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Wikepeda says: The gradient of $H$ at a point is a vector pointing in the direction of the steepest slope or grade at that point. The steepness of the slope at that point is given by the magnitude of the gradient vector.

Can one prove, without inner product that this must be so. I mean, given the partial derivatives $\frac{\partial H}{\partial x}$,$\frac{\partial H}{\partial y}$,$\frac{\partial H}{\partial z}$, etc... , can one prove that the gradient is vector pointing in the direction of the steepest slope?

My motivation is this: the gradient (as a sum of partial derivatives) existed before the concept of inner product/dot product. So I wonder how it was done.

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    $\begingroup$ Minor comment: the gradient is not a sum of partial derivatives. $\endgroup$ – littleO Jan 5 '17 at 22:16
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    $\begingroup$ The directional derivative of $f$ in the direction $u$ at $x$ is denoted $D_u f(x)$. It can be shown (by applying the chain rule) that $D_u f(x) = \langle \nabla f(x), u \rangle$ for any vector $u$. It then follows (from the Cauchy-Schwarz inequality) that if we restrict $u$ to being a unit vector, then $D_u f(x)$ is maximized when $u$ is parallel to $\nabla f(x)$. Is this proof not acceptable? Why avoid inner products when the directional derivative formula uses an inner product? $\endgroup$ – littleO Jan 5 '17 at 22:21
  • $\begingroup$ @littleO There are other definitions of directional derivatives that don’t. $\endgroup$ – amd Jan 6 '17 at 8:10
  • $\begingroup$ Have a look at this question and its answers. In several of them, the dot product enters as a consequence of basic definitions of directional derivatives and the gradient. $\endgroup$ – amd Jan 6 '17 at 8:13
  • $\begingroup$ little i thought tha proof of CSwarz also uses dot product $\endgroup$ – jan roelens Jan 6 '17 at 9:03
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Here is my intuition about the whole thing.

I think of the gradient as the vector perpendicular to a surface.

That is the equation for my surface is $f(x,y,z) = k$

for any $(x_0, y_0, z_0)$ satisfying the above.

$\frac {\partial f}{\partial x} (x-x_0) + \frac {\partial f}{\partial y} (y-y_0) + \frac {\partial f}{\partial y} (z-z_0) = 0$

Is the plane parallel to the surface and $\nabla f$ is normal to this plane.

So, now we add an extra degree of freedom and relax the requirement that $f(x,y,z) = k$

Any component of motion parallel to the plane I had prior to relaxing the constraint is not in the direction of maximal change in $f.$

$\nabla f$ points in the direction of maximal change in $f.$

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  • $\begingroup$ this answer does not satisfy me because gradient as perpendicular is a consequence of the definition of gradient as steepest hill and also consequence of dot product. I'm looking for the origin of the concept $\endgroup$ – jan roelens Jan 6 '17 at 9:22
  • $\begingroup$ What do you have against the dot product? The dot product (in this sense, as opposed to Hilbert spaces, etc.) is a direct fall-out of Euclidean geometry. Galileo discussed orthogonal projections of vectors. It certainly predates calculus. $\endgroup$ – Doug M Jan 6 '17 at 18:33
  • $\begingroup$ Doug, can you elaborate the Galileo discussion . I thought that dot product was a re arrangement of Hamiltons search of quatornions and became something on its own through Gibbs My point is : i'm looking for proofs of formulae for which we nowaydays use dot product , but for which the Original proof did not use the dot product . As an example Gauss first fundamental form in differential geometry. And also the gradiënt being or, perpendicular to level curve or being direction of maximal change. $\endgroup$ – jan roelens Jan 14 '17 at 21:08

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