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What's the tangent space of $M=\{(x,x^3,e^{x-1}): x \in \Bbb{R}\}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^\infty$.

I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M=\{(x,y,z) \in \Bbb{R}^3: x^2+y^2-z^2=1\}$. The tangent space of $M$ in this case $= \ker(\mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.

Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)

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  • $\begingroup$ Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point? $\endgroup$ – mastrok Mar 12 '18 at 17:35
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From the comment above by @mastrok.


Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $$\{ (p,v) : p = (1,1,1) \text{ and } v = (\lambda, 3\lambda, \lambda), \lambda \in \Bbb{R} \}.$$

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