I have a curve where $ t\in R^{+}_0$
$$ \Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T $$
I need to express tangent vector T in standard spherical coordinates in terms of normalized 'frame vectors' $\hat h_i $.
I started from denoting spherical cordinates and comparing with the given data $$x=r \sin\Phi \cos\lambda=2t^3 - 2t=2t(t^2-1)$$ $$y=r \sin\Phi \sin\lambda=4t^2$$ $$z=r \cos\Phi=t^3+t=t(t+1) $$
Tangent vector is defined as: $$T=\dot r |h_r| \hat h_r +\dot \phi |h_\phi| \hat h_\phi +\dot \lambda |h_\lambda| \hat h_\lambda $$
I don't know how to determine $r, \phi, \lambda$. Any tips?
\begin{align} \mathbf{r} &= \begin{pmatrix} 2t^3-2t \\ 4t^2 \\ t^3+t \end{pmatrix} \\ r &= \sqrt{(2t^3-2t)^2+(4t^2)^2+(t^3+t)^2} \\ &= t(t^2+1)\sqrt{5} \tag{$t>0$} \\ \dot{r} &= (3t^2+1)\sqrt{5} \\ \cos \theta &= \frac{z}{r} \\ &= \frac{1}{\sqrt{5}} \\ \sin \theta &= \frac{2}{\sqrt{5}} \\ \dot{\theta} &= 0 \\ \tan \phi &= \frac{y}{x} \\ \phi &= \tan^{-1} \frac{2t}{t^2-1} \\ \dot{\phi} &= -\frac{2}{t^2+1} \\ \mathbf{v} &= \dot{r} \, \mathbf{e}_r+ r\dot{\theta} \, \mathbf{e}_{\theta}+ r\dot{\phi} \sin \theta \, \mathbf{e}_{\phi} \\ &= \sqrt{5}(3t^2+1) \, \mathbf{e}_r-4t\, \mathbf{e}_{\phi} \\ \mathbf{T} &= \frac{(3t^2+1)\sqrt{5} \, \mathbf{e}_r- 4t\, \mathbf{e}_{\phi}}{\sqrt{45t^4+46t^2+5}} \end{align}