Chapter 11: Parametric Equations and Polar Coordinates - Section 11.1 - Parametrizations of Plane Curves - Exercises 11.1 - Page 647: 32

$x=\cot^2 \theta$;$y= \cot \theta$; $0\leq \theta \leq \dfrac{\pi}{2}$

Let us consider that $y=\sqrt x$ Since, $y=x \tan \theta$ Then, we have $\sqrt x= \cot \theta \implies x=\cot^2 \theta$ Now, $y=(\cot^2 \theta) (\tan \theta)$ Thus, $y=\dfrac{1}{\tan \theta}(\tan \theta)= \cot \theta$