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exam1_sol

# exam1_sol - Step 1 T(0.1 = 13 0.1-0.5(13(10 5sin(2π*0 =...

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CE 335 Exam 1 Solutions 1] (a) Only the first three derivatives of f(x) are nonzero. We have f(x) = x 3 - 2x 2 + 5x - 1 f'(x) = 3x 2 - 4x + 5 f''(x) = 6x - 4 f'''(x) = 6 So the Taylor series about zero is f(0 + h) = f(0) + hf'(0) + h 2 f''(0)/2 + h 3 f'''(0)/6 = -1 + 5h - 2h 2 + h 3 which, not coincidentally, looks much the same as the original polynomial (but written with the lowest- order terms first). (b) f(1) ≈ -1 + 5(1) = 4. (c) The true answer is f(1) = 1 - 2 + 5 - 1 = 3 The absolute error is therefore (4 - 3) = 1 The relative error is (4 - 3)/3 = 1/3 or 33%. 2] (a) The while loop will run 3 times, exiting when x = 3 so that the condition is no longer satisfied. (b) 9 6 3 9 3] (a) Differential equation: T' = -k(T - (10 + 5sin(2πt))) Euler's method: T(t + Δt) = T(t) + Δt(-k(T(t) - (10 + 5sin(2πt))))

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Unformatted text preview: Step 1: T(0.1) = 13 + 0.1*(-0.5(13 - (10 + 5sin(2π*0)))) = 12.85 °C Step 2: T(0.2) = 12.85 + 0.1*(-0.5(12.85 - (10 + 5sin(2π*0.1)))) = 12.8544 °C. (b) Because Euler's method is equivalent to using only the first two terms in the Taylor series, the error may be approximated by the next term in the Taylor series for T(t + Δt), which is (Δt) 2 T''(t)/2. In this case, T''(t) = dT ' dt = d dt − k T − 10 5sin 2πt = -kT' +10πkcos(2πt) = k 2 (T - (10 + 5sin(2πt))) + 10πkcos(2πt) = 0.75 + 5π (for k = 0.5 / d, T = 13 °C, t = 0 d) ≈ 16 °C / d 2 E ≈ (Δt) 2 T''(t)/2 ≈ 0.08 °C (for Δt = 0.1 d)...
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