Rucete ✏ AP Precalculus In a Nutshell
10. Parametric Functions — Practice Questions
This chapter explains how parametric equations describe curves and motion in the plane using a parameter t.
(Multiple Choice — Click to Reveal Answer)
1. For x(t) = t, y(t) = t², which Cartesian equation describes the curve?
(A) y = x²
(B) y = √x
(C) x = y²
(D) y = 2x
Answer
(A) — Eliminate t: t = x ⇒ y = x².
2. For x(t) = 3cos t, y(t) = 3sin t (0 ≤ t ≤ 2π), which graph is traced?
(A) Line through the origin
(B) Circle of radius 3 centered at the origin
(C) Parabola opening upward
(D) Ellipse with semiaxes 3 and 1
Answer
(B) — x² + y² = 9.
3. If x(t) = 2t + 1 and y(t) = −t + 4, which point corresponds to t = 2?
(A) (3, 2)
(B) (5, 2)
(C) (5, 3)
(D) (−3, 2)
Answer
(B) — x(2)=5, y(2)=2.
4. The slope dy/dx for a parametric curve is given by:
(A) (dx/dt) / (dy/dt)
(B) (dy/dt) / (dx/dt)
(C) d²y/dt²
(D) (dx/dt)(dy/dt)
Answer
(B) — Provided dx/dt ≠ 0.
5. For x(t) = t², y(t) = t³, what is dy/dx at t = 1?
(A) 1/2
(B) 3/2
(C) 3/2 t (at t=1) = 3/2
(D) 3t/2 (at t=1) = 3/2
Answer
(B) — dx/dt = 2t, dy/dt = 3t² ⇒ dy/dx = (3t²)/(2t) = (3/2)t ⇒ at t=1 → 3/2.
6. The speed |v(t)| of a parametric 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 velocity vector.
7. For x(t) = t, y(t) = 2t + 3, which is true?
(A) Parabola
(B) Circle
(C) Line with slope 2 and y-intercept 3
(D) Line with slope 1/2
Answer
(C) — y = 2x + 3.
8. For x(t) = cos t, y(t) = 2 sin t, which Cartesian equation holds?
(A) x² + y² = 1
(B) x² + (y/2)² = 1
(C) (x/2)² + y² = 1
(D) x² − y² = 1
Answer
(B) — cos²t + sin²t = 1 ⇒ x² + (y²/4) = 1.
9. For x(t) = e^t, y(t) = e^t, eliminate t.
(A) y = x²
(B) y = x
(C) xy = 1
(D) y = ln x
Answer
(B) — Both equal e^t.
10. For x(t) = t² − 1, y(t) = 2t, the point when t = −1 is:
(A) (0, −2)
(B) (0, 2)
(C) (−2, 0)
(D) (2, −2)
Answer
(A) — x=1−1=0, y=−2.
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 (with 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) — Undefined slope when denominator zero (with numerator nonzero).
13. For x(t) = t³, y(t) = t², what is dy/dx at t ≠ 0?
(A) (2t)/(3t²) = 2/(3t)
(B) (3t²)/(2t) = (3/2)t
(C) 2t³
(D) 3t²
Answer
(A) — dx/dt=3t², dy/dt=2t ⇒ dy/dx = (2t)/(3t²).
14. For x(t) = 1 + 2t, y(t) = 3 − t, the slope dy/dx is:
(A) −1/2
(B) −2
(C) 2
(D) 1/2
Answer
(B) — dy/dt = −1, dx/dt = 2 ⇒ −1/2? Careful: dy/dx = (−1)/2 = −1/2. Correct choice is (A).
15. For x(t) = sin t, y(t) = cos t, what is dy/dx at t = 0?
(A) 0
(B) −∞
(C) −1/t (undefined)
(D) 0/1
Answer
(B) — dy/dt = −sin t (0), dx/dt = cos t (1) ⇒ slope 0? Wait: y = cos t, x = sin t. dy/dt = −sin t = 0, dx/dt = cos t = 1 ⇒ dy/dx = 0. Correct answer is (A).
16. The arc length on t ∈ [a, b] of (x(t), y(t)) 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 arc length formula.
17. For x(t) = t², y(t) = t³, speed at t = 2 is:
(A) √( (4)² + (12)² )
(B) √( (2t)² + (3t²)² ) at t=2 = √(4² + 12²) = √160
(C) √( (2)² + (3)² )
(D) 2 + 12
Answer
(B) — dx/dt=2t, dy/dt=3t² → √(4² + 12²)=√160.
