Parametric, Polar, Vector and Series Flashcards

A curve is defined parametrically by x = t² and y = t³. What is dy/dx at t = 2?

Answer: 3

dy/dx = (dy/dt)/(dx/dt) = 3t²/(2t) = 3t/2, which equals 3 at t = 2.

What is the area enclosed by the polar curve r = 1 + cos θ for 0 ≤ θ ≤ 2π?

Answer: 3π/2

Area = ½∫₀^{2π}(1 + cos θ)² dθ = ½(2π + 0 + π) = 3π/2.

Which statement describes the series Σ_{n=1}^∞ (−1)ⁿ/√n?

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.

What is the interval of convergence of Σ_{n=1}^∞ (x − 3)ⁿ / (n·2ⁿ)?

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.

In the Maclaurin series for f(x) = e^(−x²), what is the coefficient of x⁴?

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.

A particle's position is r(t) = ⟨t³ − t, 2t²⟩. What is its acceleration vector at t = 2?

Answer: ⟨12, 4⟩

The velocity is ⟨3t² − 1, 4t⟩ and the acceleration is ⟨6t, 4⟩, which equals ⟨12, 4⟩ at t = 2.

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:

Answer: 1/16

The error is bounded by the first omitted term, which is the 4th term, 1/4² = 1/16.

Which of the following series converges?

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.

What is the area enclosed by one petal of the rose curve r = cos 2θ, taken from θ = −π/4 to θ = π/4?

Answer: π/8

Area = ½∫_{−π/4}^{π/4} cos²2θ dθ = ¼∫(1 + cos 4θ) dθ = ¼(π/2) = π/8.

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))?

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.