Velocity Vector Fields: Differentiating a Vector Function to Scalar

[dyn]

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

 

[Orodruin]

My question is ; how does differentiating a vector function Φt with respect to t result in a scalar function ?

It doesn't.

 

[BiGyElLoWhAt]

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

 

[Orodruin]

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

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

 

[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.**

 

[dyn]

It doesn't.

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.

 

[Orodruin]

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.

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

 

[dyn]

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

 

[Orodruin]

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

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.