A more elegant way to visualize the curve traced out by a vector function? *** To clarify, I am looking for algebraic manipulations that will result in a more elegant way to graph the shape by hand***
I am attempting to draw a curve for t ≥ 0, using any "projections" or "additions" that help explain the curve.
Curve: $r(t) = \cos(t^3)i + \sin(t^3)j + t^3k.$
Therefore we have:
$$\begin{align}
x &= \cos(t^3)\\
y &= \sin(t^3) \\
z &= t^3
\end{align}$$
I have crunched the numbers and generated a few tables for values for the $xy$, $xz$, and $yz$ planes – which I have sketched.
I am curious though, this process was quite tedious and I was wondering if anyone could think of a more elegant way to visualize this curve by hand?
My thoughts:
I have used 3D graphing software to visualize the curve, I understand that it's a helix. I was thinking that perhaps I could re-parameterize the VVF representing the curve without losing the shape?
Update:
Upon reading emacs drives me nuts's answer
Algebraically, to solve y in terms of x (to graph projection on the xy-plane)::
$x=cos(t^3)$
$t^3=arccos(x)$
$y=sin(t^3)$
$y=sin(arccos(x))$
From trig identity: $sin(arccos(x))=\sqrt{(1−x^2)}$
so, $y=\sqrt{(1−x^2)}$
Which is recognized as the equation of a circle with radius 1 (unit circle): $x^2+y^2=1$
 A: 
I am curious though, this process was quite tedious and I was wondering if anyone could think of a more elegant way to visualize this curve by hand?

It's a helix, obviously.  First observe that $t\mapsto t^3$ does not change the shape of the curve as it (smoothly) maps $\Bbb R$ to $\Bbb R$.  This means the curve is the same like
$$f(t) = (x(t), y(t), z(t)) = (\cos t, \sin t, t)$$
When projected to the $xy$-plane, we see that the curve traces out circles at constant (angular) velocity.  And when projected to the $z$-axis, it's just a linear movement with constant speed. Thus $f(t)$ traces out a spiral around the $z$-axis that extends to $\pm\infty$.
As already said, $f(t^3)$ is the same curve.  The difference is that it's traced with different velocity:  The farther the curve is away from the $xy$-plane, the faster it is traced.
The speed in which $f(t)$ is traced is
$$|f'(t^3)| = |3t^2(-\sin t^3, \cos t^3, 1)| = 3\sqrt{2}t^2 = 3\sqrt{2}|z|^{2/3}$$
because $\sin^2t + \cos^2 t = 1$.  And the speed of $f(t)$ is constant:
$$|f'(t)| = |(-\sin t, \cos t, 1)| = \sqrt{2}$$
In either case, the arguments of the $\sin$ and $\cos$ term equals $z$.  Therefore, the curve can be expressed as
$$(\cos z, \sin z, z)$$
which also shows that it winds a circle of radius 1 around the $z$-axis.
