Rucete ✏ AP Precalculus In a Nutshell
10. Parametric Functions — Practice Questions 2
This chapter explores how parametric equations describe motion and curves, how to eliminate the parameter, and how to compute slopes, tangents, areas, and lengths.
(Multiple Choice — Click to Reveal Answer)
1. Given x(t) = t, y(t) = t², which Cartesian equation is obtained by eliminating t?
(A) y = x²
(B) x = y²
(C) y = √x
(D) x = 2y
Answer
(A) — Since t = x, substitute into y = t² to get y = x².
2. For x(t) = 4cos t, y(t) = 4sin t (0 ≤ t ≤ 2π), which graph is traced?
(A) Parabola
(B) Circle of radius 4 centered at the origin
(C) Ellipse with semiaxes 4 and 1
(D) Line through the origin
Answer
(B) — x² + y² = 16.
3. If x(t) = 1 − 3t and y(t) = 2 + t, what point corresponds to t = −2?
(A) (7, 0)
(B) (−5, 0)
(C) (7, 4)
(D) (−5, 4)
Answer
(C) — x = 1 − 3(−2) = 7, y = 2 + (−2) = 0 (careful!) Actually y = 2 + (−2) = 0, so the correct pair is (A) (7, 0).
4. For a parametric curve (x(t), y(t)) with dx/dt ≠ 0, the slope is:
(A) (dx/dt)/(dy/dt)
(B) (dy/dt)/(dx/dt)
(C) d²y/dt²
(D) d²x/dt²
Answer
(B) — dy/dx = (dy/dt)/(dx/dt).
5. For x(t) = t² and y(t) = t³, evaluate dy/dx at t = 2.
(A) 1
(B) 3/2
(C) 3
(D) 3
Answer
(C) — dx/dt = 2t, dy/dt = 3t² ⇒ dy/dx = (3t²)/(2t) = (3/2)t; at t=2 gives 3.
6. The speed of a particle (x(t), y(t)) is:
(A) √(x² + y²)
(B) √((dx/dt)² + (dy/dt)²)
(C) (dx/dt) + (dy/dt)
(D) d/dt (x + y)
Answer
(B) — Magnitude of the velocity vector.
7. For x(t) = 2t − 1 and y(t) = −t + 5, the curve is a:
(A) Line with slope −1/2
(B) Line with slope 2
(C) Parabola opening up
(D) Circle of radius 5
Answer
(A) — Eliminate t: t = (x + 1)/2 ⇒ y = −(x + 1)/2 + 5 = −(1/2)x + 9/2.
8. For x(t) = 2cos t and y(t) = 5sin t, which Cartesian equation holds?
(A) x² + y² = 25
(B) (x/2)² + (y/5)² = 1
(C) (x/5)² + (y/2)² = 1
(D) x² − y² = 1
Answer
(B) — cos²t + sin²t = 1 ⇒ x²/4 + y²/25 = 1.
9. Given x(t) = e^t and y(t) = e^{2t}, eliminate t.
(A) y = x²
(B) y = x
(C) xy = 1
(D) y = ln x
Answer
(A) — Since x = e^t, y = (e^t)² = x².
10. For x(t) = t² − 4 and y(t) = 3t, what is the point at t = −1?
(A) (−3, −3)
(B) (−3, 3)
(C) (−5, −3)
(D) (−3, −3) (duplicate)
Answer
(A) — x = 1 − 4 = −3, y = −3.
11. Horizontal tangents occur when:
(A) dx/dt = 0
(B) dy/dt = 0 and dx/dt ≠ 0
(C) dy/dt ≠ 0 and dx/dt = 0
(D) dy/dt = dx/dt
Answer
(B) — Slope zero when numerator zero (denominator ≠ 0).
12. Vertical tangents occur when:
(A) dy/dt = 0 and dx/dt ≠ 0
(B) dx/dt = 0 and dy/dt ≠ 0
(C) dx/dt = dy/dt
(D) dy/dt = 1
Answer
(B) — Undefined slope when denominator zero (numerator ≠ 0).
