Chapter 10.7, Problem 21AYU

To determine

The graph of curve of the parametric equation $x=2\cos t,y=3\sin t;-\pi \le t\le 0$, show its orientation. Find the rectangular equation of curve.

Answer to Problem 21AYU

Solution:

The graph of the parametric equations $x=2\cos t,y=3\sin t;-\pi \le t\le 0$ is,

The rectangular equation of the parametric equation $x=2\cos t,y=3\sin t;-\pi \le t\le 0$ is $\frac{x^2}{4}+\frac{y^2}{9}=1,y\le 0$.

Explanation of Solution

Given information:

The parametric equation is $x=2\cos t,y=3\sin t;-\pi \le t\le 0$.

Explanation:

To draw the given parametric equation, plug some values of $t$ in the given equation and find few points on the curve.

The interval for $t$ is $-\pi \le t\le 0$.

$\begin{array}{cccc}t& x\left(t\right)& y\left(t\right)& \left(x,y\right)\\ -\pi & -2& 0& \left(-2,0\right)\\ -\frac{\pi }{2}& 0& -3& \left(0,-3\right)\\ 0& 2& 0& (2,0)\end{array}$

The orientation is the curve traced out in a certain direction by the corresponding succession of points $\left(x,y\right)$.

The arrow shows the orientation along the curve as $t$ varies from $-\pi \le t\le 0$.

As $t$ takes values from $-\pi \le t\le 0$, the corresponding orientation is from $\left(-2,0\right)$ to $\left(2,0\right)$

Use points in the table to sketch the curve.

The graph of the parametric equation $x=2\cos t,y=3\sin t;-\pi \le t\le 0$ is,

The orientation of the given curve is counter-clockwise.

Let $x=2\cos t,y=3\sin t$

$\Rightarrow \frac{x}{2}=\cos t,\frac{y}{3}=\sin t$

By using Pythagorean identity ${\sin }^{2}t+{\cos }^{2}t=1$,

$\Rightarrow {\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2=1$

$\Rightarrow \frac{x^2}{2^2}+\frac{y^2}{3^2}=1$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1$

The equation $\frac{x^2}{4}+\frac{y^2}{9}=1$ is similar to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which represents the ellipse.

For the interval $-\pi \le t\le 0$ it represent lower portion of the ellipse.

The rectangular equation of the parametric equation $x=2\cos t,y=3\sin t;-\pi \le t\le 0$ is $\frac{x^2}{4}+\frac{y^2}{9}=1,y\le 0$.