AP Calculus AB/BC · Unit 7 · Topic 7.3
Sketching Slope Fields
Draw short segments whose slopes equal the differential equation at sampled points.
1. Topic Focus
Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.
This topic: Draw short segments whose slopes equal the differential equation at sampled points.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For y′=x−y, all points on y=x have horizontal segments.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
A slope field, also called a direction field, is a grid of short line segments for a first-order differential equation
At the point \((x_0,y_0)\), draw a segment whose slope is \(F(x_0,y_0)\). The segment records the tangent direction that any solution through that point must have. It does not give the value of the solution, and it is not a displacement arrow.
How to sketch a slope field by hand
- Mark the requested grid. Keep the same coordinate scale throughout.
- Evaluate the right side. At each point, calculate \(m=F(x,y)\).
- Translate number into orientation. Use horizontal segments for \(m=0\), rising segments for \(m>0\), and falling segments for \(m<0\).
- Represent magnitude. A larger \(|m|\) means a steeper segment. Keep segment lengths approximately equal so length is not mistaken for rate.
- Use repeated patterns. Once points on the same isocline are recognized, copy the same orientation accurately.
- Audit several points. Recalculate one point in every visibly different region before finishing.
Reading slope values
| Value of \(F(x,y)\) | Segment | Meaning |
|---|---|---|
| \(0\) | horizontal | a solution through that point has a horizontal tangent |
| \(1\) | rises one for one | approximately a \(45^\circ\) orientation on equally scaled axes |
| \(-1\) | falls one for one | the tangent slopes downward as \(x\) increases |
| large positive or negative magnitude | very steep | orientation approaches vertical but keeps the correct sign |
| undefined | no slope segment | the differential equation assigns no finite slope there |
Isoclines and nullclines
An isocline is a set of points where the assigned slope is a constant \(m\):
Every field segment along that set has the same orientation. The special case \(F(x,y)=0\) is a nullcline, so its segments are horizontal. A nullcline is not automatically a solution curve. A horizontal line \(y=c\) is an equilibrium solution only when \(F(x,c)=0\) for every relevant \(x\).
Patterns worth recognizing
| Differential equation | Visible field pattern |
|---|---|
| \(y'=f(x)\) | all segments in a vertical column match |
| \(y'=g(y)\) | all segments in a horizontal row match |
| \(y'=x+y\) | equal slopes repeat on diagonals \(x+y=m\) |
| \(y'=xy\) | both axes have zero slope; signs alternate by quadrant |
| \(y'=g(y)\) with \(g(c)=0\) | the horizontal row \(y=c\) is an equilibrium row |
What a slope field does and does not show
The field encodes local derivative information without requiring an explicit solution formula. A denser grid improves the visual picture but does not change the differential equation. The next topic uses the field to reason about complete solution curves; here the main goal is to calculate and draw the local segments correctly.
6. Detailed Worked Example and Error Check
Example 1: Build a field from a slope table
Sketch \(y'=x+y\) on the grid \(x,y\in\{-1,0,1\}\). Evaluate the formula at every ordered pair:
| \(y\backslash x\) | \(-1\) | \(0\) | \(1\) |
|---|---|---|---|
| \(1\) | \(0\) | \(1\) | \(2\) |
| \(0\) | \(-1\) | \(0\) | \(1\) |
| \(-1\) | \(-2\) | \(-1\) | \(0\) |
The nullcline is \(y=-x\). Segments are horizontal there, positive above it, and negative below it. More generally, the isocline of slope \(m\) is \(y=m-x\), so equal orientations repeat along parallel diagonals.
Example 2: A curved nullcline
For \(y'=x^2-y\), horizontal segments satisfy
Below the parabola \(y=x^2\), the derivative is positive; above it, the derivative is negative. Because \(x\) is squared, the field is symmetric across the \(y\)-axis. The parabola is a nullcline, not a solution: its own slope is \(2x\), whereas the assigned field slope on it is \(0\).
Example 3: An autonomous equation
Consider \(y'=y(2-y)\). Since the right side depends only on \(y\), each horizontal row has one repeated slope.
