Explanation Given: The trajectory of moving object is r ( t ) = 〈 20 cos t , 20 sin t , 30 t 〉 . Formula used: Suppose r ( t ) is a smooth parameterized curve. The acceleration vector in terms of tangential and normal components a = a N N + a T T where the normal component a N = a ⋅ N and the tangential component a T = a ⋅ T . The unit tangent vector T ( t ) is v ( t ) | v ( t ) | , where v ( t ) = r ′ ( t ) . The principal unit normal vector N ( t ) = T ′ ( t ) | T ′ ( t ) | . Calculation: Differentiate r ( t ) to compute v ( t ) . v ( t ) = 〈 ( 20 cos t ) ′ , ( 20 sin t ) ′ , ( 30 t ) ′ 〉 = 〈 20 ( − sin t ) , 20 cos t , 30 〉 = 〈 − 20 sin t , 20 cos t , 30 〉 Use magnitude formula to obtain the value of | v ( t ) | . | v ( t ) | = | 〈 − 20 sin t , 20 cos t , 30 〉 | = ( − 20 sin t ) 2 + ( 20 cos t ) 2 + ( 30 ) 2 = 400 sin 2 t + 400 cos 2 t + 900 = 400 ( sin 2 t + cos 2 t ) + 900 On further simplification, | v ( t ) | = 400 ( 1 ) + 900 = 1300 = 10 13 Use unit tangent vector formula to compute the component of T ( t ) . T ( t ) = 〈 − 20 sin t , 20 cos t , 30 〉 10 13 = 〈 − 20 sin t 10 13 , 20 cos t 10 13 , 30 10 13 〉 = 〈 − 2 sin t 13 , 2 cos t 13 , 3 13 〉 = 1 13 〈 − 2 sin t , 2 cos t , 3 〉 Differentiate T ( t ) to compute the component of T ′ ( t ) . T ′ ( t ) = 1 13 〈 ( − 2 sin t ) ′ , ( 2 cos t ) ′ , 3 ′ 〉 = 1 13 〈 − 2 cos t , 2 ( − sin t ) , 0 〉 = 1 13 〈 − 2 cos t , − 2 sin t , 0 〉 Use magnitude formula to compute the value of | T ′ ( t ) |

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ISBN: 9780134856971

Author: Briggs

Publisher: PEARSON

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Chapter 14.5, Problem 40E

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To find: The tangential and normal components of the acceleration for the trajectory r(t)=〈20cost,20sint,30t〉 .

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Chapter 14.5, Problem 39E

Chapter 14.5, Problem 41E

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The tangential and normal components of the acceleration for the trajectory r ( t ) = 〈 20 cos t , 20 sin t , 30 t 〉 .

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To find: The tangential and normal components of the acceleration for the trajectory r(t)=〈20cost,20sint,30t〉 .

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