Slope Fields: Sketching and Reasoning

A slope field for $\frac{dy}{dx} = f(x, y)$ is a set of short line segments, one at each point of a grid, each with slope $f(x, y)$ at its point. A segment has no arrowhead: it only shows the steepness and the direction (rising, falling or level) of a solution through that point.

On the AP exam the usual slips are: segments drawn with the wrong sign, or with steepness that does not grow with the size of the slope (slope 2 must be steeper than slope 1); a field matched by $x$ when its slopes are constant along horizontal rows, so they depend on $y$; and a solution curve drawn cutting across the segments instead of running along them. When you sketch at given points, every zero slope must be clearly level.

1. Sketching the field at given points. Draw the slope field for $\frac{dy}{dx} = x(y - 1)$ at the nine points with $x$ in $\{-1, 0, 1\}$ and $y$ in $\{0, 1, 2\}$.

Work out the slope at each point:

The whole middle column ($x = 0$) and the whole middle row ($y = 1$) have slope 0: five level segments. The four corners have slope 1 or $-1$: segments at 45°, rising where the slope is $1$ and falling where it is $-1$.

2. Matching a field to its equation. The slope field below is for one of these equations: $\frac{dy}{dx} = x + 1$, $\frac{dy}{dx} = y + 1$, $\frac{dy}{dx} = -y - 1$, $\frac{dy}{dx} = x + y$. Which one?

Look along a row: in each horizontal row the segments are all the same, while they change from row to row. So $\frac{dy}{dx}$ depends on $y$ only, which leaves $y + 1$ and $-y - 1$; $x + 1$ would be constant down columns, and $x + y$ along neither.

Now the signs. The level segments are along $y = -1$, where both remaining choices are 0. Above that line the segments rise, so the slope is positive for $y > -1$: at $(0, 1)$, $y + 1 = 2$ is positive and $-y - 1$ is negative. And the segments get steeper farther from $y = -1$, as $y + 1$ does.

3. A solution through a point, and where it goes. For $\frac{dy}{dx} = \frac{1}{2}y(3 - y)$, sketch the solutions through $(0, 1)$ and through $(0, 4)$. Find $\lim_{x \to \infty} y$ for each.

The slope depends on $y$ only, and it is 0 when $y = 0$ or $y = 3$: the constant functions $y = 0$ and $y = 3$ are solutions, and the field is level along those two lines.

Start at $(0, 1)$: the slope is $1$, positive, so the solution rises. It keeps rising while $0 < y < 3$, but as $y$ nears 3 the slopes shrink toward 0, so the curve levels off under the line $y = 3$ without crossing it (two solutions cannot cross where the field gives one slope). Start at $(0, 4)$: the slope is $-2$, negative, so that solution falls toward $y = 3$ from above.

4. Where the slopes are zero, positive and negative. For $\frac{dy}{dx} = x(y - 1)$, describe the points where the slope is 0, where it is positive and where it is negative.

The product $x(y - 1)$ is 0 when $x = 0$ or $y = 1$: along the $y$-axis and along the horizontal line $y = 1$ (dashed below). These two lines cut the plane into four regions, and the sign of a product is positive when its two factors have the same sign:

• $x > 0$ and $y > 1$: positive (at $(2, 3)$ the slope is $4$);
• $x < 0$ and $y < 1$: positive (at $(-2, -1)$ it is $4$);
• $x > 0$ and $y < 1$, or $x < 0$ and $y > 1$: negative (at $(2, -1)$ it is $-4$).

5. Concavity and a relative minimum from the equation. Let $y = f(x)$ be the solution of $\frac{dy}{dx} = x + y - 1$ with $f(1) = 0$. Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$, and decide whether $f$ has a relative minimum, a relative maximum or neither at $x = 1$.

Differentiate the equation with respect to $x$. The derivative of $x$ is 1, and the derivative of $y$ is $\frac{dy}{dx}$:

At the point $(1, 0)$: $\frac{dy}{dx} = 1 + 0 - 1 = {0}$ and $\frac{d^2y}{dx^2} = 1 + 0 = {1}$. The first derivative is 0 and the second is positive, so by the second derivative test $f$ has a relative minimum at $x = 1$. The solution through $(1, 0)$ on the field shows it: level at the point, curving up on both sides.