Segments rise for \(0<y<2\) and fall for \(y<0\) or \(y>2\). The two horizontal zero-slope rows are equilibrium solutions because the rate remains zero for every \(x\) on each line.
Example 4: Product pattern by quadrant
For \(y'=xy\), either coordinate being zero makes the slope zero. In Quadrants I and III, \(xy>0\), so segments rise. In Quadrants II and IV, \(xy<0\), so segments fall. Moving farther from either axis increases \(|xy|\), making segments steeper.
Example 5: A field controlled mainly by \(x\)
For
the denominator is always positive, so the sign is determined by \(x\). Segments fall on the left, are horizontal on the \(y\)-axis, and rise on the right. For fixed \(x\), increasing \(|y|\) enlarges the denominator and flattens the segment.
Example 6: A singular line
For \(y'=1/(x-1)\), every vertical column has a common slope because the formula contains no \(y\). Slopes are negative for \(x<1\), positive for \(x>1\), and become very steep near \(x=1\). At \(x=1\) the differential equation is undefined, so no finite-slope segment should be drawn.
Example 7: Match a field to an equation
Suppose a field has horizontal segments on both axes, rising segments in Quadrants I and III, and falling segments in Quadrants II and IV. This sign pattern matches \(y'=xy\). It cannot match \(y'=x\), whose slopes repeat by columns, or \(y'=y\), whose slopes repeat by rows.
Common sketching errors
- Drawing arrows with different lengths instead of equal-length tangent segments.
- Using the point's \(y\)-coordinate as the slope without evaluating \(F(x,y)\).
- Making every segment in a row identical when the equation also depends on \(x\).
- Connecting the segments into a solution curve before the field is complete.
- Drawing a horizontal segment where the formula is undefined.
- Assuming every nullcline is itself a solution.
7. AP Reasoning Routine
Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
For each item, calculate or describe the slope-field segments. No differential equation needs to be solved.
(a) For \(y'=x-y\), find the slopes at \((0,0)\), \((0,1)\), \((1,0)\), and \((1,1)\).
(b) Describe the repeated pattern for \(y'=2x\) and locate all horizontal segments.
(c) Describe the field for \(y'=3-y\), including its equilibrium row and signs above and below it.
(d) For \(y'=x+y\), find the isocline on which every segment has slope \(2\).
(e) For \(y'=-xy\), determine the sign of the slope in each quadrant and on the axes.
(f) For \(y'=x^2-y\), find the nullcline and determine the sign above and below it.
(g) For \(y'=1/(y-1)\), state where the field is undefined and describe the sign on each side.
(h) A field has identical slopes across each horizontal row, with zero slope at \(y=0\), positive slope above, and negative slope below. Which better matches it: \(y'=x\) or \(y'=y\)? Explain.
(i) What segment should be drawn at \((-1,2)\) for \(y'=2x-y\)?
(j) A student makes steep slopes longer than shallow slopes. Explain why this can make a slope field misleading and give the correction.
Check the solution
(a) The slopes are \(0,-1,1,0\), respectively. Thus \(y=x\) is the zero-slope line.
(b) Slopes repeat down each vertical column. Every point on \(x=0\) has slope \(0\); columns to the right rise and columns to the left fall.
(c) Slopes repeat across each horizontal row. The row \(y=3\) is horizontal and is an equilibrium. Slopes are positive below \(3\) and negative above \(3\).
(d) Set \(x+y=2\), giving the line \(y=2-x\).
(e) Because the slope is \(-xy\), it is negative in Quadrants I and III, positive in Quadrants II and IV, and zero on both axes.
(f) The nullcline is \(y=x^2\). Slopes are negative above the parabola and positive below it.
(g) The field is undefined on \(y=1\). Slopes are negative below that line and positive above it; their magnitude grows near the line.
(h) It matches \(y'=y\), because dependence only on \(y\) creates matching rows. The signs and zero row also agree.
(i) The slope is \(2(-1)-2=-4\), so draw a short, steep downward segment centered at \((-1,2)\).
(j) Segment length should not encode magnitude because the slope is already encoded by orientation. Use approximately equal-length segments and make larger \(|m|\) appear steeper, not longer.