MATH U101 · Quiz 2 · Tue 29 Sep · Modules 4–6 §14.3–14.7

Same Skeleton, Different Skin

A quiz question isn't a new problem. It's one of nine routines with something changed on the outside. This page names the routines, shows what a "twist" does to each one, and drills the one skill that matters in the room: looking at an unfamiliar question and saying which routine it is.

Start here · what Quiz 1 actually showed

The fear: "I understand the tutorial solutions, but the quiz will be twisted, so what's the point?" Here is what the 8 Sep paper did, from the instructor's own key (post-mortem):

So "twisted" in practice means one routine with one changed step. You can't predict the question, but you can learn to recognise the routine, and that is what this page trains. The same routines (plus Lagrange) are the Modules 4–6 half of the open-book mid-sem on 5 Oct, where working earns marks. Quizzes also count best 2 of 3, so none of this work is wasted whatever happens on Tuesday.

The 40-minute plan for Tuesday

In the room · in this order
  1. Minutes 0–3: name, don't solve. Read every question. In the margin, write the routine's name next to each one (for example "tree", "∇F × ∇G", "edges + corners"). This is the Part 2 drill below.
  2. Start with the question you named fastest. An early answer settles your nerves for the rest of the paper.
  3. On scratch paper, write the routine's first line before touching the numbers (for example dy/dx = −Fx/Fy or Duf = ∇f · u). The twist, if there is one, shows up at step 2 or 3, not before.
  4. If a step doesn't fit, ask the three twist questions. Is it backwards, handing me what I normally compute? Is it a condition, where "horizontal", "parallel", "zero" or "perpendicular" turns into an equation to solve? Is it a bad point, where a piecewise function means I use the definition instead of the formula?
  5. Check, then copy once. Run the routine's 10-second check (the table below). Only then write in the answer box. Overwriting scored zero on Quiz 1.
  6. Budget: about 2 minutes a mark (Quiz 1 was 20 marks in 40 minutes). When a question has used twice its budget, write your best answer and move on. Come back only if time is left.
Routine10-second check before you write the answer
ImplicitIs the point on the curve? (Substitute: it must give 0.) Is the denominator F(dependent letter)?
Chain-rule treeOne product per path, and all paths counted. Partials evaluated at the middle variables' values, not at t.
Directional derivative|u| = 1? The answer must lie between −|∇f| and |∇f|.
Tangent plane / normal lineDoes your plane pass through the given point? (Substitute it in.)
∇F × ∇GThe direction's dot product with each gradient is 0.
Second-derivative testD = fxxfyy − fxy2 (with the square). D < 0 is a saddle whatever fxx says.
Closed regionCorners in the table? Critical points outside the region thrown away? Max ≥ every other value in the table?
Scope · what's in on Tuesday

Quiz 2 covers §14.3 §14.4 §14.5 §14.6* §14.7. The star on §14.6 means partial coverage: the handout excludes error estimation, but linearisation is in. Lagrange multipliers (§14.8) are not in Quiz 2, so card 10 is there for the mid-sem and the drill leaves it out unless you tick it. One thing is still open: the §14.1–14.2 row (limits) is partly highlighted. Ask in class or check the course announcements. If limits are in, walkthroughs W0a and W0b cover them; this page doesn't drill them.

The six twists (all of them)

Over three past mid-sems, Quiz 1 and Tutorials 6–7, every "unfamiliar" question was a known routine plus one or more of these. Learn the names: naming a twist makes it ordinary.

TwistWhat changesWhat you do
new skinDifferent function, numbers, point or region shapeNothing different. Run the routine.
backwardsYou're given what you usually compute (a rate, a direction, a slope) and asked for what you usually start fromWrite the routine's equation anyway; the unknown just moves to the other side.
condition"horizontal", "parallel to", "perpendicular", "rate is zero", "fastest"Translate the word into an equation (tangent plane horizontal ⇒ ∇F ∥ k), then solve.
storyNo ∂ signs at all: a box, a hiker, a lizard, a hot plateName the function and the variables, then it's a routine.
bad pointA piecewise function, asked at its special pointUse the limit definition (difference quotient), never the formula.
two stepsThe answer to one routine feeds anotherDo them in order; each step is a routine you know.

The routines, in syllabus order: partials · implicit · chain-rule tree · along a curve · directional derivative · tangent plane · two surfaces · critical points · closed region · Lagrange · then the drill.

