These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. All functions and data are invented models. Assumptions: no calculator for Questions 1 to 6; a graphing calculator is allowed for Question 7 (answers to 3 decimal places). This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
In a weather model, P(h) is the air pressure, in kilopascals (kPa), at an altitude of h metres. The model gives P′(1500) = −0.011. Which statement is the best interpretation?
- (A) At an altitude of 1500 metres, the air pressure is 0.011 kPa.
- (B) At an altitude of 1500 metres, the air pressure is decreasing at a rate of 0.011 kPa per metre of altitude.
- (C) Between altitudes 0 and 1500 metres, the air pressure falls by 0.011 kPa per metre on average.
- (D) At an altitude of 1500 metres, the altitude is decreasing at a rate of 0.011 metres per kPa.
Answer and explanation
Answer: (B). P′ has units kPa ÷ metres. A negative derivative means P is decreasing as h increases. The statement names the input value, the quantity, the direction and the rate with units.
- (A) confuses the rate P′(1500) with the amount P(1500).
- (C) describes an average rate over an interval. P′(1500) is the rate at one altitude only.
- (D) turns the units upside down and treats altitude as the output.
Question 2 (multiple choice · core)
A workshop’s cost of making q kites is C(q) = 500 + 12q − 0.02q² dollars, for 0 ≤ q ≤ 250. What is C′(100), and what are its units?
- (A) 8 dollars per kite
- (B) 10 dollars per kite
- (C) 15 dollars per kite
- (D) 1500 dollars
Answer and explanation
Answer: (A). C′(q) = 12 − 0.04q, so C′(100) = 12 − 4 = 8. Units: dollars ÷ kites. At a production level of 100 kites, the cost is increasing at a rate of 8 dollars per kite.
- (B) differentiates 0.02q² as 0.02q, forgetting to multiply by the power 2.
- (C) is the average cost per kite, C(100)/100 = 1500/100. That is not a derivative.
- (D) is the total cost C(100), an amount, not a rate.
Question 3 (multiple choice · core)
R(t) is the total rainfall, in millimetres, collected in a rain gauge t hours after a storm begins. Selected values are shown.
| t (hours) | 0 | 2 | 5 | 6 | 9 |
|---|---|---|---|---|---|
| R(t) (mm) | 0 | 3.0 | 10.5 | 12.0 | 15.6 |
Using the data, what is the best estimate of R′(5.5)?
- (A) 1.5 mm per hour
- (B) 2.5 mm per hour
- (C) 11.25 mm
- (D) 0.667 hours per mm
Answer and explanation
Answer: (A). Use the closest values either side of t = 5.5: R′(5.5) ≈ (12.0 − 10.5)/(6 − 5) = 1.5 mm per hour. At t = 5.5 hours, rain is collecting at approximately 1.5 mm per hour.
- (B) uses t = 2 and t = 5, (10.5 − 3.0)/3 = 2.5. That interval does not contain 5.5.
- (C) averages the two R values. That estimates the amount R(5.5), not the rate.
- (D) is the reciprocal, 1/1.5. The units are upside down.
Question 4 (multiple choice · core)
W(t) is the mass, in kilograms, of a pollutant in a lake t days after a clean-up begins. At t = 4, W′(4) = −3 and W″(4) = 0.5. Which statement is true at t = 4?
- (A) The mass of pollutant is decreasing, and it is decreasing more slowly as time passes.
- (B) The mass of pollutant is decreasing, and it is decreasing faster as time passes.
- (C) The mass of pollutant is increasing at a rate of 0.5 kg per day.
- (D) There are 3 kg of pollutant in the lake.
Answer and explanation
Answer: (A). W′(4) < 0, so the mass is decreasing. W″(4) > 0, so W′ is increasing: it is negative but moving towards 0. The size of the rate is shrinking, so the decrease is slowing down.
- (B) reads W″ > 0 the wrong way. For a negative rate, a positive second derivative means the decrease slows.
- (C) treats W″ as if it were the rate of change of W. W″ is the rate of change of the rate, in kg per day².
- (D) treats the derivative W′(4) = −3 as the amount W(4).
Question 5 (constructed response · core)
A metal rod is 50 cm long. The mass of the part of the rod from its left end to a point x cm along it is m(x) = 3x + 0.2x² grams, for 0 ≤ x ≤ 50.
(a) Find m′(20). Give units. (b) Interpret m′(20) in context. (c) Find the average rate of change of m over 0 ≤ x ≤ 50, and say what it means for the rod. (d) Use m″ to explain how the density of the rod changes from the left end to the right end.
Worked solution and suggested Marlbridge rubric
(a) m′(x) = 3 + 0.4x, so m′(20) = 3 + 8 = 11 grams per cm.
(b) At the point 20 cm from the left end, the mass is increasing at a rate of 11 grams per centimetre of rod. In other words, the rod’s linear density there is 11 g/cm.
(c) m(50) = 150 + 500 = 650 g and m(0) = 0. Average rate = (650 − 0)/(50 − 0) = 13 grams per cm. This is the average density of the whole rod: total mass 650 g spread over 50 cm.
