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Unformatted text preview: Miller, Kierste – Homework 1 – Due: Jan 24 2007, 3:00 am – Inst: Gary Berg 1 This printout should have 15 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 Determine the value of f (2) when f ( t ) = 6 t 5 , f ( 1) = 6 . Correct answer: 0 . Explanation: The most general antiderivative of f has the form F ( t ) = 3 t 2 5 t + C , where C is an arbitrary constant whose value is determined by the condition F ( 1) = 6. For then F ( 1) = 3 + 5 + C = 6 , i . e ., C = 2 . Consequently, f ( t ) = 3 t 2 5 t 2 , and so at t = 2 , f (2) = +0 . keywords: antiderivatives, affine function, particular values 002 (part 1 of 1) 10 points Consider the following functions: ( A ) F 1 ( x ) = cos2 x 4 , ( B ) F 2 ( x ) = cos 2 x 2 , ( C ) F 3 ( x ) = sin 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. F 1 and F 3 only 2. F 1 only 3. F 3 only 4. F 2 only 5. none of them 6. F 1 and F 2 only 7. all of them 8. F 2 and F 3 only correct Explanation: By trig identities, cos2 x = 2cos 2 x 1 = 1 2sin 2 x, while sin2 x = 2sin x cos x. But d dx sin x = cos x, d dx cos x = sin x. Consequently, by the Chain Rule, ( A ) Not antiderivative. ( B ) Antiderivative. ( C ) Antiderivative. keywords: antiderivative, trig function, dou ble angle formula, T/F, 003 (part 1 of 1) 10 points Find f ( t ) when f ( t ) = cos 1 3 t 2sin 2 3 t and f ( π 2 ) = 4. 1. f ( t ) = 5sin 1 3 t 3cos 2 3 t + 3 2. f ( t ) = 3sin 1 3 t + cos 2 3 t + 2 Miller, Kierste – Homework 1 – Due: Jan 24 2007, 3:00 am – Inst: Gary Berg 2 3. f ( t ) = 3sin 1 3 t + 3cos 2 3 t + 1 correct 4. f ( t ) = 5cos 1 3 t 3sin 2 3 t + 3 5. f ( t ) = 3cos 1 3 t + sin 2 3 t + 2 6. f ( t ) = 3cos 1 3 t + 3sin 2 3 t + 1 Explanation: The function f must have the form f ( t ) = 3sin 1 3 t + 3cos 2 3 t + C where the constant C is determined by the condition f ‡ π 2 · = 3sin π 6 + 3cos π 3 + C = 4 . But by known trig values sin π 6 = cos π 3 = 1 2 , so 3 + C = 4. Consequently, f ( t ) = 3sin 1 3 t + 3cos 2 3 t + 1 . keywords: antiderivatives, trigonometric functions 004 (part 1 of 1) 10 points Find the value of f ( 1) when f 00 ( t ) = 9 t 6 and f (1) = 4 , f (1) = 5 . Correct answer: 9 . Explanation: The most general antiderivative of f 00 has the form f ( t ) = 9 2 t 2 6 t + C where C is an arbitrary constant. But if f (1) = 4, then f (1) = 9 2 6 + C = 4 , i . e . C = 11 2 . Thus f ( t ) = 9 2 t 2 6 t + 11 2 , from which it follows that f ( t ) = 3 2 t 3 3 t 2 + 11 2 t + D , where the constant D is determined by the condition f (1) = 3 2 3+ 11 2 + D = 5 , i . e . D = 1 ....
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This note was uploaded on 05/08/2008 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas.
 Spring '08
 Cepparo

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