Teacher Daily Cards · Week 3
Can math solve where to move? — one page per session, generated from the current curriculum.
How to read these cards. Scan the header icons:
🔧 hardware-dependent ·
★ Gate/Milestone. In the run of show, each teacher cue is its own line;
Say: /
Ask: mark exact moves.
▽ below / ◆ at / △ above are the three differentiation tracks, and
⏱ If behind today tells you what to cut when the clock is short.
When it goes sideways, work down this ladder. You do not have to invent a fix mid-period. 1. The day's ⏱ If behind today line on the card — it names exactly what to cut, and on gate days what must NOT be cut. 2. The week's If it goes sideways card at the back of this pack, when the problem is bigger than the clock. 3. The Boost Card on the table, when it is one student stuck rather than the room. 4. Message Liz after class, not during. If your situation is not named anywhere, every recovery move here follows one principle: protect the thing being assessed, and let everything else move.
Week 3 · Day 11 🔧The reverse problem
A motion is two points. To plan it, solve the arm’s angles at each end.
Competencies this lesson builds
- ◆ PATTERN (Notice what repeats · Computational Thinking) — Inverse kinematics is forward kinematics run backwards — same relationship, opposite direction. Seeing that is the whole day.
- ○ LIMITS (Recognize constraints · Mathematical Modeling) — Elbow-up and elbow-down both reach the point. Students commit to one shape and hold it at BOTH endpoints, or the angles won't chain.
Warm-up · Daily Driver (on the board): Going from position to angles is called inverse ___. (one word)
Main work block · run of show (45 min) — one cue per line
- 0–5 · Entry Task
- kinematics.
- Name the week’s priority skill.
- 5–15 · Aim it by guessing
- Students use the on-screen aimer: move two angle sliders to land on the END point.
- Count tries.
- The frustration IS the setup — guessing is slow; the math will be exact.
- 15–30 · Watch the demo
- The app now animates this on step 4, so it runs itself and a broken arm no longer blocks the lesson — demo on the front arm as well if you have one, since seeing the real elbow drop lands harder.
- One target, reached TWICE — elbow-up and elbow-down.
- Same hand position, different elbow.
- Say: “We’ll pick elbow-up for both endpoints so the motion stays one smooth shape — and once you pick, you stay: switching shapes between the two endpoints can force the arm through a large, awkward reconfiguration instead of the smooth sweep you designed.” Have them write ‘elbow-up, both ends’ on their card now — this choice rides through Days 12–15.
- 30–35 · Distance to the target
- Introduce r = √(x² + y²) — the straight-line reach from the base.
- Have them compute r for (10, 8): r = √(100 + 64) = √164 ≈ 12.8 cm.
- This is the number next class’s Law of Cosines needs.
- 35–42 · Predict + discuss
- Can two angle pairs hit the same point?
- (Yes.) Set up the two-endpoint idea: you’ll solve IK once per end.
- 42–45 · Exit + preview
- Elbow-up vs down = different angles, same tip.
- Preview Day 12: “a new tool, for triangles with no right angle — then we point it at this arm.”
Students produce → To plan a back-and-forth motion, how many times do you solve inverse kinematics?
Three tracks — below / at / above- ▽ Below — Aimer only — land the end point by guessing; name elbow-up vs. down from the picture.
- ◆ At — Aimer + compute r for the end point; predict the two-solution answer.
- △ Above — Estimate θ₂′ from the triangle by eye first, then check it against next class's Law of Cosines.
⏱ If behind today (35 min, or yesterday ran over): Aimer + demo only; move the r calculation to Day 13's warm-up.
Students will ask · what to sayAsk · “Which elbow do we use?”
Say: For OUR worked example, elbow-up clears Mr. Rodriguez’s rail. For their own project it is a design decision: pick the branch that satisfies their constraint and keeps θ₁ inside the shoulder’s 20°–150° travel, then use that same branch at both endpoints. The consistency is the rule; which branch is theirs to argue.
Watch for- Students thinking there’s one ‘right’ answer. Two solutions is the point — the design picks one.
- Mixing shapes across the motion — elbow-up at one endpoint, elbow-down at the other. The sweep between the two poses then goes somewhere nobody planned, and it surfaces on Days 15–16 with no obvious cause. Make them commit to elbow-up at BOTH ends today and write it down.
Boost Cards: BC5, BC6, BC1 · arm setup (BC7)
Substitute today? Run it as a paper day — warm-up, work block, and exit ticket exactly as written above.
