Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 894: 30


$y=\ln \sqrt x\\z=\dfrac{1}{\sqrt x}\\z=e^{-y}$

Work Step by Step

We are given that $x=t^2; y=\ln t; z=1/t$ Simplify these given terms. $y=\ln (t)=\ln [\sqrt {t^2}]=\ln \sqrt x$ From $z=\dfrac{1}{t}$ This gives: $z=\dfrac{1}{\sqrt {t^2}}=x^{-1/2}$ Also, $z=\dfrac{1}{t}$ This gives: $z=\dfrac{1}{e^y} \implies z=e^{-y}$ Therefore, $y=\ln \sqrt x\\z=\dfrac{1}{\sqrt x}\\z=e^{-y}$
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