Rucete AP Precalculus In a Nutshell
10. Parametric Functions — Practice Questions 3
This chapter explores parametric equations for motion and conics, including elimination of the parameter, slopes, tangency conditions, areas, and arc lengths.
(Multiple Choice — Click to Reveal Answer)
1. If x(t) = t and y(t) = t² − 3t, the corresponding Cartesian equation is:
(A) y = x² − 3x
(B) x = y² − 3y
(C) y = (x − 3)²
(D) x = (y − 3)²
Answer
(A) — Eliminate t by setting t = x and substitute into y.
2. For x = 5cos t, y = 2sin t (0 ≤ t ≤ 2π), the graph is:
(A) A circle of radius 5
(B) A circle of radius 2
(C) An ellipse centered at the origin
(D) A parabola opening right
Answer
(C) — x²/25 + y²/4 = 1.
3. Given x = 1 + 3t and y = −2 + t, the point at t = −1 is:
(A) (−2, −1)
(B) (−2, −3)
(C) (−2, −3)? Check: y(−1) = −3; x(−1) = −2, so (−2, −3)
(D) (−2, −1)
Answer
(B) — x(−1) = −2, y(−1) = −3.
4. The slope dy/dx of (x(t), y(t)) with dx/dt ≠ 0 is:
(A) (dx/dt)/(dy/dt)
(B) (dy/dt)/(dx/dt)
(C) d²y/dt²
(D) d²x/dt²
Answer
(B) — Chain rule.
5. For x = t², y = t³, evaluate dy/dx at t = 2.
(A) 1
(B) 3/2
(C) 3
(D) 6
Answer
(C) — 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 (xy)
Answer
(B) — Magnitude of velocity.
7. For x = 2t − 1, y = −t + 4, the line’s slope is:
(A) −1/2
(B) −2
(C) 2
(D) 1/2
Answer
(A) — dy/dx = (−1)/2.
8. For x = 3cos t, y = 4sin t, which equation holds?
(A) x²/9 + y²/16 = 1
(B) x² + y² = 1
(C) x²/16 + y²/9 = 1
(D) x² − y² = 1
Answer
(A) — cos²t + sin²t = 1.
9. If x = e^t and y = e^{3t}, eliminate t.
(A) y = x³
(B) y = x
(C) xy = 1
(D) y = ln x
Answer
(A) — y = (e^t)³ = x³.
10. For x = t² − 9 and y = 2t, the point at t = 3 is:
(A) (0, 6)
(B) (0, −6)
(C) (−9, 6)
(D) (9, 6)
Answer
(A) — x=9−9=0, y=6.
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) — Numerator zero, denominator nonzero.
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) — Denominator zero, numerator nonzero.
13. For x = t³, y = t² (t ≠ 0), dy/dx equals:
(A) 2/(3t)
(B) 3/(2t)
(C) 2t³
(D) 3t²
Answer
(A) — (2t)/(3t²) = 2/(3t).
14. For x = 1 + 2t and y = 5 − t, dy/dx is:
(A) −1/2
(B) −2
(C) 2
(D) 1/2
Answer
(A) — (−1)/2.
15. For x = sin t, y = cos t, dy/dx at t = π/6 is:
(A) −√3
(B) −1/√3
(C) 0
(D) √3
Answer
(B) — dy/dx = (−sin t)/(cos t) = −tan t = −1/√3.
16. Arc length of (x(t), y(t)) on [a, b] is:
(A) ∫ab √(x² + y²) dt
(B) ∫ab √((x′)² + (y′)²) dt
(C) ∫ab x′y′ dt
(D) ∫ab (x′ + y′) dt
Answer
(B)
17. For x = t², y = t³, speed at t = −2 is:
(A) √( (−4)² + (−12)² ) = √160
(B) √(4 + 12) = √16
(C) √(4 + 36) = √40
(D) √(16 + 144) = √160
Answer
(D) — x′=2t, y′=3t² → √(4t² + 9t⁴); at t=−2 gives √(16 + 144).
