What is the geometrical meaning of the total differential?

517 Views Asked by At

Could anybody give me a geometrical explanation of the total differential, if there is such? For me(a non-mathematician) it just looks like the generalization of the derivative to more dimensions but in those higher dimensions it hasn't got a geometrical meaning. Am i right? Thank you.

1

There are 1 best solutions below

1
On

Just to make sure we talk about the same thing: the differential of a (differentiable) map $\mathbb{R}^n\rightarrow \mathbb{R}$ is defined as

$$df = \frac{\partial f}{\partial x^1} dx^1 + \dots+ \frac{\partial f}{\partial x^n} dx^n$$ There are more complex cases (vector valued maps or complex functions) which I'll ignore here.

I assume you know what $\frac{\partial f}{\partial x^i} $ is, so the question remains what $dx^i$ is. Actually, if you do it stringently, it's just the same, it's the total differential of the coordinate function $x^i$ which maps a vector $v$ with coordinates $v= \sum v^l e_l$ to the ith component, i.e. to the coefficient of $e_i$. So $x^i(v)= v^i$.

Now in general the differential of a map is the first order (linear) approximation of that map, which in case of the coordinate functions (which are linear maps) is just the map itself: $dx^i(v) = v^i$. This looks a bit tautological and in Euclidean space it, to some extent, is, but in more general spaces it is important to keep track of the additional information at which point this is evaluated. You don't just look at some vector $v$ but to some vector $v$ attached to some point of the space, and also the differential is restricted to the 'tangent' space in that point. In Euclidean space you can, up to some extent, ignore this additional complexity, at the cost of being less precise which can then cause confusion.

Back to the (geometrical) meaning of the total differential: with the excurse about coordinate functions in mind you can now easily verify that $$df(v) =\frac{\partial f}{\partial x^1} dx^1 (v) + \dots+ \frac{\partial f}{\partial x^n} dx^n (v) = \sum_l \frac{\partial f}{\partial x^l}v^l $$ is just the directional derivative of $f$ in direction $v$ (in some point $p$, say), which you can also write as $$\frac{d}{dt}f(p+tv))|_{t=0}$$ which is just the rate of change of $f$ in direction $v$ in $p$ or the slope of the tangent to the graph of $f$ in $p$ in direction $v$ (similar as in the case of real functions).

Yet another way you may have already encountered to write this is down is $$\langle \nabla f(p),v\rangle$$ the scalar product of the gradient of $f$ (in $p$) whith $v$, which has, of course the same geometrical meaning. The difference is that in the first case it is expressed by using a linear map ($df(p)$), in the second case by a vector ($\nabla f(p)$). I linear algebra you may have learned that linear functionals and vectors are in a one to one correspondence to each other, and in case you have a scalar product this correspondence is defined through the mapping of a vector $X$ to the linear map $v\mapsto \langle X,v\rangle$. This is exactly the correspondence between $\nabla f$ and $df$ mentioned earlier.

So, as far as I'm concerned, it has a very precise geometrical meaning (and is used throughout a branch of mathematics which is called differential geometry) to analyze the geometry of surfaces, curves and more general objects called manifolds.

(My apologies to all those who may think this is not exact enough, similarly to everyone for whom this is still to complicated. And to all those who have already seen this a thousand times) ;-)