No arm needed: defer the Arm Check to the next arm day, or have students measure from the reference video (same motion, ruler in frame) — not yet filmed, so until it is, demo the motion by hand with a ruler or use the Day-11 aimer in the app, which simulates the same landing. The prediction and the error reasoning are unchanged — only where the measured number comes from.
Week 3 · Day 12Law of Cosines
Meet the Law of Cosines — the tool for a triangle with no right angle.
Competencies this lesson builds
- ◆ CHOOSE (Pick the right tool · Computational Thinking) — Three sides, no right angle → Law of Cosines. SOHCAHTOA cannot do this triangle, and students should be able to say why.
- ◆ REPRESENT (Convert between forms · Computational Thinking) — The formula is rearranged to solve for cos(θ₂′) — the unknown they actually want.
- ★ JUSTIFY (Explain with evidence · Problem Solving) — A negative cosine means an obtuse elbow. Students explain why a bent elbow still lands the tip on the exact same point.
Warm-up · Daily Driver (on the board): The arm’s own triangle — base to elbow, elbow to tip, and the line closing it — does it always contain a right angle? (yes/no)
Note: This day is the tool, not the arm. No θ₂′, no r², no target — plain triangles only. The payoff for holding that line is that Day 13 becomes pure substitution.
Main work block · run of show (45 min) — one cue per line
- 0–5 · Entry Task
- No.
- The angle between the two links changes as the arm moves and is not generally 90°, and every tool they own so far needs a right triangle.
- That is the whole reason for the day: the Law of Cosines works when the triangle is not right.
- Skip the specific values here — this day is the tool, not the arm.
- Do NOT say 144° here — that is θ₂′, the joint command, which is this corner’s supplement, and it is next class’s answer (BC5).
- 5–20 · Teacher-led · what the tool is
- c² = a² + b² − 2ab · cos γ.
- Sell it as Pythagoras with one extra piece: at γ = 90°, cos γ = 0, the last term vanishes, and Pythagoras falls out.
- Acute γ makes c² smaller than a² + b²; obtuse makes it larger.
- Then the reverse form, cos γ = (a² + b² − c²)/(2ab), which is the one the arm needs.
- 20–28 · Teacher-led · when NOT to reach for it
- It is powerful and slow.
- If a right angle is sitting there, SOHCAHTOA is one step and this is four.
- Give them two or three quick “which tool?” calls before they touch a worksheet.
- 28–40 · Guided practice
- “Law of Cosines: the Tool” worksheet — plain triangles, both directions.
- The 9-12-15 problem is worth pausing on: cos γ comes out exactly 0, so the tool detects a right triangle by itself.
- 40–45 · Exit + preview
- Right triangle, hypotenuse and one acute angle, want the opposite side — which tool?
- (SOHCAHTOA.) Preview Day 13: “now we point it at Mr.
- Rodriguez’s motion and get the elbow angle.”
Students produce → worked problems on “Law of Cosines: the Tool”
Three tracks — below / at / above- ▽ Below — Follow the derivation on the diagram and name which tool each triangle needs (right angle → SOHCAHTOA, no right angle → Law of Cosines). Then the forward direction only, with cos γ given.
- ◆ At — Run it both directions on plain triangles: two sides + the included angle for the third side, and three sides for the angle.
- △ Above — Sketch or prove WHY the Law of Cosines generalizes SOHCAHTOA — what does c² = a² + b² − 2ab·cos(C) become when C = 90°? (It collapses to the right-triangle / Pythagorean case.)
⏱ If behind today (35 min, or yesterday ran over): Teach the formula and the Pythagoras collapse; send the plain-triangle worksheet home. Do not cut the “when NOT to use it” beat — reaching for the slow tool on a right triangle is the mistake that costs time on Day 13.
Students will ask · what to sayAsk · “Why isn’t the arm’s triangle a right triangle?”
Say: The two links meet at the elbow at any angle the servo puts them at — nothing holds that corner at 90°. That is exactly when you need the Law of Cosines.
Ask · “My textbook writes the formula the other way round.”
Say: It writes cos C = (a² + b² − c²)/(2ab), which is the same rule. Next class’s arm version looks flipped because the angle we want is the elbow’s supplement, not the triangle’s interior angle. Same law, one extra step.
Watch for- Reaching for the Law of Cosines when a right angle is already there. It is not wrong, just slow — and on a timed check it costs them.
