Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.5 Motion in 3-Space - Exercises - Page 746: 52

Answer

Because the angle between ${\bf{w}}$ and ${\bf{N}}$ is obtuse, the vector ${\bf{w}}$ cannot be the acceleration vector.

Work Step by Step

Suppose ${\bf{w}}$ is the acceleration vector of a particle moving along the circle as is shown in Figure 14. Let $v\left( t \right)$ be the speed of the particle. By Eq. (1), we have ${\bf{w}} = {w_{\bf{T}}}{\bf{T}} + {w_{\bf{N}}}{\bf{N}}$, where ${w_{\bf{T}}} = v'\left( t \right)$ ${\ }$ and ${\ }$ ${w_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$ Evaluate ${\bf{w}}\cdot{\bf{N}} = \left( {{w_{\bf{T}}}{\bf{T}} + {w_{\bf{N}}}{\bf{N}}} \right)\cdot{\bf{N}}$ ${\bf{w}}\cdot{\bf{N}} = {w_{\bf{N}}} = \kappa \left( t \right)v{\left( t \right)^2}$ Since the angle between ${\bf{w}}$ and ${\bf{N}}$ is obtuse, the left-hand side is negative value. However, $\kappa \left( t \right) = \frac{1}{R}$, where $R$ is the radius of the circle, is always positive. So, the right-hand side is a nonnegative value. Thus, it is a contradiction. Hence, we conclude that the vector ${\bf{w}}$ cannot be the acceleration vector.
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