Part 1 · Ten skeletons, each in three skins

For each routine: the skeleton (the steps that never change), where you already met it, and three twisted versions. Before opening each twist, name the part of the skeleton it changes. The working is inside.

1 · Partial derivatives §14.3

  1. For fx, freeze every other letter and differentiate with the one-variable rules.
  2. Second partials: differentiate again. fxy means x first, then y.
  3. At the special point of a piecewise function: fx(a, b) = limh→0 [f(a + h, b) − f(a, b)] / h.

Already met: Quiz 1 Q3 · Tutorial 6 Q12(a) · Module 4 notes §2–3

new skin For f = xexy, find fxy.

Same: freeze and differentiate. Changed: it's two rounds, in a stated order.

Work it through
  1. fx = exy + xyexy (product rule with y frozen).
  2. Now ∂/∂y with x frozen: xexy + (xexy + xy · xexy).
  3. fxy = (2x + x2y)exy.

bad point f = (x3 − y3)/(x2 + y2) away from the origin, f(0, 0) = 0. Find fx(0, 0) and fy(0, 0).

Same: a partial derivative. Changed: the formula doesn't hold at the origin, so use step 3.

Work it through
  1. f(h, 0) = h3/h2 = h, so [f(h, 0) − 0]/h = 1 → 1: fx(0, 0) = 1.
  2. f(0, h) = −h, so the quotient is −1: fy(0, 0) = −1.
  3. The trap: applying the quotient rule and then substituting (0, 0) gives 0/0, which is meaningless.

backwards For which constants a does w = e−t sin(ax) satisfy the heat equation wt = wxx?

Same: compute partials. Changed: you're told they must match, and a is the unknown.

Work it through
  1. wt = −e−t sin(ax); wx = ae−t cos(ax), wxx = −a2e−t sin(ax).
  2. Equal for every x, t ⇔ a2 = 1: a = ±1 (a = 0 also works, but only because it makes w = 0).

2 · Implicit differentiation §14.4

  1. Move everything to one side: F(x, y) = 0 or F(x, y, z) = 0.
  2. dy/dx = −Fx/Fy; ∂z/∂x = −Fx/Fz. The denominator is always the partial with respect to the dependent letter.
  3. Check that the point is on the surface, then substitute.

Already met: Tutorial 6 Q1–Q2 · walkthrough W1 · Module 4 notes §6

story Find the slope of the tangent line to the curve x3 + y3 = 6xy at (3, 3).

Same: everything. Changed: it says "slope of the tangent line" instead of "dy/dx".

Work it through
  1. F = x3 + y3 − 6xy; on the curve: 27 + 27 − 54 = 0 ✓.
  2. Fx = 3x2 − 6y = 9, Fy = 3y2 − 6x = 9 ⇒ slope = −9/9 = −1.

new skin Find ∂x/∂z at (1, −1, −3) if xz + y ln x − x2 + 4 = 0.

Same: the formula. Changed: now x is the dependent letter, so Fx goes in the denominator.

Work it through
  1. On the surface: −3 + (−1)(0) − 1 + 4 = 0 ✓.
  2. ∂x/∂z = −Fz/Fx, with Fz = x = 1 and Fx = z + y/x − 2x = −3 − 1 − 2 = −6.
  3. ∂x/∂z = −1/(−6) = 1/6.

condition At which points on x2 + xy + y2 = 3 is the tangent line horizontal?

Same: −Fx/Fy. Changed: no point is given. "Horizontal" becomes the equation Fx = 0, solved together with the curve.

Work it through
  1. dy/dx = 0 ⇔ Fx = 2x + y = 0 (with Fy ≠ 0), so y = −2x.
  2. Into the curve: x2 − 2x2 + 4x2 = 3x2 = 3, so x = ±1.
  3. (1, −2) and (−1, 2). Check: Fy = x + 2y = ∓3 ≠ 0 ✓.

3 · The chain-rule tree §14.4

  1. Draw the tree: output on top, middle variables below it, independent variables at the bottom.
  2. Each path from the top down to your variable contributes one product; add the products.
  3. To evaluate: first find the middle variables' values, then substitute.

Already met: Tutorial 6 Q3–Q5 · walkthrough W2 · Module 4 notes §5

story A cylinder's radius grows at 2 cm/s while its height shrinks at 3 cm/s. How fast is the volume changing when r = 5, h = 10?

