Chapter 10, Problem 65RE

To determine

Tofind:the rectangular equation and graph the curve using parametric equation.

Answer to Problem 65RE

The rectangular equation is ${\left(\frac{\text{x}}{\text{3}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{y-2}}{\text{4}}\right)}^{\text{2}}\text{=1}$ and the graph here will be of ellipse.

Explanation of Solution

Given:

  $\text{x=3sint, y=4cost+2}$ ; $\text{0}\le \text{t}\le \text{2}\pi$ .

Concept used:

The parametric equation has equation for each variable whereas rectangular equation has one equation

In rectangular equation it composed of variable like $\text{x and y}$ which can be graphed on a regular cartesian plane.

To convert parametric to rectangular $\left(\text{t}\right)$ should be eliminate so that function will be in the form of $\text{x and y}$ which consist of only one independent variable.

Calculation:

According to the given parametric equation:

  $\text{x=3sint}$

  $\frac{\text{x}}{3}\text{=sint}$ …………… $\left(\text{i}\right)$

  $\text{y=4cost+2}$

  $\frac{\text{y-2}}{4}\text{=cost}$ …………….. $\left(\text{ii}\right)$

Squaring and Adding equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$ :

  ${\left(\frac{\text{x}}{\text{3}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{y-2}}{\text{4}}\right)}^{\text{2}}\text{=1}$ .

The equation is in the form of $\frac{{\text{x}}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{+}\frac{{\text{y}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ which is the equation of ellipse.

So, the graph will be elliptical shaped.

The graph of the equation ${\left(\frac{\text{x}}{\text{3}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{y-2}}{\text{4}}\right)}^{\text{2}}\text{=1}$ :

  

Here from the graph:

At $\text{x=-2}\text{.598 and 2}\text{.598; y=0}\text{.5}$ .

At $\text{x=0 ;y=6 and -2}$ .

Hence, the rectangular equation is ${\left(\frac{\text{x}}{\text{3}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{y-2}}{\text{4}}\right)}^{\text{2}}\text{=1}$ and the graph here will be of ellipse.