AP Calc BC Parametric, Polar, Vector and Series — Questions and Answers
Question 1: A curve is defined parametrically by x = t² and y = t³. What is dy/dx at t = 2?
Correct answer: 3
dy/dx = (dy/dt)/(dx/dt) = 3t²/(2t) = 3t/2, which equals 3 at t = 2.
For a parametric curve, the slope of the tangent line is dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. Here dy/dt = 3t² and dx/dt = 2t, so dy/dx = 3t/2. Substituting t = 2 gives 3. A common mistake is to evaluate dy/dt alone (12) or to skip the division.
Question 2: What is the area enclosed by the polar curve r = 1 + cos θ for 0 ≤ θ ≤ 2π?
- π/2
- π
- 3π/2 (Correct answer)
- 3π
Correct answer: 3π/2
Area = ½∫₀^{2π}(1 + cos θ)² dθ = ½(2π + 0 + π) = 3π/2.
The area enclosed by a polar curve is A = ½∫r² dθ. Expanding (1 + cos θ)² gives 1 + 2cos θ + cos²θ. Over 0 to 2π, the integral of 1 is 2π, the integral of 2cos θ is 0, and the integral of cos²θ is π. Half of the total 3π is 3π/2.
Question 3: Which statement describes the series Σ_{n=1}^∞ (−1)ⁿ/√n?
- It converges conditionally. (Correct answer)
- It converges absolutely.
- It diverges because its terms do not approach 0.
- It diverges by the alternating series test.
Correct answer: It converges conditionally.
The terms 1/√n decrease to 0, so the alternating series test shows convergence. The series of absolute values, Σ1/√n, is a p-series with p = 1/2 ≤ 1 and diverges.
The alternating series test needs the terms to decrease in absolute value and approach 0. Both hold for 1/√n, so the series converges. Absolute convergence would require Σ1/√n to converge, but it is a p-series with p = 1/2, which diverges. A series that converges but does not converge absolutely converges conditionally.
Question 4: What is the interval of convergence of Σ_{n=1}^∞ (x − 3)ⁿ / (n·2ⁿ)?
- (1, 5)
- (1, 5]
- [1, 5]
- [1, 5) (Correct answer)
Correct answer: [1, 5)
The ratio test gives radius 2, so the series converges on (1, 5). At x = 1 it is the alternating harmonic series, which converges. At x = 5 it is the harmonic series, which diverges.
The ratio test gives |x − 3|/2 < 1, so the center is 3 and the radius is 2, which leaves the open interval (1, 5). At x = 5 the terms are 1/n, giving the divergent harmonic series. At x = 1 the terms are (−1)ⁿ/n, giving the convergent alternating harmonic series. The interval is therefore [1, 5).
Question 5: In the Maclaurin series for f(x) = e^(−x²), what is the coefficient of x⁴?
- 1/24
- −1/2
- 1/2 (Correct answer)
- 1/4
Correct answer: 1/2
Substitute u = −x² into eᵘ = 1 + u + u²/2 + …. The u²/2 term is (−x²)²/2 = x⁴/2, so the coefficient is 1/2.
The Maclaurin series for eᵘ is Σ uⁿ/n!. Substituting u = −x² gives 1 − x² + x⁴/2 − x⁶/6 + …. The x⁴ term comes from n = 2, where (−x²)²/2! = x⁴/2. The coefficient is therefore 1/2. Using 1/24 would come from wrongly applying 1/4! to the x⁴ term of eˣ.
Question 6: A particle's position is r(t) = ⟨t³ − t, 2t²⟩. What is its acceleration vector at t = 2?
- ⟨12, 4⟩ (Correct answer)
- ⟨6, 4⟩
- ⟨11, 8⟩
- ⟨12, 8⟩
Correct answer: ⟨12, 4⟩
The velocity is ⟨3t² − 1, 4t⟩ and the acceleration is ⟨6t, 4⟩, which equals ⟨12, 4⟩ at t = 2.
For a vector-valued position function, velocity is the derivative of position and acceleration is the derivative of velocity, both taken componentwise. Differentiating once gives v(t) = ⟨3t² − 1, 4t⟩, and differentiating again gives a(t) = ⟨6t, 4⟩. At t = 2 this is ⟨12, 4⟩. The vector ⟨11, 8⟩ is the velocity at t = 2, not the acceleration.
Question 7: The alternating series Σ_{n=1}^∞ (−1)^(n+1)/n² is approximated by the sum of its first 3 terms. By the alternating series error bound, the absolute error is at most:
- 1/9
- 1/25
- 1/4
- 1/16 (Correct answer)
Correct answer: 1/16
The error is bounded by the first omitted term, which is the 4th term, 1/4² = 1/16.
For a convergent alternating series whose terms decrease to 0 in absolute value, the error after S_N is at most |a_{N+1}|. Here N = 3, so the first omitted term is a₄ = 1/16. The partial sum uses 1/9, but that term was included and does not set the bound. The 1/25 option would be the bound after four terms.
Question 8: Which of the following series converges?
- Σ_{n=1}^∞ 1/√n
- Σ_{n=2}^∞ 1/(n(ln n)²) (Correct answer)
- Σ_{n=2}^∞ 1/(n ln n)
- Σ_{n=1}^∞ n/(2n + 1)
Correct answer: Σ_{n=2}^∞ 1/(n(ln n)²)
By the integral test, ∫₂^∞ dx/(x(ln x)²) = 1/ln 2 is finite, so the series converges. The other three series diverge.
With u = ln x, ∫dx/(x(ln x)²) becomes ∫du/u² = −1/u, which has a finite limit as x → ∞, so the series converges. The same substitution applied to 1/(x ln x) gives ln(ln x), which grows without bound, so that series diverges. Σ1/√n is a p-series with p = 1/2 and diverges. The terms n/(2n + 1) approach 1/2, not 0, so that series diverges by the nth-term test.
Question 9: What is the area enclosed by one petal of the rose curve r = cos 2θ, taken from θ = −π/4 to θ = π/4?
- π/8 (Correct answer)
- π/4
- π/2
- π/16
Correct answer: π/8
Area = ½∫_{−π/4}^{π/4} cos²2θ dθ = ¼∫(1 + cos 4θ) dθ = ¼(π/2) = π/8.
Apply the polar area formula A = ½∫r² dθ over the interval where one petal is traced. The petal centered on the positive x-axis is traced as θ goes from −π/4 to π/4, where r ≥ 0. The identity cos²2θ = (1 + cos 4θ)/2 turns the integral into ¼∫(1 + cos 4θ) dθ. The cosine term integrates to 0 over this interval, leaving ¼ · (π/2) = π/8.
Question 10: The Maclaurin series for arctan x is Σ_{n=0}^∞ (−1)ⁿ x^(2n+1)/(2n+1) for |x| ≤ 1. What is the sum of Σ_{n=0}^∞ (−1)ⁿ / (3ⁿ(2n + 1))?
- π/6
- π√3/6 (Correct answer)
- π/3
- π√3/3
Correct answer: π√3/6
Setting x = 1/√3 gives x^(2n+1) = 3^(−n)/√3, so arctan(1/√3) = π/6 equals (1/√3) times the given sum. The sum is therefore √3 · π/6 = π√3/6.
With x = 1/√3, we have x^(2n+1) = (1/√3)(1/3ⁿ). The arctan series then reads arctan(1/√3) = (1/√3) Σ (−1)ⁿ/(3ⁿ(2n+1)). Since arctan(1/√3) = π/6, multiplying both sides by √3 gives the sum as π√3/6. The factor of 1/√3 is easy to forget, which leads to the distractor π/6.
