Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 854: 30

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

Here, we have $x=t^2; y=\ln t; z=1/t$ Now, $y=\ln t=\ln \sqrt {t^2}=\ln \sqrt x$ From $z=\dfrac{1}{t}$, we have $z=\dfrac{1}{\sqrt t^2}=\dfrac{1}{\sqrt x}$ Also, $z=\dfrac{1}{t} \implies z=\dfrac{1}{e^y}$ or, $z=e^{-y}$ Hence, we have $y=\ln \sqrt x$ ; $z=\dfrac{1}{\sqrt x}$ and $z=e^{-y}$