18. Eliminating t from x=2cos t, y=7sin t gives:
(A) x²/4 + y²/49 = 1
(B) x² + y² = 1
(C) x²/49 + y²/4 = 1
(D) x² − y² = 1
Answer
(A)
19. For a cycloid x = t − sin t, y = 1 − cos t, the speed equals:
(A) √(2 − 2cos t)
(B) √(2 + 2cos t)
(C) √(1 − cos t)
(D) √(sin²t + cos²t)
Answer
(A) — √((1−cos t)² + sin²t) = √(2 − 2cos t).
20. For x = t and y = t³ − 4t, the Cartesian form is:
(A) y = x³ − 4x
(B) x = y³ − 4y
(C) y = x − 4x³
(D) x = y − 4y³
Answer
(A)
21. The second derivative d²y/dx² for parametrics equals:
(A) (d/dt(dy/dx))/(dx/dt)
(B) (d/dt(dx/dt))/(dy/dt)
(C) d²y/dt²
(D) d²x/dt²
Answer
(A)
22. The signed area A = ∫ y dx over t ∈ [a, b] is:
(A) ∫ab y(t) dt
(B) ∫ab y(t) x′(t) dt
(C) ∫ab x(t) y′(t) dt
(D) ∫ab x′(t) y′(t) dt
Answer
(B) — dx = x′ dt.
23. For x = t² − 4 and y = 3t + 2, a vertical tangent occurs at:
(A) t = −2 only
(B) t = 0 only
(C) t = 2 only
(D) t = 0 (since x′=2t=0) and y′ ≠ 0
Answer
(B) — x′=2t; vertical when x′=0 and y′≠0 ⇒ t=0.
24. For x = 3 − 4t and y = 1 + 2t, speed is:
(A) √(16 + 4) = √20
(B) √(9 + 1) = √10
(C) 6
(D) 4 − 2 = 2
Answer
(A)
25. For x = cos t, y = sin t on t ∈ [π, 2π], the path length is:
(A) π/2
(B) π
(C) 2π
(D) 1
Answer
(B) — Half a unit circle.
26. For x = t² − 2t and y = t³ − 3t, horizontal tangents occur at:
(A) t = 0 only
(B) t = ±1
(C) t = ±√1, 0
(D) t = ±1 with x′ ≠ 0
Answer
(B) — y′=3t² − 3 = 0 ⇒ t=±1; x′=2t − 2 ≠ 0 at ±1.
27. For x = t − sin t, y = 1 − cos t, x′(t) equals:
(A) 1 − cos t
(B) 1 + cos t
(C) sin t
(D) −sin t
Answer
(A)
28. For x = e^t and y = e^{−2t}, compute dy/dx at t = 0.
(A) −2
(B) −1
(C) −1/2
(D) −e^{−3t} at 0 = −1
Answer
(A) — dy/dx = (−2e^{−2t})/(e^t) = −2e^{−3t}; at 0 gives −2.
29. For x = t² − 1, y = t³ − t, compute d²y/dx² at t = 1.
(A) 1
(B) 1/2
(C) 0
(D) −1/2
Answer
(A) — dy/dx = (3t² − 1)/(2t); (d/dt)(dy/dx) = 3/2 + 1/(2t²); divide by x′=2t ⇒ 3/(4t)+1/(4t³). At t=1 → 1.
30. For x = 4cos t, y = 3sin t on [0, 2π], the exact arc length is:
(A) ∫₀^{2π} √(16sin²t + 9cos²t) dt
(B) 2π√25
(C) 14
(D) ∫₀^{2π} √(16cos²t + 9sin²t) dt
Answer
(A) — L = ∫ √((−4sin t)² + (3cos t)²) dt.
