Explanation Assume that the object accelerates which implies d 2 s d t 2 > 0 and κ ≠ 0 . Recall the formula for the acceleration a = a T T + a N N . The tangential component of the acceleration a T = d 2 s d t 2 . Since d 2 s d t 2 > 0 , the tangential component of the acceleration is positive. The normal component of the acceleration a N = κ | v | 2 . Since κ ≠ 0 , the normal component is positive. This implies the acceleration is a linear combination of the unit tangent vector with positive coefficient and the principal unit normal vector with positive coefficients. Therefore, the acceleration vector lies in between T and N . The graph in the case of d 2 s d t 2 > 0 is sketched below in Figure 1,

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Question

Chapter 11.9, Problem 78E

To determine

To show: The acceleration vector is in between the unit tangent vector and the principal unit normal vector if the objects accelerates and the acceleration vector does not lie between the unit tangent vector and the principal unit normal vector if the objects decelerates.

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Chapter 11.9, Problem 77E

Chapter 11.9, Problem 79E

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