Slope Fields

A differential equation like $\dfrac{dy}{dx} = x - y$ tells you the slope of a solution curve at every point, even if you can’t solve it. A slope field draws those slopes as tiny line segments, so you can see the shape of every solution at once. Slope fields show up on almost every AP Calculus exam, usually as “sketch the slope field” or “sketch the solution through this point”.

Key ideas

What a slope field is

For a differential equation $\dfrac{dy}{dx} = F(x, y)$, pick a grid of points. At each point $(x, y)$, work out the slope $F(x, y)$ and draw a short segment with that slope, centred on the point.

• Slope $0$: a flat (horizontal) segment.
• Positive slope: the segment rises from left to right. Bigger values are steeper.
• Negative slope: the segment falls from left to right.
• Undefined slope (for example, dividing by $0$): a vertical segment, or leave the point empty.

Sketching a slope field

1. Make a table of slopes at the grid points you are asked about.
2. Look for patterns before drawing everything:
   • If $\dfrac{dy}{dx}$ depends only on $x$ (like $\dfrac{dy}{dx} = x$), every segment in the same vertical column is parallel.
   • If $\dfrac{dy}{dx}$ depends only on $y$ (like $\dfrac{dy}{dx} = y$), every segment in the same horizontal row is parallel.
   • Set $\dfrac{dy}{dx} = 0$ to find where the segments are flat.
3. Draw each segment short, so neighbouring segments don’t touch.

Sketching a solution curve

A solution curve must be tangent to the segments everywhere it goes. To sketch the particular solution through a point:

• Start at the given point and follow the direction of the nearby segments, both to the right and to the left.
• Move smoothly between segments; don’t jump from one segment to the next in corners.
• Solution curves of these equations never cross each other, so don’t let your curve cut across another solution.

Reading information from a slope field

• The solution is increasing where the segments rise ($\dfrac{dy}{dx} \gt 0$) and decreasing where they fall.
• A horizontal segment means a possible maximum, minimum, or flat point.
• If $\dfrac{dy}{dx} = 0$ all along a horizontal line $y = c$, then the constant function $y = c$ is itself a solution, called an equilibrium solution. For example, $y = 0$ solves $\dfrac{dy}{dx} = 3y$.
• Concavity comes from the second derivative. Differentiate the equation implicitly, remembering that $y$ depends on $x$, then substitute $\dfrac{dy}{dx}$ back in.

Worked examples

Example 1: Building a slope field

For $\dfrac{dy}{dx} = x - y$, find the slope at each point with $x$ and $y$ in $\{-1, 0, 1\}$, and describe the segments.

Solution. Compute $x - y$ at each point:

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

The slope is $0$ at $(-1, -1)$, $(0, 0)$, and $(1, 1)$: in fact, it is $0$ everywhere on the line $y = x$. Above that line ($y \gt x$), the slopes are negative; below it, they are positive and get steeper as you move down and to the right. The full slope field is in the figure below.

Example 2: A solution curve through a point

Sketch the solution of $\dfrac{dy}{dx} = x - y$ that passes through $(0, 1)$, and describe its behaviour.

The slope field for $\frac{dy}{dx} = x - y$, the solution through $(0, 1)$ (blue), and the straight-line solution $y = x - 1$ (green, dashed).

Solution. At $(0, 1)$ the slope is $0 - 1 = -1$, so the curve heads down to the right. As it moves right it reaches the line $y = x$, where the segments are flat, so it has a low point there. After that, it is below $y = x$, where slopes are positive, so it rises.

Going right, the curve gets closer and closer to the line $y = x - 1$. That line is a solution too: its slope is $1$, and $x - y = x - (x - 1) = 1$. ✓ Since solution curves don’t cross, the blue curve can approach this line but never meet it. Going left from $(0, 1)$, the slopes are more and more negative, so the curve climbs steeply as $x$ decreases.

Example 3: Matching slope fields

Match each slope field A, B, and C to one of the equations $\dfrac{dy}{dx} = x$, $\dfrac{dy}{dx} = y$, and $\dfrac{dy}{dx} = x + y$.

