Use this page after studying Unit 4, Contextual Applications of Differentiation, or after the Unit 4 diagnostic. The unit is shared by Calculus AB and Calculus BC; every question is for both courses. These are original Marlbridge practice questions, not past exam questions, with invented data. The rubrics are a suggested Marlbridge rubric, not official scoring. No calculator.
Big ideas of the unit
- A derivative is a rate with units. Its units are (units of f) ÷ (units of input), and the input need not be time (Topic 4.1).
- A full interpretation names four things: input value, quantity, direction and rate with units. f(a) is an amount; f′(a) is a rate (Topic 4.3).
- Motion is the same idea twice. v = x′ and a = v′. The sign of v gives the direction; v and a with the same sign means speeding up (Topic 4.2).
- The second derivative says how a rate is changing: a falling quantity with a positive second derivative is falling more slowly (Topic 4.3).
- Related rates come from the chain rule. Differentiate an always-true equation with respect to t; each changing quantity brings its own rate (Topic 4.4).
- Build first, substitute last. Choose the equation, remove any variable whose rate you do not know, and substitute instant values only after differentiating (Topic 4.5).
- Close to a point, a curve behaves like its tangent line. L(x) estimates nearby values; the sign of f″ says whether the estimate is too high or too low (Topic 4.6).
- L’Hospital’s Rule is local linearity for quotients. When top and bottom both tend to 0 (or both to ∞), compare their derivatives instead (Topic 4.7).
Key relationships and methods
| You see or need | What it means or what to do | Topics |
|---|---|---|
| “Interpret f′(a)” | Input value, quantity, increasing/decreasing, rate with units | 4.1, 4.3 |
| Rate from a table | Difference quotient using the closest values either side | 4.1, 4.3 |
| Rest, direction, speeding up | v = 0; v changes sign; compare signs of v and a | 4.2 |
| Several quantities change together | Equation at all times → differentiate with respect to t | 4.4 |
| Product or quotient of changing quantities | Product or quotient rule, a rate on every changing factor | 4.4 |
| Extra variable with unknown rate | Remove it first (similar triangles, a fixed ratio) | 4.5 |
| Estimate a value near a | L(x) = f(a) + f′(a)(x − a) | 4.6 |
| Too high or too low? | f″ > 0: line below, underestimate; f″ < 0: overestimate | 4.6 |
| Limit gives 0/0 or ∞/∞ | Check both limits, then lim f′/g′ | 4.7 |
Question 1 (multiple choice · mixed)
W(t) is the volume of water, in litres, in a leaking tank after t minutes. W(5) = 300, W′(5) = −12 and W″(t) > 0 for 5 ≤ t ≤ 6. The tangent line at t = 5 is used to estimate W(5.5). Which statement is best supported?
- (A) The leak is slowing down, and the estimate 294 is too low.
- (B) The leak is speeding up, and the estimate 294 is too high.
- (C) The leak is slowing down, and the estimate 294 is too high.
- (D) The leak is speeding up, and the estimate 306 is too low.
Answer and explanation
Answer: (A). W′(5) = −12, so water leaves at 12 litres per minute. W″ > 0 means W′ is increasing towards 0, so the leak is slowing down. The estimate is W(5.5) ≈ 300 − 12(0.5) = 294. W″ > 0, so the graph bends upward and the tangent line lies below it: 294 is too low.
- (B) reads W″ > 0 as a faster leak and reverses the concavity rule.
- (C) reverses the concavity rule.
- (D) reads W″ > 0 as a faster leak and adds the change, as if the tank were filling.
Topics: 4.1, 4.3, 4.6.
Question 2 (multiple choice · mixed)
A rectangle on a screen has one corner at the origin and the opposite corner P on the curve y = 12 − x², with x > 0 and lengths in centimetres. P slides along the curve so that its x-coordinate increases at 0.5 cm per second. How fast is the area of the rectangle changing when x = 1?
