Q. 28 - Calculus

Question

Sketch the parametric curve by eliminating the parameter $x=\cos 2t,y=-\sin 2t,t\in [0,2\pi ]$

Step-by-Step Solution

Answer

The required graph is

Step 1: Given information

The parametric curve is $x=\cos 2t,y=-\sin 2t,t\in [0,2\pi ]$

Step 2: Calculation

Consider the parametric equations $x=\cos 2t,y=-\sin 2t,t\in [0,2\pi ].$

The objective is to sketch the parametric curve by eliminating the parameter.

By squaring the two parametric equations the parameter $t$ is eliminated.

Take $x=\cos 2t$

Squaring on both sides of the equation.

$x^2={\cos }^{2}2t \dots \left(1\right)$

Now take $y=-\sin 2t$

Squaring on both sides of the equation.

$y^2={\sin }^{2}2t \dots \left(2\right)$

To eliminate the parameter add the equations $\left(1\right)$ and $\left(2\right)$ that is $x^2={\cos }^{2}2t,y^2={\sin }^{2}2t.$

Then,

$$\begin{gathered} x^2+y^2={\cos }^{2}2t+{\sin }^{2}2t \\ {x}^2+y^2=1 \end{gathered}$$

Here the equation is a circle with center $(0,0)$ and radius $r=1$.

To draw the graph of the equation assume $x=-1,0,1$

Substitute $x=-1$ in the equation $x^2+y^2=1.$

Then,

$$\begin{gathered} (-1{)}^2+y^2=1 \\ 1+y^2=1 \\ y=0 \end{gathered}$$

So, $(x, y)=(-1,0)$

Step 3: Further calculation

Substitute $x=0$in the equation $x^2+y^2=1$Then,

$$\begin{gathered} 0^2+y^2=1 \\ {y}^2=1 \\ y=\pm 1 \end{gathered}$$

Then $(x, y)=(0,1)(0,-1)$

Substitute $x=1$in the equation $x^2+y^2=1.$Then,

$$\begin{gathered} 1^2+y^2=1 \\ {y}^2=1-1 \\ {y}^2=0 \end{gathered}$$

So, $(x, y)=(0,0)$

The graphical representation by using the points $(-1,0)(1,0)(0,0)$is as follows,

Therefore, the required equation after eliminating the parameter is $x^2+y^2=1.$