Question

How do you find the Cartesian equation of the curve with parametric equation \[x=2\cos \left( 3t \right)\] and $y=2\sin \left( 3t \right)$, and determine the domain and range of the corresponding relation?

Answer 1

Hint: To find the Cartesian equation square and add the given parametric equation. Take the reference of the range of $\cos \left( t \right)$ and $\sin \left( t \right)$ to find the domain and range of the parametric equation \[x=2\cos \left( 3t \right)\] and $y=2\sin \left( 3t \right)$.

Complete step by step solution:
The Cartesian equation of the curve with parametric equation \[x=2\cos \left( 3t \right)\] and $y=2\sin \left( 3t \right)$ can be obtained by squaring and adding the given parametric equations.
Squaring both the sides of the parametric equation \[x=2\cos \left( 3t \right)\], we get
\[\Rightarrow {{\left( x \right)}^{2}}={{\left( 2\cos \left( 3t \right) \right)}^{2}}\]
\[\Rightarrow {{x}^{2}}=4{{\cos }^{2}}\left( 3t \right)\]……….(1)
Again, squaring both the sides of the parametric equation $y=2\sin \left( 3t \right)$, we get
$\Rightarrow {{\left( y \right)}^{2}}={{\left( 2\sin \left( 3t \right) \right)}^{2}}$
$\Rightarrow {{y}^{2}}=4{{\sin }^{2}}\left( 3t \right)$……….(2)
Adding equation (1) and equation (2), we get
$\begin{align}
  & eq(1)+eq(2) \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4{{\cos }^{2}}\left( 3t \right)+4{{\sin }^{2}}\left( 3t \right) \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4\left( {{\cos }^{2}}\left( 3t \right)+{{\sin }^{2}}\left( 3t \right) \right) \\
\end{align}$
As we know, ${{\cos }^{2}}\left( 3t \right)+{{\sin }^{2}}\left( 3t \right)=1$
So, now our equation becomes
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}=4\times 1 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4 \\
\end{align}$
This is the required Cartesian equation.
Coming to domain and range
As we know the range of cosine function is $-1\le \cos \left( t \right)\le 1$
So, the range of $\cos \left( 3t \right)$ will be $-1\le \cos \left( 3t \right)\le 1$
Hence, the domain of ‘x’ will be $-2\le x\le 2$ as \[x=2\cos \left( 3t \right)\]
Again, as we know the range of sine function is $-1\le \sin \left( t \right)\le 1$
So, the range of $\sin \left( 3t \right)$ will be $-1\le \sin \left( 3t \right)\le 1$
Hence, the domain of ‘y’ will be $-2\le x\le 2$ as $y=2\sin \left( 3t \right)$
This is the required solution.

Note: To convert the given parametric equation to the Cartesian equation, we have to eliminate the ‘t’ from both the equations. This can be done by squaring and adding the parametric equations. The Cartesian equation we obtained ${{x}^{2}}+{{y}^{2}}=4$, can be written as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 2 \right)}^{2}}$. This is the equation of the circle with center (0,0) and radius ‘2’.