- Losing the minus sign in front of 2ab · cos γ.
Boost Cards: BC5, BC6, BC1
Substitute today? Run it as a paper day — warm-up, work block, and exit ticket exactly as written above.
There is no arm today, so nothing to defer.
Week 3 · Day 13The elbow angle θ₂′ · and criteria
Run the Law of Cosines on the motion for the elbow angle θ₂′ — then turn “close enough” into a number.
Competencies this lesson builds
- ○ LIMITS (Recognize constraints · Mathematical Modeling) — “Accurate” isn't testable. “Within 1 cm at each end — yes/no” is. Every criterion gets a number and a test.
- ★ DESIGN (Create for user needs · Problem Solving) — The criteria come from what the user actually needs, and they choose the threshold — too tight and too loose both cost something.
- ○ TEST (Prediction vs. reality · Mathematical Modeling) — A criterion is what makes tomorrow's arm run pass or fail. Set it before you run, not after.
Warm-up · Daily Driver (on the board): For target (10, 8) with L₁ = 18 and L₂ = 22, what is r²? (r² = x² + y²)
Note: This is Week 3's steepest climb, and where weaker Algebra 2 students struggle. Circulate at the tables, not the front — the bottom third needs you at the elbow today.
Main work block · run of show (45 min) — one cue per line
- 0–5 · Entry Task
- 164.
- This is step 1 of the solve.
- 5–16 · Teacher-led · point the tool at the arm
- They met the Law of Cosines last class, so this is substitution, not a new idea. r² = 164, then cos(θ₂′) = (164 − 324 − 484)/792 = −0.813, then arccos → 144.4°.
- Name the flip once: the numerator is reversed because θ₂′ is the supplement of the triangle’s interior angle (BC5).
- 16–30 · Guided practice
- Law of Cosines elbow worksheet.
- Then their own target’s θ₂′.
- Watch the sign — a negative cosine means an obtuse (bent) elbow.
- 30–42 · Teacher-led + practice · vague → measurable
- Gear change; say so out loud.
- Every criterion gets a number and a yes/no test.
- Write each one as a MAXIMUM — “no more than 1.5 cm” — not as a band.
- A band like “1–2 cm” fails a landing that is closer than 1 cm, which is the opposite of what accuracy means, and an exact value like 1.0000 cm can only ever fail.
- Then User-Defined Scales for opinions: “easy to use” → Agree=3, Neutral=2, Difficult=1.
- Distribute BC8.
- Students sharpen the three criteria from their Day 5 Brief — including the accuracy target they will measure against on Day 16.
- 42–45 · Exit + preview
- A measurable criterion has a number (or a UDS) plus a yes/no test.
- Preview Day 14: “the shoulder angle θ₁.”
Students produce → worked problems on “Criteria vs. Constraints” and “Law of Cosines: the Elbow Angle θ₂′” (2 sheets — the second is extra practice in exam wording, print it for anyone who finishes)
Three tracks — below / at / above- ▽ Below — Worked example + fill-the-blank — the formula is set up with r² = 164 and the arm lengths in; they simplify the fraction and take arccos.
- ◆ At — Standard problems: set up and solve cos(θ₂′) for their own target.
- △ Above — Solve both endpoints, then sharpen a genuinely subjective criterion into a User-Defined Scale and defend the cut points.
⏱ If behind today (35 min, or yesterday ran over): One worked solve + one solo with BC5 open; worksheet to homework. Do not cut the sign discussion (a negative cosine means θ₂′ is greater than 90°) and do not cut criteria — Day 16 measures against it.
Students will ask · what to sayAsk · “My criterion is a feeling — is it invalid?”
Say: No — build a User-Defined Scale. Turn the opinion into 1–3 and you can score it. That's BC8.
Ask · “Why a maximum instead of an exact number?”
Say: An exact value (1.0000 cm) is nearly impossible to hit, so it can only ever fail. A ceiling — no more than 1.5 cm — passes anything closer, which is what accuracy means. Watch for students writing a band like “1–2 cm”: that fails a landing 0.4 cm out.
Ask · “My cos came out bigger than 1.”
Say: Then the target is unreachable — cosine can’t exceed 1. Check your arithmetic first, then whether the point is past the 40 cm reach. The step after the sign question makes this point: cos = +1 IS r = 40 and cos = −1 IS r = 4, so the cosine runs out of room exactly where the arm does.
