# Tangent to reparameterized curve

1. Sep 5, 2010

### monea83

Given is a curve $$\gamma$$ from $$\mathbb{R} \rightarrow M$$ for some manifold M. The tangent to $$\gamma$$ at $$c$$ is defined as

$$(\gamma_*c)g = \frac{dg \circ {\gamma}}{du}(c)$$

Now, the curve is to be reparameterized so that $$\tau = \gamma \circ f$$, with f defining the reparametrization. (f' > 0 everywhere)

The book I'm reading claims that $$\tau_* = f' \cdot \gamma_* \circ f$$, however I do not quite see how this result is derived.

Using the chain rule, I get

$$(\tau_*c)g = \frac{dg \circ \gamma \circ f}{du}(c) =\frac{dg \circ \gamma \circ f}{df} \cdot \frac{df}{du}(c)$$

The second part of the rhs is obviously f', but how is the first part equal to $$\gamma_* \circ f$$?

2. Sep 5, 2010

### quasar987

Your notation is really bad. Try this:

The tangent to $$\gamma$$ at $$\gamma(u_0)$$ is the tangent vector $$\gamma_{*,u_0}$$ defined by

$$(\gamma_{*,u_0})g := \frac{d(g \circ {\gamma})}{du}(u_0)$$

So, for $$\tau:= \gamma \circ f$$ a reparametrization, the chain rule yields

$$(\tau_{*,t_0})g = \frac{d(g \circ \gamma \circ f)}{dt}(t_0) =\frac{d(g \circ \gamma)}{du}(f(t_0)) \frac{df}{dt}(t_0)=f'(t_0)(\gamma_{*,f(t_0)})g$$

That is to say,

$$\tau_{*,t_0}=f'(t_0)\cdot \gamma_{*,f(t_0)}$$

for all t_0.

Or, even more compactly,

$$\tau_*=f'\cdot \gamma_{*}\circ f$$

I highly recommend the book Introduction to Smooth Manifolds by John Lee.

Last edited: Sep 5, 2010
3. Sep 5, 2010

### monea83

Thank you very much, I think I can work it out now!

When you say my notation is bad, are you referring to my application of the chain rule (which I believe is flawed), or to the notation in which the problem was posed (which was taken from "Tensor analysis on manifolds", Bishop & Goldberg)?