Explanation Given: The vector -valued functions r ( t ) = 〈 2 t + 1 , sin π t , 4 〉 ; t 0 = 4 . Definition used: Let r ( t ) = 〈 f ( t ) , g ( t ) , h ( t ) 〉 be the vector-valued function. Its derivative can be given by r ′ ( t ) = 〈 f ′ ( t ) , g ′ ( t ) , h ′ ( t ) 〉 where f , g and h are differentiable functions. Derivative rule: (1) Constant rule: d d t ( c ) = 0 (2) d d t ( cos a t ) = − a sin a t [ a is constant ] (3) d d t ( sin a t ) = a cos a t [ a is constant ] Calculation: Compute the derivative of the vector-valued function r ( t ) = 〈 2 t + 1 , sin π t , 4 〉 ; t 0 = 4 as follows, r ′ ( t ) = d d t ( r ( t ) ) = d d t 〈 2 t + 1 , sin π t , 4 〉 = 〈 d d t ( 2 t + 1 ) ( 1 2 ) , d d t ( sin π t ) , d d t ( 4 ) 〉 Apply the constant rule and derivative rules (2) and (3), r ′ ( t ) = 〈 ( ( 1 2 ) ( 2 ) ( 2 t + 1 ) ( 1 2 − 1 ) ) , ( π ) ( cos π t ) , ( 0 ) 〉 = 〈 ( ( 2 t + 1 ) ( − 1 2 ) ) , π cos π t , 0 〉 = 〈 ( 1 2 t + 1 ) , π cos π <

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ISBN: 9780135873311

Author: Briggs

Publisher: PEARSON CO

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Chapter 14.2, Problem 82E

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To find: An equation of the line parallel to the curve at t=t0 .

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Chapter 14.2, Problem 81E

Chapter 14.2, Problem 83E

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Chapter 14 Solutions

EP CALCULUS:EARLY TRANS.-MYLABMATH ACC.

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An equation of the line parallel to the curve at t = t 0 .

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To find: An equation of the line parallel to the curve at t=t0 .

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