Chapter 10, Problem 1RE

To determine

To calculate: The parametric equations for a portion of the circle, $x^2+y^2=2$ , that lies outside the first quadrant, oriented clockwise and plot the curve using the graphing utility.

Answer to Problem 1RE

The parametric equations for the given curve are $x=\sqrt{2}\cos t$ and $y=-\sqrt{2}\sin t$ over the interval, $\left(0\le t\le \frac{3\pi }{2}\right)$ .

Explanation of Solution

Calculation:

Consider that the given curve, $x^2+y^2=2$ , is a circle of radius $\sqrt{2}$ .

Use sine and cosine to parameterize the given curve.

Let $x=\sqrt{2}\cos t$ and $y=-\sqrt{2}\sin t$ .

Use the trigonometric equation, ${\sin }^2\theta +{\cos }^2\theta =1$ .

To verify, substitute the parametric values of $x$ and $y$ on the left-hand side of the equation of the circle.

$\begin{array}{c}{x}^2+y^2={\left(\sqrt{2}\cos t\right)}^2+{\left(-\sqrt{2}\sin t\right)}^2\\ =2{\cos }^{2}t+2{\sin }^{2}t\\ =2\left({\cos }^{2}t+{\sin }^{2}t\right)\\ =2\end{array}$

Thus, the parametric equations for the given curve are $x=\sqrt{2}\cos t$ and $y=-\sqrt{2}\sin t$ .

Since the curve does not lie in the first quadrant, the interval is $\left(0\le t\le \frac{3\pi }{2}\right)$ .

The graph of the given curve with the parametric equations, $x=\cos t$ and $y=\sin t$ , over the interval, $\left(\pi \le t\le \frac{3\pi }{2}\right)$ , is shown below.

The graph represents a portion of the circle, $x^2+y^2=2$ , that lies outside the first quadrant.