4.9-solutions

# 4.9-solutions - tran(pt4954 4.9 campisi(54970 This...

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tran (pt4954) – 4.9 – campisi – (54970) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the value oF f (0) when f ( t ) = 3 sin 2 t , f p π 2 P = 2 . 1. f (0) = 5 2. f (0) = 1 correct 3. f (0) = 2 4. f (0) = 4 5. f (0) = 3 Explanation: Since d dx cos mt = m sin mt , For all m n = 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) = 2. But cos 2 t v v v t = π/ 2 = cos π = 1 . Thus f p π 2 P = 3 2 + C = 2 , and so f ( t ) = 3 2 cos 2 t + 1 2 . Consequently, f (0) = 1 . 002 10.0 points ±ind the value oF f ( 1) when f ′′ ( t ) = 9 t 2 and f (1) = 5 , f (1) = 4 . Correct answer: 4. Explanation: The most general anti-derivative oF f ′′ has the Form f ( t ) = 9 2 t 2 2 t + C where C is an arbitrary constant. But iF f (1) = 5, then f (1) = 9 2 2 + C = 5 , i . e . C = 5 2 . Thus f ( t ) = 9 2 t 2 2 t + 5 2 , From which it Follows that f ( t ) = 3 2 t 3 t 2 + 5 2 t + D , where the constant D is determined by the condition f (1) = 3 2 1 + 5 2 + D = 4 , i . e . D = 1 . Consequently, f ( t ) = 3 2 t 3 t 2 + 5 2 t , and so at t = 1, f ( 1) = 4 . 003 10.0 points ±ind the unique anti-derivative F oF f ( x ) = 3 e 5 x 3 e 2 x + 5 e 3 x e 2 x For which F (0) = 0. 1. F ( x ) = e 3 x 3 x e 3 x

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tran (pt4954) – 4.9 – campisi – (54970) 2 2. F ( x ) = 3 5 e 5 x 3 x e 5 x 2 5 3. F ( x ) = e 3 x 3 x e 5 x correct 4. F ( x ) = e 3 x + 3 x + e 3 x 5. F ( x ) = 3 5 e 5 x + 3 x + e 3 x 2 5 6. F ( x ) = e 3 x + 3 x + e 5 x 2 Explanation: After division, 3 e 5 x 3 e 2 x + 5 e 3 x e 2 x = 3 e 3 x 3 + 5 e 5 x . Since d dx e αx = αe αx , it thus follows that F ( x ) = e 3 x 3 x e 5 x + C where the constant C is determined by the condition F (0) = 0. For then F (0) = 1 1 + C = 0 . Consequently, F ( x ) = e 3 x 3 x e 5 x .
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## This note was uploaded on 09/20/2011 for the course CALCULUS 7234832 taught by Professor Campsisi during the Spring '11 term at University of Texas.

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4.9-solutions - tran(pt4954 4.9 campisi(54970 This...

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