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

Calculus III (Maths 201–DDB)

Answers

1. ∑_{n=0}^{∞}(−1)^{n} \frac{x^{4n+1}}{(4n + 1)(2n + 3)!} ≃ \frac{1}{12} − \frac{1}{19200} ± \frac{1}{23224320} = 0.08328125 ± 4 × 10^{−8}

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

(c) π/6 = arcsin(\frac{1}{2}) ≃ \frac{1}{2} + \frac{1}{48} + \frac{3}{1280} − \frac{15}{2^{10}·6·7} = \frac{112579}{215040} = 0.524

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 = −3, x = −9 at t = 0; y-int: y = 0 at t = −3, y = 18 at t = 3. \frac{dy}{dx} = 1 + \frac{3}{2t}, \frac{d^{2}y}{dx^{2}} = −\frac{3}{4t^{3}} . HT at t = −\frac{3}{2}, (−\frac{27}{4}, −\frac{9}{4}); VT at t = 0, (−9, 0). Graph: See Fig 1b.

(b) area = ∫_{−3}^{0}(t^{2} + 3t)2t dt = \frac{27}{2} length = ∫_{−3}^{0} \sqrt{(2t)^{2} + (2t + 3)^{2}} dt

6. Graph: see Fig 2b area = \frac{1}{2} ∫_{2π/3}^{4π/3}(1 + 2 cos θ)^{2} dθ = π − \frac{3}{2} \sqrt{3}

length = 2 ∫_{0}^{2π/3} \sqrt{4 sin^{2} θ + (1 + 2 cos θ)^{2}} dθ