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. BC-only material. Assumptions: angles in radians; no calculator for Questions 1–5 and 7; in Question 6 a calculator is allowed and decimal answers should be given to 3 decimal places. Notation: x′(t) = dx/dt and y′(t) = dy/dt. All contexts and data are fictional.
Question 1 (multiple choice · foundation)
A curve is given by x = t³ + 2t and y = t² − 5t. What is the slope of the tangent line to the curve at t = 1?
- (A) −3/5
- (B) −5/3
- (C) −3
- (D) −4/3
Answer and explanation
Answer: (A). dx/dt = 3t² + 2 = 5 and dy/dt = 2t − 5 = −3 at t = 1. So dy/dx = (dy/dt) ÷ (dx/dt) = −3/5.
- (B) divides the other way, (dx/dt) ÷ (dy/dt) = 5/(−3). That is dx/dy, not dy/dx.
- (C) gives dy/dt alone. That is the rate of change of y with respect to t, not the slope of the curve.
- (D) divides the coordinates, y ÷ x = (−4)/3 at t = 1. A ratio of coordinates is not a slope.
Question 2 (multiple choice · core)
An ellipse is given by x = 4 cos t, y = 3 sin t. Which is an equation of the tangent line to the ellipse at t = π/4?
- (A) y − 3√2/2 = −¾(x − 2√2)
- (B) y − 3√2/2 = ¾(x − 2√2)
- (C) y − 3√2/2 = −4/3(x − 2√2)
- (D) y − 2√2 = −¾(x − 3√2/2)
Answer and explanation
Answer: (A). At t = π/4 the point is (4 · √2/2, 3 · √2/2) = (2√2, 3√2/2). dx/dt = −4 sin t and dy/dt = 3 cos t, so dy/dx = (3 cos t)/(−4 sin t). At t = π/4, sin t = cos t, so the slope is −¾. A negative slope makes sense: at t = π/4 the point is on the upper right part of the ellipse, where the curve falls to the right.
- (B) loses the minus sign from d/dt [cos t] = −sin t.
- (C) uses (dx/dt) ÷ (dy/dt), the reciprocal of the slope.
- (D) swaps the x- and y-coordinates of the point.
Question 3 (multiple choice · stretch)
A curve is given by x = t² − 4t and y = t³ − 12t. For which value(s) of t does the curve have a horizontal tangent line?
- (A) t = −2 only
- (B) t = −2 and t = 2
- (C) t = 2 only
- (D) No values of t
Answer and explanation
Answer: (A). dy/dt = 3t² − 12 = 0 gives t = ±2. Now check dx/dt = 2t − 4 at each.
- t = −2: dx/dt = −8 ≠ 0, so the tangent is horizontal, at the point (12, 16).
- t = 2: dx/dt = 0 as well. Both derivatives are zero, so dy/dx = 0/0 gives no conclusion. In fact, for t ≠ 2, dy/dx = 3(t − 2)(t + 2) / (2(t − 2)) = 3(t + 2)/2, which approaches 6 as t → 2. The tangent direction there is not horizontal.
Why the others are wrong:
- (B) stops at dy/dt = 0 and does not check dx/dt.
- (C) chooses the value where dx/dt = 0, which is the condition for a vertical tangent (and here both derivatives are zero).
- (D) is wrong because t = −2 satisfies both conditions.
Question 4 (multiple choice · core)
A curve is given by x = t² + t and y = √(t + 3), for t ≥ −3. What is the slope of the curve at the point (2, 2)?
- (A) 1/12
- (B) −1/6
- (C) √5/50
- (D) 12
Answer and explanation
Answer: (A). Find t first. x = 2 gives t² + t − 2 = 0, so t = 1 or t = −2. Then y(1) = √4 = 2 ✓ but y(−2) = √1 = 1 ✗. So the point (2, 2) belongs only to t = 1. At t = 1: dx/dt = 2t + 1 = 3 and dy/dt = 1/(2√(t + 3)) = ¼. So dy/dx = (¼) ÷ 3 = 1/12.
