Motivation for tangent space
The isomorphism between tangents and derivations is surprising! As is the fact it is a linear map. However with calculus one is not far from that definition already.
Motivating example
Let us take (informally, because cases where [ilmath]\theta=\tfrac{\pi}{2}[/ilmath] and [ilmath]r=0[/ilmath] must be treated carefully) the manifold of the plane. The reader should be familiar with polar coordinates (giving things as an angle and a distance from the origin, rather than [ilmath]x[/ilmath] and [ilmath]y[/ilmath])
We will have two ways of looking at points, as [ilmath](x,y)[/ilmath] - traditionally, and [ilmath](r,\theta)[/ilmath] where:
- [ilmath](r,\theta)\mapsto(r\cos(\theta),r\sin(\theta))[/ilmath]
- [ilmath](x,y)\mapsto\left(\sqrt{x^2+y^2},\arctan(\frac{y}{x})\right)[/ilmath]
The line
Take the line [ilmath]y=mx+c[/ilmath], where [ilmath]m[/ilmath] is the gradient and [ilmath]c[/ilmath] is the intercept with the y axis, writing this we see the line can be given as:
| Form | First coord | Second coord |
|---|---|---|
| [ilmath]x,y[/ilmath] | [ilmath]x=t[/ilmath] | [ilmath]y=mt+c[/ilmath] |
| [ilmath]r,\theta[/ilmath] | [ilmath]r=\sqrt{t^2(m^2+1)+2mct+c^2}[/ilmath] | [ilmath]\theta=\arctan\left(m+\frac{c}{t}\right)[/ilmath] |
| Pure forms | ||
| Form | map | |
| Cartesian | [ilmath]y=mx+c[/ilmath] | |
| Polar[1] | [math]r=|c|\sqrt{\frac{(m^2+1)}{(\tan(\theta)-m)^2}+\frac{2m}{\tan(\theta)-m}+1}[/math] | |
These formulas are easily found from substitution.
TODO: Add picture, talk about tangents between the two!
