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. Assumptions for every question: no calculator, exact answers, and every changing quantity is a differentiable function of time t. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
A square array of solar panels is being extended so that it stays square. At the moment when its side length is 6 m, the side length is increasing at 0.5 m per day. At what rate is the area of the array increasing at that moment?
- (A) 0.25 m² per day
- (B) 1 m² per day
- (C) 6 m² per day
- (D) 12 m² per day
Answer and explanation
Answer: (C). A = s², so dA/dt = 2s · ds/dt = 2(6)(0.5) = 6 m² per day.
- (A) squares the rate, (0.5)² = 0.25. The rate is not squared; the side is.
- (B) computes 2 · ds/dt and forgets the factor s.
- (D) computes 2s = 12 and forgets the factor ds/dt. That is dA/ds, not dA/dt.
Question 2 (multiple choice · core)
A rectangular animal pen is reshaped, but its area is always 20 m². Its length x and width y change with time. At the moment when x = 4 m and y = 5 m, the length is increasing at 2 m per minute. What is dy/dt at that moment?
- (A) −2.5 m per minute
- (B) −1.6 m per minute
- (C) 0 m per minute
- (D) 2.5 m per minute
Answer and explanation
Answer: (A). At all times xy = 20. Differentiate with the product rule: x · dy/dt + y · dx/dt = 0. Substitute: 4 · dy/dt + 5(2) = 0, so dy/dt = −10/4 = −2.5 m per minute. The width must shrink to keep the area fixed.
- (B) swaps the roles of x and y: −(dx/dt)(x)/y = −8/5.
- (C) uses (dx/dt)(dy/dt) = 0 instead of the product rule, then concludes dy/dt = 0.
- (D) has the wrong sign. If the length grows and the area is fixed, the width cannot also grow.
Question 3 (multiple choice · core)
A cylindrical piece of dough rises and spreads, so both its radius r and its height h change with time. Its volume is V = πr²h. Which expression gives dV/dt?
- (A) 2πr · (dr/dt) · (dh/dt)
- (B) πr² · (dh/dt)
- (C) 2πrh · (dr/dt) + πr² · (dh/dt)
- (D) 2πr · (dr/dt) + dh/dt
Answer and explanation
Answer: (C). V is the product of r² and h, and both change. Product rule: dV/dt = π[(d/dt of r²) · h + r² · dh/dt] = π[2r · (dr/dt) · h + r² · dh/dt].
- (A) multiplies the rates together instead of using the product rule.
- (B) treats r as a constant. That would only be right if the radius were fixed.
- (D) differentiates each factor separately and adds, which is not a valid rule.
Question 4 (multiple choice · core)
Quantities x and y depend on time, and z = x/y. At a certain instant, x = 6, y = 3, dx/dt = 1 and dy/dt = 2. What is dz/dt at that instant?
- (A) −1
- (B) 1/3
- (C) 1/2
- (D) 1
Answer and explanation
Answer: (A). Quotient rule: dz/dt = (y · dx/dt − x · dy/dt)/y² = (3 × 1 − 6 × 2)/9 = (3 − 12)/9 = −1.
- (B) is (dx/dt)/y = 1/3. It ignores the change in y.
- (C) is (dx/dt)/(dy/dt) = 1/2. Rates do not divide like that.
- (D) reverses the order in the numerator, (6 × 2 − 3 × 1)/9 = 1, which flips the sign.
Question 5 (constructed response · core)
A cylindrical candle has a fixed radius of 3 cm. As it burns, its height h decreases at a constant 0.4 cm per hour. The volume of wax is V = πr²h.
(a) Explain which quantities in V = πr²h are constants and which change with time. (b) Find dV/dt at the moment when h = 10 cm. Give units. (c) A student substitutes h = 10 before differentiating and gets dV/dt = 0. Explain the error. (d) Does your answer to (b) depend on the height? Explain.
Worked solution and suggested Marlbridge rubric
(a) π is a constant. The radius r = 3 cm is fixed for all time, so it is a constant too. The height h changes with time, so V changes with time.
(b) With r = 3, V = 9πh at all times. Differentiate: dV/dt = 9π · dh/dt. Since h decreases, dh/dt = −0.4. So dV/dt = 9π(−0.4) = −3.6π cm³ per hour (about −11.3 cm³ per hour). At that moment the wax volume is decreasing at 3.6π cm³ per hour.
(c) Substituting h = 10 first gives V = 90π, a constant, whose derivative is 0. But h = 10 is true only at one instant; the height is still changing. The equation must be differentiated while h is still a variable.
