This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: kodeih (njk359) – HW02 – Henry – (54974) 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 Find all functions g such that g ′ ( x ) = 4 x 2 + 5 x + 3 √ x . 1. g ( x ) = 2 √ x ( 4 x 2 + 5 x − 3 ) + C 2. g ( x ) = √ x ( 4 x 2 + 5 x + 3 ) + C 3. g ( x ) = 2 √ x ( 4 x 2 + 5 x + 3 ) + C 4. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x − 3 parenrightbigg + C 5. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C cor rect 6. g ( x ) = √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 5 x 1 / 2 + 3 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 + 6 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x 2 , ( B ) F 2 ( x ) = cos 2 x 2 , ( C ) F 3 ( x ) = − cos 2 x 4 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. none of them 2. F 1 only 3. F 1 and F 2 only 4. F 2 only 5. F 2 and F 3 only 6. all of them 7. F 3 only 8. F 1 and F 3 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 ) Antiderivative. ( B ) Not antiderivative. ( C ) Antiderivative. kodeih (njk359) – HW02 – Henry – (54974) 2 003 10.0 points Find f ( t ) when f ′ ( t ) = 4 3 cos 1 3 t − 4 sin 2 3 t and f ( π 2 ) = 2. 1. f ( t ) = 4 cos 1 3 t + 6 sin 2 3 t − 3 2. f ( t ) = 8 cos 1 3 t − 6 sin 2 3 t + 1 3. f ( t ) = 4 cos 1 3 t + 2 sin 2 3 t − 1 4. f ( t ) = 4 sin 1 3 t + 2 cos 2 3 t − 1 5. f ( t ) = 8 sin 1 3 t − 6 cos 2 3 t + 1 6. f ( t ) = 4 sin 1 3 t + 6 cos 2 3 t − 3 correct Explanation: The function f must have the form f ( t ) = 4 sin 1 3 t + 6 cos 2 3 t + C where the constant C is determined by the condition f parenleftBig π 2 parenrightBig = 4 sin π 6 + 6 cos π 3 + C = 2 . But by known trig values sin π 6 = cos π 3 = 1 2 , so 5 + C = 2. Consequently, f ( t ) = 4 sin 1 3 t + 6 cos 2 3 t − 3 . 004 10.0 points Find f ( x ) on ( − π 2 , π 2 ) when f ′ ( x ) = 4 + 2 tan 2 x and f (0) = 4. 1. f ( x ) = 2 + 4 x + 2 sec x 2. f ( x ) = 4 − 2 x − 2 tan x 3. f ( x ) = 4 + 2 x + 2 tan 2 x 4. f ( x ) = 6 − 2 x − 2 sec x 5. f ( x ) = 4 + 2 x + 2 tan x correct 6. f ( x ) = 2 + 4 x + 2 sec 2 x Explanation: The properties d dx (tan x ) = sec 2 x, tan 2 x = sec 2 x − 1 , suggest that we rewrite...
View
Full
Document
This note was uploaded on 09/25/2010 for the course M 54980 taught by Professor Henry during the Fall '10 term at University of Texas.
 Fall '10
 HENRY

Click to edit the document details