- (A) Increasing at 5.5 cm² per second
- (B) Increasing at 4.5 cm² per second
- (C) Increasing at 9 cm² per second
- (D) Decreasing at 1 cm² per second
Answer and explanation
Answer: (B). The area is A = xy, and y = 12 − x² at all times, so A = 12x − x³. Differentiate with respect to t: dA/dt = (12 − 3x²) · dx/dt. At x = 1: dA/dt = 9 × 0.5 = 4.5 cm² per second. (Product rule check: dy/dt = −2x · dx/dt = −1, and 11(0.5) + 1(−1) = 4.5.)
- (A) keeps only y · dx/dt and forgets that the height y is also changing.
- (C) is dA/dx, with the rate dx/dt left out.
- (D) keeps only x · dy/dt.
Topics: 4.1, 4.4, 4.5.
Question 3 (multiple choice · mixed)
Functions f and g have continuous derivatives. The line y = 3x is tangent to the graph of f at x = 0, and the line y = 5x is tangent to the graph of g at x = 0. What is lim (x → 0) f(x)/g(x)?
- (A) 3/5
- (B) 5/3
- (C) 0
- (D) It cannot be determined without formulas for f and g.
Answer and explanation
Answer: (A). Both tangent lines pass through the origin, so f(0) = g(0) = 0; their slopes give f′(0) = 3 and g′(0) = 5. f and g are continuous, so the form is 0/0. By L’Hospital’s Rule, the limit is lim f′(x)/g′(x) = 3/5, since the derivatives are continuous and g′(0) ≠ 0. Local linearity says the same thing: near 0, f(x)/g(x) ≈ 3x/5x.
- (B) inverts the ratio.
- (C) treats 0/0 as 0.
- (D) misses that the tangent lines give all the values the rule needs.
Topics: 4.6, 4.7.
Question 4 (constructed response · mixed)
A particle moves along a horizontal line. Its velocity, in metres per second, is v(t) = 3 − 6 sin(πt/6) for 0 ≤ t ≤ 6 seconds. At t = 4 its position is x(4) = 2 m.
(a) Find a(t). Give a(2) with units. (b) Find the times when the particle changes direction. Justify. (c) Is the particle speeding up or slowing down at t = 2? At t = 4? Justify. (d) Use the tangent line to the graph of x at t = 4 to estimate x(4.1). Is the estimate too high or too low? Justify.
Worked solution and suggested Marlbridge rubric
(a) a(t) = −6 cos(πt/6) × (π/6) = −π cos(πt/6). a(2) = −π cos(π/3) = −π/2 m/s².
(b) v = 0 when sin(πt/6) = 1/2. With 0 ≤ πt/6 ≤ π, πt/6 = π/6 or 5π/6, so t = 1 or t = 5. v(0) = 3 > 0, v(2) = 3 − 3√3 < 0 and v(6) = 3 > 0, so v changes sign at both and the particle changes direction at t = 1 and t = 5.
(c) At t = 2: v(2) = 3 − 3√3 < 0 and a(2) = −π/2 < 0. Same signs: speeding up. At t = 4: v(4) = 3 − 6 sin(2π/3) = 3 − 3√3 < 0 and a(4) = −π cos(2π/3) = π/2 > 0. Opposite signs: slowing down.
(d) The slope of x at t = 4 is v(4), so x(4.1) ≈ 2 + (3 − 3√3)(0.1) = 2.3 − 0.3√3 (about 1.78 m). x″ = a > 0 on (3, 6), so the graph of x bends upward and the tangent line lies below it: too low.
| Point | What earns it |
|---|---|
| 1 | a(t) = −π cos(πt/6) and a(2) = −π/2 m/s² with units |
| 1 | t = 1 and t = 5 from v = 0 |
| 1 | Sign change of v shown at both times |
| 1 | Both conclusions in (c), each with the signs of v and a stated |
| 1 | x(4.1) ≈ 2.3 − 0.3√3, using v(4) as the slope |
| 1 | Underestimate, justified by x″ = a > 0 |
Total: 6 points. Topics: 4.1, 4.2, 4.6.