Three slope fields, each drawn for $-2 \le x \le 2$ and $-2 \le y \le 2$.

Solution. Look for patterns first.

• In B, every vertical column has parallel segments, so the slope depends only on $x$. The segments are flat along the $y$-axis ($x = 0$), rise to the right of it, and fall to the left. That is $\dfrac{dy}{dx} = x$.
• In C, every horizontal row has parallel segments, so the slope depends only on $y$. It is flat along the $x$-axis and positive above it. That is $\dfrac{dy}{dx} = y$.
• In A, the slopes change in both directions. They are flat along the line $y = -x$, which is exactly where $x + y = 0$. So A is $\dfrac{dy}{dx} = x + y$.

Example 4: Concavity from the equation

A solution of $\dfrac{dy}{dx} = x - y$ passes through $(0, 1)$. Is it increasing or decreasing there? Is it concave up or concave down there?

Solution. The slope at $(0, 1)$ is $0 - 1 = -1 \lt 0$, so the solution is decreasing.

For concavity, differentiate both sides with respect to $x$. Since $y$ is a function of $x$, the derivative of $y$ is $\dfrac{dy}{dx}$:

$$\frac{d^2y}{dx^2} = 1 - \frac{dy}{dx} = 1 - (x - y) = 1 - x + y$$

At $(0, 1)$: $\dfrac{d^2y}{dx^2} = 1 - 0 + 1 = 2 \gt 0$, so the solution is concave up there. This matches the figure: the blue curve is falling but bending upward at $(0, 1)$.

Common mistakes

Making segments too long. Long segments cross each other and make the field unreadable. Keep them short and centred on each grid point.

Plugging the point in backwards. At $(1, 2)$, $x = 1$ and $y = 2$. For $\dfrac{dy}{dx} = x - y$ the slope is $1 - 2 = -1$, not $2 - 1 = 1$.

Drawing the solution curve through only one side. When asked for the solution through a point, sketch it both to the left and to the right of the point (within the region shown), not just in one direction.

Letting the curve cut across segments. The curve should run along the segments, tangent to them, never at an angle across them. It also should not cross another solution, such as an equilibrium line.

Forgetting the chain rule in the second derivative. When you differentiate $x - y$ with respect to $x$, the derivative of $y$ is $\dfrac{dy}{dx}$, not $1$ and not $0$. Then substitute the original equation for $\dfrac{dy}{dx}$.

Mixing up “depends only on $x$” with rows. If the slope depends only on $x$, the segments match down each column (same $x$). If it depends only on $y$, they match across each row (same $y$).

Practice

1. (Warm-up) For $\dfrac{dy}{dx} = xy$, find the slope of the segment at each point: $(1, 2)$, $(-1, 3)$, $(0, 5)$, and $(2, -1)$.

Solution

• $(1, 2)$: $1 \cdot 2 = 2$
• $(-1, 3)$: $(-1)(3) = -3$
• $(0, 5)$: $0 \cdot 5 = 0$ (flat)
• $(2, -1)$: $2(-1) = -2$

2. (Warm-up) For $\dfrac{dy}{dx} = y - 2$, where are the segments horizontal? Where do they have positive slope?

Solution

The slope is $0$ when $y - 2 = 0$, so the segments are horizontal all along the line $y = 2$. (That line is an equilibrium solution.)

The slope is positive when $y - 2 \gt 0$, that is, at every point above the line $y = 2$.

3. (Core) In a slope field, all the segments in each horizontal row are parallel. The segments are flat along $y = 1$ and $y = -1$, and they have positive slope between those lines. Which equation could it be?

• (a) $\dfrac{dy}{dx} = 1 - x^2$
• (b) $\dfrac{dy}{dx} = 1 - y^2$
• (c) $\dfrac{dy}{dx} = y^2 - 1$

Solution

Parallel segments across each row means the slope depends only on $y$, which rules out (a).

Both (b) and (c) are $0$ at $y = \pm 1$. Test a point between the lines, say $y = 0$: (b) gives $1 - 0 = 1 \gt 0$ and (c) gives $0 - 1 = -1 \lt 0$. So the answer is (b).

