Explanation Given information: r ( t ) = ( e t cos t ) i + ( e t sin t ) j + 2 e t k (1) Formula used: Write the general expression to calculate the velocity. v = d r d t (2) Here, d r d t is the rate of change in displacement with respect to time. Write the general expression to calculate the acceleration. a = d v d t (3) Here, d v d t is the rate of change in velocity with respect to time. Write the general expression for the acceleration vector. a = a T T + a N N (4) Take magnitude on both sides of the equation (4). | a | = a T 2 + a N 2 (5) Write the general expression to calculate the tangential scalar component of acceleration. a T = d d t | v | (6) Calculation: Substitute equation (1) in equation (2). v = d d t ( ( e t cos t ) i + ( e t sin t ) j + 2 e t k ) v = ( e t cos t − e t sin t ) i + ( e t sin t + e t cos t ) j + 2 e t k (7) Take magnitude on both sides of the above equation. | v | = ( e t cos t − e t sin t ) 2 + ( e t sin t + e t cos t ) 2 + ( 2 e t ) 2 = e 2 t cos 2 t + e 2 t sin 2 t − 2 e 2 t cos t sin t + e 2 t cos 2 t + e 2 t sin 2 t + 2 e 2 t cos t sin t + 2 e 2 t = 2 e 2 t cos 2 t + 2 e 2 t sin 2 t + 2 e 2 t = 2 e 2 t ( cos 2 t + sin 2 t ) + 2 e 2 t | v | = 2 e 2 t + 2 e 2 t { ∵ sin 2 θ + cos 2 θ = 1 } = 4 e 2 t | v | = 2 e t (8) Substitute equation (8) in equation (6). a T = d d t ( 2 e t ) = 2 e t At t = 0 , the tangential scalar component of acceleration a T is, a T ( 0 ) = 2 e 0 = 2 { ∵ e 0 = 1 } Substitute equation (7) in equation (3). a = d d t ( ( e t cos t − e t sin t ) i + ( e t sin t + e t cos t ) j + 2 e t k ) = ( e t cos t − e t sin t − e t sin t − e t cos t ) i + ( e t sin t + e t cos t + e t cos t − e t sin t ) j + 2 e t k = ( − 2 e t sin t ) i + ( 2 e t cos t ) j + 2 e t k Take magnitude on both sides of the above equation

Thomas' Calculus: Early Transcendentals (14th Edition)

14th Edition

ISBN: 9780134439020

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 13.5, Problem 6E

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Find the acceleration vector a in the form a=aTT+aNN at the value of t without finding T and N for the given vector r(t)=(etcost)i+(etsint)j+2etk, t=0.

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Chapter 13.5, Problem 5E

Chapter 13.5, Problem 7E

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Thomas' Calculus: Early Transcendentals (14th Edition)

Ch. 13.1 - In Exercises 1–4, find the given limits. 1. Ch. 13.1 - In Exercises 1–4, find the given limits. 2. Ch. 13.1 - In Exercises 1–4, find the given limits. 3. Ch. 13.1 - In Exercises 1–4, find the given limits. 4. Ch. 13.1 - Motion in the Plane In Exercises 5–8, r(t) is the...Ch. 13.1 - Motion in the Plane In Exercises 5–8, r(t) is the...Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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Find the acceleration vector a in the form a = a T T + a N N at the value of t without finding T and N for the given vector r ( t ) = ( e t cos t ) i + ( e t sin t ) j + 2 e t k , t = 0 .

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Find the acceleration vector a in the form a=aTT+aNN at the value of t without finding T and N for the given vector r(t)=(etcost)i+(etsint)j+2etk, t=0.

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Chapter 13 Solutions