1. ## [SOLVED] The Gradient Vector

I have to show that if one of the following statements is true, then the other statement is also true.
Let f be a scalar field and let P be a point in the domain of f.

There is some Cartesian system in which f is represented by a function which has a linear approximation at P.

For every choice of Cartesian system, f is represented by a function which has linear approximation at P.

I don't really know what either statement means by itself, so it is difficult to show how one implies the other. In particular, what does it mean to choose a different Cartesian system (that sounds like just moving the origin around). Further, I don't know what the book is saying in an earlier paragraph when they say "the gradient vector can be visualized without reference to a particular coordinate system"

The only thing that I do know is that to have a linear approximation at a point means that the function is both continuous at the point and has partial derivatives at the point (there might be cases where this is not enough to say that a linear approximation exists however).

Thanks for helpful responses

2. For some function $f \in C^1, \nabla f = (\frac{\partial f}{dx},\frac{\partial f}{dy},\frac{\partial f}{dz})$ in rectangular coordinates. From the definition of derivate can be shown that $\nabla f$ it's a good aproximation of f. Further reading, Marsden & Tromba Vector Calculus, section 2.3 differentiation.

3. I don't know what to say

4. A)There is some Cartesian system in which f is represented by a function which has a linear approximation at P.

B)For every choice of Cartesian system, f is represented by a function which has linear approximation at P.

If we prove that B is true, then A is true.

What book are you talking about? It will be helpful to know, for useful help lol. Sorry for the non-helping previous post.

5. I don't like the way this question is worded- there is no statement of how good the approximation is. "Having a linear approximation" doesn't mean anything by itself. Any function can be "approximated", perhaps with large error, by a linear or even constant function.

But if that linear approximation has the property that E(h), the error within distance h of the point, goes to 0 faster than h itself, then the function is differentiable at that point. And being differentiable is a property of the function, not the coordinate system. If a function is differentiable at a point in one coordinate system, it is differentiable there in any coordinate system.