Reasoning Using Slope Fields

AP Calculus · Differential Equations

Reasoning Using Slope Fields

Increasing, concave, asymptotic — all read straight off the ticks, with nothing solved.

key idea$\text{A solution curve is the curve that stays \emph{tangent} to the field at every point}$

Reading the graph

Once a field is drawn, it answers questions no formula is needed for: where a solution rises, where it turns, what it approaches in the long run, and which equation produced it. The rule for tracing is simple — enter at the given point and never cross a tick at the wrong angle.

Two solution curves of $\dfrac{dy}{dx}=x-y$, one through $(0,1)$ and one through $(0,-2.2)$. Each threads the field tangentially. Both are drawn toward the same slanted channel — and neither ever crosses the other, because a single tick at a point fixes one and only one direction there.

Question	What to look at	How to read it
Is the solution increasing at $P$?	the tick at $P$	rising tick means $y'\gt 0$
Where are the extrema?	the flat ticks	the curve turns where it meets a horizontal tick
Is the curve concave up?	how ticks change along the curve	steepening upward means $y''\gt 0$
What happens as $x\to\infty$?	the far-right ticks	flattening toward a level means a horizontal asymptote
Which equation made this field?	flat ticks $+$ row/column patterns	test $f(x,y)=0$ against the flat set
Can two solutions cross?	—	no — one tick per point gives one direction

Every one of these is answered without solving anything. That is the whole point of a slope field: qualitative behaviour, extracted directly from the equation.

$\dfrac{dy}{dx}=\dfrac{y}{2}$

$\dfrac{dy}{dx}=1-y$

$\dfrac{dy}{dx}=-\dfrac{x}{y}$

Three fields with very different long-run stories. On the left, ticks steepen away from $y=0$ — solutions run off to $\pm\infty$. In the middle, ticks push toward $y=1$ from both sides, so every solution flattens onto that line. On the right there is no equilibrium at all; the ticks curl, and the solutions close into circles.

Rising ticksthe solution is increasing there

Flat ticksa possible maximum or minimum

Ticks flattening rightwarda horizontal asymptote

Worked examples

EX 1

Long-run behaviour

$\dfrac{dy}{dx}=1-y$

A field has ticks rising below $y=1$, flat along $y=1$, and falling above it. Describe the solution through $(0,3)$ as $x\to\infty$.

Locate the starting point

$(0,3)$ lies above the flat row.

Read the tick there

Above $y=1$ the ticks fall, so the solution decreases immediately.

Follow it down

As it descends toward $y=1$, the ticks flatten — the descent slows.

Ask whether it crosses

It cannot. $y=1$ is itself a constant solution, and two solutions never cross.

Conclude

The curve decreases and approaches $y=1$ asymptotically from above, never reaching it.

EX 2

Match a field to an equation

eliminate by structure

A field is symmetric about the $x$-axis, has vertical ticks along $y=0$, and horizontal ticks along $x=0$. Which of $\dfrac{dy}{dx}=x+y$, $\dfrac{dy}{dx}=-\dfrac{x}{y}$, $\dfrac{dy}{dx}=y^{2}$ is it?

Use the vertical ticks

Vertical means the slope is undefined, so the right-hand side must have $y$ in a denominator.

Eliminate two candidates

$x+y$ and $y^{2}$ are defined everywhere, so neither can produce a vertical tick.

Test the survivor

$-\dfrac{x}{y}$ is undefined at $y=0$ ✓ and equals $0$ when $x=0$ ✓.

Confirm the symmetry

Replacing $y$ by $-y$ negates the slope, which is exactly the reflection symmetry described.

Answer

$\dfrac{dy}{dx}=-\dfrac{x}{y}$, whose solution curves are circles centred at the origin.

EX 3

Concavity from the field

second derivative without a formula

Along a solution of $\dfrac{dy}{dx}=x-y$ the ticks steepen as $x$ increases. What does that say, and how can it be confirmed?

Translate ‘steepen upward’

The slope $y'$ is increasing along the curve, so $y''\gt 0$.

Name the consequence

The solution is concave up on that stretch.

Confirm algebraically

Differentiate the equation: $y''=1-y'$.

Substitute the equation back in

$y''=1-(x-y)=1-x+y$.

Reconcile

$y''\gt 0$ exactly where $y\gt x-1$ — and that is precisely the region where the field is seen to steepen.

EX 4

Locate an extremum

flat tick plus a sign change

A solution of $\dfrac{dy}{dx}=x-y$ passes through $(1,0)$. Classify what happens there.

Compute the slope

$y'=1-0=1\neq0$, so $(1,0)$ is not a critical point.

Find where the curve does flatten

$y'=0$ requires $y=x$ — the curve must reach the line $y=x$ to turn.

Test the sign on each side

Below $y=x$, $y'\gt 0$; above it, $y'\lt 0$.

Classify

A rise followed by a fall is a local maximum wherever the curve meets $y=x$.

The general lesson

A flat tick alone is not an extremum. You still need the slope to change sign across it.

Watch out

Solution curves never cross

One tick per point means one direction per point. Two curves meeting would need two different slopes at the same place.

A flat tick is not automatically an extremum

Check that the field's slopes change sign as the curve passes through. A field can flatten and then resume in the same direction.

Follow the ticks, do not connect the dots

A traced solution is tangent to the field everywhere along it, not merely at the grid points it happens to pass near.

An equilibrium is a barrier

If $f(x,y)=0$ along a horizontal line, that line is itself a solution and no other solution can cross it. That single observation settles most long-run questions.

Course map. Topic 7.4 · LO FUN-7.C — estimate solutions to differential equations (EK FUN-7.C.3).