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blueexam2sol.07f

# blueexam2sol.07f - blue MATH 32A Exam 2 LAST NAME FIRST...

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blue MATH 32A Exam 2 November 29, 2007 LAST NAME FIRST NAME ID NO. Your TA: To receive credit, you must write your answer in the space provided . DO NOT WRITE BELOW THIS LINE 1 (25 pts) 5 (25 pts) 2 (25 pts) 6 (25 pts) 3 (25 pts) 7 (25 pts) 4 (25 pts) 8 (25 pts) TOTAL

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2 PROBLEM 1 (25 Points) A miniature rocket leaves the origin at time t = 0 with initial velocity (units are meters and seconds) v 0 = ( 40 , 40 ) m/s 2 x y The rocket’s acceleration vector a ( t ) has a horizontal com- ponent ( 8 , 0 ) supplied by the rocket engines and a verti- cal component due to gravity of magnitude g m/ s 2 in the downward direction. Use the approximate value g = 10. (A) Find the rocket’s position r ( t ) at time t . Answer : (B) How far from the origin does the rocket land? Answer :
3 Solution: (A) We have r ′′ ( t ) = ( 8 , 10 ) . Therefore r ( t ) = t ( 8 , 10 ) + ( 40 , 40 ) We have r (0) = ( 0 , 0 ) since the rocket begins at the origin. Thus r ( t ) = 1 2 t 2 ( 8 , 10 ) + t ( 40 , 40 ) (B) The rocket’s y -coordinate is 5 t 2 +40 t = 5 t ( t 8), so the rocket lands (crashes) at time t = 8 s. The x -coordinate is x = 4 t 2 + 40 t , so the distance from the origin when the rocket lands is x (8) = 4(8 2 ) + 40(8) = 576 m

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4 PROBLEM 2 (25 Points) Calculate the curvature of r ( t ) = ( t, t 1 , t 2 ) at t = 1. Answer : Solution: Use the formula κ = || r ′′ ( t ) × r ( t ) || || r ( t ) || 3 We have r ( t ) = ( 1 , t 2 , 2 t ) r ′′ ( t ) = ( 0 , 2 t 3 , 2 ) r ′′ ( t ) × r ( t ) = 6 t 2 i + 2 j 2 t 3 k At t = 1 we have r = ( 1 , 1 , 2 ) r ′′ × r = 6 i + 2 j 2 k and κ = || 6
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blueexam2sol.07f - blue MATH 32A Exam 2 LAST NAME FIRST...

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