M408D Quest Homework 5-solutions

# M408D Quest Homework 5-solutions - hernandez(ah29758 M408D...

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hernandez (ah29758) – M408D Quest Homework 5 – pascaleff – (54550) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine the degree 2 Taylor polynomial T 2 ( x ) centered at x = 1 for the function f when f ( x ) = radicalbig 8 + x 2 . 1. T 2 ( x ) = 3 - 1 3 ( x + 1) + 8 27 ( x + 1) 2 2. T 2 ( x ) = 3 + 1 3 ( x - 1) + 8 27 ( x - 1) 2 3. T 2 ( x ) = 3 - 1 3 ( x - 1) + 4 27 ( x - 1) 2 4. T 2 ( x ) = 3 + 1 3 ( x + 1) + 8 27 ( x + 1) 2 5. T 2 ( x ) = 3 + 1 3 ( x - 1) + 4 27 ( x - 1) 2 correct 6. T 2 ( x ) = 3 - 1 3 ( x + 1) + 4 27 ( x + 1) 2 Explanation: For a function f the degree 2 Taylor poly- nomial centered at x = 1 is given by T 2 ( x ) = f (1) + f (1)( x - 1) + 1 2 f ′′ (1)( x - 1) 2 . Now when f ( x ) = radicalbig 8 + x 2 , f ( x ) = x 8 + x 2 , while f ′′ ( x ) = 8 + x 2 - x 2 8 + x 2 8 + x 2 = 8 (8 + x 2 ) 3 / 2 . But then f (1) = 3 , f (1) = 1 3 , f ′′ (1) = 8 27 . Consequently, T 2 ( x ) = 3 + 1 3 ( x - 1) + 4 27 ( x - 1) 2 . 002(part1of2)10.0points (i) Compute the degree 2 Taylor polynomial for f centered at x = 1 when f ( x ) = x . 1. T 2 ( x ) = 1 - 1 4 ( x - 1) - 1 4 ( x - 1) 2 2. T 2 ( x ) = 1 - 1 2 ( x - 1) + 1 8 ( x - 1) 2 3. T 2 ( x ) = 1 - 1 4 ( x - 1) + 1 8 ( x - 1) 2 4. T 2 ( x ) = 1 + 1 4 ( x - 1) + 1 4 ( x - 1) 2 5. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 4 ( x - 1) 2 6. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 8 ( x - 1) 2 correct Explanation: The degree 2 Taylor polynomial centered at x = 1 for a general f is given by T 2 ( x ) = f (1)+ f (1) ( x - 1)+ f ′′ (1) 2! ( x - 1) 2 . Now when f ( x ) = x , f ( x ) = 1 2 x , f ′′ ( x ) = - 1 4 x x , in which case, f (1) = 1 , f (1) = 1 2 ,

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hernandez (ah29758) – M408D Quest Homework 5 – pascaleff – (54550) 2 while f ′′ (1) 2! = - 1 8 . Consequently, T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 8 ( x - 1) 2 . 003(part2of2)10.0points (ii) What estimate does Taylor’s Inequality provide for the error R 2 ( x ) = x - T 2 ( x ) in using the degree 2 Taylor polynomial T 2 ( x ) you derived in part (i) as an approximation to x on the interval [1 , 1 . 15]? 1. | R 2 ( x ) | ≤ 23 . 193 × 10 5 2. | R 2 ( x ) | ≤ 16 . 893 × 10 5 3. | R 2 ( x ) | ≤ 18 . 993 × 10 5 4. | R 2 ( x ) | ≤ 21 . 093 × 10 5 correct 5. | R 2 ( x ) | ≤ 25 . 293 × 10 5 Explanation: Taylor’s Inequality says that if T 2 ( x ) is the degree 2 Taylor polynomial for f centered at x = a and if | f (3) ( x ) | ≤ M , then the error R 2 ( x ) = f ( x ) - T 2 ( x ) satisfies the inequality | R 2 ( x ) | ≤ 1 3! M | x - a | 3 . We apply this estimate with f ( x ) = x , a = 1 . In this case, by the calculations in part (i), f (3) ( x ) = d dx f ′′ ( x ) = d dx parenleftBig - 1 4 x x parenrightBig = 3 8 x 2 x . Since we need to estimate R 2 ( x ) on the inter- val [1 , 1 . 15], we require the estimate | f (3) ( x ) | ≤ M to hold for all x in [1 , 1 . 15]. But f (3) ( x ) is decreasing and positive on (0 , ), so | f (3) ( x ) | = f (3) ( x ) f (3) (1) . for all x 1. Thus 1 x 1 . 15 = | f (3) ( x ) | ≤ 3 8 . On the other hand, 1 x 1 . 15 = | x - 1 | ≤ 3 20 . Consequently, | R 2 ( x ) | ≤ 1 3! 3 8 parenleftBig 3 20 parenrightBig 3 21 . 093 × 10 5 . 004 10.0points Lance Armstrong is moving with speed 30 feet/sec and acceleration 10 feet/sec 2 when he crosses the starting line of a stage in the Tour de France. Use a degree two Taylor poly- nomial T 2 ( t ) to estimate how far he travels in the next second.
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