<s>[INST] Tangent vector of curve $ \Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T $ expressed in spherical coordinates 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?
 [/INST] \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}
</s>