dx} = y - x$ immediately, since one gives $1$ and the other gives $-1$; testing $(0, 0)$ distinguishes neither, because both give $0$.

Three well-chosen points is usually a proof. Zero points, and a decision made on whether the picture "looks exponential", is the failure this topic exists to prevent. Fields for very different equations look alike at a glance, and they stop looking alike the moment a coordinate pair goes in.

Given a field and asked what it says, work from the same features you would use to draw one.

1. Find the horizontal segments. Their locations are the solution set of $f(x, y) = 0$. If they lie along a horizontal line $y = c$, the slope depends on $y$ there; along a vertical line $x = c$, on $x$.
2. Check a steep region against a shallow one. A field that steepens with height suggests $y$ appears; one that steepens as you move right suggests $x$ does.
3. Look for vertical segments. They mean a denominator vanished, which points at a quotient.

A horizontal line of zero segments deserves special attention because it is often an equilibrium solution: the constant function $y = c$ solves the equation exactly. That is Topic 7.4's subject and it is visible here.

Finally, a warning about scale. Fields are drawn on whatever window fits the page, and a slope of $2$ drawn on axes with different horizontal and vertical scales does not look like a slope of $2$. Read the axes before judging steepness by eye, and prefer the substitution.