- (B) uses t = −2, which gives x = 2 but y = 1, so it is a different point, (2, 1).
- (C) substitutes the x-coordinate as if it were t, using t = 2: (1/(2√5)) ÷ 5 = 1/(10√5) = √5/50.
- (D) divides the other way, (dx/dt) ÷ (dy/dt).
Question 5 (calculation · core)
A curve is given by x = eᵗ + 1 and y = t eᵗ.
(a) Show that dy/dx = 1 + t. (b) Find the equation of the tangent line at t = 0. (c) Find the exact coordinates of the point where the tangent line is horizontal. (d) Explain why the curve has no vertical tangent lines.
Worked solution and suggested Marlbridge rubric
(a) dx/dt = eᵗ. By the product rule, dy/dt = eᵗ + t eᵗ = eᵗ(1 + t). So dy/dx = eᵗ(1 + t) ÷ eᵗ = 1 + t.
(b) At t = 0: x = 2, y = 0, slope = 1. Tangent line: y = x − 2.
(c) dy/dt = eᵗ(1 + t) = 0 gives t = −1 (eᵗ is never 0). There dx/dt = e⁻¹ ≠ 0 ✓. The point is (1 + 1/e, −1/e), about (1.368, −0.368).
(d) A vertical tangent needs dx/dt = 0, but dx/dt = eᵗ > 0 for every t.
| Point | What earns it |
|---|---|
| 1 | dy/dt = eᵗ(1 + t) by the product rule, divided by dx/dt = eᵗ to give 1 + t |
| 1 | Tangent line y = x − 2, with the point (2, 0) and slope 1 from the same t |
| 1 | t = −1 from dy/dt = 0, with a check that dx/dt ≠ 0 |
| 1 | Exact point (1 + 1/e, −1/e) |
| 1 | dx/dt = eᵗ is never 0, so no vertical tangent |
Total: 5 points. Optional check: eliminating t gives y = (x − 1) ln(x − 1), and dy/dx = ln(x − 1) + 1 = t + 1. ✓ Common error: writing dy/dt = eᵗ (forgetting the product rule), which gives dy/dx = 1 and a “straight line”.
Question 6 (constructed response · calculator · core)
A remote-controlled boat moves on a lake. Its position at time t seconds is x(t) = 10 ln(t + 1) metres east of a jetty and y(t) = 12 + 5 sin(0.8t) metres north of the jetty, for 0 ≤ t ≤ 6.
(a) Find dy/dx at t = 2. (b) Interpret your answer to (a) in context. (c) Explain why the boat’s path has no vertical tangent line for 0 ≤ t ≤ 6. (d) Find all times in 0 < t < 6 at which the path has a horizontal tangent line, and the boat’s position at each.
Worked solution and suggested Marlbridge rubric
(a) dx/dt = 10/(t + 1) and dy/dt = 4 cos(0.8t). At t = 2: dx/dt = 10/3 ≈ 3.333 and dy/dt = 4 cos(1.6) ≈ −0.117. So dy/dx = (4 cos 1.6) ÷ (10/3) = 1.2 cos 1.6 ≈ −0.035.
(b) At t = 2 the path slopes slightly downward as x increases: the boat’s northward position decreases by about 0.035 m for each metre it moves east. (Since dx/dt > 0, the boat is moving east, so it is heading east and very slightly south.)
(c) A vertical tangent needs dx/dt = 0. But dx/dt = 10/(t + 1) > 0 for all t in [0, 6], so this never happens.
(d) dy/dt = 4 cos(0.8t) = 0 when 0.8t = π/2 or 3π/2 (the next solution, 5π/2, gives t ≈ 9.817, outside the interval). So t = 5π/8 ≈ 1.963 and t = 15π/8 ≈ 5.890. At both, dx/dt > 0 by part (c), so both are horizontal tangents.
