| { |
| "description": "All 18 assessment items (two counterbalanced nine-item forms) and 10 assisted problems. The same form can be used at pre-test or post-test according to participant assignment. Choice indices are zero-based canonical source indices.", |
| "problems": [ |
| { |
| "id": "A1", |
| "skill": "differentiating a product constraint", |
| "type": "choice", |
| "prompt": "Two positive quantities \\(x\\) and \\(y\\) both change with time and satisfy \\(xy=24\\) throughout the motion. Which equation follows by differentiating with respect to time \\(t\\)?", |
| "choices": [ |
| "\\(x\\frac{dy}{dt}+y\\frac{dx}{dt}=0\\)", |
| "\\(\\frac{dx}{dt}\\frac{dy}{dt}=0\\)", |
| "\\(x\\frac{dy}{dt}+y\\frac{dx}{dt}=24\\)", |
| "\\(x\\frac{dy}{dt}+y=0\\)" |
| ], |
| "key": 0, |
| "explanation": "Use the product rule; the derivative of the constant 24 is zero.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A2", |
| "skill": "reverse chain rule", |
| "type": "number", |
| "prompt": "For a vase, water volume \\(V\\) in cm³ depends on water depth \\(h\\) in cm. At the current depth, \\(dV/dh=40\\) cm³ per cm: this describes the local volume change per unit change in depth. Water enters at \\(dV/dt=10\\) cm³/s, with no outflow. How fast is the depth rising at this instant?", |
| "unit": "cm/s", |
| "key": 0.25, |
| "explanation": "h′=V′/(dV/dh)=10/40=1/4.", |
| "tolerance": 1e-08, |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A3", |
| "skill": "one-variable geometric model", |
| "type": "choice", |
| "prompt": "A pile of sand keeps the shape of a square-based pyramid as it grows: it is always similar to a pyramid with height \\(4\\) m and base side \\(2\\) m. Its height is \\(h\\) and its base side is \\(s\\). The volume formula is \\(V=\\frac{1}{3}s^2h\\). Which expression gives the pile's volume using only \\(h\\)?", |
| "choices": [ |
| "\\(V=\\frac{h^3}{3}\\)", |
| "\\(V=\\frac{4h^3}{3}\\)", |
| "\\(V=\\frac{h^3}{12}\\)", |
| "\\(V=\\frac{h^2}{12}\\)" |
| ], |
| "key": 2, |
| "explanation": "s/h=1/2, so V=(1/3)(h/2)²h=h³/12.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A4", |
| "skill": "solving with a hidden second value", |
| "type": "number", |
| "prompt": "A rectangle keeps a constant area of \\(48\\) cm² while its length \\(l\\) and width \\(w\\) both change. At one instant, \\(l=8\\) cm and the length is increasing at \\(4\\) cm/s. Find \\(dw/dt\\) at that instant. Use a negative value if the width is decreasing.", |
| "unit": "cm/s", |
| "key": -3, |
| "explanation": "w=48/8=6; 8w′+6(4)=0, so w′=−3.", |
| "tolerance": 1e-08, |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A5", |
| "skill": "constants versus variables", |
| "type": "choice", |
| "prompt": "A cone-shaped tank with its point at the bottom is \\(6\\) m tall and has a top radius of \\(3\\) m. Water is draining out. Consider:\n(i) the tank's height, 6 m\n(ii) the tank's top radius, 3 m\n(iii) the depth \\(h\\) of the water\n(iv) the radius \\(r\\) of the water's surface\n\nA student plans to differentiate the volume relationship with respect to time. Which quantities can be replaced by their fixed numerical values before differentiating?", |
| "choices": [ |
| "(i) and (ii) only", |
| "(i), (ii), and (iv), because the surface radius is fixed by the tank", |
| "All four: substituting their values at that instant is always valid", |
| "None of them; everything must stay symbolic until after differentiating" |
| ], |
| "key": 0, |
| "explanation": "Only the tank dimensions are constant. Water depth and water-surface radius vary.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A6", |
