566 Partial derivatives
as valuable as the one that the rightmember of (11.242) is the scalar
product of two vectors. Thus(11.25)du=( i +au +au k).(dxi +d y +d% k).
dt ax ay az J at dt dt JThe remarkable feature of this product is the fact that, for each t, the
first vector depends only upon the first partial derivatives of u at the
point P(x,y,z) and the second vector is (when it is not 0) simply a forward
tangent to C at P.
Information is ready to gush from (11.25), and we make progress by
learning about the first vector which is called the gradient of u at P and
is denoted by Vu so that(11.26) Vu = ax i + ay j + az k.
The symbol V, an inverted delta, is read "del" and Vu is read "del u."
We must always remember that Vu is a vector. For many purposes, it
is very helpful to consider V itself to be an operator,(11.261) V = ax l +as i+ az k,
which can be applied to a scalar function u having continuous partial
derivatives to produce the gradient vector Vu. In case Vu = 0, this is
the whole story and there is nothing more to be learned. Henceforth
we suppose that Vu 0. Let the last vector in (11.25) be a unit vector v
so that du/dt is the directional derivative of u at P in the direction of v.
Then(11.262) dt = Ivul cos 0,
where JVul is the length of the gradient Vu and 0 is the angle between the
vectors Vu and v. Since -1 <= cos 0 < 1 and cos 0 = 1, it follows that
the direction of Vu is the direction in which the directional derivative of u
at P attains its maximum value and that the length of Vu is this maximum.
This is the fundamental intrinsic property of the gradient of u at P.
Since cos -r = -1, the direction opposite to that of Vu is the direction
in which the directional derivative of u at P attains its minimum value,
and -IVul is the minimum. Some applications are easy to understand.
If u is temperature and Mr. Walker is at a place that is too cold to suit
him, he will walk in the direction of the gradient of u, and the length of
the gradient will tell him the rate (in degrees per meter, for example)
at which his position becomes more comfortable.
Since cos r/2 = 0, directional derivatives in directions orthogonal to
the gradient Vu will be 0, and this can make us think about level surfaces