Same: a tree with two paths. Changed: you name the function yourself (V = πr2h), and "shrinks" means a minus sign.

Work it through
  1. dV/dt = Vr r′ + Vh h′ = 2πrh(2) + πr2(−3).
  2. = 200π − 75π = 125π cm3/s (growing).

new skin w = f(x − y, y − x) for some differentiable f. Show that wx + wy = 0.

Same: the tree. Changed: f is never given, so the answer stays in the symbols fu, fv.

Work it through
  1. Name the middle: u = x − y, v = y − x.
  2. wx = fu(1) + fv(−1); wy = fu(−1) + fv(1).
  3. Add: everything cancels, giving 0 ✓.

new skin z = f(x, y), x = g(t), y = h(t), with g(3) = 2, h(3) = 7, g′(3) = 5, h′(3) = −4, fx(2, 7) = 6, fy(2, 7) = −8. Find dz/dt at t = 3.

Same: two paths. Changed: the data is a list of values, and some are decoys. You need fx, fy at (g(3), h(3)) = (2, 7).

Work it through
  1. dz/dt = fx(2, 7) g′(3) + fy(2, 7) h′(3) = 6 · 5 + (−8)(−4) = 62.

4 · A function along a curve §14.4

  1. On the curve r(t), f becomes a function of t alone.
  2. df/dt = fx x′ + fy y′ (+ fz z′) = ∇f · r′(t).
  3. Extremes along the curve: set df/dt = 0, then classify with d2f/dt2 or by comparing values.

Already met: Tutorial 6 Q6–Q7 · walkthrough W3

new skin The temperature is T = x2 − y2. Where on the unit circle x = cos t, y = sin t is it hottest and coldest?

Same: steps 1–3. Changed: T itself is given, so you can substitute directly (the chain rule gives the same result).

Work it through
  1. T = cos2t − sin2t = cos 2t, so dT/dt = −2 sin 2t = 0 at t = 0, π/2, π, 3π/2.
  2. Hottest T = 1 at (±1, 0); coldest T = −1 at (0, ±1).

story A bug crawls along r(t) = (t, t2, 1) (time in seconds) through the temperature field T = xyz. How fast is the bug's temperature changing at t = 1?

Same: ∇T · r′. Changed: it looks like a directional-derivative question, but "per second" means use r′ itself, not a unit vector.

Work it through
  1. At t = 1 the bug is at (1, 1, 1); ∇T = (yz, xz, xy) = (1, 1, 1); r′ = (1, 2t, 0) = (1, 2, 0).
  2. dT/dt = 1 + 2 + 0 = 3 degrees per second.
  3. Compare: "rate per unit distance in the direction of motion" would be DuT = 3/√5. Per second uses r′; per unit distance uses the unit u.

backwards All you know is fx = y and fy = x. On the line x = t, y = 2 − t, where can f have an extreme value?

Same: step 2. Changed: f is never given (just like Tutorial 6 Q6), so you can't substitute and must use the partials.

Work it through
  1. df/dt = y(1) + x(−1) = (2 − t) − t = 2 − 2t, which is 0 at t = 1.
  2. d2f/dt2 = −2 < 0: a maximum along the line, at (1, 1).

5 · Directional derivative and gradient §14.5

  1. ∇f at the point.
  2. Make the direction a unit vector: u = v/|v|.
  3. Duf = ∇f · u = |∇f| cos φ. Fastest increase is along ∇f at rate |∇f|; fastest decrease is along −∇f; the rate is zero when u ⟂ ∇f.
  4. If a value c is given, solve ∇f · u = c together with u12 + u22 = 1.

Already met: Tutorial 6 Q8–Q12 · walkthrough W4 · 2023 mid-sem Q2(a) · Module 5 notes §2

new skin Find the rate of change of f = x2 + 3xy at P(1, 2) in the direction towards Q(4, 6).

Same: steps 1–3. Changed: the direction is hidden in "towards Q": v = Q − P.

Work it through
  1. ∇f = (2x + 3y, 3x) = (8, 3). v = (3, 4), length 5, u = (3/5, 4/5).
  2. (24 + 12)/5 = 36/5.

new skin Find the derivative of f = x2y at (1, 2) in the direction making an angle π/3 with the positive x-axis.

Same: everything. Changed: the direction is given as an angle, so u = (cos θ, sin θ), which is already a unit vector.

Work it through
  1. ∇f = (2xy, x2) = (4, 1); u = (1/2, √3/2).
  2. 2 + √3/2 ≈ 2.87.

condition At (1, 2), is there a direction in which f = x2 + y2 changes at rate 3? At rate 10?

Same: |∇f| cos φ. Changed: it's a yes/no question, answered without solving anything.

Work it through
  1. ∇f = (2, 4), |∇f| = √20 = 2√5 ≈ 4.47. Since cos φ runs from −1 to 1, every rate lies in [−2√5, 2√5].
  2. Rate 3: yes (two directions, one on each side of ∇f). Rate 10: impossible.

6 · Tangent plane, normal line, linearisation §14.6*

  1. Write the surface as a level surface F(x, y, z) = c. For a graph z = f(x, y), take F = f(x, y) − z.
  2. The normal is n = ∇F at the point.
  3. Plane: Fx(x − x0) + Fy(y − y0) + Fz(z − z0) = 0. Normal line: (x0, y0, z0) + t∇F.
  4. For a graph, the plane is the linearisation: L = f(a, b) + fx(x − a) + fy(y − b).

Already met: Tutorial 7 Q1 · walkthrough W5 · 2023 mid-sem Q2(b) · 2024 Q3(b) · Module 5 notes §5

condition At which points of x2 + y2 + z2 − 2x − 4y = 4 is the tangent plane parallel to the xy-plane?

Same: the normal is ∇F. Changed: no point is given. "Parallel to the xy-plane" means ∇F ∥ k, i.e. Fx = Fy = 0. (This is the 2023 mid-sem's twist.)

Work it through
  1. Fx = 2x − 2 = 0, Fy = 2y − 4 = 0 ⇒ x = 1, y = 2.
  2. Back into the surface: 1 + 4 + z2 − 2 − 8 = 4 ⇒ z2 = 9: (1, 2, 3) and (1, 2, −3).
  3. Sense check: the surface is the sphere with centre (1, 2, 0) and radius 3, and these are its top and bottom points ✓.

story Without a calculator, approximate √(3.022 + 3.972).

Same: step 4. Changed: no surface is mentioned. "Approximate" means linearise f = √(x2 + y2) at the nearby nice point (3, 4).

Work it through
  1. f(3, 4) = 5, fx = x/f = 3/5, fy = 4/5.
  2. L = 5 + 0.6(0.02) + 0.8(−0.03) = 5 + 0.012 − 0.024 = 4.988 (true value 4.98812…).

two steps Where does the normal line to z = x2 + y2 at (1, 1, 2) meet the xy-plane?

Same: steps 1–3 give the line. Changed: one extra step: set z = 0 on the line.

Work it through
  1. F = x2 + y2 − z, ∇F = (2x, 2y, −1) = (2, 2, −1). Line: (1 + 2t, 1 + 2t, 2 − t).
  2. z = 0 ⇒ t = 2 ⇒ (5, 5, 0).

7 · Two surfaces at once §14.6

  1. Write both as level surfaces, F = c1 and G = c2, and find both gradients at the point.
  2. Tangent line to the curve where they meet: direction ∇F × ∇G.
  3. Angle between the surfaces = angle between their normals: cos φ = ∇F · ∇G / (|∇F| |∇G|).

Already met: Tutorial 7 Q2–Q3 · walkthrough W5 · Module 5 notes, curve of intersection

new skin Find the tangent line at (1, 1, 1) to the curve where the ellipsoid x2 + 2y2 + 3z2 = 6 meets the plane x = 1.

Same: step 2. Changed: one surface is a plane, and its gradient is just its normal (1, 0, 0).

Work it through
  1. ∇F = (2x, 4y, 6z) = (2, 4, 6); ∇G = (1, 0, 0).
  2. Cross product: (4·0 − 6·0, 6·1 − 2·0, 2·0 − 4·1) = (0, 6, −4), or (0, 3, −2).
  3. (1, 1 + 3t, 1 − 2t). Check: x stays 1, so the line lies in the plane ✓.

new skin At what angle do the sphere x2 + y2 + z2 = 3 and the surface z = xy meet at (1, 1, 1)?

