HW 9 - Tran Tony Homework 9 Due 3:00 am Inst Samuels This...

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Tran, Tony – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Samuels 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the linearization of f ( x ) = 1 5 + x at x = 0. 1. L ( x ) = 1 5 1 - 1 5 x · 2. L ( x ) = 1 5 1 + 1 10 x · 3. L ( x ) = 1 5 + 1 5 x 4. L ( x ) = 1 5 1 + 1 10 x · 5. L ( x ) = 1 5 - 1 5 x 6. L ( x ) = 1 5 1 - 1 10 x · correct Explanation: The linearization of f is the function L ( x ) = f (0) + f 0 (0) x . But for the function f ( x ) = 1 5 + x = (5 + x ) - 1 / 2 , the Chain Rule ensures that f 0 ( x ) = - 1 2 (5 + x ) - 3 / 2 . Consequently, f (0) = 1 5 , f 0 (0) = - 1 10 5 , and so L ( x ) = 1 5 1 - 1 10 x · . keywords: linearization, square root function, differentials 002 (part 1 of 1) 10 points Use linear approximation with a = 4 to estimate the number 3 . 7 as a fraction. 1. 3 . 7 1 7 8 2. 3 . 7 1 9 10 3. 3 . 7 1 17 20 4. 3 . 7 1 33 40 5. 3 . 7 1 37 40 correct Explanation: For a general function f , its linear approxi- mation at x = a is defined by L ( x ) = f ( a ) + f 0 ( a )( x - a ) and for values of x near a f ( x ) L ( x ) = f ( a ) + f 0 ( a )( x - a ) provides a reasonable approximation for f ( x ). Now set f ( x ) = x, f 0 ( x ) = 1 2 x . Then, if we can calculate a easily, the linear approximation a + h a + h 2 a provides a very simple method via calculus for computing a good estimate of the value of a + h for small values of h . In the given example we can thus set a = 4 , h = - 3 10 .
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Tran, Tony – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Samuels 2 For then 3 . 7 1 37 40 . keywords: linear approximation, square roots 003 (part 1 of 1) 10 points A cube with sides 3 inches long is covered with a fiberglass coating . 01 inches thick. Es- timate the volume of the fiberglass shell. 1. fiberglass vol 23 50 cu.ins. 2. fiberglass vol 12 25 cu.ins. 3. fiberglass vol 27 50 cu.ins. correct 4. fiberglass vol 1 2 cu.ins. 5. fiberglass vol 13 25 cu.ins. Explanation: A cube with side length x has volume V ( x ) = x 3 . If the length of each side is changed by an amount Δ x , then the approxi- mate change, Δ V , in volume is given by Δ V V 0 ( x x = 3 x 2 Δ x . Now a . 01 inch thick fiberglass coating on each face will increase the side length by 1 50 = (2 × . 01) inches . When the side length of the cube is 3 inches, therefore, the volume of this fiberglass shell will be approximately Δ V = fiberglass vol 27 50 cu. ins. . keywords: cube, volume, estimate 004 (part 1 of 1) 10 points The radius of a circle is estimated to be 14 inches, with a maximum error in measure- ment of ± 0 . 02 inches. Use differentials to estimate the maximum error in calculating the area of the circle using this estimate. 1. Max error ≈ ± 1 . 7672 sq.ins 2. Max error ≈ ± 1 . 7512 sq.ins 3. Max error ≈ ± 1 . 7592 sq.ins correct 4. Max error ≈ ± 1 . 7832 sq.ins 5. Max error ≈ ± 1 . 7752 sq.ins Explanation: The area of a circle having radius x is given by A = πx 2 . By differentials, therefore, Δ A Δ x dA dx = 2 πx Δ x.
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