AP Calculus AB/BC · Unit 7 · Topic 7.4
Reasoning Using Slope Fields
Match solution curves, equilibrium behavior, and increasing or concave regions to a slope field.
1. Topic Focus
Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.
This topic: Match solution curves, equilibrium behavior, and increasing or concave regions to a slope field.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A solution cannot cross an equilibrium line when uniqueness conditions hold.
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 turns the local rule \(y'=F(x,y)\) into qualitative information about entire solution curves. A solution through \((x_0,y_0)\) must pass through that point smoothly and remain tangent to the nearby field segments as \(x\) moves both forward and backward.
A reasoning checklist
- Locate the initial point. It selects one curve from the family represented by the field.
- Read the sign of \(y'\). Positive segments mean the solution increases; negative segments mean it decreases.
- Track changing steepness. Slopes becoming more positive indicate concave up behavior; slopes becoming less positive or more negative indicate concave down behavior.
- Find zero-slope sets. Decide whether each is merely a nullcline or an equilibrium solution.
- Respect barriers and singularities. A solution cannot pass through a point where the equation is undefined, and equilibrium rows are barriers when standard uniqueness conditions hold.
- Describe long-term behavior cautiously. State what the visible field supports, such as approaching a level, moving away, or remaining bounded.
Increasing, decreasing, and horizontal tangents
| Field along the solution | Conclusion |
|---|---|
| \(F(x,y)>0\) | the solution is increasing |
| \(F(x,y)<0\) | the solution is decreasing |
| \(F(x,y)=0\) | the solution has a horizontal tangent at that point |
| sign changes from \(+\) to \(-\) | a local maximum is possible |
| sign changes from \(-\) to \(+\) | a local minimum is possible |
A solution may cross a curved nullcline. At the crossing it has a horizontal tangent, and the sign on the two sides determines whether the point is a local extremum. By contrast, a horizontal nullcline \(y=c\) is an equilibrium only if \(F(x,c)=0\) for every relevant \(x\).
Equilibrium solutions and stability
For an autonomous equation \(y'=g(y)\), solve \(g(c)=0\) to find constant solutions \(y=c\). A sign chart or phase line classifies nearby motion:
| Direction below \(c\) | Direction above \(c\) | Classification |
|---|---|---|
| up toward \(c\) | down toward \(c\) | stable: nearby solutions approach |
| down away from \(c\) | up away from \(c\) | unstable: nearby solutions separate |
| up or down | the same direction as below | semistable: approaches from one side only |
Concavity from the field
Concavity depends on how the slope changes along the chosen solution, not merely from left to right across a fixed row. If tangent slopes increase as the curve advances, then \(y''>0\); if they decrease, then \(y''<0\). When algebra is available, differentiating the differential equation confirms the visual reasoning:
For an autonomous equation \(y'=g(y)\), this becomes \(y''=g'(y)g(y)\). The formula is optional support; a clear slope-field argument can establish the same qualitative conclusion.
Can solution curves cross?
If \(F\) and the conditions ensuring local uniqueness are satisfied near a point, two distinct solutions cannot cross there: an intersection would give the same initial point two different solution curves. In typical AP examples with smooth right-hand sides, this means equilibrium curves and already-drawn solution curves act as barriers. Do not apply this claim across a singularity or when uniqueness is not guaranteed.
From one curve to a family
The field represents a family of solution curves. Different initial values select different members, but every valid member must follow the same local segments. Curves may share qualitative behavior, such as approaching one stable equilibrium, without being vertical translations of one another.
6. Detailed Worked Example and Error Check
Example 1: Cooling toward equilibrium
Consider \(y'=2-y\). The equilibrium is \(y=2\). Above it, slopes are negative; below it, slopes are positive.
- If \(y(0)=5\), the solution decreases toward \(2\).
- If \(y(0)=0\), the solution increases toward \(2\).
- If \(y(0)=2\), the solution remains constant.
The upper solution is concave up because its negative slopes become less negative. The lower solution is concave down because its positive slopes become less positive. Algebra confirms this: \(y''=-y'\).
Example 2: Two equilibria with different stability
For \(y'=y(y-3)\), the equilibria are \(y=0\) and \(y=3\). A sign chart gives
Below \(0\), solutions rise; between \(0\) and \(3\), they fall; above \(3\), they rise. Therefore \(y=0\) attracts from both sides and is stable, while \(y=3\) repels from both sides and is unstable.
Example 3: A semistable equilibrium
For \(y'=(y-1)^2\), the only equilibrium is \(y=1\), and every non-equilibrium slope is positive. Solutions below \(1\) rise toward the equilibrium, while solutions above \(1\) rise away from it. The equilibrium is semistable.
