This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Villarreal, Natalie – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Louiza Fouli 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 √ 3 + x at x = 0. 1. L ( x ) = 1 √ 3 ‡ 1 + 1 6 x · 2. L ( x ) = 1 √ 3 + 1 3 x 3. L ( x ) = 1 3 ‡ 1 + 1 6 x · 4. L ( x ) = 1 3 ‡ 1 1 3 x · 5. L ( x ) = 1 √ 3 1 3 x 6. L ( x ) = 1 √ 3 ‡ 1 1 6 x · correct Explanation: The linearization of f is the function L ( x ) = f (0) + f (0) x. But for the function f ( x ) = 1 √ 3 + x = (3 + x ) 1 / 2 , the Chain Rule ensures that f ( x ) = 1 2 (3 + x ) 3 / 2 . Consequently, f (0) = 1 √ 3 , f (0) = 1 6 √ 3 , and so L ( x ) = 1 √ 3 ‡ 1 1 6 x · . keywords: linearization, square root function, differentials 002 (part 1 of 1) 10 points Use linear approximation with a = 25 to estimate the number √ 24 . 8 as a fraction. 1. √ 24 . 8 ≈ 4 19 20 2. √ 24 . 8 ≈ 4 24 25 3. √ 24 . 8 ≈ 4 49 50 correct 4. √ 24 . 8 ≈ 4 97 100 5. √ 24 . 8 ≈ 4 47 50 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 = 25 , h = 2 10 . Villarreal, Natalie – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Louiza Fouli 2 For then √ 24 . 8 ≈ 4 49 50 . keywords: linear approximation, square roots 003 (part 1 of 1) 10 points A cube with sides 4 inches long is covered with a fiberglass coating . 02 inches thick. Es timate the volume of the fiberglass shell. 1. fiberglass vol ≈ 49 25 cu.ins. 2. fiberglass vol ≈ 48 25 cu.ins. correct 3. fiberglass vol ≈ 99 50 cu.ins. 4. fiberglass vol ≈ 97 50 cu.ins. 5. fiberglass vol ≈ 19 10 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 . 02 inch thick fiberglass coating on each face will increase the side length by 1 25 = (2 × . 02) inches . When the side length of the cube is 4 inches, therefore, the volume of this fiberglass shell will be approximately Δ V = fiberglass vol ≈ 48 25 cu. ins. ....
View
Full
Document
This note was uploaded on 04/15/2011 for the course MATH 408K taught by Professor Gualdani during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Gualdani

Click to edit the document details