Converting Parametric Form to Cartesian Equation

[Alex23]

r(t)=(3cost)i+(sint+1)j

Find a Cartesian equation for the curve traced out by this function.

My work :

x=3cost
y=sint+1

sint = y-1 >> t= arcsin(y-1)

Plug that in for t in the x equation. x=3cos(arcsin(y-1))

I don't know what to do from here or if I'm going in the right direction or not. I'm not looking for the answer here. Just trying to learn how to do this question. All help is greatly appreciated.

 

[soroban]

Hello, Alex23!

Find a Cartesian equation for the curve traced out by this function:
. \(\displaystyle r(t) \:=\3\cos t)i + (\sin t + 1)j\)

\(\displaystyle \text{We have: }\:\begin{Bmatrix}x &=& 3\cos t && \Rightarrow && \dfrac{x}{3} &=& \cos t & [1] \\ y &=& 1 + \sin t && \Rightarrow && y-1 &=& \sin t & [2] \end{Bmatrix}\)

\(\displaystyle \begin{Bmatrix}\text{Square [1]:} & \dfrac{x^2}{9} &=& \cos^2\!t \\ \text{Square [2]:} & (y-1)^2 &=& \sin^2\!t \end{Bmatrix}\)

\(\displaystyle \text{Add: }\;\dfrac{x^2}{9} + (y-1)^2 \:=\:\underbrace{\cos^2\!t + \sin^2\!t}_{\text{This is 1}} \)

\(\displaystyle \text{Therefore: }\:\dfrac{x^2}{9} + \dfrac{(y-1)^2}{1} \;=\;1\) . . . An ellipse!

 

[Alex23]

That make sense now. I was trying to solve for x for some reason.

Thanks a lot.