Where $t$ ranges from $0$ to $\infty$.
After a bit of manipulation, I got $x^2 + y^2 = t^2$ but I have no idea how to sketch this since $t$ isn't a constant.
Where $t$ ranges from $0$ to $\infty$.
After a bit of manipulation, I got $x^2 + y^2 = t^2$ but I have no idea how to sketch this since $t$ isn't a constant.
Your curve is given by
$$ \alpha(t) = (t \cos t, t \sin t) = t (\cos t, \sin t). $$
As $t$ runs from $0$ to $2\pi$ the curve $(\cos t, \sin t)$ traces a circle of radius one around the origin. Multiplying by $t$ scales the point $(\cos t, \sin t)$ appropriately so that at time $t$ the resulting point makes an angle $t$ with the positive $x$-axis but lies on the circle of radius $t$ instead of the circle on radius one. Hence, you get a spiral that starts (when $t = 0$) at the origin and spirals counter clockwise as $t$ increases from $0$ to $2\pi$ and reaches $(1,0)$ when $t = 2\pi$ and then the pattern continues.
If you are have access to any mathematical software you can visual what's going on quite nicely. I've plotted the two situations which are described in the above answer.
For a curve given by $x(t)=t \cos(t)$ and $y(t)= t\sin(t)$ in the range $0\leq t \leq 5\pi$ (just for visual purposes) you get the following:
Then notice for $x(t)= \cos(t)$ and $y(t)= \sin(t)$ in the range $0\leq t \leq 2\pi$ you get the following:

Notice for the second graph as long as the range is $\geq 2\pi$ we will always get a circle. It won't spiral as in the first case because there is no factor of $t$ to scale the graph. Anyway - I always find visual aids fantastic help in remembering what's going on.