How do you follow a solution through a slope field?
You will be able to: Use an initial condition and derivative signs to describe a solution and its concavity.
How do you follow a solution through a slope field?
A quantity adjusts toward a target level of 2. Below the target it rises; above the target it falls. The rule y′=2−y tells us how the same field guides many different starting levels.
A useful starting point: What can repeated rows and columns reveal about the rate rule? →
Words and symbols before equations
- Initial point
- The point (0,y₀) through which the selected solution passes.
- Equilibrium solution
- The constant solution y=2 in this field.
- Concavity
- Whether slope increases or decreases as x increases.
- Asymptote
- A line approached by a curve in a limiting direction.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. y′=2−y; exact solution 2+(y₀−2)e^(−x), starting at x=0. Teal dashed line is the equilibrium y=2; navy is the selected solution.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Increasing and concave down; approaches 2 from below.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The rate is positive below y=2 and negative above it. Thus a solution starting below 2 rises, and one starting above 2 falls. y=2 itself is constant.
The exact family y=2+(y₀−2)e^(−x) supports the picture. For finite x, e^(−x)>0, so a nonconstant solution stays on its starting side of 2. As x→∞ it approaches 2.
Differentiate the equation: y″=−y′=y−2. Below 2 the rising curve is concave down; above 2 the falling curve is concave up.
For this smooth rate rule, solution curves cannot cross because an initial point has a unique solution. Do not assume uniqueness for every possible nonsmooth differential equation.
A worked example, step by step
Describe the solution of y′=2−y, y(0)=0, for x≥0.
- At (0,0), y′=2: the curve initially rises.
- As long as y<2, y′>0, but y″=y−2<0, so the rise slows.
- The solution is y=2−2e^(−x); differentiating verifies the equation and the initial value.
- It remains below 2 at finite x and tends to 2 from below as x→∞.
An increasing function can be concave down. A finite graph cannot by itself prove a long-term limit.
If y(0)=4, does the solution cross y=2?
Compare with an explanation
No. y−2=2e^(−x) stays positive at every finite x.
Predict. Change one thing. Explain.
Try initial values 0,2 and 4. Compare direction, concavity and the horizontal equilibrium. Explain the long-term behavior using e^(−x), not just the plotted window.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Increasing and concave down; approaches 2 from below.
Signs and exact limit
y=2+(-2)e^(−x).
Initial slope=2.
At x=4, y=1.96337; this is not the infinite-time limit.
As x→∞, e^(−x)→0, so y→2.
Original model; rounded readouts and finite sampled graphs. y′=2−y; exact solution 2+(y₀−2)e^(−x), starting at x=0. Teal dashed line is the equilibrium y=2; navy is the selected solution.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the rate units, derivative signs, initial condition, domain or solution check. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAnalyze y′=3−y with y(0)=5: initial slope, direction, concavity and limiting value.
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Compare with the answer and four-point rubric
- 1 point: Initial slope is 3−5=−2.
- 1 point: The solution y=3+2e^(−x) stays above 3 and decreases.
- 1 point: y″=y−3>0, so it is concave up.
- 1 point: As x→∞, e^(−x)→0 and y→3 from above.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Can rising curves be concave down?
Yes; their positive slopes can decrease.
RECALL 2What selects one curve in this field?
An initial value.
RECALL 3Why can these particular curves not cross?
The smooth rate rule gives a unique solution through each initial point.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you follow a solution through a slope field?
- y=2+(y₀−2)e^(−x).
- y″=y−2; sign depends on position relative to equilibrium.
Remember: An increasing function can be concave down. A finite graph cannot by itself prove a long-term limit.
Conditions: Original model; rounded readouts and finite sampled graphs. y′=2−y; exact solution 2+(y₀−2)e^(−x), starting at x=0. Teal dashed line is the equilibrium y=2; navy is the selected solution.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.C · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.4, FUN-7.C. CED and Fall 2026 clarifications checked September 17, 2026. All nine BC topics are included, with Euler’s method and logistic interpretation. The correction to FUN-7.B.2 is reflected: an equation may have infinitely many solutions. Focused explanations, examples, visual models and practice are original Refresh Kid work.
Rate and amount units, signs, initial data, lost equilibria, solution intervals and model assumptions are explicit. Slope fields and Euler polygons are finite illustrations; exact algebra supports identities and limits. The logistic explicit formula supports the visualization; required interpretation can be done from the rate law without deriving that formula. The smooth examples have unique solutions; no universal uniqueness claim is made.
Organic Chemistry Tutor video titles and destinations were checked, not the full videos. Khan Academy’s destination was checked; its JavaScript lesson contents were not fully readable by the research tool. OpenStax sections 4.1–4.4 supplied conceptual cross-checks. No provider questions, artwork or scripts were copied. No affiliation or endorsement is implied.
GitHub’s 3D website collection informed optional spatial inspection and camera controls. The original tank reuses self-hosted Three.js with its MIT license retained. The 2×2 dm base connects water depth to volume: 1 dm³=1 L. This tank uses a feedback-controlled pump, not a gravity-drain law. Its 3D height and labeled 2D graphs use the same exact exponential formula. Rotation changes only the view, and complete explanations remain available without WebGL.
Independent teacher review and student usability testing remain pending. Technical checks do not certify mathematical accuracy, full accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. This multi-unit archive is not assigned as a complete Unit 7 task.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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