18. Eliminating t from x = 2cos t, y = 5sin t yields:
(A) x²/4 + y²/25 = 1
(B) x² + y² = 1
(C) x²/25 + y²/4 = 1
(D) 2x² + 5y² = 1
Answer
(A) — (x/2)² + (y/5)² = 1.
19. For x(t) = t − sin t, y(t) = 1 − cos t (cycloid), what is x′(t)?
(A) 1 − cos t
(B) 1 + cos t
(C) 1 − sin t
(D) cos t
Answer
(A) — d/dt (t − sin t) = 1 − cos t.
20. For x(t) = t, y(t) = t³, which best describes the curve near the origin?
(A) Line
(B) Parabola opening right
(C) Cubic y = x³
(D) y = x³ with steepening slope near 0
Answer
(C) — Eliminate t: y = x³.
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) — Chain rule.
22. Area under a parametric curve y(x) from t=a to t=b is:
(A) ∫ y dt
(B) ∫ x dy
(C) ∫ y dx = ∫ab y(t) x′(t) dt
(D) ∫ x dx
Answer
(C) — Parametric area formula.
23. For x= t² − 4, y= 3t + 1, which t gives vertical tangent?
(A) t = 0
(B) t = 2
(C) t = −2
(D) t = ±2
Answer
(A) — x′=2t, vertical when x′=0 and y′≠0 ⇒ t=0.
24. For x= 2 + 4t, y= 5 − t, what is the speed?
(A) √(4² + (−1)²) = √17
(B) 4 − 1 = 3
(C) √(4 − 1)
(D) 5
Answer
(A) — √(x′² + y′²) = √(16 + 1).
25. For x= cos t, y= sin t on t ∈ [0, π], the path length is:
(A) π/2
(B) π
(C) 2π
(D) 1
Answer
(B) — Half a unit circle ⇒ length π.
26. For x= t², y= t³ − 3t, find where the curve has horizontal tangents.
(A) t = 0 only
(B) t = ±1
(C) t = 0, ±1
(D) None
Answer
(B) — y′ = 3t² − 3 = 3(t² − 1) ⇒ zeros at t = ±1 with x′ = 2t ≠ 0.
27. For x= t − sin t, y= 1 − cos t, the speed is:
(A) √(2 − 2cos t)
(B) √(2 − 2sin t)
(C) √(2 + 2cos t)
(D) √(1 − cos t)
Answer
(A) — x′=1 − cos t, y′= sin t ⇒ √((1 − cos t)² + sin² t) = √(2 − 2cos t).
28. For x= e^t, y= e^{−t}, compute dy/dx at t = 0.
(A) −1
(B) 1
(C) 0
(D) −e^{−2t} at t=0 = −1
Answer
(A) — dy/dt = −e^{−t}, dx/dt = e^{t} ⇒ dy/dx = −e^{−2t} ⇒ at 0 is −1.
29. For x(t) = t² − 1 and y(t) = t³ − t, compute d²y/dx² at t = 1.
(A) 1
(B) 0
(C) 1/2
(D) −1/2
Answer
(A) — dy/dx = (3t² − 1)/(2t) = (3/2)t − (1/(2t)). Then (d/dt)(dy/dx) = 3/2 + 1/(2t²). Divide by x′ = 2t: d²y/dx² = [(3/2) + 1/(2t²)]/(2t) = 3/(4t) + 1/(4t³). At t = 1 this is 3/4 + 1/4 = 1.
30. For x(t) = 4cos t and y(t) = 2sin t on t ∈ [0, 2π], which expression equals the exact arc length?
(A) ∫02π √(16sin²t + 4cos²t) dt
(B) ∫02π √(16cos²t + 4sin²t) dt
(C) 2π√5
(D) 8π
Answer
(A) — L = ∫ √(x′(t)² + y′(t)²) dt = ∫ √((−4sin t)² + (2cos t)²) dt = ∫ √(16sin²t + 4cos²t) dt. (No elementary closed form.)
31. For projectile parametrics x= v₀ cosα · t, y= v₀ sinα · t − (1/2)gt², the range (landing where y=0, t>0) is:
(A) v₀² sin2α / g
(B) 2v₀² sin2α / g
(C) v₀² sin²α / g
(D) v₀² / g
Answer
(A) — Time to land 2v₀ sinα / g; multiply by v₀ cosα.
32. For x= t³ − 3t, y= t⁴ − 2t², choose the condition indicating a cusp candidate.
(A) x′=0, y′≠0
(B) x′≠0, y′=0
(C) x′=0 and y′=0 at same t
(D) x′y′ < 0
Answer
(C) — Simultaneous vanishing of first derivatives is a cusp candidate.
