Chapter 0.4, Problem 47E

To determine

To find: The parametrization for the witch by expressing the coordinates of $P$ in terms of $t$ , the radian measure of the angle that segment $OA$ makes with the positive $x$ -axis.

Answer to Problem 47E

  $x=2\cot t$ and

  $y=2{\sin }^{2}t$ for all $t\in (0,\pi )$

Explanation of Solution

Given information: The graph is:

  

Calculation:

  $OQ$ is a line segment on the $y$ -axis, and $OQ\perp QA$ is parallel to the $x$ -axis at $y=2$ . $OQA$ is a right triangle as a result. Because the line $y=2$ is parallel to the $x$ -axis, take note that $\angle OAQ=t$ . Identify the relationships in this right triangle as follows:

  $\frac{AQ}{AO}=\cos t$

And,

  $\begin{array}{c}\frac{AQ}{OQ}\\ =\frac{AQ}{2}\\ =\cot t\to x\\ =AQ\\ =2\cot t\end{array}$

Hence, $AB\cdot AO=A{Q}^2$ and

  $\frac{AQ}{AO}=\cos t$

  $\begin{array}{c}AB=\frac{A{Q}^2}{AO}\\ =AQ\cdot \cos t\\ =x\cos t\end{array}$

Since, $AB=x\cos t$ and

  $x=2\cot t$

Since,

  $\begin{array}{c}y=2-AB\sin t\\ =2-x\cos t\sin t\\ =2-2\cot t\cos t\sin t\\ =2-2\frac{\cos t}{\sin t}\cos t\sin t\\ =2-2{\cos }^{2}t\\ =2\left(1-{\cos }^{2}t\right)\\ =2{\sin }^{2}t\end{array}$

Since the curve is on the upper half plane, we want the interval of $t$ to be $(0,\pi )$

Therefore, the parametrization is $x=2\cot t$ and

  $y=2{\sin }^{2}t$ for all $t\in (0,\pi )$