How do step size and direction affect Euler’s method?
You will be able to: Compare Euler approximations and justify error direction in a specified example.
How do step size and direction affect Euler’s method?
You can follow a bending road more closely with many short straight moves than with one long move. For y′=y, the exact curve e^x bends upward, so forward tangent steps stay below it.
A useful starting point: How does Euler’s method turn local slopes into a prediction? →
Words and symbols before equations
- Endpoint error
- Approximation minus exact value at the target.
- Forward step
- A positive h.
- Backward step
- A negative h.
- Concave up
- A graph lying above its tangent lines on a concave-up interval.
What this picture assumes
Original model; rounded readouts and finite sampled graphs. y′=y, y(0)=1, target x=1. Euler step h=1/n. Orange straight segments approximate the navy exact curve e^x. The table lists every step; underestimation is specific to this equation and forward steps.
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.
- n=2, h=0.5: Euler y(1)≈2.25; exact e≈2.71828; signed error=-0.468282. Orange approximation lies below the navy exact curve.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For the displayed problem, n equal forward steps to x=1 give h=1/n and yₙ=(1+1/n)^n. The exact solution is e^x; the endpoint is e.
Every forward tangent step starts below or on e^x and the equation y′=y preserves this underestimation. Increasing n gives closer endpoint values here. This example does not prove that every equation, step size or method is stable.
Moving backward requires negative h. Starting at (0,1) for y′=y, a step h=−0.25 gives y≈0.75 at x=−0.25.
Error direction must be supported by the equation and interval. A coarse approximation can fail badly; for y′=−2y and h=1, Euler changes 1 to −1 even though the exact positive solution e^(−2x) never becomes negative.
A worked example, step by step
Compare Euler endpoint estimates for y′=y, y(0)=1 at x=1 using n=1,2 and 4.
- With n=1, h=1 and y₁=2.
- With n=2, h=0.5 and y₂=1.5²=2.25.
- With n=4, h=0.25 and y₄=1.25⁴=2.44140625.
- All are below e≈2.71828; their errors shrink in magnitude for this tested sequence.
Smaller steps often improve a well-behaved calculation, but an attractive polygon does not certify accuracy. Never ignore the sign of h.
What happens to the first update if h is negative?
Compare with an explanation
The slope contribution is multiplied by a negative step, moving backward in x.
Predict. Change one thing. Explain.
Compare 1,2,4 and 16 steps. Record the endpoint and signed error. Explain the underestimation using the convex exact curve and the recurrence, and state why this evidence is specific to this problem.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
n=2, h=0.5: Euler y(1)≈2.25; exact e≈2.71828; signed error=-0.468282. Orange approximation lies below the navy exact curve.
| Step | Current x | Current y | Slope | Next y |
|---|---|---|---|---|
| 1 | 0 | 1 | 1 | 1.5 |
| 2 | 0.5 | 1.5 | 1.5 | 2.25 |
Original model; rounded readouts and finite sampled graphs. y′=y, y(0)=1, target x=1. Euler step h=1/n. Orange straight segments approximate the navy exact curve e^x. The table lists every step; underestimation is specific to this equation and forward steps.
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′=−2y, y(0)=1, compare one h=1 step with two h=0.5 steps to x=1 and interpret against the exact e^(−2).
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Compare with the answer and four-point rubric
- 1 point: One step gives 1+1(−2)=−1.
- 1 point: Two half steps give 1→0→0.
- 1 point: The exact endpoint e^(−2)≈0.1353 is positive.
- 1 point: Both coarse approximations misrepresent positivity; step refinement and accuracy checks are needed.
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 1How is signed endpoint error defined here?
Approximate value minus exact value.
RECALL 2Can coarse Euler steps violate physical positivity?
Yes, even if the exact solution stays positive.
RECALL 3What justifies an error-direction claim?
The equation, recurrence and shape on the stated interval, not a universal slogan.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do step size and direction affect Euler’s method?
- For y′=y, y(0)=1: yₙ=(1+1/n)^n at x=1.
- Endpoint error=yₙ−e.
Remember: Smaller steps often improve a well-behaved calculation, but an attractive polygon does not certify accuracy. Never ignore the sign of h.
Conditions: Original model; rounded readouts and finite sampled graphs. y′=y, y(0)=1, target x=1. Euler step h=1/n. Orange straight segments approximate the navy exact curve e^x. The table lists every step; underestimation is specific to this equation and forward steps.
Refresh Kid · AP Calculus BC Unit 7 · Objectives FUN-7.C · Review edition
Framework, scope and review status
Mapped to College Board CED Topic 7.5, 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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