How do short line segments encode a differential equation?
You will be able to: Calculate and sketch tangent directions for y′=F(x,y).
How do short line segments encode a differential equation?
Imagine a tiny sign at each point of a map telling a curve which way to tilt there. For y′=x−y, the sign at (1,0) tilts upward while the sign at (1,2) tilts downward.
A useful starting point: How can you check a proposed solution without solving the equation? →
Words and symbols before equations
- Slope field
- Short tangent-direction segments at sampled (x,y) points.
- Slope m
- Change in y per unit change in x along the tangent.
- Horizontal segment
- A segment with slope zero.
- Solution curve
- A differentiable curve whose tangent follows the given slope at each point.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. Short equal-screen-length segments encode dy/dx in coordinate units; axis scales differ. The orange segment marks the selected point. A finite field does not prove a global 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.
- At (1,0), slope=1. Positive tilt. Segment direction represents dy/dx, not the amount y.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
At a grid point insert both coordinates into F. For (1,0), m=1−0=1; for (1,2), m=−1. These are slopes, not heights of the solution.
Draw short segments centered on each point. Equal segment lengths help compare direction; the length is an artistic choice and does not measure a rate magnitude.
On y=x, the equation gives slope zero. Above that diagonal slopes are negative; below it slopes are positive. A zero-slope set need not itself be a solution curve.
The field does not pick a starting point. An initial condition selects a curve when the equation has the appropriate uniqueness properties. The smooth fields in these examples have unique local solutions.
A worked example, step by step
Sketch the field directions at (0,0), (0,1), (1,0) and (1,1) for y′=x−y.
- At (0,0), m=0: draw a horizontal segment.
- At (0,1), m=−1: draw a downward-tilted segment.
- At (1,0), m=1: draw an upward-tilted segment.
- At (1,1), m=0 again. The diagonal y=x is a zero-slope set, not a line that the solution follows.
The field gives y′, not y. A horizontal segment at one point does not mean the whole solution is constant.
Is y=x a solution of y′=x−y?
Compare with an explanation
No. Its derivative is 1, but the right side along y=x is 0.
Predict. Change one thing. Explain.
Select x−y and move the marked point across y=x. Predict the slope sign before reading it. Then compare x-only and y-only equations to explain repeated columns and rows.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At (1,0), slope=1. Positive tilt. Segment direction represents dy/dx, not the amount y.
Read patterns, then substitute
Both coordinates affect the rate; zero slopes lie on y=x.
An initial condition selects a solution curve.
Finite field samples support interpretation, not a global proof.
Original model; rounded readouts and finite sampled graphs. Short equal-screen-length segments encode dy/dx in coordinate units; axis scales differ. The orange segment marks the selected point. A finite field does not prove a global 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 questionFor y′=x+y, give the slopes at (0,1), (1,−1) and (−1,−1), and identify all points with zero slope.
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Compare with the answer and four-point rubric
- 1 point: At (0,1), slope is 1.
- 1 point: At (1,−1), slope is 0.
- 1 point: At (−1,−1), slope is −2.
- 1 point: The zero-slope set is y=−x; it is not itself a solution because its slope is −1.
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 1Which coordinates enter F(x,y)?
The coordinates of the grid point, not a guessed solution value elsewhere.
RECALL 2What do short segment lengths mean?
They are drawing choices; direction encodes slope.
RECALL 3Is a zero-slope set always a solution?
No; its own derivative must also satisfy the equation.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do short line segments encode a differential equation?
- At (a,b), m=F(a,b).
- For y′=x−y: m=0 along y=x.
Remember: The field gives y′, not y. A horizontal segment at one point does not mean the whole solution is constant.
Conditions: Original model; rounded readouts and finite sampled graphs. Short equal-screen-length segments encode dy/dx in coordinate units; axis scales differ. The orange segment marks the selected point. A finite field does not prove a global 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 7.3, 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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