Example 4: Crossing a non-equilibrium nullcline
For \(y'=x-y\), the nullcline is \(y=x\). It is not an equilibrium because it is not a horizontal constant solution. Above the line, \(x-y<0\), so solutions decrease; below it, they increase. A solution that crosses from above the line to below it has a horizontal tangent and changes from decreasing to increasing, producing a local minimum.
Example 5: Logistic behavior without solving
For
the equilibria are \(P=0\) and \(P=100\). For \(0<P<100\), the population increases and remains between the equilibrium barriers. Field segments become steepest near \(P=50\), so a positive solution is concave up below \(50\), concave down above \(50\), and approaches \(100\).
Example 6: Choosing a curve from a family
Suppose the field is generated by \(y'=-2y\) and a curve must pass through \((0,3)\). The correct curve begins above the equilibrium \(y=0\), decreases, remains above the axis, and flattens toward it. A curve that crosses the axis or initially rises contradicts the field.
Example 7: Analytic confirmation of concavity
For \(y'=x+y\), differentiate along a solution:
At \((0,0)\), the solution has \(y'=0\) and \(y''=1>0\). Thus it has a horizontal tangent and is locally concave up, consistent with a local minimum.
Example 8: Bounds from equilibrium barriers
For \(y'=y(4-y)\), solutions with \(0<y(0)<4\) have positive slope. They cannot cross \(y=0\) or \(y=4\) in a smooth uniqueness setting because both are equilibrium solutions. Therefore the solution remains bounded between \(0\) and \(4\) while increasing toward \(4\).
Common reasoning errors
- Drawing a curve that passes through the initial point but is not tangent to nearby segments.
- Calling every zero-slope curve an equilibrium solution.
- Inferring concavity from the sign of \(y'\); increasing does not automatically mean concave up.
- Allowing a solution to cross a smooth equilibrium barrier.
- Classifying stability from only one side of an equilibrium.
- Claiming a precise limiting value beyond what the visible field or equation supports.
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.
Use slope-field or sign reasoning; do not solve the differential equations explicitly.
(a) For \(y'=-3y\) with \(y(0)=4\), describe monotonicity, concavity, bounds, and long-term behavior.
(b) For \(y'=y(y-2)\), identify and classify all equilibria.
(c) Classify the equilibrium of \(y'=(y+1)^2\) and describe motion on each side.
(d) For \(y'=x-y\), identify the nullcline, state the sign above and below it, and explain why the nullcline is not an equilibrium solution.
(e) For \(y'=1-y^2\), identify each equilibrium and classify its stability.
(f) For \(P'=0.2P(1-P/600)\) with \(0<P(0)<600\), describe monotonicity, the concavity change, and the limiting level suggested by the field.
(g) A field has horizontal segments on \(y=4\), positive slopes below, and negative slopes above. Classify \(y=4\) and describe a solution through \((0,7)\).
(h) Could a solution of \(y'=y(2-y)\) starting at \(y(0)=1\) cross the equilibrium \(y=2\) in a region where uniqueness holds? Explain.
(i) Along a solution curve, tangent slopes change from \(-3\) to \(-1\) to \(0\) as \(x\) increases. State monotonicity and concavity on that portion.
(j) For \(y'=x+y\), determine \(y'\) and \(y''\) at \((0,0)\), then classify the local shape.
Check the solution
(a) Above \(y=0\), slopes are negative, so the solution decreases. Since \(y''=-3y'=9y>0\), it is concave up. It remains between \(0\) and \(4\) and approaches the stable equilibrium \(0\).
(b) Equilibria are \(y=0\) and \(y=2\). Slopes are positive below \(0\), negative between \(0\) and \(2\), and positive above \(2\). Thus \(y=0\) is stable and \(y=2\) is unstable.
(c) \(y=-1\) is semistable. Slopes are positive on both sides: solutions below rise toward \(-1\), while solutions above rise away.
(d) The nullcline is \(y=x\). Slopes are negative above it and positive below it. The line is not an equilibrium because an equilibrium must be a constant horizontal solution \(y=c\).
(e) Equilibria are \(y=-1\) and \(y=1\). Slopes are negative below \(-1\), positive between them, and negative above \(1\). Thus \(y=-1\) is unstable and \(y=1\) is stable.
(f) The population increases while it lies between \(0\) and \(600\). It is concave up below \(300\), concave down above \(300\), and approaches \(600\).
(g) Segments on both sides point toward \(y=4\), so it is stable. The solution through \((0,7)\) decreases, remains above \(4\), and flattens as it approaches \(4\).
(h) No. The line \(y=2\) is an equilibrium solution and acts as a barrier under uniqueness; the solution from \(y=1\) increases toward it without crossing.
(i) The slopes remain negative, so the solution is decreasing. They increase from \(-3\) toward \(0\), so the curve is concave up.
(j) At \((0,0)\), \(y'=0\). Differentiating gives \(y''=1+y'=1\), so the solution has a horizontal tangent and is locally concave up, indicating a local minimum.