| "skill": "instant-specific rate", |
| "type": "number", |
| "prompt": "A \\(5\\) m ladder leans against a vertical wall. Its foot slides away from the wall at a constant \\(1\\) m/s. Let \\(x\\) be the distance from the foot to the wall and \\(y\\) the height of the top, so \\(x^2+y^2=25\\). At the instant when \\(x=3\\) m, a student correctly finds \\(dy/dt=-3/4\\) m/s. Find \\(dy/dt\\) exactly one second later. Give a fraction or a decimal rounded to two decimal places.", |
| "unit": "m/s", |
| "key": -1.3333333333333333, |
| "explanation": "One second later x=4 and y=3. y′=−xx′/y=−4/3.", |
| "tolerance": 0.005, |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A7", |
| "skill": "linear in rates, nonlinear in state", |
| "type": "choice", |
| "prompt": "A cube's edge length \\(s\\) is increasing, so its volume \\(V=s^3\\) is increasing too. Consider two separate changes, each made on its own:\n(P) The edge grows twice as fast, with the current edge length unchanged.\n(Q) The current edge length is twice as large, with the growth rate unchanged.\n\nHow does \\(dV/dt\\) change in each case? Choose the option that is correct for both P and Q.", |
| "choices": [ |
| "(P) doubles; (Q) doubles", |
| "(P) doubles; (Q) becomes four times as large", |
| "(P) becomes four times as large; (Q) becomes four times as large", |
| "(P) becomes four times as large; (Q) doubles" |
| ], |
| "key": 1, |
| "explanation": "V′=3s²s′ is linear in s′ and quadratic in s.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A8", |
| "skill": "rate dependence on state", |
| "type": "two-tier", |
| "prompt": "Water flows into a tank at a constant positive rate, with no outflow. The volume \\(V\\) in liters and depth \\(h\\) in cm satisfy \\(V=h^2+4h\\). Answer both parts.", |
| "parts": [ |
| { |
| "id": "answer", |
| "prompt": "Part 1. Compare how fast the depth is rising when \\(h=6\\) with how fast it is rising when \\(h=2\\). At \\(h=6\\), the depth is rising…", |
| "choices": [ |
| "at the same rate.", |
| "one half as fast.", |
| "one third as fast.", |
| "one fifth as fast." |
| ], |
| "key": 1 |
| }, |
| { |
| "id": "reason", |
| "prompt": "Part 2. Which statement best justifies your answer to Part 1?", |
| "choices": [ |
| "A constant inflow produces a constant rate of depth increase.", |
| "The depth rate equals the volume rate divided by \\(dV/dh\\), and \\(dV/dh\\) grows as \\(h\\) grows.", |
| "The depth rate is inversely proportional to the depth.", |
| "The depth rate is inversely proportional to the volume." |
| ], |
| "key": 1 |
| } |
| ], |
| "explanation": "h′=V′/(2h+4). The denominator rises from 8 to 16, so the depth rate halves. Both parts must be correct for one point.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point only when both parts are correct; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "A9", |
| "skill": "first error in a worked solution", |
| "type": "choice", |
| "prompt": "A \\(13\\) m ladder leans against a vertical wall. Its foot slides away from the wall at \\(2\\) m/s. A student wants the rate of change of the top's height when the foot is \\(5\\) m from the wall. Here is the student's work:\n\nLine 1: \\(x^2+y^2=169\\), and when \\(x=5\\), \\(y=12\\).\nLine 2: Differentiating: \\(2x\\frac{dx}{dt}+2y\\frac{dy}{dt}=0\\).\nLine 3: Substituting: \\(2(5)+2(12)\\frac{dy}{dt}=0\\).\nLine 4: \\(\\frac{dy}{dt}=-10/24=-5/12\\) m/s.\n\nIn which line does the first error occur?", |
| "choices": [ |
| "Line 2", |
| "Line 3", |
| "Line 4", |
| "There is no error." |
| ], |
| "key": 1, |
