1 v t 7 t 1 i 4 t k 2 v t i 7 j 4 t k 3 v t 7 i j 4 t

This preview shows page 7 - 9 out of 9 pages.

1. v ( t ) = (7 t 1) i + 4 t k 2. v ( t ) = i 7 j + 4 t k 3. v ( t ) = 7 i j + 4 t kcorrect 4. v ( t ) = 7 t i t j + 2 t 2 k 5. v ( t ) = 7 i j + 4 k Explanation: The velocity vector, v ( t ), of the particle satisfies the equation d v dt = a ( t ) = 4 k . So v ( t ) = 4 t k + C where C is a constant vector such that v (0) = C = 7 i j . Thus v ( t ) = 7 i j + 4 t k . keywords: Stewart5e, vector function, veloc- ity vector, acceleration, initial velocity 014 10.0points Determine the position vector, r (1), at time t = 1 for a particle moving in 3-space and having acceleration a ( t ) = 2 k when its initial velocity and position are given by v (0) = i + j 4 k , r (0) = 3 i + 5 j respectively. 1. r (1) = 6 i + 4 j + 5 k 2. r (1) = 4 i 6 j 5 k 3. r (1) = 6 i + 4 j 5 k 4. r (1) = 4 i + 6 j 5 kcorrect 5. r (1) = 4 i 6 j + 5 k 6. r (1) = 6 i 4 j + 5 k Explanation: Since a ( t ) = d v dt = 2 k , we see that v ( t ) = 2 t k + C where C is a constant vector such that v (0) = C = i + j 4 k . Thus v ( t ) = d r dt = i + j (2 t + 4) k . But then r ( t ) = t i + t j ( t 2 + 4 t ) k + D
hill (cgh723) – HW07 – neitzke – (55325) 8 where D is a constant vector such that r (0) = D = 3 i + 5 j . Consequently, r ( t ) = ( t + 3) i + ( t + 5) j ( t 2 + 4 t ) k , and so at t = 1 r (1) = 4 i + 6 j 5 k . 015 10.0points Determine the minimum speed of a particle moving in 3-space with position function r ( t ) = ( t 2 , 5 t, t 2 20 t ) . 1. min speed = 17 units/sec 2. min speed = 16 units/sec 3. min speed = 15 units/sec correct 4. min speed = 18 units/sec 5. min speed = 19 units/sec Explanation: The velocity of the particle is given by r ( t ) = ( 2 t, 5 , 2 t 20 ) , so the particle has speed | r ( t ) | = radicalBig 4 t 2 + 25 + (2 t 20) 2 . This has a minimum when d | r ( t ) | dt = 8 t + 4(2 t 20) radicalbig 4 t 2 + 25 + (2 t 20) 2 = 0 , i.e. , when t = 5. Consequently, the particle has minimum when min speed = | r (5) | = 15 units/sec . 016 10.0points Find the tangential component a T of the acceleration vector of a particle moving in the plane with position function r ( t ) = (9 t 3 t 3 ) i + 9 t 2 j . 1. a T = 18 t + 18 2. a T = 18 t correct 3. a T = 18 t 4. a T = 9 t 5. a T = 9 9 t 2 Explanation: The tangential component of the accelera- tion vector of a particle moving in the plane with position function r ( t ) is given by a T = r ′′ ( t ) · T ( t ) = r ′′ ( t ) · r ( t ) bardbl r ( t ) bardbl . But for the given r ( t ), r ( t ) = (9 9 t 2 ) i + 18 t j , bardbl r ( t ) bardbl = radicalBig 9 2 ( t 2 + 1) 2 = 9( t 2 + 1) , while r ′′ ( t ) = 18 t i + 18 j . Consequently, T ( t ) = (9 9 t 2 ) i + 18 t j 9( t 2 + 1) = (1 t 2 ) i + 2 t j t 2 + 1 , while r ′′ ( t ) · T ( t ) = 18 t (1 t 2 ) + 2 t (18) . Thus a T = 18 t . 017 10.0points Find the normal component, a N , of the acceleration vector of a particle moving in 3-space with position function r ( t ) = 5 cos 5 t i + 5 sin 5 t j + 6 t k .
hill (cgh723) – HW07 – neitzke – (55325) 9 1. a N = 25 2. a N = 125 cos 5 t + 125 sin 5 t 3. a N = 125 correct 4. a N = 25 cos 5 t + 25 sin 5 t 5. a N = 5 Explanation: The normal component of the acceleration vector of a particle moving in 3-space with position function r ( t ) is given by a N = bardbl r ( t ) × r ′′ ( t ) bardbl bardbl r ( t ) bardbl . But, when r ( t ) = 5 cos 5 t i + 5 sin 5 t j + 6 t k , we see that r ( t ) = 25 sin 5 t i + 25 cos 5 t j + 6 k , while r ′′ ( t ) = 125 cos 5 t i 125 sin 5 t j . In this case r ( t ) × r ′′ ( t ) = 750 sin 5 t i 750 cos 5 t j + 3125 k . Thus bardbl r ( t ) × r ′′ ( t ) bardbl = radicalbig 750 2 sin 2 5 t + 750 2 cos 2 5 t + 3125 2 , while bardbl r ( t ) bardbl = radicalbig 625 sin 2 5 t + 625 cos 2 5 t + 36 = 661 .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture