is [imath]f:\mathbb{R}\rightarrow \mathbb{R}^2, f(t)=(t^2, t^3)[/imath] an embedding?

[MathNugget]

I am using this definition:
An embedding is a function [math]f:M\rightarrow N[/math], [imath]m=dim(M)\leq n=dim(N)[/imath] (dim is dimension) that satifies [imath]rg (d_xf)=m, \forall x\in M[/imath] (the differential is injective).

I suppose I have to calculate [imath]\frac{df_1}{dt}, \frac{df_2}{dt}[/imath], with [imath]f(t)=(f_1(t), f_2(t))[/imath] ?
Also, rg here is the rank of the matrix... there supposedly should be a matrix I get out of this...

 

[blamocur]

I suppose I have to calculate df1dt,df2dt\frac{df_1}{dt}, \frac{df_2}{dt}dtdf1,dtdf2, with f(t)=(f1(t),f2(t))f(t)=(f_1(t), f_2(t))f(t)=(f1(t),f2(t)) ?

You suppose correctly. What is the resulting differential in your example?

 

[MathNugget]

You suppose correctly. What is the resulting differential in your example?

Wouldn't it be [imath](2t, 3t^2)=(\frac{df_1}{dt}, \frac{df_2}{dt})[/imath]. The rank would be 0 in 0... so it's not an embedding, innit?

 

[blamocur]

Wouldn't it be [imath](2t, 3t^2)=(\frac{df_1}{dt}, \frac{df_2}{dt})[/imath]. The rank would be 0 in 0... so it's not an embedding, innit?

That would be my answer too.

 

[MathNugget]

That would be my answer too.

Finally, we're getting some easy tasks I can comprehend. I'm starting to regain my hope for Monday. Thanks for help!!