| "explanation": "Line 3 drops x′=2. Correct substitution gives y′=−20/24=−5/6. Line 4 correctly follows from the erroneous Line 3.", |
| "form": "A", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B1", |
| "skill": "differentiating a product constraint", |
| "type": "choice", |
| "prompt": "The pressure \\(P\\) and volume \\(V\\) of a trapped gas both change with time and satisfy \\(PV=300\\) throughout the change. Which equation follows by differentiating with respect to time \\(t\\)?", |
| "choices": [ |
| "\\(\\frac{dP}{dt}\\frac{dV}{dt}=0\\)", |
| "\\(P\\frac{dV}{dt}+V\\frac{dP}{dt}=300\\)", |
| "\\(P\\frac{dV}{dt}+V\\frac{dP}{dt}=0\\)", |
| "\\(P\\frac{dV}{dt}+V=0\\)" |
| ], |
| "key": 2, |
| "explanation": "Use the product rule; the derivative of the constant 300 is zero.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B2", |
| "skill": "reverse chain rule", |
| "type": "number", |
| "prompt": "In a simplified heating model, a room's stored energy \\(E\\) in kJ depends on its temperature \\(T\\) in °C. At the current temperature, \\(dE/dT=15\\) kJ per °C: this describes the local energy change per unit change in temperature. Energy enters at \\(dE/dt=3\\) kJ/min, with no energy loss. How fast is the temperature rising at this instant?", |
| "unit": "°C/min", |
| "key": 0.2, |
| "explanation": "T′=E′/(dE/dT)=3/15=1/5.", |
| "tolerance": 1e-08, |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B3", |
| "skill": "one-variable geometric model", |
| "type": "choice", |
| "prompt": "A pile of gravel keeps the shape of a cone as it grows: it is always similar to a cone with height \\(3\\) m and base radius \\(1\\) m. Its height is \\(h\\) and its base radius is \\(r\\). The volume formula is \\(V=\\frac{1}{3}\\pi r^2h\\). Which expression gives the pile's volume using only \\(h\\)?", |
| "choices": [ |
| "\\(V=\\frac{\\pi h^3}{27}\\)", |
| "\\(V=\\frac{\\pi h^3}{3}\\)", |
| "\\(V=\\frac{\\pi h^2}{27}\\)", |
| "\\(V=3\\pi h^3\\)" |
| ], |
| "key": 0, |
| "explanation": "r/h=1/3, so V=(1/3)π(h/3)²h=πh³/27.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B4", |
| "skill": "solving with a hidden second value", |
| "type": "number", |
| "prompt": "A trapped gas at constant temperature obeys \\(PV=120\\), with \\(P\\) in kPa and \\(V\\) in liters. At one instant, \\(V=6\\) liters and the volume is increasing at \\(3\\) liters/min. Find \\(dP/dt\\) at that instant. Use a negative value if the pressure is decreasing.", |
| "unit": "kPa/min", |
| "key": -10, |
| "explanation": "P=120/6=20; 20(3)+6P′=0, so P′=−10.", |
| "tolerance": 1e-08, |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B5", |
| "skill": "constants versus variables", |
| "type": "choice", |
| "prompt": "A tank has the shape of an inverted square-based pyramid with its point at the bottom. It is \\(8\\) m deep and its top is a square of side \\(4\\) m. Water is draining out. Consider:\n(i) the tank's depth, 8 m\n(ii) the side of the tank's top, 4 m\n(iii) the depth \\(h\\) of the water\n(iv) the side \\(s\\) of the square water surface\n\nA student plans to differentiate the volume relationship with respect to time. Which quantities can be replaced by their fixed numerical values before differentiating?", |
| "choices": [ |
| "None of them; everything must stay symbolic until after differentiating", |
| "All four: substituting their values at that instant is always valid", |
| "(i) and (ii) only", |
| "(i), (ii), and (iv), because the surface side is fixed by the tank" |
| ], |
| "key": 2, |
| "explanation": "Only the tank dimensions are constant. Water depth and the water-surface side vary.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B6", |
