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University of Texas M 408D Rusin HW 09

University of Texas M 408D Rusin HW 09 - byrne(cmb3744 HW...

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byrne (cmb3744) – HW 09 – rusin – (55565) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. I forgot to say in class: we are SKIP- PING all the material on curvature, torsion of curves, etc. You’re off the hook! Meanwhile, the new material we are starting (functions of two or more variables) will give you lots of stuff to calculate and a lot less stuff to draw. Read up on partial derivatives and ask the TAs for help on Monday as you dive into the second half of the assignment below. We’ll discuss this on Tuesday too, of course. Good luck! 001 10.0points Determine the position vector, r ( t ), of a particle having acceleration a ( t ) = 6 k when its initial velocity and position are given by v (0) = i + j 4 k , r (0) = 2 i + 3 j respectively. 1. r ( t ) = ( t + 2) i ( t 3) j (3 t 2 4 t ) k 2. r ( t ) = ( t 2) i + ( t + 3) j (3 t 2 + 4 t ) k 3. r ( t ) = ( t + 2) i ( t 3) j (3 t 2 + 4 t ) k 4. r ( t ) = ( t + 2) i + ( t + 3) j (3 t 2 4 t ) k 5. r ( t ) = ( t 2) i + ( t + 3) j (3 t 2 4 t ) k 6. r ( t ) = ( t + 2) i + ( t + 3) j (3 t 2 + 4 t ) k correct Explanation: Since a ( t ) = d v dt = 6 k , we see that v ( t ) = 6 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 (6 t + 4) k . But then r ( t ) = t i + t j (3 t 2 + 4 t ) k + D where D is a constant vector such that r (0) = D = 2 i + 3 j . Consequently, r ( t ) = ( t + 2) i + ( t + 3) j (3 t 2 + 4 t ) k . 002 10.0points Determine the minimum speed of a particle moving in 3-space with position function r ( t ) = ( t 2 , 6 t, t 2 24 t ) . 1. min speed = 19 units/sec 2. min speed = 15 units/sec 3. min speed = 16 units/sec 4. min speed = 17 units/sec 5. min speed = 18 units/sec correct Explanation: The velocity of the particle is given by r ( t ) = ( 2 t, 6 , 2 t 24 ) , so the particle has speed | r ( t ) | = radicalBig 4 t 2 + 36 + (2 t 24) 2 . This has a minimum when d | r ( t ) | dt = 8 t + 4(2 t 24) radicalbig 4 t 2 + 36 + (2 t 24) 2 = 0 ,
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byrne (cmb3744) – HW 09 – rusin – (55565) 2 i.e. , when t = 6. Consequently, the particle has minimum when min speed = | r (6) | = 18 units/sec . 003 10.0points A ball is thrown at an angle of 42 to the ground. If the ball lands 111 meters away, what was the initial speed of the ball? ( Assume g = 9 . 8 meters/sec.) 1. v 0 33 . 1 m / s correct 2. v 0 34 . 7 m / s 3. v 0 33 . 4 m / s 4. v 0 33 . 7 m / s 5. v 0 34 . 0 m / s Explanation: We can assume that the snowball is moving in the xy -plane and that it is thrown from the origin. Then its position function is given by r ( t ) = 1 2 gt 2 j + t v 0 , where g is the acceleration due to gravity and v 0 is the initial velocity. But the ball is thrown at an elevation of 42 and lands 111 meters away. Thus we have v 0 = | v 0 | ((cos 42) i + (sin 42) j ) , so the equation becomes r ( t ) = 1 2 gt 2 j + t | v 0 | ((cos 42) i + (sin 42) j ) = t | v 0 | (cos 42) i + ( t | v 0 | sin 42 1 2 gt 2 ) j . Now, the ball hits the ground when t | v 0 | sin 42 1 2 gt 2 = 0 . i.e. , after t = 2( | v 0 | sin 42) /g seconds, and it will then be a distance of | v 0 | 2 sin 84 g .
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