Same: step 1. Changed: it asks for an angle, so use the dot product of the two gradients instead of the cross product.

Work it through
  1. ∇F = (2, 2, 2); G = xy − z, ∇G = (y, x, −1) = (1, 1, −1).
  2. cos φ = (2 + 2 − 2)/(2√3 · √3) = 2/6 = 1/3: φ = arccos(1/3) ≈ 70.5°.

condition Are the sphere x2 + y2 + z2 = 3 and the plane x + y + z = 3 tangent at (1, 1, 1)?

Same: the gradients. Changed: "tangent to each other" means the normals are parallel (their cross product is 0).

Work it through
  1. ∇F = (2, 2, 2) = 2(1, 1, 1) = 2∇G, so the normals are parallel.
  2. Yes, they are tangent at that point (both points lie on both surfaces ✓).

8 · Critical points and the second-derivative test §14.7

  1. Solve fx = 0 and fy = 0 together (factor, or substitute one into the other).
  2. D = fxxfyy − fxy2 at each point.
  3. D > 0 and fxx > 0: local min. D > 0 and fxx < 0: local max. D < 0: saddle. D = 0: the test says nothing.

Already met: Tutorial 7 Q5–Q6 · walkthrough W6 · Module 6 notes §2

new skin Classify the critical points of f = x3 − 3xy + y3.

Same: steps 1–3. Changed: unlike Tutorial 7 Q5, the two equations are coupled, so substitute one into the other.

Work it through
  1. fx = 3x2 − 3y = 0 ⇒ y = x2. fy = 3y2 − 3x = 0 ⇒ x = y2 = x4 ⇒ x(x3 − 1) = 0. Points: (0, 0), (1, 1).
  2. fxx = 6x, fyy = 6y, fxy = −3. At (0, 0): D = −9. At (1, 1): D = 36 − 9 = 27, fxx = 6.
  3. Saddle at (0, 0); local min f = −1 at (1, 1).

new skin Classify the critical point of f = x4 + y4.

Same: step 1. Changed: D = 0, so step 3 fails and you must argue from the function itself.

Work it through
  1. The only critical point is (0, 0), where all three second partials are 0, so D = 0.
  2. But f ≥ 0 = f(0, 0) everywhere: a minimum. (Compare x3 + y3: also D = 0, but f(t, 0) = t3 takes both signs near 0, so it is neither.)

story Find the point of the plane x + 2y + z = 4 closest to the origin.

Same: step 1. Changed: you build f yourself: the squared distance, with z eliminated using the plane. (Minimise the square to avoid a root.)

Work it through
  1. g = x2 + y2 + (4 − x − 2y)2.
  2. gx = 4x + 4y − 8 = 0, gy = 4x + 10y − 16 = 0 ⇒ y = 4/3, x = 2/3, z = 2/3.
  3. (2/3, 4/3, 2/3), distance 2√6/3. Lagrange solves it too; see card 10.

9 · Absolute extrema on a closed region §14.7

  1. Sketch the region and label every edge and corner.
  2. Interior critical points: keep only those strictly inside.
  3. Each edge: substitute its equation to get a one-variable problem, then find its critical points.
  4. Corners (the edges' endpoints).
  5. One table of values: the biggest is the absolute max, the smallest the absolute min.

Already met: Tutorial 7 Q4 · walkthrough W7 · 2023 mid-sem Q3 · 2024 Q2(a) · Module 6 notes §3

Three regions: a disk, a diamond and a square, each with its candidate points marked (a) round edge (b) four edges (c) point outside
Each region is a different shape, but the candidate list is always the same kind of thing: interior points, points on each edge, corners. Red marks the absolute max and green the absolute min. In (c) the only critical point, (3, 0), lies outside the square and is thrown away.

new skin (a) Absolute max and min of f = x2 + y2 − 2x on the disk x2 + y2 ≤ 4.

Same: steps 2–5. Changed: the edge is round. On it, y2 = 4 − x2, with x restricted to [−2, 2]. There are no corners, but the ends of that x-interval play their part.

Work it through
  1. Interior: fx = 2x − 2 = 0, fy = 2y = 0 ⇒ (1, 0), f = −1.
  2. Edge: f = 4 − 2x on −2 ≤ x ≤ 2, which is a straight line, so its extremes are at the ends: f(−2, 0) = 8, f(2, 0) = 0.
  3. Max 8 at (−2, 0); min −1 at (1, 0).

new skin (b) Absolute extrema of f = 2x2 + y − 3xy on |x| + |y| ≤ 1. This is the 2024 mid-sem Q2(a).

