MHB Cant quite see this grad relation

[ognik]

Book states: $\nabla f\left(u, v\right) =0 = \frac{\partial f}{\partial u}\nabla u + \frac{\partial f}{\partial v} \nabla v, \therefore \nabla u$ and $ \nabla v $ are parallel

1. I know $d f\left(u, v\right) = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v} dv$, but how can we just replace the d<...> by $\nabla <... >$ like that? seems a little circular?

2. Can't anyway see how that makes them parallel?
I get $ =\frac{\partial f}{\partial u}\left( \pd{u}{x}, \pd{u}{y},\d{u}{z}\right) + \frac{\partial f}{\partial v}\left( \pd{v}{x}, \pd{v}{y},\d{v}{z}\right)
= \left( \pd{f}{x}, \pd{f}{y},\d{f}{z}\right) + \left( \pd{f}{x}, \pd{f}{y},\d{f}{z}\right)$ so both tterms are equal and parallel, but why does that make the $\nabla$ parts parallel?

 

[I like Serena]

Book states: $\nabla f\left(u, v\right) =0 = \frac{\partial f}{\partial u}\nabla u + \frac{\partial f}{\partial v} \nabla v, \therefore \nabla u$ and $ \nabla v $ are parallel

1. I know $d f\left(u, v\right) = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v} dv$, but how can we just replace the d<...> by $\nabla <... >$ like that? seems a little circular?

It's a generalization for multiple dimensions.
The chain rule for multiple dimensions says:
$$\d f x = \pd f u \pd u x + \pd f v \pd v x$$

2. Can't anyway see how that makes them parallel?
I get $ =\frac{\partial f}{\partial u}\left( \pd{u}{x}, \pd{u}{y},\d{u}{z}\right) + \frac{\partial f}{\partial v}\left( \pd{v}{x}, \pd{v}{y},\d{v}{z}\right)
= \left( \pd{f}{x}, \pd{f}{y},\d{f}{z}\right) + \left( \pd{f}{x}, \pd{f}{y},\d{f}{z}\right)$ so both tterms are equal and parallel, but why does that make the $\nabla$ parts parallel?

Are they really parallel?
And does $\pd f u \pd u x = \pd f x$ hold?

Let's pick an example.
Suppose we pick $f(u,v)=u^2+v^2-1, \quad u(x,y)=x+y, \quad v(x,y)=x-y$
How would that work out?Oh, and I've moved your thread to Calculus, which is where Partial Derivatives and Multi-variable Calculus belongs.
Please check the descriptions of the forums on the main page.

 

[ognik]

It's a generalization for multiple dimensions.
The chain rule for multiple dimensions says:
$$\d f x = \pd f u \pd u x + \pd f v \pd v x$$.

So $ \pd{f}{x}, \pd{f}{y} = \nabla f = \left( \pd{f}{u}\pd{u}{x}+\pd{f}{v}\pd{v}{x}, \pd{f}{u}\pd{u}{y}+\pd{f}{v}\pd{v}{y} \right) = \pd{f}{u} \left( \pd{u}{x},\pd{u}{y} \right) + \pd{f}{v}\left( \pd{v}{x},\pd{v}{y} \right) ...$?

Are they really parallel?.

I couldn't see why they should be, but the book says if $\nabla f = 0$, then they are, I can't see this is a general case?

And does $\pd f u \pd u x = \pd f x$ hold?.

didn't understand why you asked, its missing the $\pd{f}{v}...$ part?

Oh, and I've moved your thread to Calculus, which is where Partial Derivatives and Multi-variable Calculus belongs.
Please check the descriptions of the forums on the main page.

Thanks, I was honestly undecided 'cos in my book this is a section on vectors, just looking at grad, curl etc...

 

[I like Serena]

So $ \pd{f}{x}, \pd{f}{y} = \nabla f = \left( \pd{f}{u}\pd{u}{x}+\pd{f}{v}\pd{v}{x}, \pd{f}{u}\pd{u}{y}+\pd{f}{v}\pd{v}{y} \right) = \pd{f}{u} \left( \pd{u}{x},\pd{u}{y} \right) + \pd{f}{v}\left( \pd{v}{x},\pd{v}{y} \right) ...$?

Correct.

I couldn't see why they should be, but the book says if $\nabla f = 0$, then they are, I can't see this is a general case?

I actually overlooked that it was given that $\nabla f = 0$.
In that case they have to be parallel, since the only way 2 vectors can sum up to zero is if they are equal and opposite.
Since the partial derivatives are only scalars, that implies that one vector must be a multiple of the other. That is, parallel.

didn't understand why you asked, its missing the $\pd{f}{v}...$ part?

It's not true, which should become clear if we work out the example I posted.

Thanks, I was honestly undecided 'cos in my book this is a section on vectors, just looking at grad, curl etc...

If you look at the descriptions of the forums, it should be clear where it should go, regardless of the section in your book.