HW08-solutions

HW08-solutions - qazi(kaq87 – HW08 – berg –(55290 1...

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Unformatted text preview: qazi (kaq87) – HW08 – berg – (55290) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the velocity of a particle with the given position function r ( t ) = 9 e 4 t i + 3 e − 2 t j . 1. v ( t ) = 36 e t i- 6 e − t j 2. v ( t ) = 9 e 4 t i- 3 e − 2 t j 3. v ( t ) = 13 e 4 t i + 5 e − 2 t j 4. v ( t ) = 13 e 4 t i- 5 e − 2 t j 5. v ( t ) = 36 e 4 t i- 6 e − 2 t j correct Explanation: The velocity of a particle with position function r ( t ) = x ( t ) i + y ( t ) j is given by r ′ ( t ) = x ′ ( t ) i + y ′ ( t ) j . Thus, since r ′ ( t ) = 36 e 4 t i- 6 e − 2 t j , we have that v ( t ) = 36 e 4 t i- 6 e − 2 t j . keywords: 002 10.0 points Find the acceleration of a particle having position function r ( t ) = 5 sin t i + 6 t j + 6 cos t k in 3-space. 1. a ( t ) = 5 cos t i- 6 sin t k 2. a ( t ) =- 5 cos t i + 6 sin t k 3. a ( t ) = 5 sin t i- 6 cos t k 4. a ( t ) =- 5 sin t i + 6 cos t k 5. a ( t ) =- 5 sin t i- 6 cos t k correct Explanation: The acceleration of a particle having posi- tion function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k is given by r ′′ ( t ) = x ′′ ( t ) i + y ′′ ( t ) j + z ′′ ( t ) k . Thus, since r ′ ( t ) = 5 cos t i + 6 j- 6 sin t k we have that a ( t ) = r ′′ ( t ) =- 5 sin t i- 6 cos t k . keywords: Stewart5e, vector function, posi- tion function, acceleration, curve in 3-space 003 10.0 points A particle moving in 3-space has position function r ( t ) = (5 √ 2) t i + e 5 t j + e − 5 t k . What is its speed at time t ? 1. speed = e 5 t + e − 5 t 2. speed = 5( e 5 t + e − 5 t ) correct 3. speed = radicalbig 10 + e 10 t + e − 10 t 4. speed = radicalbig 10 + 5 e t + 5 e − t 5. speed = 5( √ 2 + e 5 t- e − 5 t ) Explanation: qazi (kaq87) – HW08 – berg – (55290) 2 A particle with position function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k , has velocity r ′ ( t ) = x ′ ( t ) i + y ′ ( t ) j + z ′ ( t ) k , and speed = | r ′ ( t ) | . Now when r ( t ) = (5 √ 2) t i + e 5 t j + e − 5 t k , we see that r ′ ( t ) = (5 √ 2) i + 5 e 5 t j- 5 e − 5 t k , while | r ′ ( t ) | 2 = 50 + 25 e 5 t + 25 e − 5 t = 25( e 5 t + e − 5 t ) 2 ....
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HW08-solutions - qazi(kaq87 – HW08 – berg –(55290 1...

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