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HW02-solutions - nunez(djn358 HW02 Radin(54915 This...

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nunez (djn358) – HW02 – Radin – (54915) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind all Functions g such that g ( x ) = 2 x 2 + 5 x + 1 x . 1. g ( x ) = 2 x ( 2 x 2 + 5 x 1 ) + C 2. g ( x ) = 2 x p 2 5 x 2 + 5 3 x + 1 P + C cor- rect 3. g ( x ) = x ( 2 x 2 + 5 x + 1 ) + C 4. g ( x ) = 2 x p 2 5 x 2 + 5 3 x 1 P + C 5. g ( x ) = 2 x ( 2 x 2 + 5 x + 1 ) + C 6. g ( x ) = x p 2 5 x 2 + 5 3 x + 1 P + C Explanation: AFter division g ( x ) = 2 x 3 / 2 + 5 x 1 / 2 + x 1 / 2 , so we can now fnd an antiderivative oF each term separately. But d dx p ax r r P = ax r 1 For all a and all r n = 0. Thus 4 5 x 5 / 2 + 10 3 x 3 / 2 + 2 x 1 / 2 = 2 x p 2 5 x 2 + 5 3 x + 1 P is an antiderivative oF g . Consequently, g ( x ) = 2 x p 2 5 x 2 + 5 3 x + 1 P + C with C an arbitrary constant. 002 10.0 points Consider the Following Functions: ( A ) F 1 ( x ) = cos 2 x 4 , ( B ) F 2 ( x ) = sin 2 x 2 , ( C ) F 3 ( x ) = cos 2 x 2 . Which are anti-derivatives oF f ( x ) = sin x cos x ? 1. F 3 only 2. F 2 and F 3 only 3. F 1 and F 3 only 4. F 2 only 5. F 1 only 6. all oF them 7. none oF them 8. F 1 and F 2 only correct 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 . But d dx sin x = cos x, d dx cos x = sin x . Consequently, by the Chain Rule, ( A ) Anti-derivative. ( B ) Anti-derivative. ( C ) Not anti-derivative.
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nunez (djn358) – HW02 – Radin – (54915) 2 003 10.0 points Find f ( t ) when f ( t ) = cos 1 3 t 2 sin 2 3 t and f ( π 2 ) = 2. 1. f ( t ) = 3 sin 1 3 t + cos 2 3 t 2. f ( t ) = 5 cos 1 3 t 3 sin 2 3 t + 1 3. f ( t ) = 3 sin 1 3 t + 3 cos 2 3 t 1 correct 4. f ( t ) = 3 cos 1 3 t + sin 2 3 t 5. f ( t ) = 5 sin 1 3 t 3 cos 2 3 t + 1 6. f ( t ) = 3 cos 1 3 t + 3 sin 2 3 t 1 Explanation: The function f must have the form f ( t ) = 3 sin 1 3 t + 3 cos 2 3 t + C where the constant C is determined by the condition f p π 2 P = 3 sin π 6 + 3 cos π 3 + C = 2 . But by known trig values sin π 6 = cos π 3 = 1 2 , so 3 + C = 2. Consequently, f ( t ) = 3 sin 1 3 t + 3 cos 2 3 t 1 . 004 10.0 points Find f ( x ) on ( π 2 , π 2 ) when f ( x ) = 7 + 5 tan 2 x and f (0) = 4. 1. f ( x ) = 9 2 x 5 sec x 2. f ( x ) = 4 + 2 x + 5 tan x correct 3. f ( x ) = 1 + 7 x + 5 sec x 4. f ( x ) = 4 + 2 x + 5 tan 2 x 5. f ( x ) = 4 2 x 5 tan x 6. f ( x ) = 1 + 7 x + 5 sec 2 x Explanation: The properties d dx (tan x ) = sec 2 x, tan 2 x = sec 2 x 1 , suggest that we rewrite f ( x ) as f ( x ) = 2 + 5 sec 2 x, for then the most general anti-derivative of f is f ( x ) = 2 x + 5 tan x + C, with C an arbitrary constant. But if f (0) = 4, then f (0) = C = 4 .
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HW02-solutions - nunez(djn358 HW02 Radin(54915 This...

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