Instructor: Dr. R.A.G. Seely Test 1_{(version A)}

Calculus III (Maths 201–DDB)

Answers

1. ∑_{n=0}^{∞}(−1)^{n} \frac{x^{4n+1}}{(4n + 1)(2n + 2)!} ≃ \frac{1}{4} − \frac{1}{3840} ± \frac{1}{3317760} = 0.2497396 ± 3 × 10^{−7}

2. (a) ∑_{n=0}^{∞}(−1)^{n}x^{2n} , IofC: (−1, 1) (b) ∑_{n=0}^{∞}(−1)^{n} \frac{x^{2n+1}}{2n + 1} , IofC: [−1, 1]

(c) π/4 = arctan 1 ≃ 1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + \frac{1}{9} = \frac{263}{315} = 0.83 (±\frac{1}{11} = ±0.09)

3. T_{3}(x) = 2[1+\frac{1}{160}(x−32)−\frac{1}{12800}(x−32)^{2}+\frac{3}{2048000}(x−32)^{3}] so T_{3}(33) = \frac{2060643}{1024000} = 2.01234668. Error: Take M = |f^{(4)}(31)| = 1.7 × 10^{−6} and hence |R_{3}(33)| ≤ 7.2 × 10^{−8} (i.e. good to 7 decimal places). So ^{5} \sqrt{33} = 2.01234668 ± 1.7 × 10^{−6}

4. (a) 1 (b) \frac{1}{10}

5. (a) x-int: x = 0 at t = −2, x = −4 at t = 0; y-int: y = 0 at t = −2, y = 8 at t = 2. \frac{dy}{dx} = 1 + \frac{1}{t}, \frac{d^{2}y}{dx^{2}} = −\frac{1}{2t^{3}} . HT at t = −1, (−3, −1); VT at t = 0, (−4, 0). Graph: See Fig 1a.

(b) area = ∫_{−2}^{0}(t^{2} + 2t)2t dt = \frac{8}{3} length = ∫_{−2}^{0} \sqrt{(2t)^{2} + (2t + 2)^{2}} dt

6. Graph: see Fig 2a area = \frac{1}{2} ∫_{π/6}^{5π/6}(1 − 2 sin θ)^{2} dθ = π − \frac{3}{2} \sqrt{3}

length = 2 ∫_{−π/2}^{π/6} \sqrt{4 cos^{2} θ + (1 − 2 sin θ)^{2}} dθ