31. Projectile: x = v₀cosα·t, y = v₀sinα·t − (1/2)gt². The range is:
(A) v₀² sin2α / g
(B) 2v₀² sin2α / g
(C) v₀² sin²α / g
(D) v₀² / g
Answer
(A)
32. For x = t³ − 3t and y = t⁴ − 2t², which condition characterizes a cusp candidate?
(A) x′ = 0 and y′ = 0 at the same t
(B) x′ ≠ 0 and y′ = 0
(C) x′ = 0 and y′ ≠ 0
(D) x′y′ < 0
Answer
(A) — Simultaneous zero derivatives.
33. For x = ln t and y = t (t > 0), dy/dx at t = e is:
(A) e
(B) 1/e
(C) 1
(D) e²
Answer
(A) — dy/dx = (1)/(1/t) = t; at t=e gives e.
34. For x = cos t − cos 2t, y = sin t − sin 2t, compute y′(t).
(A) cos t − 2cos 2t
(B) −sin t + 2sin 2t
(C) cos t − 2sin 2t
(D) −sin t − 2cos 2t
Answer
(A) — d/dt(sin t)=cos t; d/dt(−sin 2t)=−2cos 2t.
35. For x = t² + 4 and y = t³ − 3t, the curve is increasing in x when:
(A) t > 0
(B) t < 0
(C) t ≠ 0
(D) All real t
Answer
(A) — x′ = 2t > 0 ⇔ t > 0.
36. For x = t² − 6t and y = 3t + 1, find the point and classify the tangent at t = 3.
Answer
(−9, 10); vertical tangent — x′=2t − 6=0 at t=3 while y′=3≠0.
37. Eliminate t from x = 2 + 3t and y = −1 + 5t to obtain y in terms of x.
Answer
y = −1 + (5/3)(x − 2) — t = (x − 2)/3, substitute in y.
38. For x = t and y = t² − 4t + 1, find all horizontal tangents.
Answer
t = 2 — y′=2t − 4=0; x′=1≠0.
39. For x = R cos t, y = R sin t, give speed and total length on [0, 2π].
Answer
Speed R; length 2πR.
40. For x = e^{2t}, y = e^{t}, compute dy/dx and evaluate at t = 0.
Answer
dy/dx = (e^{t})/(2e^{2t}) = (1/2)e^{-t}; at t=0 ⇒ 1/2.
41. For x = t² − 1, y = t³ − 3t, find all t where the tangent is vertical.
Answer
t = 0 — x′=2t=0 and y′(0)=−3≠0.
42. Compute d²y/dx² at t = 2 for x = t² and y = t³.
Answer
3/8 — dy/dx=(3/2)t; d/dt=3/2; divide by x′=2t ⇒ 3/(4t); at 2 ⇒ 3/8.
43. For x = cos t, y = 2 + sin t on [0, 2π], compute ∫ y dx.
Answer
−π — ∫ (2 + sin t)(−sin t) dt over a full period = −∫ sin²t dt = −π.
44. Given x = t − 1/t (t > 0), y = t + 1/t, eliminate t.
Answer
y² − x² = 4 — (t + 1/t)² − (t − 1/t)² = 4.
45. For x = ln t and y = t² (t > 0), find dy/dx at t = 1 and state monotonicity of y with x there.
Answer
dy/dx = (2t)/(1/t) = 2t²; at 1 ⇒ 2 > 0, increasing.
46. For x = 1 − t², y = t³ − t, find all horizontal tangents.
Answer
t = ±1/√3 — y′=3t² − 1 = 0 with x′=−2t ≠ 0.
47. Write the arc-length integral for x = a cos t, y = b sin t on 0 ≤ t ≤ π.
Answer
L = ∫₀^{π} √(a² sin²t + b² cos²t) dt.
48. Eliminate t from x = t² + t and y = t² − t to relate 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 = 3t − t³ and y = 2 − t². Find all t where the speed is zero.
Answer
None — Need x′=3 − 3t²=0 and y′=−2t=0 simultaneously; impossible.
50. For x = t − sin t and y = 1 − cos t, compute dy/dx at t = π/3.
Answer
√3 — dy/dx = (sin t)/(1 − cos t); at π/3 → (√3/2)/(1/2) = √3.