]> 6.6 The Gradient and Directional Derivatives

## 6.6 The Gradient and Directional Derivatives

We have seen above that the 2-vector

$( ∂ f ∂ x , ∂ f ∂ y ) ( x 0 , y 0 )$

is called the gradient of $f$ at argument $( x 0 , y 0 )$ and that it is generally written as $grad ⟶ f$ or $∇ ⟶ f$ .

The equation for the tangent plane to the surface defined by $f$ at $( x 0 , y 0 )$ can be described in terms of the gradient as

$f L ( x , y ) = f ( x 0 , y 0 ) + ( grad ⟶ f ) · ( r ⟶ − r 0 ⟶ )$

From this equation we can deduce that a normal to this tangent plane is in the direction in the three dimensional space whose coordinates represent $( x , y f L )$ of $( grad ⟶ f , − 1 )$ .

The projection of this normal into the $( x , y )$ plane is the vector $grad ⟶ f$ itself.

Thus, $grad ⟶ f$ is in the direction of the projection of the normal to the tangent plane to $f$ at $( x 0 , y 0 )$ into the $( x , y )$ plane.

This relationship can be seen in the applet below.

The symbol $∇ ⟶$ is called "del". It is a strange thing called a vector operator. By itself it makes about as much sense as the noise of one hand clapping. But put next to something that the derivatives in it can act on, it makes perfect sense.

The equation for the linear approximation $f L$ to $f$ at $( x 0 , y 0 )$ allows us to compute the directional derivatives of $f$ at that point.

Suppose we seek the directional derivative of $f$ in a direction defined by unit vector $u ^$ . Then if $r ⟶ − r 0 ⟶ = s u ^$ , the directional derivative of $f$ (which is close to $f L$ near $( x 0 , y 0 )$ ) in that direction is the derivative of $f L$ with respect to $s$ .

But we have

$f L ( x , y ) = f ( x 0 , y 0 ) + ( grad ⟶ f ) · ( r ⟶ − r 0 ⟶ ) = f ( x 0 , y 0 ) + ( grad ⟶ f ) · s u ^$

so that $f L$ 's derivative with respect to $s$ , the directional derivative of $f$ in the direction of $u ^$ , is given by $( grad ⟶ f ) · u ^$ .

If $f$ were a function of more variables, say $x , y , z , t , …$ we can use exactly the same approach to describe changes, the only difference being that the tangent plane becomes the tangent hyper-plane, and there are partial derivatives in more directions. The conclusions are exactly the same:

1. The gradient vector is in the direction of the projection of the normal to the tangent hyper-plane into the hyper-plane of coordinates.

2. The directional derivative in any direction is given by the dot product of a unit vector in that direction with the gradient vector.

3. The component of the gradient vector in the direction of any axis is the partial derivative of $f$ with respect to the corresponding distance variable in that direction.

4. That partial derivative is the ordinary derivative with respect to that variable assuming all the other variables remain constant.

The upshot of all this is that the gradient vector, whose components can be computed by ordinary one dimensional differentiation for a field in any number of dimensions, is all you need to compute its directional derivative in any direction.

If these concepts seem strange to you, play around with the applet below until you feel comfortable with them. The graph on the left shows the field represented on the right restricted to the cutting half plane pictured on the right. Its linear approximation at the edge of that half plane is also shown. The slope of that linear approximation is the directional derivative of the field at that edge, in the direction of the cutting half plane.

(Of course what is shown is not exactly a half plane, being rather a rectangle. The edge of it of interest here is the one that is the axis of rotation in the third slider.)