(d) No. dV/dt = 9π · dh/dt and dh/dt is constant, so the volume decreases at 3.6π cm³ per hour at every height. The value h = 10 is not needed.
| Point | What earns it |
|---|---|
| 1 | Identifies r and π as constants and h (and V) as changing |
| 1 | dV/dt = πr² · dh/dt with dh/dt = −0.4 (negative sign) |
| 1 | −3.6π cm³ per hour with units, stated as a decrease |
| 1 | Explains that h = 10 holds only at an instant, so substituting first wrongly removes dh/dt; notes in (d) that the rate does not depend on h |
Question 6 (constructed response · core)
A small bead slides around a circular wire hoop. With the centre of the hoop at the origin and lengths in cm, the bead’s position (x, y) always satisfies x² + y² = 100.
(a) Differentiate x² + y² = 100 with respect to t. (b) At the moment when the bead is at (6, 8), dx/dt = −4 cm per second. Find dy/dt. (c) Interpret the signs of dx/dt and dy/dt in (b). (d) Later the bead is at (6, −8), again with dx/dt = −4 cm per second. Find dy/dt and explain why it differs from (b).
Worked solution and suggested Marlbridge rubric
(a) 2x · dx/dt + 2y · dy/dt = 0 (the derivative of the constant 100 is 0).
(b) Substitute: 2(6)(−4) + 2(8) · dy/dt = 0, so −48 + 16 · dy/dt = 0 and dy/dt = 3 cm per second. (Check: 6² + 8² = 100, so (6, 8) is on the hoop.)
(c) dx/dt < 0: the bead is moving left. dy/dt > 0: it is moving up. At (6, 8), on the upper right of the hoop, moving left means moving up and over the top.
(d) 2(6)(−4) + 2(−8) · dy/dt = 0, so −48 − 16 · dy/dt = 0 and dy/dt = −3 cm per second. The relationship between the rates depends on where the bead is: dy/dt = −(x/y) · dx/dt, and y has changed sign. On the lower half of the hoop, moving left means moving down.
| Point | What earns it |
|---|---|
| 1 | Correct derivative 2x · dx/dt + 2y · dy/dt = 0 |
| 1 | dy/dt = 3 cm/s at (6, 8), with values substituted after differentiating |
| 1 | Interprets both signs (moving left, moving up) |
| 1 | dy/dt = −3 cm/s at (6, −8) with a reason based on the sign of y |
Question 7 (constructed response · stretch)
In a 3D design program, a closed box has a square base of side s cm and height h cm. The designer stretches the base so that s increases at 0.1 cm per second, while the height decreases at 0.3 cm per second. Consider the instant when s = 10 cm and h = 4 cm.
(a) The volume is V = s²h. Find dV/dt at that instant. (b) The total surface area is S = 2s² + 4sh. Find dS/dt at that instant. (c) The base is growing, yet the volume is shrinking. Use your working in (a) to explain how this can happen.
Worked solution and suggested Marlbridge rubric
(a) Product rule (s² times h): dV/dt = 2s · (ds/dt) · h + s² · dh/dt. Substitute s = 10, h = 4, ds/dt = 0.1, dh/dt = −0.3:
dV/dt = 2(10)(0.1)(4) + (100)(−0.3) = 8 − 30 = −22 cm³ per second.
(b) dS/dt = 4s · ds/dt + 4(ds/dt · h + s · dh/dt) (chain rule on 2s², product rule on 4sh). Substitute:
dS/dt = 4(10)(0.1) + 4(0.1 × 4 + 10 × (−0.3)) = 4 + 4(0.4 − 3) = 4 − 10.4 = −6.4 cm² per second.
(c) The first term in (a), 8 cm³/s, is the gain from the growing base. The second term, −30 cm³/s, is the loss from the falling height. At this instant the base is large (area 100 cm²), so even a small drop in height removes a lot of volume. The loss is bigger than the gain, so the total rate is negative.
| Point | What earns it |
|---|---|
| 1 | Correct product-rule derivative of s²h, with ds/dt and dh/dt attached |
| 1 | dV/dt = −22 cm³/s with units |
| 1 | Correct derivative of S and dS/dt = −6.4 cm²/s with units |
| 1 | Explains (c) by comparing the two terms of dV/dt (gain of 8 versus loss of 30) |
A common error in (b) is to differentiate 4sh as 4 · (ds/dt) · (dh/dt). That ignores the product rule.
How did you do?
- Q1 or Q5 wrong: reread “The chain rule does the linking” and Worked example 1 in the study guide.
- Q2, Q3 or Q7 wrong: redo Worked example 2(a): the product rule with two changing quantities.
- Q4 wrong: redo Worked example 2(b) and check the order of the quotient rule.
- Q5(c) or Q6 wrong: reread “Always true, or only true now?” and differentiate before you substitute.
Then tick off the topic checklist.