Question 5 (constructed response · mixed)
Two rods, 5 m and 8 m long, are hinged at one end. A cloth covers the triangle formed by the rods and the line joining their free ends. A motor pushes the free ends apart so that their distance c increases at a steady 0.5 m per minute. If θ is the angle between the rods, c² = 89 − 80 cos θ and the cloth’s area is A = 20 sin θ m².
(a) Find c when θ = π/3. Find dθ/dt at that moment, with units. (b) Find dA/dt when θ = π/3. Interpret your answer in context. (c) Find the distance c at the instant the area stops increasing. Explain. (d) Use the tangent line to A, as a function of θ, at θ = π/3 to estimate A when θ = π/3 + 0.1. Is this an overestimate or an underestimate? Justify.
Worked solution and suggested Marlbridge rubric
(a) c² = 89 − 80(1/2) = 49, so c = 7 m. Differentiate with respect to t: 2c · dc/dt = 80 sin θ · dθ/dt. At θ = π/3: 2(7)(0.5) = 80(√3/2) · dθ/dt, so 7 = 40√3 · dθ/dt and dθ/dt = 7/(40√3) = 7√3/120 radians per minute (about 0.10).
(b) dA/dt = 20 cos θ · dθ/dt = 20(1/2)(7√3/120) = 7√3/12 m² per minute (about 1.01). When the angle is π/3, the cloth’s area is increasing at about 1.01 m² per minute.
(c) From (a), dθ/dt = c · (dc/dt)/(40 sin θ) > 0 for 0 < θ < π. So dA/dt = 20 cos θ · dθ/dt is 0 only when cos θ = 0, at θ = π/2, where it changes from positive to negative. Then c² = 89, so c = √89 m (about 9.4 m).
(d) dA/dθ = 20 cos θ, which is 10 at θ = π/3, and A(π/3) = 10√3. So A ≈ 10√3 + 10(0.1) = 10√3 + 1 m² (about 18.32). d²A/dθ² = −20 sin θ < 0 for 0 < θ < π, so the tangent line lies above the graph: an overestimate. (The actual area is about 18.23 m².)
| Point | What earns it |
|---|---|
| 1 | c = 7 m |
| 1 | 2c · dc/dt = 80 sin θ · dθ/dt, with both rates attached |
| 1 | dθ/dt = 7√3/120 radians per minute |
| 1 | dA/dt = 7√3/12 m² per minute, with an interpretation |
| 1 | θ = π/2 from cos θ = 0 (with dθ/dt > 0), and c = √89 m |
| 1 | Estimate 10√3 + 1 from the tangent line in θ |
| 1 | Overestimate, justified by d²A/dθ² < 0 |
Total: 7 points. Topics: 4.1, 4.4, 4.5, 4.6.
Question 6 (constructed response · mixed)
Let h(x) = ln(1 + 2x)/x for x > −1/2, x ≠ 0, and let h(0) = k.
(a) Find lim (x → 0) h(x). Show that L’Hospital’s Rule applies. (b) Find the value of k that makes h continuous at x = 0. (c) Find lim (x → ∞) h(x). Show that the rule applies. (d) Write the linearization of f(x) = ln(1 + 2x) at x = 0. Use it to estimate ln(1.1), and decide whether the estimate is too high or too low. (e) Use your linearization to explain the answer to (a) without the rule.
Worked solution and suggested Marlbridge rubric
(a) lim (x → 0) ln(1 + 2x) = ln 1 = 0 and lim (x → 0) x = 0. The form is 0/0, so the rule applies: lim (x → 0) [2/(1 + 2x)]/1 = 2.
(b) Continuity needs h(0) = lim (x → 0) h(x), so k = 2.