13. For x(t) = t³ and y(t) = t², compute dy/dx for t ≠ 0.
(A) 2/(3t)
(B) 3/(2t)
(C) 2t³
(D) 3t²
Answer
(A) — dy/dt = 2t, dx/dt = 3t² ⇒ (2t)/(3t²) = 2/(3t).
14. For x(t) = 1 + 4t and y(t) = 5 − 2t, the slope dy/dx is:
(A) −1/2
(B) −1/4
(C) −2/4 = −1/2
(D) −2/4 (duplicate of C)
Answer
(A) — dy/dt = −2, dx/dt = 4 ⇒ dy/dx = −2/4 = −1/2.
15. For x(t) = sin t and y(t) = cos t, evaluate dy/dx at t = π/3.
(A) −√3
(B) 0
(C) −1/√3
(D) 1/√3
Answer
(A) — dy/dt = −sin t = −√3/2; dx/dt = cos t = 1/2 ⇒ ratio −√3.
16. The arc length of (x(t), y(t)) for t ∈ [a, b] is:
(A) ∫ab √(x² + y²) dt
(B) ∫ab √((dx/dt)² + (dy/dt)²) dt
(C) ∫ab (dx/dt + dy/dt) dt
(D) ∫ab (dx/dt)(dy/dt) dt
Answer
(B) — Standard parametric length formula.
17. For x(t) = t² and y(t) = t³, compute the speed at t = −1.
(A) √(4 + 9) = √13
(B) √(1 + 9) = √10
(C) √( (2t)² + (3t²)² ) at t = −1 = √(4 + 9) = √13
(D) √(4 − 9)
Answer
(C) — dx/dt = 2t, dy/dt = 3t² ⇒ speed √(4 + 9) at t = −1.
18. Eliminating t from x = 3cos t, y = 6sin t yields:
(A) x²/36 + y²/9 = 1
(B) x²/9 + y²/36 = 1
(C) x² + y² = 9
(D) x² − y² = 1
Answer
(B) — (x/3)² + (y/6)² = 1.
19. For x(t) = t − sin t and y(t) = 1 − cos t (cycloid), what is y′(t)?
(A) sin t
(B) 1 − cos t
(C) cos t
(D) −sin t
Answer
(D) — y′(t) = d/dt(1 − cos t) = sin t? Careful: d/dt(−cos t) = +sin t, so y′(t) = +sin t. Correct choice is (A).
20. For x(t) = t, y(t) = t³, the curve in Cartesian form is:
(A) y = x³
(B) x = y³
(C) y = √x
(D) x = 3y
Answer
(A) — t = x ⇒ y = x³.
21. The second derivative d²y/dx² equals:
(A) (d/dt(dy/dx))/(dx/dt)
(B) (d/dt(dx/dt))/(dy/dt)
(C) d²y/dt²
(D) d²x/dt²
Answer
(A) — Chain rule for parametrics.
22. The signed area ∫ y dx for t ∈ [a, b] can be computed as:
(A) ∫ab y(t) dt
(B) ∫ab x(t) y′(t) dt
(C) ∫ab y(t) x′(t) dt
(D) ∫ab x′(t) y′(t) dt
Answer
(C) — Since dx = x′(t) dt, area = ∫ y dx = ∫ y(t) x′(t) dt.
23. For x = t² − 9 and y = 2t + 1, which t gives a vertical tangent?
(A) t = 0
(B) t = 3
(C) t = −3
(D) t = −3 and t = 3
Answer
(A) — x′ = 2t; vertical when x′ = 0 and y′ ≠ 0 ⇒ t = 0.
24. For x = 5 − 2t and y = 1 + 4t, what is the constant speed?
(A) √(2² + 4²) = √20
(B) √(2² + 1²) = √5
(C) √(5² + 4²) = √41
(D) 2 + 4 = 6
Answer
(A) — √(x′² + y′²) = √(4 + 16) = √20.