33. For x= ln t, y= t (t>0), find dy/dx at t=1.
(A) 1
(B) t at 1 ⇒ 1
(C) t² at 1 ⇒ 1
(D) 1/ t at 1 ⇒ 1
Answer
(A) — dy/dt=1, dx/dt=1/t ⇒ dy/dx = t ⇒ at 1 gives 1.
34. For x= cos t − cos 2t, y= sin t − sin 2t, x′(t) equals:
(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, d/dt(−cos2t)=+2sin2t.
35. For x= t² + 1, y= t³ − 3t, where is the curve increasing in x?
(A) t > 0
(B) t < 0
(C) t ≠ 0
(D) t > 0 and t < 0 (all t except 0)
Answer
(A) — x′=2t > 0 for t > 0.
36. For x= t² − 4t, y= 3t − 2, find the point and slope dy/dx at t = 2.
Answer
Point (−4, 4); slope 3/(2t − 4) undefined at t=2 ⇒ vertical tangent — x(2)=−4, y(2)=4; x′=2t − 4=0, y′=3 ⇒ vertical.
37. Eliminate t for x= 1 + 2t, y= 4 − t and give the Cartesian equation.
Answer
y = 4 − (x − 1)/2 — From x=1+2t ⇒ t=(x−1)/2; substitute into y.
38. For x= t, y= t² − 2t, find all t where the tangent is horizontal.
Answer
t = 1 — y′=2t − 2; set to 0 ⇒ t=1 with x′=1 ≠ 0.
39. For x= 2cos t, y= 2sin t, compute the speed and the total length on [0, 2π].
Answer
Speed 2 (constant); length 4π — √(x′² + y′²)=√(4sin²t + 4cos²t)=2.
40. For x= e^t, y= e^{2t}, compute dy/dx as a function of t, then evaluate at t = 0.
Answer
dy/dx = (2e^{2t})/(e^{t}) = 2e^{t}; at t=0 → 2.
41. For x= t² − 1, y= t³ − 3t, find all t with vertical tangents.
Answer
t = 0 — x′=2t; vertical when x′=0 and y′≠0; y′=3t² − 3=−3 ≠ 0 at t=0.
42. Compute d²y/dx² at t = 2 for x= t², y= t³.
Answer
d²y/dx² = ( (d/dt)(dy/dx) )/(dx/dt) = ( (d/dt)( (3t²)/(2t) ) )/(2t) = ( (3/2) )/(2t) = 3/(4t); at t=2 → 3/8.
43. For x= cos t, y= sin t on [0, π], compute the signed area ∫ y dx.
Answer
∫₀^π y x′ dt = ∫₀^π (sin t)(−sin t) dt = −∫₀^π sin²t dt = −π/2 — Magnitude π/2 (upper semicircle above x-axis, leftwards orientation gives negative sign).
44. Given x= t − 1/t (t>0), y= t + 1/t, eliminate t to get a Cartesian relation.
Answer
y² − x² = 4 — (t + 1/t)² − (t − 1/t)² = 4.
45. For x= ln t, y= t² (t>0), find dy/dx at t = 1 and state whether the curve is increasing or decreasing in y with respect to x there.
Answer
dy/dx = (2t)/(1/t) = 2t²; at t=1 → 2 > 0, so increasing.
46. For x= 1 − t², y= t³ − t, find all t with horizontal tangents.
Answer
t = ±1 — y′=3t² − 1 ⇒ zeros at t=±1 with x′=−2t ≠ 0.
47. For x= a cos t, y= b sin t, write the arc length integral L on [0, 2π] (no evaluation).
Answer
L = ∫₀^{2π} √(a² sin²t + b² cos²t) dt.
48. For x= t² + t, y= t² − t, eliminate t to obtain a Cartesian equation.
Answer
(x − y) = 2t and (x + y) = 2t² ⇒ (x + y) = (x − y)² / 2 — From t = (x − y)/2 and t² = (x + y)/2.
49. A particle has x= 3t − t³ and y= 4 − 2t². Find all t where speed is zero.
Answer
None — Speed zero requires x′=0 and y′=0: x′=3 − 3t²=3(1 − t²); y′=−4t. The system has t=0 ⇒ x′=3 ≠ 0; t=±1 ⇒ y′=∓4 ≠ 0. No common solution.
50. For x= t − sin t, y= 1 − cos t, find dy/dx at t = π/3.
Answer
dy/dx = (sin t)/(1 − cos t); at t=π/3 → (√3/2)/(1 − 1/2) = (√3/2)/(1/2) = √3.