# I Velocity vector fields

1. Apr 6, 2016

### dyn

Hi. Given a one-parameter family of maps such as
Φt : ( x , y ) → ( xet + 2et -2 , ye2t ) the velocity vector field at t=0 is given by d(Φt)/dt = (x+2) ∂/∂x + 2y ∂/∂y
My question is ; how does differentiating a vector function Φt with respect to t result in a scalar function ? Thanks

2. Apr 6, 2016

### Orodruin

Staff Emeritus
It doesn't.

3. Apr 6, 2016

### BiGyElLoWhAt

How do you go from $\frac{d}{dt}[(xe^t + 2e^t - 2)\hat{x} + (ye^{2t})\hat{y}]$ to what you have?

4. Apr 6, 2016

### Orodruin

Staff Emeritus
The derivative is evaluated at $t = 0$. The tangent vector basis is $\partial_x$ and $\partial_y$.

5. Apr 6, 2016

### BiGyElLoWhAt

Ahh. The basis was what I was missing. I suppose that makes sense, maybe. I would think that we would be looking at $\frac{\partial \hat{x}}{\partial t}|_{t=0}$ and $\frac{\partial \hat{y}}{\partial t}|_{t=0}$, however.
Not trying to hijack the thread.
**I think I see the error of my ways, now.**

Last edited: Apr 6, 2016
6. Apr 6, 2016

### dyn

The function Φ t takes 2 coordinates ( x , y ) into 2 different coordinates. I would think this makes them vectors while the velocity vector field doesn't look like a vector.

Last edited by a moderator: Apr 6, 2016
7. Apr 6, 2016

### Orodruin

Staff Emeritus
Yes it does. Remember that the tangent vector basis is $\partial_\mu$.

8. Apr 7, 2016

### dyn

So the original function is a vector with 2 components ? And the velocity vector field is a vector with one component ?

9. Apr 7, 2016

### Orodruin

Staff Emeritus
No, both have two components. It is just a different way of writing it as a linear combination of the basis instead of a collection of components.