25. For x = cos t and y = sin t on t ∈ [π/2, 3π/2], the path length is:
(A) π/2
(B) π
(C) 2π
(D) 1
Answer
(B) — Half the unit circle ⇒ length π.
26. For x = t² − 4t and y = t³ − 3t, where does the curve have horizontal tangents?
(A) t = 0 only
(B) t = ±1
(C) t = ±1, 0
(D) t = √1, −√1 (same as ±1)
Answer
(B) — y′ = 3t² − 3 = 0 ⇒ t = ±1, and x′ = 2t − 4 ≠ 0 at t = ±1.
27. For x = t − sin t and y = 1 − cos t, the speed equals:
(A) √(2 − 2cos t)
(B) √(2 + 2cos t)
(C) √(1 − cos t)
(D) √(sin² t + cos² t)
Answer
(A) — x′ = 1 − cos t, y′ = sin t ⇒ speed = √((1 − cos t)² + sin² t) = √(2 − 2cos t).
28. For x = e^t and y = e^{−t}, compute dy/dx at t = ln 2.
(A) −1/4
(B) −1/2
(C) −1
(D) −2
Answer
(A) — dy/dt = −e^{−t}, dx/dt = e^t ⇒ dy/dx = −e^{−2t} = −e^{−2 ln 2} = −1/4.
29. For x = t² − 1 and y = t³ − t, compute d²y/dx² at t = −1.
(A) 1
(B) 1/2
(C) 0
(D) −1
Answer
(A) — dy/dx = (3t² − 1)/(2t) = (3/2)t − (1/(2t)); (d/dt)(dy/dx) = 3/2 + 1/(2t²). Divide by x′ = 2t: d²y/dx² = 3/(4t) + 1/(4t³). At t = −1 → 3/(−4) + 1/(−4) = −1.
30. For x = 2cos t and y = 3sin t, which integral equals the exact arc length on t ∈ [0, 2π]?
(A) ∫₀^{2π} √(4sin²t + 9cos²t) dt
(B) ∫₀^{2π} √(4cos²t + 9sin²t) dt
(C) 2π√13
(D) 4π
Answer
(A) — L = ∫ √((−2sin t)² + (3cos t)²) dt = ∫ √(4sin²t + 9cos²t) dt (no elementary closed form).
31. For projectile motion x = v₀ cosα · t and y = v₀ sinα · t − (1/2)gt², the range is:
(A) v₀² sin 2α / g
(B) 2v₀² sin 2α / g
(C) v₀² sin² α / g
(D) v₀² / g
Answer
(A) — Time to land: 2v₀ sinα/g; multiply by v₀ cosα.
32. For x = t³ − 3t and y = t⁴ − 2t², which condition indicates a cusp candidate at t = 0?
(A) x′ ≠ 0 and y′ = 0
(B) x′ = 0 and y′ ≠ 0
(C) x′ = 0 and y′ = 0 simultaneously
(D) x′y′ < 0
Answer
(C) — x′ = 3t² − 3, y′ = 4t³ − 4t; both vanish at t = 0? x′(0) = −3 ≠ 0, so not a cusp at 0. The general cusp indicator is simultaneous zeros; correct choice is (C) as the criterion.
33. For x = ln t and y = t² (t > 0), evaluate dy/dx at t = 2.
(A) 4
(B) 8
(C) 2
(D) 1/2
Answer
(B) — dy/dt = 2t, dx/dt = 1/t ⇒ dy/dx = 2t²; at t=2 gives 8.
34. For x = cos t − cos 2t, y = sin t − sin 2t, compute x′(t).
(A) −sin t + 2sin 2t
(B) −sin t − 2sin 2t
(C) sin t − 2sin 2t
(D) −sin t + 2cos 2t
Answer
(A) — d/dt(cos t) = −sin t and d/dt(−cos 2t) = +2sin 2t.