| "skill": "instant-specific rate", |
| "type": "number", |
| "prompt": "A taut \\(10\\) m tether connects a balloon to a ground anchor. The balloon moves while the tether stays taut. Its horizontal distance \\(p\\) from the anchor increases at a constant \\(2\\) m/s, and its height is \\(q\\), so \\(p^2+q^2=100\\). At the instant when \\(p=6\\) m, a student correctly finds \\(dq/dt=-3/2\\) m/s. Find \\(dq/dt\\) exactly one second later. Give a fraction or a decimal rounded to two decimal places.", |
| "unit": "m/s", |
| "key": -2.6666666666666665, |
| "explanation": "One second later p=8 and q=6. q′=−pp′/q=−16/6=−8/3.", |
| "tolerance": 0.005, |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B7", |
| "skill": "linear in rates, nonlinear in state", |
| "type": "choice", |
| "prompt": "A spherical balloon's radius \\(r\\) is increasing, so its volume \\(V=\\frac{4}{3}\\pi r^3\\) is increasing too. Consider two separate changes, each made on its own:\n(P) The radius grows twice as fast, with the current radius unchanged.\n(Q) The current radius is twice as large, with the growth rate unchanged.\n\nHow does \\(dV/dt\\) change in each case? Choose the option that is correct for both P and Q.", |
| "choices": [ |
| "(P) becomes four times as large; (Q) doubles", |
| "(P) becomes four times as large; (Q) becomes four times as large", |
| "(P) doubles; (Q) becomes four times as large", |
| "(P) doubles; (Q) doubles" |
| ], |
| "key": 2, |
| "explanation": "V′=4πr²r′ is linear in r′ and quadratic in r.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B8", |
| "skill": "rate dependence on state", |
| "type": "two-tier", |
| "prompt": "A device's stored energy increases at a constant positive rate. Its stored energy \\(E\\) and temperature \\(T\\) satisfy \\(E=T^2+8T\\). Answer both parts.", |
| "parts": [ |
| { |
| "id": "answer", |
| "prompt": "Part 1. Compare how fast the temperature is rising when \\(T=12\\) with how fast it is rising when \\(T=4\\). At \\(T=12\\), the temperature is rising…", |
| "choices": [ |
| "one fifth as fast.", |
| "one third as fast.", |
| "one half as fast.", |
| "at the same rate." |
| ], |
| "key": 2 |
| }, |
| { |
| "id": "reason", |
| "prompt": "Part 2. Which statement best justifies your answer to Part 1?", |
| "choices": [ |
| "The temperature rate is inversely proportional to the stored energy.", |
| "The temperature rate is inversely proportional to the temperature.", |
| "The temperature rate equals the energy rate divided by \\(dE/dT\\), and \\(dE/dT\\) grows as \\(T\\) grows.", |
| "A constant energy supply produces a constant rate of temperature increase." |
| ], |
| "key": 2 |
| } |
| ], |
| "explanation": "T′=E′/(2T+8). The denominator rises from 16 to 32, so the temperature rate halves. Both parts must be correct for one point.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point only when both parts are correct; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "B9", |
| "skill": "first error in a worked solution", |
| "type": "choice", |
| "prompt": "A taut \\(10\\) m tether connects a balloon to a ground anchor. The balloon moves so that its horizontal distance \\(p\\) from the anchor increases at \\(3\\) m/s. A student wants the rate of change of its height \\(q\\) when \\(p=6\\) m. Here is the student's work:\n\nLine 1: \\(p^2+q^2=100\\), and when \\(p=6\\), \\(q=8\\).\nLine 2: Differentiating: \\(2p+2q\\frac{dq}{dt}=0\\).\nLine 3: Substituting: \\(2(6)+2(8)\\frac{dq}{dt}=0\\).\nLine 4: \\(\\frac{dq}{dt}=-12/16=-3/4\\) m/s.\n\nIn which line does the first error occur?", |
| "choices": [ |
| "There is no error.", |
