# Tangent space of smooth manifold $M=\{(x,x^3,e^{x-1}) : x \in \Bbb{R}\}$ at $(1,1,1)$

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. :)

• Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point? – mastrok Mar 12 '18 at 17:35

## 1 Answer

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 $$\{ (p,v) : p = (1,1,1) \text{ and } v = (\lambda, 3\lambda, \lambda), \lambda \in \Bbb{R} \}.$$