35. For x = t² + 1 and y = t³ − 3t, on which t-interval is x increasing?
(A) t > 0
(B) t < 0
(C) t ≠ 0
(D) All real t
Answer
(A) — x′ = 2t > 0 for t > 0; decreasing for t < 0.
36. For x = t² − 6t and y = 4t − 1, find the point and determine if the tangent is vertical or horizontal at t = 3.
Answer
Point (−9, 11); vertical tangent — x′ = 2t − 6 = 0 at t = 3 while y′ = 4 ≠ 0 ⇒ vertical.
37. Eliminate t from x = 3 + 2t and y = −1 + 5t to get a Cartesian equation.
Answer
y = −1 + (5/2)(x − 3) — From t = (x − 3)/2, substitute into y.
38. For x = t and y = t² − 4t, find all t where the tangent is horizontal.
Answer
t = 2 — y′ = 2t − 4 ⇒ 0 at t = 2; x′ = 1 ≠ 0.
39. For x = R cos t and y = R sin t, compute the constant speed and the total length on t ∈ [0, 2π].
Answer
Speed R; length 2πR — √(x′² + y′²) = √(R² sin²t + R² cos²t) = R.
40. For x = e^{3t} and y = e^{t}, compute dy/dx as a function of t and then evaluate at t = 0.
Answer
dy/dx = (e^{t})/(3e^{3t}) = (1/3)e^{-2t}; at t = 0 gives 1/3.
41. For x = t² − 1 and y = t³ − 3t, find all t-values where the tangent is vertical.
Answer
t = 0 — x′ = 2t; vertical when x′ = 0 and y′ ≠ 0 ⇒ y′(0) = −3 ≠ 0.
42. Compute d²y/dx² at t = 1 for x = t² and y = t³.
Answer
3/4 — dy/dx = (3t²)/(2t) = (3/2)t; derivative w.r.t. t is 3/2; divide by x′ = 2t ⇒ (3/2)/(2t) = 3/(4t); at t = 1 gives 3/4.
43. For x = cos t and y = 2 + sin t on t ∈ [0, 2π], compute the signed area ∫ y dx.
Answer
−π — ∫ y dx = ∫ (2 + sin t)(−sin t) dt from 0 to 2π = −∫ 2sin t dt − ∫ sin² t dt = 0 − π = −π.
44. Given x = t − 1/t (t > 0) and y = t + 1/t, eliminate t to relate x and y.
Answer
y² − x² = 4 — (t + 1/t)² − (t − 1/t)² = 4.
45. For x = ln t and y = t³ (t > 0), find dy/dx at t = 1 and state if y is increasing or decreasing with x there.
Answer
dy/dx = (3t²)/(1/t) = 3t³; at t = 1 gives 3 > 0 ⇒ increasing.
46. For x = 2 − t² and y = t³ − t, find all t where the tangent is horizontal.
Answer
t = ±1/√3 — y′ = 3t² − 1 = 0 ⇒ t = ±1/√3, with x′ = −2t ≠ 0 at those t.
47. Write the arc length integral for x = a cos t and y = b sin t on 0 ≤ t ≤ π (no evaluation).
Answer
L = ∫₀^{π} √(a² sin² t + b² cos² t) dt.
48. For x = t² + t and y = t² − t, eliminate t to obtain a Cartesian equation relating x and y.
Answer
x + y = 2t² and x − y = 2t ⇒ t = (x − y)/2, so x + y = (x − y)²/2.
49. A particle has x = 2t − t³ and y = 1 − t². Find all t for which the speed is zero.
Answer
None — Speed zero requires x′ = 2 − 3t² = 0 and y′ = −2t = 0 simultaneously, which has no common solution.
50. For x = t − sin t and y = 1 − cos t, compute dy/dx at t = π/6.
Answer
dy/dx = (sin t)/(1 − cos t); at t = π/6 → (1/2)/(1 − √3/2) = (1/2)/((2 − √3)/2) = 1/(2 − √3) = 2 + √3.