- t ≈ 1.963: position ≈ (10.864, 17.000), the furthest north the boat goes.
- t ≈ 5.890: position ≈ (19.301, 7.000), the furthest south.
| Point | What earns it |
|---|---|
| 1 | dx/dt and dy/dt correct, including the chain-rule factor 0.8 |
| 1 | dy/dx ≈ −0.035 at t = 2 |
| 1 | Interpretation: change in north position per metre east, with units, referring to t = 2 |
| 1 | No vertical tangent because dx/dt = 10/(t + 1) is never 0 on the interval |
| 1 | Both times t ≈ 1.963 and t ≈ 5.890 from dy/dt = 0, with dx/dt ≠ 0 |
| 1 | Both positions correct |
Total: 6 points. A calculator’s numerical derivative for (a) is fine. An interpretation such as “the boat moves 0.035 m per second” does not earn the (b) point: dy/dx is not a rate with respect to time.
Question 7 (constructed response · stretch)
A curve is given by x = t + 1/t and y = t − 1/t, for t > 0.
(a) Find dy/dx in terms of t, simplified so that it has no fractions inside fractions. (b) Find the point where the curve has a vertical tangent. Justify your answer. (c) Find the value of t at which the slope of the curve is 2, and write the equation of the tangent line there. (d) Show that every point of the curve satisfies x² − y² = 4. Use implicit differentiation of this equation to check your slope in (c).
Worked solution and suggested Marlbridge rubric
(a) dx/dt = 1 − 1/t² and dy/dt = 1 + 1/t². Multiply top and bottom by t²: dy/dx = (t² + 1)/(t² − 1), for t ≠ 1.
(b) dx/dt = 1 − 1/t² = 0 gives t = 1 (t > 0). There dy/dt = 2 ≠ 0, so the tangent is vertical at (2, 0).
(c) (t² + 1)/(t² − 1) = 2 gives t² + 1 = 2t² − 2, so t² = 3 and t = √3. Point: x = √3 + 1/√3 = 4√3/3, y = √3 − 1/√3 = 2√3/3. Tangent line: y − 2√3/3 = 2(x − 4√3/3), which simplifies to y = 2x − 2√3 (about y = 2x − 3.464).
(d) x² − y² = (t² + 2 + 1/t²) − (t² − 2 + 1/t²) = 4 for every t. Differentiating x² − y² = 4 implicitly: 2x − 2y (dy/dx) = 0, so dy/dx = x/y. At (4√3/3, 2√3/3): dy/dx = (4√3/3) ÷ (2√3/3) = 2. ✓
| Point | What earns it |
|---|---|
| 1 | dy/dx = (t² + 1)/(t² − 1) |
| 1 | t = 1 from dx/dt = 0, with the check dy/dt ≠ 0, and the point (2, 0) |
| 1 | t = √3 from solving dy/dx = 2 |
| 1 | Correct tangent line through (4√3/3, 2√3/3) |
| 1 | x² − y² = 4 shown by expanding |
| 1 | Implicit derivative x/y evaluated to confirm the slope 2 |
Total: 6 points. Acceptable alternative for (c): the point written as (4/√3, 2/√3). Common error in (b): saying “vertical” without checking dy/dt, or giving the answer as t = 1 without the point.
How did you do?
- Q1 or Q2 wrong: check which derivative goes on top, and your signs for trig derivatives; see “Deriving dy/dx with the chain rule” in the study guide.
- Q3 or Q5(c) wrong: reread “Horizontal and vertical tangents”, especially the both-zero case.
- Q4 or Q7(c) wrong: practise finding the value of t first, from a given point or a given slope (Worked example 1, part (c)).
- Q6 wrong: compare with Worked example 2, where dy/dx is interpreted separately from the direction of motion.
Then tick off the topic checklist.