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Unformatted text preview: Tran, Tony – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Samuels 1 This printout should have 20 questions. Multiplechoice 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) x. But for the function f ( x ) = 1 √ 5 + x = (5 + x ) 1 / 2 , the Chain Rule ensures that f ( x ) = 1 2 (5 + x ) 3 / 2 . Consequently, f (0) = 1 √ 5 , f (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 ( a )( x a ) and for values of x near a f ( x ) ≈ L ( x ) = f ( a ) + f ( a )( x a ) provides a reasonable approximation for f ( x ). Now set f ( x ) = √ x, f ( 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 . 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 ( x )Δ 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. ....
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This note was uploaded on 04/27/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz

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