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Unformatted text preview: pokharel (yp624) – HW02 – Radin – (57410) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine f ( t ) when f ′′ ( t ) = 2(6 t + 1) and f ′ (1) = 2 , f (1) = 1 . 1. f ( t ) = 6 t 3 + t 2 − 6 t + 0 2. f ( t ) = 6 t 3 − 2 t 2 + 6 t − 9 3. f ( t ) = 2 t 3 + t 2 − 6 t + 4 correct 4. f ( t ) = 6 t 3 + 2 t 2 − 6 t − 1 5. f ( t ) = 2 t 3 − t 2 + 6 t − 6 6. f ( t ) = 2 t 3 − 2 t 2 + 6 t − 5 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 6 t 2 + 2 t + C where C is an arbitrary constant. But if f ′ (1) = 2, then f ′ (1) = 6 + 2 + C = 2 , i.e., C = − 6 . From this it follows that f ′ ( t ) = 6 t 2 + 2 t − 6 . The most general antiderivative of f is thus f ( t ) = 2 t 3 + t 2 − 6 t + D , where D is an arbitrary constant. But if f (1) = 1, then f (1) = 2 + 1 − 6 + D = 1 , i.e., D = 4 . Consequently, f ( t ) = 2 t 3 + t 2 − 6 t + 4 . 002 10.0 points Find all functions g such that g ′ ( x ) = 4 x 2 + 5 x + 4 √ x . 1. g ( x ) = 2 √ x ( 4 x 2 + 5 x + 4 ) + C 2. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C cor rect 3. g ( x ) = 2 √ x ( 4 x 2 + 5 x − 4 ) + C 4. g ( x ) = √ x ( 4 x 2 + 5 x + 4 ) + C 5. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x − 4 parenrightbigg + C 6. g ( x ) = √ x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 5 x 1 / 2 + 4 x − 1 / 2 , so we can now find an antiderivative of each term separately. But d dx parenleftbigg ax r r parenrightbigg = ax r − 1 for all a and all r negationslash = 0. Thus 8 5 x 5 / 2 + 10 3 x 3 / 2 + 8 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C pokharel (yp624) – HW02 – Radin – (57410) 2 with C an arbitrary constant. 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x 2 , ( B ) F 2 ( x ) = − cos 2 x 4 , ( C ) F 3 ( x ) = cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. all of them 2. none of them 3. F 1 only 4. F 1 and F 2 only correct 5. F 2 only 6. F 1 and F 3 only 7. F 3 only 8. F 2 and F 3 only Explanation: By trig identities, cos 2 x = 2 cos 2 x − 1 = 1 − 2 sin 2 x , while sin 2 x = 2 sin x cos x ....
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin

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