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Unformatted text preview: MATH 2401A1-B1 Name: SOLUTION TEST1 (01/25/02) Please read questions carefully and show all your work. You are allowed 50 minutes on this exam. 1. (20 pts) Find d dt ( f · g ), d dt ( f × g ), d dt ( u f ), and d dt ( f ◦ u ) for the functions f ( t ) = ln t i + t j- ( t 2 + 1) k , g ( t ) = 1 t i + t 3 j- t k , u ( t ) = t 2 . Solution Since f · g = ln t t + t 4 + t ( t 2 + 1) , we have d dt ( f · g ) = 1- ln t t 2 + 4 t 3 + 3 t 2 + 1 . Since f × g = fl fl fl fl fl fl fl i j k ln t t- ( t 2 + 1) 1 t t 3- t fl fl fl fl fl fl fl = ( t 5 + t 3- t 2 ) i + ( t ln t- t- 1 t ) j + ( t 3 ln t- 1) k , we have d dt ( f × g ) = (5 t 4 + 3 t 2- 2 t ) i + (ln t + 1 t 2 ) j + (3 t 2 ln t + t 2 ) k . By the differentiation formulas we have d dt ( u f ) = u d dt f + u ( t ) f = t 2 ( 1 t i + j- 2 t k ) + 2 t (ln t i + t j- ( t 2 + 1) k ) = ( t + 2 t ln t ) i + 3 t 2 j- 2 t (2 t 2 + 1) k ; and d dt ( f ◦ u ) = f ( u ( t )) u ( t ) = ( 1 t 2 i + j- 2 t 2 k )2 t = 2 t i + 2 t j- 4 t 3 k ....
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