Chapter 10.1, Problem 9QR

To determine

To find: The equation that relates $x$ and $y$ .

Answer to Problem 9QR

The equation is: $y=+\sqrt{1-x^2}$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\cos \theta \\ y=\sin \theta (0\le \theta \le \pi )\end{array}$

Calculation:

The two equations will be known as follows:

  $\begin{array}{c}x=\cos \theta \mathrm{.....}(1)\\ y=\sin \theta \mathrm{.....}(2)\end{array}$

To generate the following equations, square both of the aforementioned equations:

  $\begin{array}{c}{x}^2={\cos }^2\theta \mathrm{.....}(3)\\ y^2={\sin }^2\theta \mathrm{......}(4)\end{array}$

Add equation $(3)$ and $(4)$ , get:

  $\begin{array}{c}{x}^2+y^2={\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow x^2+y^2=\mathrm{1.....}(5)\end{array}$

The sine function is negative in Quadrants I and II, this means that $y\ge 0$ .

So,

  $\begin{array}{c}{y}^2=1-x^2\\ \Rightarrow y=\pm \sqrt{1-x^2}\end{array}$

Hence,

  $y\ge 0$

And will reject the square root that is negative:

  $y=+\sqrt{1-x^2}$

Therefore, the required equation that relates $x$ and $y$ is $y=+\sqrt{1-x^2}$ .