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 ) = 〈 1 3 t 3 , 1 − 3 t 〉 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 ) = 〈 1 3 t 3 , 1 − 3 t 〉 ----------------(4) Therefore, velocity, v ( t ) = r ′ ( t ) = 〈 t 2 , − 3 〉 ------------(5) and acceleration, a ( t ) = v ′ ( t ) = r ″ ( t ) = 〈 2 t , 0 〉 ---------(6) also, ∥ v ∥ = t 4 + 9 --------------------------------------(7) Using equations (2), (5), (6) and (7) we determine the decomposition of a ( t ) into tangential componen...

Applied Calculus (with Infotrac) 3rd Edition By Waner, Stefan; Costenoble, Steven Published By Brooks Cole Hardcover

3rd Edition

ISBN: 9781319048518

Author: Colin Adams Jon Rogawski

Publisher: Macmillan Higher Education

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Chapter 13.5, Problem 37E

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

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Chapter 13.5, Problem 36E

Chapter 13.5, Problem 38E

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Chapter 13 Solutions

Applied Calculus (with Infotrac) 3rd Edition By Waner, Stefan; Costenoble, Steven Published By Brooks Cole Hardcover

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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

Decomposition of acceleration vector a(t) of r(t)=〈13t3,1−3t〉 into tangential and normal components at the point t=−1.

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Chapter 13 Solutions

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