2011 M480S solutions

# 2011 M480S solutions - Na (gn2849) – 4.9 – campisi –...

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Unformatted text preview: Na (gn2849) – 4.9 – campisi – (55842) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the value of f (0) when f ′ ( t ) = 3 sin2 t , f parenleftBig π 2 parenrightBig = 5 . 1. f (0) = 2 correct 2. f (0) = 5 3. f (0) = 3 4. f (0) = 4 5. f (0) = 6 Explanation: Since d dx cos mt = − m sin mt , for all m negationslash = 0, we see that f ( t ) = − 3 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 2) = 5. But cos 2 t vextendsingle vextendsingle vextendsingle t = π/ 2 = cos π = − 1 . Thus f parenleftBig π 2 parenrightBig = 3 2 + C = 5 , and so f ( t ) = − 3 2 cos 2 t + 7 2 . Consequently, f (0) = 2 . 002 10.0 points Find the value of f ( − 1) when f ′′ ( t ) = 15 t − 6 and f ′ (1) = 6 , f (1) = 1 . Correct answer: − 13. Explanation: The most general anti-derivative of f ′′ has the form f ′ ( t ) = 15 2 t 2 − 6 t + C where C is an arbitrary constant. But if f ′ (1) = 6, then f ′ (1) = 15 2 − 6 + C = 6 , i . e . C = 9 2 . Thus f ′ ( t ) = 15 2 t 2 − 6 t + 9 2 , from which it follows that f ( t ) = 5 2 t 3 − 3 t 2 + 9 2 t + D , where the constant D is determined by the condition f (1) = 5 2 − 3 + 9 2 + D = 1 , i . e . D = − 3 . Consequently, f ( t ) = 5 2 t 3 − 3 t 2 + 9 2 t − 3 , and so at t = − 1, f ( − 1) = − 13 . 003 10.0 points Find the unique anti-derivative F of f ( x ) = e 5 x + 4 e 2 x + 3 e − 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x + 4 x − 1 3 e − 3 x Na (gn2849) – 4.9 – campisi – (55842) 2 2. F ( x ) = 1 5 e 5 x − 4 x + 1 3 e − 3 x − 2 5 3. F ( x ) = 1 3 e 3 x +4 x − 3 5 e − 5 x + 4 15 correct 4. F ( x ) = 1 3 e 3 x − 4 x + 3 5 e − 3 x − 4 15 5. F ( x ) = 1 5 e 5 x + 4 x − 3 5 e − 5 x − 2 5 6. F ( x ) = 1 3 e 3 x − 4 x + 3 5 e − 5 x − 14 15 Explanation: After division, e 5 x + 4 e 2 x + 3 e − 3 x e 2 x = e 3 x + 4 + 3 e − 5 x ....
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## This note was uploaded on 02/11/2011 for the course M 480S taught by Professor Campisi during the Spring '11 term at University of Texas.

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2011 M480S solutions - Na (gn2849) – 4.9 – campisi –...

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