Suppose a frictionless particle $P$ is sliding down the helix $$ r(t) = a\cos(\theta)i + a\sin(\theta)j+b\theta k $$ with the initial position $(a,0,0)$. The law of conservation of energy tells us that the particle's speed is $\sqrt{2gz}$ when the particle has fallen down a distance $z$.
I want to represent $\theta$ as a function of $t$.
My attempt is as following: first, speed of the particle is $$ \|r'(t)\| = \sqrt{a^2+b^2}|\theta'(t)|=-\sqrt{a^2+b^2}\theta'(t)=\sqrt{2gb\theta(t)} $$ I tried to solve this differential equation and I concluded that $$ \theta=\frac{gb}{2(a^2+b^2)}t^2 $$ but I am not sure that this is the right answer. Can anyone verify this?
Thanks in advance.
Edit: there was a typo, and I fixed it.
I'll assume you made a typo and the equation of motion actually is
$$ \textbf{r}(t) = a\cos\theta(t)\ \textbf{i} + a\sin\theta(t)\ \textbf{j} + b\theta(t)\ \textbf{k} $$
which is actually the shape of a helix
The initial condition implies $\theta(0) = 0$
The velocity vector is
$$ \textbf{r}'(t) = \big[{-a}\sin\theta(t)\ \textbf{i} + a\cos\theta(t)\ \textbf{j} + b\ \textbf{k}\big]\theta'(t) $$
From the given speed, we have
$$ |\textbf{r}'(t)| = \sqrt{a^2+b^2}\big|\theta'(t)\big|= \sqrt{2gz(t)} $$
The particle is always falling downwards, so $\theta' \ge 0$ and we can ignore the absolute values
$$ \theta'(t) = \sqrt{\frac{2gb\theta(t)}{a^2+b^2}} $$
Solving the separable equation
$$ \frac{\theta'}{2\sqrt{\theta}} = \sqrt{\frac{gb}{2(a^2+b^2)}} \implies \sqrt{\theta} = \sqrt{\frac{gb}{2(a^2+b^2)}}t + c $$
where $\theta(0)=0$ gives $c=0$, thus
$$ \theta(t) = \frac{gb}{2(a^2+b^2)}t^2 $$