Explanation Acceleration vector a ( t ) can be decompose into tangential component (tangent to the path of motion) and normal component (perpendicular to the path of motion) as a ( t ) = a T T ( t ) + a N N ( t ) ------------------------(1) where unit tangent vector, T ( t ) = r ′ ( t ) ∥ r ′ ( t ) ∥ = v ∥ v ∥ and unit normal vector, N ( t ) = T ′ ( t ) ∥ T ′ ( t ) ∥ a T = a · T = a · v ∥ v ∥ a N = a · N = ∥ a × v ∥ ∥ v ∥ Given: Position vector r ( t ) = 〈 t , 1 2 t 2 , 1 6 t 3 〉 Formula used: Decomposition of a ( t ) into tangential component T ( t ) is, a T = a · T = a · v ∥ v ∥ ----------------(2) Decomposition of a ( t ) into normal component N ( t ) is, a N = a · N = ∥ a × v ∥ ∥ v ∥ ---------------(3) Calculation: Since position vector r ( t ) = 〈 t , 1 2 t 2 , 1 6 t 3 〉 ----------------(4) Therefore, velocity, v ( t ) = r ′ ( t ) = 〈 1 , t , 1 2 t 2 〉 ------------(5) and acceleration, a ( t ) = v ′ ( t ) = r ″ ( t ) = 〈 0 , 1 , t 〉 ----------(6) also, ∥ v ∥ = 1 + t 2 + 1 4 t 4 ----------------------------------(7) Using equations (2), (5), (6) and (7) we determine the decomposition of a ( t ) into tang...

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Author: Rogawski

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Chapter 14.5, Problem 39E

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Decomposition of acceleration vector $a (t)$ of $r (t) = 〈 t, \frac{1}{2} t^{2}, \frac{1}{6} t^{3} 〉$ into tangential and normal components at the point $t = 4$ .

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Chapter 14.5, Problem 38E

Chapter 14.5, Problem 40E

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Chapter 14 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

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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Decomposition of acceleration vector a ( t ) of r ( t ) = 〈 t , 1 2 t 2 , 1 6 t 3 〉 into tangential and normal components at the point t = 4 .

Solution Summary: The author explains the decomposition of acceleration vector a (t) into tangential and normal components at the point t = 4.

Decomposition of acceleration vector $a (t)$ of $r (t) = 〈 t, \frac{1}{2} t^{2}, \frac{1}{6} t^{3} 〉$ into tangential and normal components at the point $t = 4$ .

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Chapter 14 Solutions

Decomposition of acceleration vector a ( t ) of r ( t ) = 〈 t , 1 2 t 2 , 1 6 t 3 〉 into tangential and normal components at the point t = 4 .

Solution Summary: The author explains the decomposition of acceleration vector a (t) into tangential and normal components at the point t = 4.

Decomposition of acceleration vector a(t) of r(t)=〈t,12t2,16t3〉 into tangential and normal components at the point t=4.

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Chapter 14 Solutions

Decomposition of acceleration vector $a (t)$ of $r (t) = 〈 t, \frac{1}{2} t^{2}, \frac{1}{6} t^{3} 〉$ into tangential and normal components at the point $t = 4$ .