(d) m″(x) = 0.4 g/cm², which is positive for every x. So m′ increases along the rod: from 3 g/cm at the left end to 23 g/cm at the right end. The rod gets denser from left to right.
| Point | What earns it |
|---|---|
| 1 | m′(20) = 11 with units g/cm |
| 1 | Interpretation names the position (20 cm from the left end), the quantity (mass), increasing, and the rate with units |
| 1 | Average rate 13 g/cm from (m(50) − m(0))/50, identified as average density |
| 1 | m″ = 0.4 > 0 used to conclude the density increases along the rod |
Note the input here is distance, not time. The same unit rule applies.
Question 6 (constructed response · core)
In a chemistry experiment, the concentration of a reactant A, in moles per litre (mol/L), is modelled by A = 0.80/(1 + 0.5t), where t is measured in seconds.
(a) Find [A]′(t). (b) Find [A]′(2), with units, and interpret it. (c) Chemists usually report the rate at which a reactant is used up as a positive number. What rate would they report at t = 2, and why is it positive? (d) Find [A]″(2). Is the reactant being used up faster or more slowly as time passes? Explain.
Worked solution and suggested Marlbridge rubric
(a) Write A = 0.80(1 + 0.5t)⁻¹. By the chain rule, [A]′(t) = −0.80(1 + 0.5t)⁻² × 0.5 = −0.4/(1 + 0.5t)².
(b) [A]′(2) = −0.4/(1 + 1)² = −0.4/4 = −0.1 mol/L per second. At t = 2 seconds, the concentration of A is decreasing at a rate of 0.1 mol/L per second.
(c) They would report 0.1 mol/(L·s). The “rate of using up A” describes a decrease, so it equals −[A]′(t). The minus sign is absorbed into the words “used up”, just as “decreasing at a rate of 0.1” absorbs it.
(d) [A]″(t) = 0.4/(1 + 0.5t)³, so [A]″(2) = 0.4/8 = 0.05 mol/L per second². It is positive, so [A]′ is increasing towards 0. The concentration is still falling, but more slowly: the reactant is used up more slowly as time passes.
| Point | What earns it |
|---|---|
| 1 | Correct derivative −0.4/(1 + 0.5t)², using the chain rule (or the quotient rule) |
| 1 | [A]′(2) = −0.1 with units mol/L per second, interpreted at t = 2 with “decreasing” |
| 1 | Reports 0.1 and explains the sign is carried by “used up” (rate of use = −[A]′) |
| 1 | [A]″(2) = 0.05 > 0 and concludes the decrease is slowing |
Acceptable alternative for (a): the quotient rule on 0.80/(1 + 0.5t) gives (0 − 0.80 × 0.5)/(1 + 0.5t)², the same result.
Question 7 (constructed response · stretch · calculator allowed)
The amount of a medicine in a patient’s body t hours after an injection is modelled by A(t) = 200t·e^(−0.5t) milligrams, for t ≥ 0.
(a) Find A′(1) and A′(3), each to 3 decimal places, with units. (b) Interpret A′(1) and A′(3) in context. (c) Find A″(1) to 3 decimal places and interpret it. (d) A student says: “A′(3) = −22.313, so the amount in the body at t = 4 will be exactly 22.313 mg less than at t = 3.” Use the model to test this claim and explain the difference.
Worked solution and suggested Marlbridge rubric
(a) By the product rule, A′(t) = 200e^(−0.5t) + 200t × (−0.5)e^(−0.5t) = 100e^(−0.5t)(2 − t).
- A′(1) = 100e^(−0.5) ≈ 60.653 mg per hour.
- A′(3) = −100e^(−1.5) ≈ −22.313 mg per hour.
(A calculator’s numerical derivative gives the same values.)
(b) At t = 1 hour, the amount of medicine in the body is increasing at a rate of about 60.653 mg per hour. At t = 3 hours, it is decreasing at a rate of about 22.313 mg per hour.
(c) A″(t) = 50e^(−0.5t)(t − 4), so A″(1) = −150e^(−0.5) ≈ −90.980 mg per hour². At t = 1 hour, the amount is still increasing, but the rate of increase is falling by about 90.980 mg per hour each hour.
(d) A(3) ≈ 133.878 mg and A(4) ≈ 108.268 mg, so the actual change is about −25.610 mg, not −22.313 mg. The derivative gives the rate at the single instant t = 3. During the next hour the rate changes (it becomes more negative, since A″(3) < 0), so the derivative only approximates the change. The student’s “exactly” is wrong.
| Point | What earns it |
|---|---|
| 1 | Both values A′(1) ≈ 60.653 and A′(3) ≈ −22.313 with units mg/h |
| 1 | Both interpretations with time, quantity, direction and rate (no double negatives) |
| 1 | A″(1) ≈ −90.980 interpreted as the rate of increase slowing |
| 1 | Computes the actual change ≈ −25.610 and explains that a derivative is a rate at one instant, so it gives only an approximate change |
Acceptable alternative for (d): any correct argument that the rate is not constant over 3 ≤ t ≤ 4, supported by a value such as A′(4) ≈ −27.067.
How did you do?
- Q1, Q4 or Q5(b) wrong: reread “Writing an interpretation that earns credit” and “What the second derivative adds” in the study guide.
- Q2 or Q6 wrong: redo Worked example 1, and check every derivative with the unit rule.
- Q3 wrong: redo Worked example 2 (estimating from a table).
- Q5(c) or Q7(d) wrong: review the difference between an average rate, an instantaneous rate and an actual change.
Then tick off the topic checklist.