Parametric equations, polar coordinates and vector-valued functions (BC) practice problems AP Calculus, Unit 9
BC only. Derivatives and arc length of parametric curves, motion with vectors, and slopes and areas in polar coordinates. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for parametric equations, polar coordinates and vector-valued functions (bc)
| dy/dx = (dy/dt)/(dx/dt) | parametric slope |
|---|---|
| Speed = √[(dx/dt)2 + (dy/dt)2] | speed of a moving particle |
| L = ∫ √[(dx/dt)2 + (dy/dt)2] dt | arc length |
| A = ½∫ r2 dθ | polar area |
Questions marked No calculator should be done by hand, as in Part A of the multiple-choice section. You can type fractions, such as 3/4. The full sheet is in our AP Calculus formula sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
Divide dy/dt by dx/dt.
Worked solution
- dy/dt = 3t2 and dx/dt = 2t, so dy/dx = 3t/2.
- At t = 2: 3.
Hint
Speed is the length of the velocity vector.
Worked solution
- √(32 + 42) = 5.
Hint
A = ½∫02π r2 dθ.
Worked solution
- A = ½ × 9 × 2π = 9π, the area of a circle of radius 3.
- ≈ 28.27.
Exam-style questions
Questions at the level of the AP Calculus exam. The ones marked Challenge combine more than one idea.
Hint
dy/dx = (3cos t)/(−3sin t).
Worked solution
- dy/dx = −cos t/sin t = −cot t.
- At π/4: −1.
Hint
dx/dt = 1 and dy/dt = t1/2.
Worked solution
- L = ∫03 √(1 + t) dt.
- = (2/3)[(1 + t)3/2] from 0 to 3 = (2/3)(8 − 1) = 14/3.
- ≈ 4.667.
Hint
d2y/dx2 = [d/dt(dy/dx)] ÷ (dx/dt).
Worked solution
- dy/dx = 3t/2, so d/dt(dy/dx) = 3/2.
- Divide by dx/dt = 2t = 2.
- 3/4 = 0.75.
Hint
A = ½∫02π (1 + cos θ)2 dθ.
Worked solution
- Expand: 1 + 2cos θ + cos2θ. Over a full turn, ∫cos θ = 0 and ∫cos2θ = π.
- A = ½(2π + 0 + π) = 3π/2.
- ≈ 4.712.
Hint
dy/dx = (r' sinθ + r cosθ)/(r' cosθ − r sinθ).
Worked solution
- r = θ and r' = 1. At π/2: sin = 1 and cos = 0.
- dy/dx = (1 + 0)/(0 − π/2) = −2/π.
- ≈ −0.637.
Free-response practice
One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.
Calculator-active free-response style · 5 points · about 15 minutes
A particle moves in the plane with x(t) = t2 − 3t and y(t) = 2t3/3, for t ≥ 0.
- (a) Find the velocity vector at t = 2.
- (b) Find the speed at t = 2.
- (c) At what time is the particle moving vertically?
- (d) Find the total distance travelled from t = 0 to t = 2.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
x'(t) = 2t − 3 and y'(t) = 2t2, so v(2) = 〈1, 8〉.
Part (b)
√(1 + 64) = √65 ≈ 8.062.
Part (c)
x'(t) = 0 at t = 1.5, while y'(1.5) = 4.5 ≠ 0.
Part (d)
∫02 √[(2t − 3)2 + 4t4] dt ≈ 6.951, by calculator.
Your mark: 0 of 5
Common mistakes in parametric equations, polar coordinates and vector-valued functions (bc)
- Writing dy/dx = (dx/dt)/(dy/dt), upside down.
- Forgetting to divide by dx/dt for the second derivative.
- Using the wrong limits for one loop of a polar curve.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Calculus past papers, with official scoring guidelines. · All AP Calculus units