# Basic Question on Gradients

I am having trouble understanding how the gradient of a scalar field is the direction along the $x$-$y$ plane that yields the maximum inclination. Sure it takes into account the partial derivatives along both the x and y axes but does that mean "maximum inclination"? How do the partial derivatives come into the gradient vector field?
A basic intuition would be great.

Thank You in advance.

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If you are satisfied with the answers that people have given already, you should "accept" the one that was most helpful to you by clicking the gray check mark to the left of it. – Rahul Dec 24 '11 at 10:58

For a smooth function $f: \mathbb{R}^2 \to \mathbb{R}$, the gradient at a point $(x_0, y_0)$ is perpendicular to the level curve at that point. If $c = f(x_0, y_0)$, the level curve is $f(x,y) = c$. If you move along a level curve, then the value of $f$ does not change at all. If you move perpendicular to the level curve (i.e., in the direction of the gradient), then you are moving "as fast as possible" away from the level curve. That is what is meant by "maximum inclination"; if you think of $f$ as the height of a mountain, then the gradient is the direction which is locally the steepest.
First note that for a unit vector $v\in \mathbb{R}^n$ and a function which is differentiable in $x\in \mathbb{R}^n$, the derivative in the direction $v$ (which is defined as the limit of $(f(x+vt)-f(x))/t$ as $t\to 0$) is $\nabla f(x) \cdot v$ (use the chain rule to verify this).
Assuming $\nabla f(x)\ne 0$, the question is for which unit vector $v$ the number $\nabla f(x) \cdot v$ is maximal, and it turns out that this is the case for $v=\nabla f(x)/|\nabla f(x)|$ (the direction of the gradient); in fact this is precisely the Cauchy-Schwarz inequality.
OP should try to verify this chain rule. And it's not hard to see that $v\|\nabla f$ extremizes $D_vf$ with the identity $a\cdot b =\|a\| \|b\| \cos\theta$ in mind. – anon Oct 2 '11 at 9:32