| "Line 2", |
| "Line 3", |
| "Line 4" |
| ], |
| "key": 1, |
| "explanation": "Line 2 omits the factor dp/dt when differentiating p². The correct derivative is 2p(dp/dt)+2q(dq/dt)=0; substitution gives q′=−36/16=−9/4.", |
| "form": "B", |
| "phases": [ |
| "pretest", |
| "posttest" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "One point for a correct response; otherwise zero. Missing and explicit unsure responses score zero." |
| }, |
| { |
| "id": "ms-hemisphere-mound", |
| "order": 1, |
| "skill": "state_dependence", |
| "title": "Hemispherical salt mound", |
| "prompt": "A pile of salt grows while keeping the shape of half a sphere. Its volume is \\(V=(2/3)\\pi r^3\\), where r is its radius. The radius grows at a constant 0.5 m per hour.\n\nHow many times as large is the volume’s rate of increase, \\(dV/dt\\), when \\(r=3\\) m as when \\(r=1\\) m?", |
| "fields": [ |
| { |
| "id": "ratio", |
| "label": "Ratio of the two volume rates", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 9, |
| "tolerance": 0.01 |
| }, |
| "points": 1 |
| } |
| ], |
| "worked": [ |
| "Differentiate \\(V=(2/3)\\pi r^3\\): \\(dV/dt=2\\pi r^2\\,dr/dt\\).", |
| "Use the same \\(dr/dt=0.5\\) at both instants. At \\(r=1\\), \\(dV/dt=\\pi\\); at \\(r=3\\), \\(dV/dt=9\\pi\\) m³/hour.", |
| "The ratio is \\(9\\pi/\\pi=9\\). Constant radius growth does not mean constant volume growth; each rate holds at its own instant.", |
| "The volume rate is linear in \\(dr/dt\\): tripling the radius-growth rate at a fixed radius triples \\(dV/dt\\). It is quadratic in \\(r\\): tripling radius multiplies \\(dV/dt\\) by nine. The volume itself is 27 times larger, a different ratio." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-changing-rectangle", |
| "order": 2, |
| "skill": "constraint_rates", |
| "title": "Changing rectangle", |
| "prompt": "A rectangle changes shape while its area stays at 12 cm². Initially, its length is 4 cm. The length increases at a constant 2 cm per second.\n\nHow fast is its width changing one second later? Use a negative value if the width is decreasing.", |
| "fields": [ |
| { |
| "id": "width_rate", |
| "label": "Width rate (cm/s)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": -0.6666666666666666, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "cm/s" |
| } |
| ], |
| "worked": [ |
| "The area stays fixed while both dimensions change, so \\(lw=12\\). Increasing the length therefore makes the width smaller.", |
| "One second later, \\(l=4+2(1)=6\\) cm, so \\(w=12/6=2\\) cm.", |
| "Differentiate the constant-area relation: \\(w\\,dl/dt+l\\,dw/dt=0\\). Thus \\(dw/dt=-(w/l)\\,dl/dt\\).", |
| "At this instant, \\(dw/dt=-(2/6)(2)=-2/3\\) cm/s. The negative sign means the width is decreasing.", |
| "An equally valid route is \\(w=12/l\\), giving \\(dw/dt=-12l^{-2}\\,dl/dt\\). A constant length rate does not give a constant width rate: the width rate depends on the current length." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-filling-cone", |
| "order": 3, |
| "skill": "chain_inversion", |
| "title": "Filling a cone-shaped tank", |
| "prompt": "A cone-shaped tank has its point at the bottom. The tank is 4 m deep and has a top radius of 2 m. Water flows in at a constant \\(\\pi\\) m³/min, with no outflow.\n\nThe water forms a smaller cone, with volume \\(V=(1/3)\\pi r^2h\\), where h is the water depth and r is the radius of the water surface.\n\nHow fast is the water depth rising when the water is 2 m deep?", |
| "fields": [ |
| { |
| "id": "depth_rate", |
| "label": "Depth rate (m/min)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 1, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "m/min" |