Same: every step. Changed: the absolute values hide four straight edges, one per quadrant.

Work it through
  1. Interior: fy = 1 − 3x = 0, fx = 4x − 3y = 0 ⇒ (1/3, 4/9), f = 2/9.
  2. Third-quadrant edge y = −1 − x, −1 ≤ x ≤ 0: f = 5x2 + 2x − 1, critical at x = −1/5, giving f(−1/5, −4/5) = −6/5. The other three edges and the corners are on the past-paper page. Their values top out at 2.
  3. Max 2 at (±1, 0); min −6/5 at (−1/5, −4/5). The interior point is neither.

new skin (c) Absolute extrema of f = x2 + y2 − 6x on the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

Same: every step. Changed: step 2 finds a point outside the region, so both answers are on the boundary.

Work it through
  1. Interior: fx = 2x − 6 = 0 ⇒ (3, 0), which is outside the square, so throw it away.
  2. Edges: y = 0 gives x2 − 6x, decreasing on [0, 2] (values 0 down to −8). y = 2 gives x2 − 6x + 4 (4 down to −4). x = 0 gives y2 (0 up to 4). x = 2 gives y2 − 8 (−8 up to −4). No edge has an interior critical point.
  3. Corners: (0, 0) → 0, (2, 0) → −8, (0, 2) → 4, (2, 2) → −4: max 4 at (0, 2); min −8 at (2, 0).

10 · Lagrange multipliers §14.8 · not in Quiz 2, for the mid-sem

  1. Name f (what to optimise) and the constraint g = c.
  2. ∇f = λ∇g, one equation per component, together with g = c.
  3. Eliminate λ by comparing or dividing the equations. Watch for the cases where a variable is 0.
  4. Evaluate f at every solution and compare.

Already met: Module 6 notes §4 and the Module 6 lesson. Not on Tutorials 6–7.

story Find the largest area of a rectangle, with sides parallel to the axes, inscribed in the ellipse x2 + 4y2 = 8.

Same: steps 2–4. Changed: you build f: with a corner at (x, y) in the first quadrant, the area is 4xy.

Work it through
  1. ∇(xy) = (y, x) = λ(2x, 8y): y = 2λx, x = 8λy.
  2. Substitute: x = 16λ2x ⇒ λ = 1/4 (first quadrant) ⇒ y = x/2 ⇒ 2x2 = 8 ⇒ (2, 1).
  3. Largest area 4 · 2 · 1 = 8.

new skin Find the max and min of x + 2y on the circle x2 + y2 = 5.

Same: everything.

Work it through
  1. (1, 2) = λ(2x, 2y) ⇒ y = 2x ⇒ 5x2 = 5 ⇒ x = ±1.
  2. Max 5 at (1, 2); min −5 at (−1, −2).

story The closest point of x + 2y + z = 4 to the origin, again (card 8), this time by Lagrange.

The point: some questions fit two routines. Either one earns the marks, so use whichever you name first.

Work it through
  1. (2x, 2y, 2z) = λ(1, 2, 1) ⇒ (x, y, z) = (λ/2)(1, 2, 1), a multiple of the plane's normal.
  2. Into the plane: (λ/2)(1 + 4 + 1) = 4 ⇒ λ/2 = 2/3: (2/3, 4/3, 2/3), the same point as before.

Part 2 · Name the routine (30 seconds each)

Ten questions a round, drawn from a bank of 24 questions not on the tutorial sheets (26 with Lagrange). Don't solve them. Read, pick the routine, and check. The skill being trained is the first three minutes of Tuesday.

How to do a round

For each question, choose the routine from the menu and press Check. You'll see the giveaway words, which twist it is, and the first line to write. "Finish it" has the full answer if you want it, but it's optional. The target is about 5 minutes for ten. Do two rounds tonight and one before the quiz. If you pick the same wrong routine twice, reread that card in Part 1.

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    Ask an AI well

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    "Here's Tutorial 7 Q4. Rewrite it three times with a different twist each time (a round region, a critical point outside the region, a story). Give no answers."

    "I wrote this first line for [paste question]. Is it the right routine? Don't go further than that."