(c) As x → ∞, ln(1 + 2x) → ∞ and x → ∞, so the form is ∞/∞. The rule gives lim (x → ∞) 2/(1 + 2x) = 0.
(d) f(0) = 0 and f′(x) = 2/(1 + 2x), so f′(0) = 2 and L(x) = 2x. ln(1.1) = f(0.05) ≈ 2(0.05) = 0.1. f″(x) = −4/(1 + 2x)² < 0, so the line lies above the graph: 0.1 is too high. (The true value is about 0.0953.)
(e) Near x = 0, ln(1 + 2x) ≈ 2x, so h(x) ≈ 2x/x = 2: the top behaves like its tangent line.
| Point | What earns it |
|---|---|
| 1 | Both limits in (a) stated separately as 0 |
| 1 | Limit 2 from the derivatives of top and bottom, and k = 2 |
| 1 | ∞/∞ form shown, and limit 0 at infinity |
| 1 | L(x) = 2x and the estimate 0.1 |
| 1 | Too high, justified by f″ < 0 |
| 1 | (e): links ln(1 + 2x) ≈ 2x to the limit 2 |
Total: 6 points. Topics: 4.6, 4.7.
Question 7 (constructed response · mixed)
A water tank is a vertical cylinder with radius 2 m. The volume of water in it, V(t) cubic metres, is measured at selected times t hours. V is differentiable.
| t (hours) | 0 | 3 | 4 | 8 | 10 |
|---|---|---|---|---|---|
| V(t) (m³) | 30 | 35.4 | 36.2 | 37.4 | 36.8 |
(a) Use the data to estimate V′(3.5). Give units and interpret your answer. (b) The depth of water is h m, so V = 4πh. Use (a) to estimate the rate at which the depth is changing at t = 3.5. (c) Was the depth rising faster at t = 1.5 or at t = 3.5? Use the data to support your answer. (d) At t = 9, a sensor shows that the depth is decreasing at 0.02 m per hour. Find dV/dt at t = 9, in litres per minute. (1 m³ = 1000 litres.)
Worked solution and suggested Marlbridge rubric
(a) V′(3.5) ≈ (36.2 − 35.4)/(4 − 3) = 0.8 m³ per hour. At 3.5 hours, the volume is increasing at about 0.8 m³ per hour.
(b) dV/dt = 4π · dh/dt, so dh/dt ≈ 0.8/(4π) = 1/(5π) m per hour (about 0.064 m per hour).
(c) V′(1.5) ≈ (35.4 − 30)/3 = 1.8 m³ per hour, so dh/dt ≈ 9/(20π) m per hour (about 0.14). Since dh/dt = V′ ÷ 4π, the larger V′ gives the larger dh/dt: the depth was rising faster at t = 1.5.
(d) dV/dt = 4π · dh/dt = 4π(−0.02) = −0.08π m³ per hour. That is −80π litres per hour, or −80π/60 = −4π/3 litres per minute (about −4.2).
| Point | What earns it |
|---|---|
| 1 | 0.8 from the closest values either side of 3.5, with units |
| 1 | Interpretation with the time, “increasing” and units |
| 1 | dV/dt = 4π · dh/dt and 1/(5π) m per hour |
| 1 | (c): t = 1.5, with both rates compared through the factor 1/(4π) |
| 1 | −0.08π m³ per hour, with a negative sign |
| 1 | Conversion to −4π/3 litres per minute |
Total: 6 points. Topics: 4.1, 4.3, 4.4.
How did you do?
Add up your points from Questions 4–7 (25 in total) and your correct answers to Questions 1–3. The total is only a guide, not a predicted exam score. Note which topics your lost points came from (each answer lists them), then use those topic checklists:
4.1 · 4.2 · 4.3 · 4.4 · 4.5 · 4.6 · 4.7
If many topics need work, go back to the Unit 4 diagnostic and use its “Your next step” table to choose where to start.