13.5 Directional Derivatives and Gradients
As \(\displaystyle x\) changes, we know how \(\displaystyle f(x,\, y)\) changes. The partial derivative \(\displaystyle \partial f/\partial x\) treats \(\displaystyle y\) as constant. Similarly \(\displaystyle \partial f/\partial y\) keeps \(\displaystyle x\) constant and gives the slope in the \(\displaystyle y\) direction. But east-west and north-south are not the only directions to move. We could go along a 45° line, where \(\displaystyle \Delta x\, =\, \Delta y\). In principle, before we draw axes, no direction is preferred. The graph is a surface with slopes in all directions.
On that surface, calculus looks for the rate of change (or the slope). There is a directional derivative, whatever the direction. In the 45° case we are inclined to divide \(\displaystyle \Delta f\) by \(\displaystyle \Delta x\), but we would be wrong.
Let me state the problem. We are given \(\displaystyle f(x,\, y)\) around a point \(\displaystyle P\, =\, (x_0,\, y_0)\). We are also given a direction u (a unit vector). There must be a natural definition of \(\displaystyle D_u f\) — the derivative of f in the direction u. To compute this slope at \(\displaystyle P\), we need a formula. Preferably the formula is based on \(\displaystyle \partial f/\partial x\) and \(\displaystyle \partial f/\partial y\), which we already know.
Note that the 45° direction has u \(\displaystyle \, =\, \) i\(\displaystyle /\sqrt{2}\, +\, \) j\(\displaystyle /\sqrt{2}\). This shows that the square root of \(\displaystyle 2\) is going to enter the derivative. This shows that dividing \(\displaystyle \Delta f\) by \(\displaystyle \Delta x\) is wrong. We should divide by the step length \(\displaystyle \Delta s\).