4. (Core) A solution of $\dfrac{dy}{dx} = 2x - y$ passes through $(1, 3)$. Is it increasing or decreasing at that point? Is it concave up or concave down?

Solution

Slope: $2(1) - 3 = -1 \lt 0$, so it is decreasing.

Second derivative:

$$\frac{d^2y}{dx^2} = 2 - \frac{dy}{dx} = 2 - (2x - y) = 2 - 2x + y$$

At $(1, 3)$: $2 - 2 + 3 = 3 \gt 0$, so it is concave up.

5. (Core) Consider $\dfrac{dy}{dx} = (y - 1)(y - 3)$.

• (a) Find the equilibrium solutions.
• (b) A solution passes through $(0, 2)$. Is it increasing or decreasing there?
• (c) Describe what happens to that solution as $x$ increases.

Solution

(a) The slope is $0$ for every $x$ when $y = 1$ or $y = 3$, so the equilibrium solutions are $y = 1$ and $y = 3$.

(b) At $y = 2$: $(2 - 1)(2 - 3) = -1 \lt 0$, so it is decreasing.

(c) For every $y$ between $1$ and $3$, the slope is negative, so the solution keeps decreasing. It can’t cross the equilibrium solution $y = 1$, so it levels off and gets closer and closer to $y = 1$ as $x$ increases.

6. (Core) Write a differential equation whose slope field has these features: the slope depends only on $x$, the segments are flat along the line $x = 2$, and they have negative slope to the left of $x = 2$ and positive slope to the right.

Solution

One answer is

$$\frac{dy}{dx} = x - 2$$

It depends only on $x$, is $0$ at $x = 2$, is negative for $x \lt 2$, and is positive for $x \gt 2$. (Other answers work too, such as $\dfrac{dy}{dx} = 3(x - 2)$ or $\dfrac{dy}{dx} = (x - 2)^3$.)

7. (Core) For $\dfrac{dy}{dx} = x - y$, show that the line $y = x - 1$ is a solution. Then explain why the solution through $(0, 1)$ stays above this line for every $x$.

Solution

For $y = x - 1$: the left side is $\dfrac{dy}{dx} = 1$, and the right side is $x - (x - 1) = 1$. They match, so it is a solution.

At $x = 0$, the solution through $(0, 1)$ is above the line, which passes through $(0, -1)$. Solution curves of this equation never cross, so the curve through $(0, 1)$ can never reach the line. It stays above it for every $x$ (it just gets closer and closer as $x$ increases).

8. (Challenge) For $\dfrac{dy}{dx} = x - y$, show that wherever a solution curve meets the line $y = x$, it has a relative minimum.

Solution

On the line $y = x$, the slope is $x - y = 0$, so any solution that meets the line has a horizontal tangent there.

From Example 4, $\dfrac{d^2y}{dx^2} = 1 - \dfrac{dy}{dx}$. At a point where $\dfrac{dy}{dx} = 0$, this gives $\dfrac{d^2y}{dx^2} = 1 \gt 0$.

The first derivative is $0$ and the second derivative is positive, so by the second derivative test the solution has a relative minimum at that point.

9. (Challenge) Let $y = f(x)$ be the solution of $\dfrac{dy}{dx} = x^2 + y$ with $f(0) = 1$.

• (a) Write the equation of the tangent line to the graph of $f$ at $x = 0$, and use it to approximate $f(0.2)$.
• (b) Is your approximation an overestimate or an underestimate? Justify your answer.

Solution

(a) $f'(0) = 0^2 + 1 = 1$, so the tangent line is $y = 1 + 1(x - 0) = 1 + x$. Then $f(0.2) \approx 1 + 0.2 = 1.2$.

(b) Differentiate the equation:

$$\frac{d^2y}{dx^2} = 2x + \frac{dy}{dx} = 2x + x^2 + y$$

At $(0, 1)$: $\dfrac{d^2y}{dx^2} = 0 + 0 + 1 = 1 \gt 0$. The graph of $f$ is concave up near $x = 0$, so it lies above its tangent line there. The tangent-line value $1.2$ is an underestimate.