Question about grad f on the x-y plane

[7thSon]

Suppose I have a smooth scalar function f defined on some region in the x-y plane. Its partial derivatives with respect to x and y are well-defined.

Someone explain this "proof" to me that the quantity:

\frac{dx}{df} = \frac{f_{,x}}{f_{,x}^2 + f_{,y}^2} (similar expression for dy/df)

The authors from which I read this do this by taking the the total differential of f
df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy

dividing both sides by dx:
\frac{df}{dx} = frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}

and substituting in the relation

\frac{dy}{dx} = \frac{f_{,y}}{f_{,x}}

which gives you the resulting expression.

I would be fine with this if they substituted in the normal result of the implicit function theorem, which gives you, for a level set of f,
\frac{dy}{dx} = - \frac{f_{,x}}{f_{,y}}

However, it seems like since they decided they were more interested in the normal to the level set, rather than the tangent to the level set, they substituted in the different value of dy/dx! Why are they allowed to substitute in a different value for dy/dx, or is there something that I'm missing?

Another thing I was thinking was maybe there is a dual to the idea of the dot product of grad f with a tangent vector? But instead, between a total differential and a covector [dx,dy]? Is that why they can use an expression for dy/dx that is not the tangent to the level curve?

 

[7thSon]

Maybe someone could start out by explaining to me whether or not it is correct to write, as this author does,

\frac{dx}{df} = \frac{\partial x}{\partial f} + \frac{\partial x}{\partial y} \frac{dy}{df}

Isn't \frac{\partial x}{\partial y} always zero?

 

[7thSon]

This thread has gotten a lot of views but no responses, can someone at least say because they are perplexed or maybe my initial post is too long :P

 

[HallsofIvy]

If x and y are independent variables, then, yes, dx/dy is 0. And the formula you give makes no sense. However, I suspect that this is a situation where x and y are not independent. If f(x,y) is defined for all (x, y) in the plane but we then restrict ourselves to some curve y= u(x), we say that
\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}

which leads to the results given.

 

[7thSon]

If x and y are independent variables, then, yes, dx/dy is 0. And the formula you give makes no sense. However, I suspect that this is a situation where x and y are not independent. If f(x,y) is defined for all (x, y) in the plane but we then restrict ourselves to some curve y= u(x), we say that
\frac{df}{dx}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx} which leads to the results given.

<br /> <br /> Thanks HallsOfIvy. So this makes sense that there is a restriction of y to be a function of x, or vice versa... given that the partial derivative identity i described looks like a consequence of the implicit function theorem (though strangely different for a reason that still eludes me).<br /> <br /> What I am still hoping someone could clarify are the geometric differences between<br /> <br /> \frac{dy}{dx} vs. \frac{\partial y}{\partial x}<br /> \frac{df}{dx} vs. \frac{\partial f}{\partial x} and<br /> \frac{dx}{df} vs. \frac{\partial x}{\partial f}<br /> <br /> Most confusingly, the exact derivative \frac{dy}{dx} seems to be on a level set of f, which seems to be the same case I would get if I wrote y as an implicit function of x restricted to some level contour I was considering. (i.e. why is that any different from \frac{\partial y}{\partial x}

 

[7thSon]

Ok, I think 60% of my remaining doubts will be answered if someone can answer the following question:

We know that the tangent vector

\mathbf{v} = \nabla f = \frac{\partial f}{\partial x} \mathbf{e_1} + \frac{\partial f}{\partial y} \mathbf{e_2}

points in the direction of maximum ascent of the scalar function f at some point p.

In the same way, does the covector [dx,dy] = [ f_{,x} , f_{,y} ] point in the fastest rate of incremental ascent of f in the covector space?

Am I just completely making up concepts that don't exist? :)