| } |
| ], |
| "worked": [ |
| "As water rises, its surface radius grows too. Similar cones give \\(r/h=2/4\\), so \\(r=h/2\\). The tank dimensions are fixed; the water dimensions vary.", |
| "Use this relation in the volume formula: \\(V=(1/3)\\pi(h/2)^2h=\\pi h^3/12\\).", |
| "Differentiate with respect to time: \\(dV/dt=(\\pi h^2/4)\\,dh/dt\\).", |
| "At \\(h=2\\), the volume gained per unit depth is \\(dV/dh=\\pi\\). Therefore \\(dh/dt=\\pi/\\pi=1\\) m/min.", |
| "The same volume flow raises the level more slowly when the surface is wider. Constant inflow fixes \\(dV/dt\\), not \\(dh/dt\\)." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-depth-rate-error", |
| "order": 4, |
| "skill": "state_dependence", |
| "title": "Checking a rate comparison", |
| "prompt": "Water enters a container at a constant 12 liters per minute, with no outflow. Its water volume V (liters) and depth h (cm) satisfy \\(V=2h^2+8h\\).\n\nA student writes:\n\\(dV/dt=(4h+8)\\,dh/dt\\)\n\\(dh/dt=12/(4h+8)\\)\n“The depth triples from 1 cm to 3 cm, so the depth rises one third as fast.”\n\nWhat is the correct multiplier for the depth’s rate of increase at 3 cm compared with 1 cm?", |
| "fields": [ |
| { |
| "id": "ratio", |
| "label": "Rate at 3 cm divided by rate at 1 cm", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 0.6, |
| "tolerance": 0.01 |
| }, |
| "points": 1 |
| } |
| ], |
| "worked": [ |
| "The student's differentiation is correct: \\(dV/dt=(4h+8)\\,dh/dt\\), so \\(dh/dt=12/(4h+8)\\).", |
| "The error is assuming that tripling depth triples the denominator. The fixed term 8 also matters.", |
| "At \\(h=1\\), \\(dh/dt=12/(4+8)=1\\) cm/min. At \\(h=3\\), \\(dh/dt=12/(12+8)=3/5\\) cm/min.", |
| "Divide the second rate by the first: \\((3/5)/1=3/5=0.6\\). The depth rises 0.6 times as fast, rather than one third as fast." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-boat-rope", |
| "order": 5, |
| "skill": "constraint_rates", |
| "title": "Boat and rope", |
| "prompt": "A rope attached to the front of a boat passes through a pulley on a dock. The pulley is 3 m above the rope’s attachment point on the boat. This height difference stays constant, and the rope stays straight and tight.\n\nThe rope is pulled in at a constant 1 m/s. A student claims that the boat therefore moves toward the dock at the same speed.\n\nHow fast is the boat moving toward the dock when its horizontal distance from the dock is 4 m? Give the speed as a positive number.", |
| "fields": [ |
| { |
| "id": "speed", |
| "label": "Boat's speed toward the dock (m/s)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 1.25, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "m/s" |
| } |
| ], |
| "worked": [ |
| "The horizontal distance \\(x\\) and rope length \\(L\\) both change; the 3 m height is fixed. At every instant, \\(L^2=x^2+3^2\\).", |
| "Differentiate before substituting current values: \\(2L\\,dL/dt=2x\\,dx/dt\\). The constant 9 contributes zero.", |
| "When \\(x=4\\), \\(L=\\sqrt{16+9}=5\\). The rope shortens, so \\(dL/dt=-1\\).", |
| "Thus \\(dx/dt=(L/x)(dL/dt)=(5/4)(-1)=-1.25\\) m/s. The boat's speed toward the dock is 1.25 m/s.", |
| "Equal rates would require \\(L=x\\), impossible with the fixed height. Each changing quantity carries its own rate. This answer holds at this instant: as \\(x\\) shrinks, \\(L/x\\) grows and the boat speeds up." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-power-error", |
| "order": 6, |
| "skill": "constraint_rates", |
| "title": "Power with a student error", |
