Explanation Given information: A particle follows a path r 1 ( t ) on the helical curve with parametrization r ( θ ) = 〈 c o s θ , s i n θ , θ 〉 when it is at position r ( π 2 ) its speed is 3 m s − 1 and its speed is increasing at a rate of 1 2 m s − 1 .Note: Particle’s acceleration vector does not coincide with r " ( θ ) . Formula Used: a = a T T + a N N Where a T = ‖ v ' ( t ) ‖ and a N = ‖ v ( t ) ‖ . k ( t ) Also known, the tangent vector and the unit normal vector is defined by T ( t ) = v ( t ) ‖ v ( t ) ‖ and N ( t ) = T ' ( t ) ‖ T ' ( t ) ‖ . Thus, v ( t ) = T ( t ) ‖ v ( t ) ‖ . So, by the product rule; a ( t ) = d v d t = d ( T ( t ) ‖ v ( t ) ‖ ) d t a ( t ) = T ( t ) ‖ v ' ( t ) ‖ + ‖ v ( t ) ‖ T ' ( t ) We know T ' ( t ) = ‖ v ( t ) ‖ N ( t ) . k ( t ) k ( t ) is the curvature. k ( t ) = ‖ d T d t ‖ Calculation: A particle follows a path given below, r ( θ ) = 〈 c o s θ , s i n θ , θ 〉 r ' ( θ ) = 〈 − s i n θ , c o s θ , 1 〉 θ = π 2 ; r ( π 2 ) = 〈 0 , 1 , π 2 〉 r ' ( π 2 ) = 〈 − 1 , 0 , 1 〉 Now consider the acceleration vector; a = a T T + a N N Here the second term is complicated for calculation. So, we will consider only the first term. Hence a = a T T a = a T T a = ‖ v ' ( t ) ‖

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Author: Rogawski

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Chapter 14.5, Problem 51E

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To find: The acceleration vector.

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Chapter 14.5, Problem 50E

Chapter 14.5, Problem 52E

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Chapter 14 Solutions

CALCULUS (CLOTH)

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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The acceleration vector .

Solution Summary: The author explains that the acceleration vector does not coincide with r"(theta).

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To find: The acceleration vector.

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Chapter 14 Solutions