Explanation Given data: Vector function r ( t ) = 〈 t 2 , 2 3 t 3 , t 〉 and point ( 1 , 2 3 , 1 ) . Formula used: Write the expression for tangent vector of a vector function r ( t ) ( T ( t ) ) . T ( t ) = r ′ ( t ) | r ′ ( t ) | (1) Here, r ′ ( t ) is first derivative of function r ( t ) . Write the expression for normal vector of vector function r ( t ) ( N ( t ) ) . N ( t ) = T ′ ( t ) | T ′ ( t ) | (2) Here, T ′ ( t ) is first derivative of tangent function. Write the expression for binormal vector of vector function r ( t ) ( B ( t ) ) . B ( t ) = T ( t ) × N ( t ) (3) Consider the two three-dimensional vector functions such as u ( t ) = 〈 u 1 ( t ) , u 2 ( t ) , u 3 ( t ) 〉 and v ( t ) = 〈 v 1 ( t ) , v 2 ( t ) , v 3 ( t ) 〉 . Cross product of vectors: Write the expression for cross product of vectors u ( t ) and v ( t ) ( u ( t ) × v ( t ) ) . u ( t ) × v ( t ) = | i j k u 1 ( t ) u 2 ( t ) u 3 ( t ) v 1 ( t ) v 2 ( t ) v 3 ( t ) | = 〈 [ ( u 2 ( t ) v 3 ( t ) − v 2 ( t ) u 3 ( t ) ) ] , [ − ( u 1 ( t ) v 3 ( t ) − v 1 ( t ) u 3 ( t ) ) ] , [ ( u 1 ( t ) v 2 ( t ) − v 1 ( t ) u 2 ( t ) ) ] 〉 Write the expression for magnitude of vector a ( | a | ) . | a | = a 1 2 + a 2 2 + a 3 2 Here, a 1 , a 2 and a 3 are the x , y , and z -coordinates of vector respectively. Write the vector function. r ( t ) = 〈 t 2 , 2 3 t 3 , t 〉 (4) Equate the components of r ( t ) with point ( 1 , 2 3 , 1 ) . t 2 = 1 t = 1 t = 1 2 3 t 3 = 2 3 t 3 = ( 2 3 ) ( 2 3 ) t 3 = 1 t = 1 3 t = 1 t = 1 Hence, the value of t is 1. Apply differentiation with respect to t on both sides of equation (4). r ′ ( t ) = d d t 〈 t 2 , 2 3 t 3 , t 〉 = 〈 d d t ( t 2 ) , d d t ( 2 3 t 3 ) , d d t ( t ) 〉 = 〈 2 t , 2 3 ( 3 t 2 ) , 1 〉 { ∵ d d x ( x ) = 1 , d d x ( x 2 ) = 2 x , d d x ( x n ) = n x n − 1 } = 〈 2 t , 2 t 2 , 1 〉 Find the value of | r ′ ( t ) | . | r ′ ( t ) | = ( 2 t ) 2 + ( 2 t 2 ) 2 + 1 2 = 4 t 2 + ( 2 t 2 ) 2 + 1 2 = ( 2 t 2 ) 2 + 1 2 + 2 ( 2 t 2 ) ( 1 ) = ( 2 t 2 + 1 ) 2 { ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b } | r ′ ( t ) | = 2 t 2 + 1 Substitute 〈 2 t , 2 t 2 , 1 〉 for r ′ ( t ) and 2 t 2 + 1 for | r ′ ( t ) | in equation (1), T ( t ) = 〈 2 t , 2 t 2 , 1 〉 2 t 2 + 1 = 1 2 t 2 + 1 〈 2 t , 2 t 2 , 1 〉 T ( t ) = 〈 2 t 2 t 2 + 1 , 2 t 2 2 t 2 + 1 , 1 2 t 2 + 1 〉 (5) Substitute 1 for t , T ( 1 ) = 〈 2 ( 1 ) 2 ( 1 ) 2 + 1 , 2 ( 1 ) 2 2 ( 1 ) 2 + 1 , 1 2 ( 1 ) 2 + 1 〉 = 〈 2 2 + 1 , 2 2 + 1 , 1 2 + 1 〉 = 〈 2 3 , 2 3 , 1 3 〉 Apply differentiation with respect to t on both sides of equation (5). T ′ ( t ) = d d t 〈 2 t 2 t 2 + 1 , 2 t 2 2 t 2 + 1 , 1 2 t 2 + 1 〉 = 〈 d d t ( 2 t 2 t 2 + 1 ) , d d t ( 2 t 2 2 t 2 + 1 ) , d d t ( 1 2 t 2 + 1 ) 〉 = 〈 ( ( d d t ( 2 t ) ) ( 2 t 2 + 1 ) − 2 t ( d d t ( 2 t 2 + 1 ) ) ( 2 t 2 + 1 ) 2 ) , ( ( d d t ( 2 t 2 ) ) ( 2 t 2 + 1 ) − 2 t 2 ( d d t ( 2 t 2 + 1 ) ) ( 2 t 2 + 1 ) 2 ) , ( ( − 1 ( 2 t 2 + 1 ) 2 ) ( 4 t ) ) 〉 { ∵ d d x ( u v ) = u ′ v − u v ′ v 2 , d d x ( 1 x ) = − 1 x 2 , d d x ( x 2 ) = 2 x } = 〈 ( 2 ( 1 ) ( 2 t 2 + 1 ) − 2 t ( 2 ( 2 t ) + 0 ) ( 2 t 2 + 1 ) 2 ) , ( ( 2 ( 2 t ) ) ( 2 t 2 + 1 ) − 2 t 2 ( 2 ( 2 t ) + 0 ) ( 2 t 2 + 1 ) 2 ) , − 4 t ( 2 t 2 + 1 ) 2 〉 { ∵ d d x ( x ) = 1 } T ′ ( t ) = 〈 4 t 2 + 2 − 8 t 2 ( 2 t 2 + 1 ) 2 , 8 t 3 + 4 t − 8 t 3 ( 2 t 2 + 1 ) 2 , − 4 t ( 2 t 2 + 1 ) 2 〉 = 〈 2 − 4 t 2 ( 2 t 2 + 1 ) 2 , 4 t ( 2 t 2 + 1 ) 2 , − 4 t ( 2 t 2 + 1 ) 2 〉 Find the value of | T ′ ( t ) |

9th Edition

ISBN: 9780357598511

Author: James Stewart; Daniel K. Clegg; Saleem Watson

Publisher: Cengage Learning US

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Chapter 13.3, Problem 51E

To determine

To find: The vectors T, N, and B of vector function r(t)=〈t2,23t3,t〉 at point (1,23,1) .

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Chapter 13.3, Problem 50E

Chapter 13.3, Problem 52E

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Chapter 13 Solutions

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The vectors T , N , and B of vector function r ( t ) = 〈 t 2 , 2 3 t 3 , t 〉 at point ( 1 , 2 3 , 1 ) .

To find: The vectors T, N, and B of vector function r(t)=〈t2,23t3,t〉 at point (1,23,1) .

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Chapter 13 Solutions