How to express a parametrized curve in spherical coordinates Given an helix parametrized by $\lambda$ with
$$ x(\lambda)=\cos \lambda, \quad y(\lambda)=\sin \lambda, \quad z(\lambda)=\lambda$$
How can I express this curve in spherical coordinates $(r,\theta,\phi)$?
For example, I know that the unit circle parametrized by 
$$x(t)=\sin t, \quad y(t)= \cos t$$
Can be expressed in, say, cartesian coordinates, as $x^2 + y^2 = 1$, just by using the identity $\sin^2t + \cos^2t = 1$. But, with the helix, I can't seem to find such correspondence. And I think maybe it doesn't exist, because for the same value of $x$, there are infinite values for $z$. So, I don't know if this question even makes sense. Thanks for the attention.
 A: The same way that you convert any coordinate to spherical
$$\begin{align} r(\lambda) &= \sqrt{x^2+y^2+z^2} = \sqrt{1 + \lambda^2}
\\
\theta(\lambda) &= \arctan(\frac{y}{x}) = \lambda
\\
\phi(\lambda) &= \cos^{-1}(\frac{z}{\sqrt{x^2 + y^2 + z^2}}) = \cos^{-1}(\frac{\lambda}{\sqrt{1+\lambda^2}}) \end{align}$$
A: There is a minor issue with your problem. You need two equations to describe the helix. To explain why, think about how would you write the equation of a circle in 3D. $x^2+y^2=1$ is the equation of the surface of the cylinder. You would need to specify another equation (for example $z=0$) to get a single circle.
As you can see, the equations for $x$ and $y$ can be combined into $x^2+y^2=1$, which will give the equation for the cylinder. In spherical coordinates you get $$x^2+y^2+z^2=r^2$$ and $$z=r\cos\theta$$ Using these, the equation of the cylinder in polar coordinates is $$r^2\sin^2\theta=1$$
The second equation is given by $$\frac xy=\frac{r\sin\theta\cos\phi}{r\sin\theta\sin\phi}=\frac{\cos\phi}{\sin\phi}=\frac{\cos\lambda}{\sin\lambda}$$
This can be simplified into $$\phi=\lambda=z=r\cos\theta$$
So the two equations describing the helix are $$r^2\sin^2\theta=1\\\phi=r\cos\theta$$
A: Your question is confuse because on one hand you ask to express a (3D) parametric curve in spherical coordinates, but on the other hand you give the example of a (2D) parametric curve turned to an implicit equation and not using polar coordinates.

2D case:
$$\begin{cases}x=\cos t,\\y=\sin t\end{cases}\to\begin{cases}\rho\cos\theta=\cos t,\\\rho\sin\theta=\sin t\end{cases}\to\begin{cases}\theta=t,\\\rho=1\end{cases}$$
gives the polar parametric equations,
$$x^2+y^2=1$$
is the implicit Cartesian equation, and
$$\rho=1$$ the implicit (at the same time explicit) polar equation.

3D case:
Correspondingly,
$$\begin{cases}x=\cos t,\\y=\sin t,\\z=t\end{cases}\to\begin{cases}\rho\cos\theta\sin\phi=\cos t,\\\rho\sin\theta\sin\phi=\sin t,\\\rho\cos\phi=t\end{cases}\to\begin{cases}\phi=\arcsin\dfrac1{\sqrt{1+t^2}},\\\theta=t,\\\rho=\sqrt{1+t^2}\end{cases}$$
$$\begin{cases}x=\cos z,\\y=\sin z\end{cases}$$
and
$$\begin{cases}\rho\sin\phi=1,\\\rho^2=1+\theta^2\end{cases}.$$
(A 3D curve requires a system of two implicit equations.)
