Determine the initial point and terminal point, then determine the rectangular equation of the...

Question:

Determine the initial point and terminal point, then determine the rectangular equation of the curve whose parametric equation are given,

{eq}\displaystyle x = 2 \sin (t),\ y = 4 \cos (t),\ 0 \le t \le \pi {/eq}.

Parametric to Cartesian Coordinates

Given a parametric function {eq}\textbf{r}:\left[a,b\right]\rightarrow\mathbb{R}^2 {/eq} defined as {eq}\textbf{r}(t)=(x(t),y(t)) {/eq} it can be expressed in rectangular Cartesian coordinates. Exactly how the translation from parametric to rectangular coordinates is accomplished depends on the exact expressions of the parametric equations. Moreover, the parametric equation includes an initial point corresponding to {eq}t=a {/eq} and a terminal point corresponding to {eq}t=b {/eq}. This determines the extent of the resulting curve and its orientation.

Answer and Explanation:

The question is restated with slightly different notation. Determine the initial point, terminal point and the rectangular coordinate equation of the curve whose parametric equations are {eq}x=2\sin(t) {/eq}, {eq}y=4\cos(t) {/eq} for {eq}0\leq t\leq\pi {/eq}.

When {eq}t=0 {/eq} we have

{eq}\begin{eqnarray*} x &=& 2\sin(0) \\ &=& 0 \\ y &=& 4\cos(0) \\ &=& 4. \end{eqnarray*} {/eq}

When {eq}t=\pi {/eq} we have

{eq}\begin{eqnarray*} x &=& 2\sin(\pi) \\ &=& 0 \\ y &=& 4\cos(\pi) \\ &=& -4. \end{eqnarray*} {/eq}

Consequently, the initial point is {eq}(0,4) {/eq} and the terminal point is {eq}(0,-4) {/eq}. Lets also find the point that corresponds to {eq}t=\frac{\pi}{2} {/eq}, namely,

{eq}\begin{eqnarray*} x &=& 2\sin\left(\frac{\pi}{2}\right) \\ &=& 2 \\ y &=& 4\cos\left(\frac{\pi}{2}\right) \\ &=& 0. \end{eqnarray*} {/eq}

This point is {eq}(2,0) {/eq}. The curve begins at {eq}(0,4) {/eq}, passes through {eq}(2,0) {/eq} and ends at {eq}(0,-4) {/eq}. Consequently, the curve has a clockwise orientation.

Now consider expressing the parametric equations in the rectangular coordinate system. Note that

{eq}\begin{eqnarray*} x &=& 2\sin(t) \\ \frac{x}{2} &=& \sin(t) \\ y &=& 4\cos(t) \\ \frac{y}{4} &=& \cos(t). \end{eqnarray*} {/eq}

Consequently

{eq}\begin{eqnarray*} \left(\frac{x}{2}\right)^2+\left(\frac{y}{4}\right)^2 &=& \sin^2(t)+\cos^2(t) \\ &=& 1. \end{eqnarray*} {/eq}

Therefore, the equation in rectangular coordinates is

{eq}\begin{eqnarray*} \left(\frac{x}{2}\right)^2+\left(\frac{y}{4}\right)^2 &=& \frac{x^2}{2^2}+\frac{y^2}{4^2} \\ &=& 1. \end{eqnarray*} {/eq}

This is an ellipse centered at the origin with semi-major axis 4 along the {eq}y {/eq}-axis and semi-minor axis 2 along the {eq}x {/eq}-axis. Since the parametrization begins at {eq}(0,4) {/eq} and ends at {eq}(0,-4) {/eq} it is only half the ellipse oriented clockwise.


Learn more about this topic:

Loading...
Converting Between Parametric & Rectangular Forms

from Precalculus: High School

Chapter 24 / Lesson 4
51K

Related to this Question

Explore our homework questions and answers library