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Calculus AB · Unit 7 · 7.3
AP Calculus · Differential Equations
Sketching Slope Fields
Plot the equation itself: at each point, a short tick whose slope is what the equation says it is.
key idea$\frac{dy}{dx}=f(x,y)\ \text{assigns a \emph{slope} to every point of the plane}$
Reading the graph
A slope field draws, at a grid of points, a short segment whose slope is what the differential equation says $\dfrac{dy}{dx}$ is there. No solving is involved — you are plotting the equation itself. Solution curves are the curves that stay tangent to the segments everywhere they pass.
The field for $\dfrac{dy}{dx}=x+y$. Segments are flat along the line $y=-x$, because that is exactly where $x+y=0$; they steepen upward to the right of it and downward to the left. Reading where the slope is zero is always the fastest way into a field.| Point $(x,y)$ | $\dfrac{dy}{dx}=x+y$ | Segment drawn |
|---|
| $(0,0)$ | $0$ | horizontal |
| $(1,1)$ | $2$ | steep, rising |
| $(1,-1)$ | $0$ | horizontal |
| $(-2,1)$ | $-1$ | falling at $45^{\circ}$ |
| $(2,2)$ | $4$ | very steep, rising |
Sketching is nothing more than filling in this table at a handful of lattice points and drawing what the third column says. Slope, not value — the number you compute is the steepness of the tick, never a height.
Three fields, three signatures. If $f$ contains only $x$, every segment in a vertical column is parallel. If it contains only $y$, every segment in a horizontal row is parallel — and any $y$ making $f=0$ gives a constant solution, marked here in amber. If both variables appear, neither pattern holds and you must compute point by point.what a tick means
- The number you compute is a slope. Draw a tick at that steepness; never plot it as a height.
- Segments are drawn at equal length — only their direction carries information.
- $f(x,y)=0$ locates every horizontal tick, and these are the fastest features to find first.
- For an autonomous equation, a constant $y$ with $f=0$ is an equilibrium solution — a horizontal line that is itself a solution.
Worked examples
EX 1
Sketch by lattice points
$\dfrac{dy}{dx}=x-y$Describe the segments of the field for $\dfrac{dy}{dx}=x-y$ at the nine lattice points with $x,y\in\{-1,0,1\}$.
1
Find the flat ticks first
$x-y=0$ on the line $y=x$, so $(-1,-1)$, $(0,0)$ and $(1,1)$ all get horizontal segments.
2
Compute below that line
$(0,-1)$: slope $1$. $(1,-1)$: slope $2$. $(1,0)$: slope $1$. All rising.
3
Compute above it
$(0,1)$: slope $-1$. $(-1,1)$: slope $-2$. $(-1,0)$: slope $-1$. All falling.
4
State the pattern
Below $y=x$ the ticks rise, above it they fall, and on it they are flat.
5
Predict solution behaviour
Curves are pushed toward the line $y=x$ from both sides — visible before any solving.
EX 2
Identify the equation from a field
read the signatureA field's segments are identical along every horizontal row, and are flat exactly at $y=2$. Which of $\dfrac{dy}{dx}=x-2$, $\dfrac{dy}{dx}=y-2$, $\dfrac{dy}{dx}=xy$ is it?
1
Use the row test
Identical along a row means the slope does not change as $x$ varies. So $f$ depends on $y$ alone.
2
Eliminate
$x-2$ varies with $x$, and $xy$ varies with both. Both are out.
3
Check the remaining candidate
$y-2=0$ exactly when $y=2$ — matching the flat row.
4
Answer
$\dfrac{dy}{dx}=y-2$.
5
The general method
Two questions settle almost every matching problem: which variable does the field respond to, and where are the flat ticks.
EX 3
Sketch a solution curve through a point
follow the ticksOn the field for $\dfrac{dy}{dx}=-\dfrac{x}{y}$, describe the solution curve through $(3,4)$.
1
Compute the slope there
$-\dfrac34$ — the curve leaves the point falling gently.
2
Follow the field
Move a little right and the ticks tilt further downward; move left and up, they flatten.
3
Locate the flat ticks
$-\dfrac{x}{y}=0$ requires $x=0$, so the curve is horizontal as it crosses the $y$-axis.
4
Note where the field is vertical
At $y=0$ the slope is undefined and the ticks stand vertical — the curve turns there.
5
Name the shape
Horizontal at the top, vertical at the sides: the curve is a circle, here of radius $5$.
EX 4
Slope versus value
the standard confusionFor $\dfrac{dy}{dx}=2$, a student draws dots at height $2$ across the field. What went wrong?
1
Read the equation
It says the derivative is $2$ everywhere, not that $y$ is $2$.
2
Draw the correct field
Every segment, at every point, has slope $2$ — a field of identical parallel ticks.
3
Identify the solutions
Straight lines $y=2x+C$, one through each starting point.
4
Contrast with the student's picture
Height $2$ everywhere would be the horizontal line $y=2$, whose derivative is $0$ — the opposite claim.
5
The check
Ask what the tick is doing, not where it sits. A field carries no height information at all.
Watch out
Compute a slope, draw a tick
The value of $f(x,y)$ is a steepness. Plotting it as a height is the single most common slope-field error.
All segments have the same length
Length is drawing convention, not data. A large slope shows as a steeper tick, never a longer one.
Find the flat ticks first
Solving $f(x,y)=0$ gives the whole horizontal skeleton of the field in one step, and usually identifies the equation immediately.
Undefined slope means a vertical tick
Where the right-hand side divides by zero, the segment stands upright and no solution curve can be differentiable there.