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 - 14.1 Vector-Valued Functions - Exercises - Page 711: 39

Answer

$$r(t)=\lt1, 2+2\cos t,5+2\sin t\gt.$$

Work Step by Step

We know that the sphere of radius 2, centered at $(1,2,5)$, has the equation $$(x-1)^2+(y-2)^2+(z-5)^2=4.$$ Its projection on the yz-plane is $$(y-2)^2+(z-5)^2=4.$$ So we get the parametrization $$x=1, \quad y=2+2\cos t, \quad z=5+2\sin t.$$ That is $$r(t)=\lt1, 2+2\cos t,5+2\sin t\gt.$$
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