Ask · “Why isn’t this a right triangle?”
Say: The two links meet at the elbow at any angle — the corner isn’t 90°. That’s exactly when you need the Law of Cosines.
Watch for- Order of operations in the numerator (r² − L₁² − L₂²). Have them compute L₁²=324, L₂²=484 first.
- The gear change at 30 minutes. Criteria is the shortest block on the day and the easiest to let slide — but Day 16 measures against whatever they write here.
Boost Cards: BC8, BC5, BC6, BC1
Substitute today? Run it as a paper day — warm-up, work block, and exit ticket exactly as written above.
There is no arm today, so nothing to defer.
Week 3 · Day 14 🔧Solving for θ₁
Find θ₁, the shoulder angle, at each endpoint — then you have both poses.
Competencies this lesson builds
- ◆ REPRESENT (Convert between forms · Computational Thinking) — θ₁ = α + β decomposes one angle into “aim at the target” plus “the lean the bent elbow forces.”
- ○ MODEL (Name the quantities · Mathematical Modeling) — Two solved poses ARE the motion — sweeping the joints between them is the path.
- ★ JUSTIFY (Explain with evidence · Problem Solving) — Students explain why aiming (α) alone misses: the shoulder and elbow angles are linked, not independent.
Warm-up · Daily Driver (on the board): You know θ₂′ = 144.4° at the end point. Is that enough on its own to say where the tip lands? (yes/no)
Main work block · run of show (45 min) — one cue per line
- 0–5 · Entry Task
- No — two joints place the tip. θ₂′ bends the elbow; θ₁ aims the whole arm, and the same 144.4° elbow can point anywhere until θ₁ is fixed.
- That is today.
- 5–20 · Teacher-led · θ₁ = α + β
- θ₁ aims the whole arm.
- Two pieces: α = arctan(y/x) points straight at the target; β is the lean from the inner triangle.
- Work (10,8): α=38.7°, β=89.5° → θ₁=128.2°.
- 20–35 · Guided practice
- Shoulder-angle worksheet.
- Students find α for their own target, then combine with β.
- 35–42 · Arm Check
- Use the GO button — enter (10, 8) and press GO; the app solves IK and the end-effector should land on (10, 8).
- Measure the miss in cm.
- 42–45 · Exit + preview
- θ₁ is built from the aim α and the lean β.
- Preview Day 15: “IK Solve Card, Part A — solve both endpoints on paper.”
Students produce → worked problems on “The Shoulder Angle θ₁ = α + β”
Three tracks — below / at / above- ▽ Below — α = arctan(y/x) worked; β SCAFFOLDED — give them d = 12.8 and the BC6 step-3 formula with the numbers already substituted, so they only take the arccos. Never just hand them 89.5°: β changes with every target, and that is the whole point of the day.
- ◆ At — Find α AND derive β themselves (BC6 step 3: cos β = (L₁² + d² − L₂²)/(2·L₁·d)), then combine for θ₁.
- △ Above — θ₁ = α + β is the ELBOW-UP pose. Work out the elbow-down formula, then test it on (10, 8): θ₁ = α − β = 38.7° − 89.5° = −50.8°, which falls outside the shoulder's 20°–150° travel. So this target has two correct solutions and the arm can only hold one. Which targets have two the arm CAN hold?
⏱ If behind today (35 min, or yesterday ran over): The math (α, β, θ₁) is the must-do; make the Arm Check a single teacher demo instead of every group taking a turn.
Students will ask · what to sayAsk · “Is θ₁ just arctan(y/x)?”
Say: That’s only α — the aim. You add β, the lean from the arm’s own triangle, to get the real shoulder angle.
Watch for- Stopping at α and calling it θ₁. Both pieces are required.
Boost Cards: BC5, BC6, BC1 · arm setup (BC7)
Substitute today? Run it as a paper day — warm-up, work block, and exit ticket exactly as written above.
No arm needed: defer the Arm Check to the next arm day, or have students measure from the reference video (same motion, ruler in frame) — not yet filmed, so until it is, demo the motion by hand with a ruler or use the Day-11 aimer in the app, which simulates the same landing. The prediction and the error reasoning are unchanged — only where the measured number comes from.
Week 3 · Day 15IK Solve Card · Part A (the math)
IK Solve Card, Part A — solve BOTH endpoints of your motion on paper (no arm run yet).