| "prompt": "The electrical power supplied to a component is \\(P=IV\\), where I is current in amperes and V is voltage in volts.\n\nAt one instant, the current is 3 A and the power is 36 W. The current is increasing at 0.5 A/s, while the voltage is decreasing at 1 V/s.\n\nA student calculates \\(dP/dt=12(0.5)=6\\) W/s. What is the correct value of \\(dP/dt\\)? Use a negative value if the power is decreasing.", |
| "fields": [ |
| { |
| "id": "power_rate", |
| "label": "dP/dt (W/s)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 3, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "W/s" |
| } |
| ], |
| "worked": [ |
| "Recover voltage: \\(V=P/I=36/3=12\\) V. The student correctly found this value.", |
| "Both \\(I\\) and \\(V\\) change, so the product rule requires \\(dP/dt=V\\,dI/dt+I\\,dV/dt\\).", |
| "The signed rates are \\(dI/dt=0.5\\) and \\(dV/dt=-1\\).", |
| "Therefore \\(dP/dt=12(0.5)+3(-1)=6-3=3\\) W/s.", |
| "The student dropped \\(I\\,dV/dt\\), treating voltage as fixed. If voltage instead fell at 2 V/s, the power would momentarily have zero rate even though both factors changed." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-trapezoid-trough", |
| "order": 7, |
| "skill": "chain_inversion", |
| "title": "Trapezoidal trough", |
| "prompt": "A long, open water tank (a trough) is 6 m long. Viewed from one end, it is a trapezoid: 1 m wide at the bottom, 3 m wide at the top, and 1 m deep. Both sides slope outward equally, and this shape is the same along the whole tank.\n\nWater flows in at 3 m³/min, with no outflow. How fast is the water depth rising when the water surface is 0.75 m above the bottom?", |
| "fields": [ |
| { |
| "id": "depth_rate", |
| "label": "Depth rate (m/min)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 0.2, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "m/min" |
| } |
| ], |
| "worked": [ |
| "The trough dimensions are fixed, but water depth \\(h\\) and surface width \\(w\\) vary. The water-surface width is not always 3 m.", |
| "Each side slopes outward 1 m per metre of depth, so \\(w=1+2h\\). The wet trapezoid has area \\((1+w)h/2=h+h^2\\), hence \\(V=6(h+h^2)\\).", |
| "The local volume change per unit depth is \\(dV/dh=6+12h\\), which is 15 m³ per metre at \\(h=0.75\\).", |
| "Divide the volume rate by this derivative: \\(dh/dt=(dV/dt)/(dV/dh)=3/15=0.2\\) m/min.", |
| "At \\(h=0.25\\), \\(dV/dh=9\\) and \\(dh/dt=1/3\\). Tripling depth reduces the rate to three fifths, not one third, because \\(dV/dh\\) has a constant term." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-two-cars-revised", |
| "order": 8, |
| "skill": "constraint_rates", |
| "title": "Two cars", |
| "prompt": "Two straight roads meet at a right angle. Car A is 8 km east of the intersection and travels east at 60 km/h. Car B is 6 km north of the intersection and travels south toward it at 20 km/h.\n\nA student claims that the distance between the cars changes at \\(60+20=80\\) km/h. What is the actual rate of change of the distance between them at this instant? Use a positive value if the distance is increasing.", |
| "fields": [ |
| { |
| "id": "distance_rate", |
| "label": "Rate of change of the distance (km/h)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 36, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "km/h" |
| } |
| ], |
| "worked": [ |
| "Let \\(x\\) be A's distance east, \\(y\\) B's distance north, and \\(D\\) their separation. At every instant, \\(D^2=x^2+y^2\\).", |
| "Differentiate: \\(2D\\,dD/dt=2x\\,dx/dt+2y\\,dy/dt\\). Both cars contribute.", |
| "Car A moves east, so \\(dx/dt=60\\). Car B moves south toward the intersection, so \\(dy/dt=-20\\).", |
