Exam1_practice_solutions

Exam1_practice_solutions - MATH 2401 Sections F4& F5...

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Unformatted text preview: MATH 2401 September 16, 2009 Sections F4 & F5 PRACTICE EXAM 1 Name: gtID # : Note: There are ve problems in this exam, each worth 20 points. Write out your solutions neatly and explain your work. Calculators are not allowed. Basic formulas: Unit tangent and principal normal: T ( t ) = r ( t ) k r ( t ) k , N ( t ) = T ( t ) k T ( t ) k Arc length: L ( C ) = Z b a k r ( t ) k dt Curvature: κ = d T ds = k d T /dt k ds/dt = k v × a k k v k 3 κ = | x ( t ) y 00 ( t )- y ( t ) x 00 ( t ) | [[ x ( t )] 2 + [ y ( t )] 2 )] 3 / 2 Components of acceleration: a = a T T + a N N a T = d 2 s dt 2 = T · a = v · a k v k , a N = κ ds dt 2 k T × a k = k v × a k k v k 1 Problem 1 (20 points). Determine the unit tangent, principal normal and the osculating plane of the space curve r ( t ) = 2 t i- cos 2 t j + sin 2 t k at t = π/ 8 . Solution: We have r ( t ) = 2 i + 2 sin 2 t j + 2 cos 2 t k = 2 ( i + sin 2 t j + cos 2 t k ) , and thus k r ( t ) k = 2 p 1 + sin 2 2 t + cos 2 2 t = 2 √ 2 , so the unit tangent vector is T = T ( π/ 8) = r ( π/ 8) k r ( π/ 8) k = 1 √ 2 ( i + sin 2 t j + cos 2 t k ) t = π/ 8 = 1 √ 2 i + 1 √ 2 j + 1 √ 2 k . Furthermore, we have T ( t ) = 1 √ 2 (2 cos 2 t j- 2 sin 2 t k ) = √ 2 (cos 2 t j- sin 2 t k ) , which gives the principal normal N...
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