HW06-solutions

# HW06-solutions - husain(aih243 HW06 Gilbert(56215 This...

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husain (aih243) – HW06 – Gilbert – (56215) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Richard removes a turkey From the oven aFter it reaches a temperature oF 166 F and places on a table in his dining room where the temperature is 85 F . AFter 10 minutes the temperature oF the turkey is 158 F , while aFter 20 minutes it is 152 F . Use linear ap- proximation to predict the temperature oF the turkey aFter halF an hour. 1. temp 143 F 2. temp 147 F 3. temp 144 F 4. temp 146 F correct 5. temp 145 F Explanation: IF T ( t ) is the temperature oF the turkey aFter t minutes, then the given data tell us that T (0) = 166 , T (10) = 158 , T (20) = 152 , while the Fact that the room temperature is 85 F means that T ( t ) gets closer to 85 F as t increases. Now by linear approximation T ( t ) T (20) + T (20)( t 20) , ( t > 20) , with T (20) = lim t 20 T ( t ) T (20) t 20 . But T (20) T (10) T (20) 10 20 = 158 152 10 = 3 5 degrees/min . Consequently, temp = T (30) 146 F . 002 10.0 points ±ind the linearization oF f ( x ) = 1 3 + x at x = 0. 1. L ( x ) = 1 3 p 1 + 1 6 x P 2. L ( x ) = 1 3 p 1 1 3 x P 3. L ( x ) = 1 3 + 1 3 x 4. L ( x ) = 1 3 p 1 1 6 x P correct 5. L ( x ) = 1 3 p 1 + 1 6 x P 6. L ( x ) = 1 3 1 3 x 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 p 1 1 6 x P .

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husain (aih243) – HW06 – Gilbert – (56215) 2 003 10.0 points Use linear approximation with a = 16 to estimate the number 15 . 4 as a fraction. 1. 15 . 4 3 39 40 2. 15 . 4 3 15 16 3. 15 . 4 3 77 80 4. 15 . 4 3 37 40 correct 5. 15 . 4 3 19 20 Explanation: For a general function f , its linear approxi- mation at x = a is de±ned 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 = 16 , h = 6 10 . For then 15 . 4 3 37 40 . 004 10.0 points Estimate the value of 17 1 / 4 using di²eren- tials. 1. 17 1 / 4 63 32 2. 17 1 / 4 31 16 3. 17 1 / 4 2 4. 17 1 / 4 33 16 5. 17 1 / 4 65 32 correct Explanation: Set f ( x ) = x 1 / 4 . Then df dx = 1 4 x 3 / 4 . By di²erentials, therefore, we see that f ( a + Δ x ) f ( a ) dx v v v x = a Δ x = Δ x 4 a 3 / 4 . Thus, with a = 16 and Δ x = 1, 17 1 / 4 2 1 32 . Consequently, 17 1 / 4 65 32 . 005 10.0 points At the H-E-B stores throughout Texas the daily demand (in pounds) for candy at \$ x per pound is given by D = 8000 100 x 2 , 1 x 7 .
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HW06-solutions - husain(aih243 HW06 Gilbert(56215 This...

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