| "Recover the separation: \\(D=\\sqrt{8^2+6^2}=10\\) km.", |
| "Then \\(dD/dt=(8(60)+6(-20))/10=(480-120)/10=36\\) km/h.", |
| "Adding the speeds would describe motion along one line, not this geometry. If B drove south at 80 km/h, the separation would momentarily have zero rate even though both cars moved." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-wedge-pool", |
| "order": 9, |
| "skill": "chain_inversion", |
| "title": "Wedge pool with gauge", |
| "prompt": "A pool is 20 m long and 10 m wide. Its floor slopes evenly: it meets the top edge at the shallow end and is 2 m below the top edge at the deep end.\n\nWater flows in at 10 m³/min, with no outflow. A sensor at the deep end displays \\(G=50h\\), where h is the water depth in metres at that end.\n\nHow fast is the sensor reading changing when \\(h=1\\) m?", |
| "fields": [ |
| { |
| "id": "gauge_rate", |
| "label": "Sensor reading rate (units/min)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 5, |
| "tolerance": 0.01 |
| }, |
| "points": 1, |
| "unit": "units/min" |
| } |
| ], |
| "worked": [ |
| "The pool dimensions are fixed, but depth \\(h\\) and the length under water vary.", |
| "The floor rises 2 m over 20 m. At depth \\(h\\), the wet length is \\(10h\\); the triangular cross-section has area \\(5h^2\\), giving \\(V=50h^2\\).", |
| "Differentiate: \\(dV/dh=100h\\). At \\(h=1\\), the local volume change per unit depth is 100 m³ per metre.", |
| "Divide to obtain the depth rate: \\(dh/dt=10/100=0.1\\) m/min.", |
| "Then use \\(G=50h\\): \\(dG/dt=50\\,dh/dt=5\\) units/min. Divide to go from volume to depth, then multiply to go from depth to the gauge.", |
| "As the depth approaches 2 m, \\(dV/dh\\) doubles and the gauge rate halves. Even a gauge linear in \\(h\\) can have a rate that depends on depth." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| }, |
| { |
| "id": "ms-spherical-tank", |
| "order": 10, |
| "skill": "integrated", |
| "title": "Spherical tank with gauge", |
| "prompt": "A spherical water tank has a fixed radius of 3 m. When the water is h metres deep, its volume is \\(V=(\\pi/3)h^2(9-h)\\). Water flows in at a constant \\(2\\pi\\) m³/min, with no outflow. A sensor displays \\(G=20h^2\\).\n\nA student claims that constant water flow means the sensor reading must rise at a constant rate. What is \\(dG/dt\\) when \\(h=2\\) m?", |
| "fields": [ |
| { |
| "id": "gauge_rate", |
| "label": "dG/dt at h = 2 (units/min)", |
| "type": "number", |
| "required": true, |
| "inputMode": "decimal", |
| "validator": { |
| "kind": "number", |
| "expected": 20, |
| "tolerance": 0.05 |
| }, |
| "points": 1, |
| "unit": "units/min" |
| } |
| ], |
| "worked": [ |
| "The tank radius of 3 m is fixed. Water depth \\(h\\), volume \\(V\\), and gauge reading \\(G\\) vary.", |
| "Expand \\(V=3\\pi h^2-(\\pi/3)h^3\\), so \\(dV/dh=6\\pi h-\\pi h^2=\\pi h(6-h)\\). At \\(h=2\\), this is \\(8\\pi\\).", |
| "The depth rate is \\(dh/dt=(2\\pi)/(8\\pi)=1/4\\) m/min.", |
| "Since \\(G=20h^2\\), \\(dG/dt=40h\\,dh/dt=40(2)(1/4)=20\\) units/min.", |
| "Constant inflow fixes the volume rate, not the depth rate, since \\(dV/dh\\) varies. Even constant depth growth would not fix the gauge rate because \\(G\\) is quadratic. At \\(h=1\\) the gauge rate is 16; at \\(h=3\\) it is \\(80/3\\), about 26.7." |
| ], |
| "form": null, |
| "phases": [ |
| "assisted" |
| ], |
| "maximum_score": 1, |
| "scoring_rule": "Recorded numeric answer check; see field validator and tolerance. The consolidated participant score is the best recorded submission, or null if none is scored." |
| } |
| ] |
| } |
|
|
