Velocity of a curve meaning

Question

This is from p.69 of John Lee's Introduction to Smooth Maniflds. Here $J$ is just an interval of $\mathbb{R}$. I am quite suspicious that this definition agrees with the usual of definition of velocity for curves whose codomains are Euclidean spaces, but entirely certain. For example. for a curve $f(t)=(t,0)$, Am I justified to write $f'(t)$ whose definition is given above just as $(1,0)$? I am very curious.

Answer 1

Yes you are.

As you can see in the later part of the chapter, if you a mapping F from $M\rightarrow N$ then to compute the differential $DF_p$ you just compute the Jacobian of the represantation $[D(ψ^-1Fφ)_p]$ where $ψ,φ$ ar charts in $M,N$ repsectively.

Now, in your case, you have a mapping $f$ from $\mathbb{R^2}\rightarrow \mathbb{R}$ and therefore we can take $ψ=φ=id$ therefore $[D(ψ^{-1}fφ)_p]=[Df_p]$.

One good thing to do when first studing smooth manifold theory is to always check if the new definitions don't change what we already now from usual vector calculus.

P.S. Yes for curves $f(t)=(f_1(t),...,f_n(t))$ from $\mathbb{R}\rightarrow \mathbb{R^n} $ we have that $f'(t)=(f'_1(t),...,f'_n(t))$. Note that what we are saying is just a special case of the general case of the Jacobian matrix.

Answer 2

Your example is correct.

Let $U \subset \mathbb R^2$, and ${\bf X} : U \to \mathbb R^3$.

Let $I \subset \mathbb R$, and $\gamma : I \to U$.

The composite ${\bf X} \circ \gamma$ is given by $({\bf X} \circ \gamma) \ : \ I \stackrel{\gamma}{\longrightarrow} U \stackrel{{\bf X}}{\longrightarrow} \mathbb R^3$.

For any $t \in I$, we get $(u(t),v(t)) \in U$ and then $\alpha(t):=({\bf X} \circ \gamma)(t) = {\bf X}(u(t),v(t)) \in \mathbb R^3$.

Using the Chain Rule gives $\dot{\alpha}(t)=\dot u{\bf X}_u +\dot v{\bf X}_v$.

This is a vector in $\mathbb R^3$, based at ${\bf X}(u(t),v(t))$ and tangent to ${\bf X}(U)$.

As an example:

Let ${\bf X}(u,v) = (u,v,u^2-v^2)$ and $\gamma(t) = (\cos t, \sin t)$. In this case $U=\mathbb R^2$ and $I = [0,2\pi)$.

$({\bf X} \circ \gamma)(t) = (\cos t, \sin t, \cos^2t - \sin^2t)$.

Simplifying gives $({\bf X} \circ \gamma)(t) = (\cos t, \sin t, \cos 2t)$.

Hence $\dot \alpha(t) = (-\sin t, \cos t, -2\sin 2t)$.