Another question: From my understanding, we can think of the directional derivative as the rate at which a function changes at a point in some velocity $\dot{x}$. This definition, however, implies that the directional derivative will scale linearly with the norm of the direction vector. I.e. if we have two vectors $\dot{x_{1}}$ and $\dot{x_{2}}$ where $\dot{x_{2}} = 2\cdot \dot{x_{1}}$ then the directional derivative w.r.t $\dot{x_{2}}$ will be $2 \times$ the directional derivative w.r.t. $\dot{x_{1}}$.
My question is: doesn't it make more sense to think of the directional derivative as just the rate of change of our function in some unit direction? In other words, why don't we project the gradient in the direction of a unit vector $\dot{x} / |\dot{x}|$ so that we don't also scale it by $|\dot{x}|$ when we project? If we're trying to determine the rate at which our function changes in some direction then why would we want to consider the norm of that direction too?
Edit: I think I'm just overthinking this. The magnitude of the velocity implicitly encapsulates the chain rule when we define the directional derivative as the dot product of the velocity with the gradient.
kmcrane
@dvernet: Yes, sorry; that's just a typo. In TeX those symbols are "phi" and "varphi"; just different typographic variants on the same Greek letter. But yes, they should be the same.
kmcrane
@dvernet: Right. When you're first trying to get your head around the directional derivative, it's perhaps ok to just think about directions (unit vectors), and the fact that moving in different directions will result in different amounts of change. But of course, moving at different speeds in those directions will also affect the rate of change. So in general you'd like to be able to talk about the derivative with respect to any vector, not just unit vectors. Also, working with general vectors means that the directional derivative is linear in the direction of differentiation. Otherwise it would be nonlinear, since you would have to divide by the norm.
Should this be $d\varphi / dt$?
Another question: From my understanding, we can think of the directional derivative as the rate at which a function changes at a point in some velocity $\dot{x}$. This definition, however, implies that the directional derivative will scale linearly with the norm of the direction vector. I.e. if we have two vectors $\dot{x_{1}}$ and $\dot{x_{2}}$ where $\dot{x_{2}} = 2\cdot \dot{x_{1}}$ then the directional derivative w.r.t $\dot{x_{2}}$ will be $2 \times$ the directional derivative w.r.t. $\dot{x_{1}}$.
My question is: doesn't it make more sense to think of the directional derivative as just the rate of change of our function in some unit direction? In other words, why don't we project the gradient in the direction of a unit vector $\dot{x} / |\dot{x}|$ so that we don't also scale it by $|\dot{x}|$ when we project? If we're trying to determine the rate at which our function changes in some direction then why would we want to consider the norm of that direction too?
Edit: I think I'm just overthinking this. The magnitude of the velocity implicitly encapsulates the chain rule when we define the directional derivative as the dot product of the velocity with the gradient.
@dvernet: Yes, sorry; that's just a typo. In TeX those symbols are "phi" and "varphi"; just different typographic variants on the same Greek letter. But yes, they should be the same.
@dvernet: Right. When you're first trying to get your head around the directional derivative, it's perhaps ok to just think about directions (unit vectors), and the fact that moving in different directions will result in different amounts of change. But of course, moving at different speeds in those directions will also affect the rate of change. So in general you'd like to be able to talk about the derivative with respect to any vector, not just unit vectors. Also, working with general vectors means that the directional derivative is linear in the direction of differentiation. Otherwise it would be nonlinear, since you would have to divide by the norm.