Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.4 - Curvature and Normal Vectors of a Curve - Exercises 13.4 - Page 766: 21


$(x-\dfrac{\pi}{2} )^2 +y^2=1$

Work Step by Step

When $r(t)$ is a smooth curve, then the curvature can be defined as: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|...(1)$ where $T=\dfrac{v}{|v|} $, a unit tangent vector. We have: $v=\dfrac{dr}{dt}= i+ j \cos (t)$ and $|v|=\sqrt {(1)^2+(\cos t)^2}=\sqrt {1+\cos^2 t}$ So, $T=\dfrac{v}{|v|}=\dfrac{i+j \cos t }{\sqrt {1+\cos^2 t}}$ Equation (1) becomes: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|=\dfrac{1}{\sqrt {1+\cos^2 t}} \times \dfrac{\sin t}{\cos^2 t +1}=\dfrac{\sin t \sqrt {cos^2 t+1}}{(\cos^2 t+1)^2}$ $\kappa (\dfrac{\pi}{2})=\dfrac{\sin (\dfrac{\pi}{2}) \sqrt {cos^2 (\dfrac{\pi}{2})+1}}{(\cos^2 (\dfrac{\pi}{2})+1)^2}=1$ Therefore, the radius of curvature is: $\rho=\dfrac{1}{\kappa}=1$ and the center is $(\dfrac{\pi}{2}, 0)$. Now, the solution is: $(x-\dfrac{\pi}{2} )^2 +y^2=1$
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