Competencies this lesson builds
- ○ LIMITS (Recognize constraints · Mathematical Modeling) — Step 0 is reachability: compute r at each endpoint and confirm 8 ≤ r ≤ 40 BEFORE solving. Catching this at the arm is too late.
- ○ MODEL (Name the quantities · Mathematical Modeling) — Both endpoints get named quantities: r², cos(θ₂′), θ₂′, θ₁ — the full pose, twice.
- ◆ REPRESENT (Convert between forms · Computational Thinking) — The solved angles get written as a path plan: how each joint sweeps from the start pose to the end pose.
Warm-up · Daily Driver (on the board): A target gives r = 5 cm. Reachable on the real arm? (yes/no)
Note: Part A is MATH ONLY — no arm run today (that's Part B, Day 16). 45 minutes is plenty for both endpoints. Students who need it draft it from the Days 12–13 refreshers.
Main work block · run of show (45 min) — one cue per line
- 0–5 · Entry Task
- No — the practical floor is 8 cm (the elbow bends only to 160°).
- Warm reachability.
- 5–14 · Frame Part A · Step 0 first
- Put Step 0 on the board: for each endpoint, r = √(x² + y²), confirm 8 ≤ r ≤ 40.
- (Geometry says 4 cm, but the elbow's 160° bend makes 8 cm the real floor — catch it here, not at the arm next class.) Then hand-solve each endpoint: r² → cos(θ₂′) → θ₂′ → θ₁.
- Two poses + the sweep.
- NO arm run today.
- Show the W3D5 exemplars.
- 14–40 · Solve both endpoints
- Students solve both endpoints on paper.
- BC5/BC6 on the tables.
- Circulate for sign errors and the elbow-up choice at BOTH ends.
- 40–45 · Submit Part A + preview
- Submit solved angles — a checkpoint, not the gate yet.
- Preview Day 16: “run it on the arm with GO, measure the error — that completes and tags the card.”
Students produce → worked problems on “IK Solve Card — Full Pipeline”
Three tracks — below / at / above- ▽ Below — BC6 skeleton (r → cos → θ₂′ → θ₁ boxes) for ONE endpoint; they fill the blanks.
- ◆ At — Both endpoints, full pipeline, plus the sweep between the poses.
- △ Above — EC1 / EC2 — solve the other IK solution, or add a wrist link.
⏱ If behind today (35 min, or yesterday ran over): One full endpoint solved in class; the second endpoint becomes homework or Day 16's warm-up.
Students will ask · what to sayAsk · “Do I solve both endpoints the same way?”
Say: Exactly the same pipeline, run twice — that's the pattern. The numbers differ; the steps don't.
Ask · “Why 8 cm, not 4?”
Say: 4 cm is pure geometry (arm folded straight back). The real elbow only bends to 160°, so a point closer than ~8 cm can't be formed on the arm — it fails there even though the math says reachable.
Watch for- Solving one endpoint and stopping. Part A needs BOTH poses + the sweep between them.
- Mixing elbow-up at one end and elbow-down at the other — the θ₁ values won't chain into one motion. Commit to elbow-up at both.
Boost Cards: BC5, BC6, BC1, EC1, EC2
Substitute today? Run it as a paper day — warm-up, work block, and exit ticket exactly as written above.
There is no arm today, so nothing to defer.
Week 3 · ContingenciesIf it goes sideways
Three recovery moves for this week — preserve rigor, ease the crunch. Reach for this card when the problem is bigger than the clock; if it is only the clock, the day's ⏱ If behind today line is the faster fix.
- Students can't finish both endpoints on Day 15 — Prioritize ONE full endpoint solve (r → cos θ₂′ → θ₂′ → θ₁) with a written explanation; the second endpoint becomes Day 16's warm-up. Rigor preserved, crunch eased.
- The arm is down on Day 16 — Day 16 is a run. If hardware's out, students submit Part A (both endpoints solved) and you demo one run when it's back — the card still tags on the math, with the measured box deferred.
- Lost time / calculator chaos — Pre-write L₁² = 324, L₂² = 484 and the reachability check on the board so students start at the plug-in, not the setup.
- The app or the internet is down — Days 12, 13, 14 and 15 print in full with answer keys, and Day 11 is a teacher demo on an arm that does not need the internet. Day 13's criteria half is the one with no worksheet, so run criteria from the board against the class's own Briefs, using its Catch-Up Card.
Generated from the live curriculum · npm run gen:printables