Cartesian equation of $ \vec{r}(t)=\left(e^{\omega t}(\cos(\omega t)+\sin(\omega t)),2\omega e^{\omega t} \cos(\omega t)\right) $

Question

I have this parametric equation: $$ \vec{r}(t)=\left(e^{\omega t}(\cos(\omega t)+\sin(\omega t)),2\omega e^{\omega t} \cos(\omega t)\right) $$ and I have to obtain the Cartesian equation. Any suggestions please?

Answer 1

It can be done but only implicitly and the expression is pretty ugly... $$ \begin{cases} \frac{y}{2\omega} &= e^{\omega t}\cos(\omega t)\\ x - \frac{y}{2\omega} &= e^{\omega t}\sin(\omega t)\\ \end{cases} \implies \begin{cases} e^{\omega t} &= \sqrt{\left(\frac{y}{2\omega}\right)^2 + \left(x - \frac{y}{2\omega}\right)^2}\\ \tan(\omega t) &= \frac{2\omega x}{y} - 1 \end{cases}\\ \implies \frac{2\omega x}{y} = 1 + \tan\left\{\frac12\log\left[\left(\frac{y}{2\omega}\right)^2 + \left(x - \frac{y}{2\omega}\right)^2\right]\right\} $$ Update

As pointed out by B.S., above implicit representation is plagued with artifact. This is caused by the use of ratio like $\frac{x}{y}$ which killed the information about the signs and by the discontinuity of the $\tan$ function. A more well behaved implicit representation is given by:

$$R(x,y) - \frac{y}{2\omega}\cos\log R(x,y) - \left(x - \frac{y}{2\omega}\right)\sin\log R(x,y) = 0$$ where $\quad\displaystyle R(x,y) = \sqrt{\left(\frac{y}{2\omega}\right)^2 + \left(x - \frac{y}{2\omega}\right)^2}$.

Up to what I have checked, this reproduce the same curve as the parametric parametrization.

Please note that even this implicit representation is correct, it is always non-negative and only becomes 0 on the curve. The implicit plot of some CAS (e.g Maxima) cannot handle this type of representation properly without some tweaking. If you want to plot this out for comparison and don't mind a reflected copy of the curve. You can use following implicit representation instead:

$$\frac{y}{2\omega}\sin\log R(x,y) - \left(x - \frac{y}{2\omega}\right)\cos\log R(x,y) = 0$$

Answer 2

Hint: Let $x=e^{\omega t}(\cos(\omega t)+\sin(\omega t))$ and $y=2\omega e^{\omega t} \cos(\omega t)$ and try to eliminate $t$