In each of the following examples a scalar field is defined by the given function for all points in the plane for which the expression on the right is defined. In each example determine the set of points at which is continuous.
Let if Show that but that
Let
whenever
Show that
but that
does not tend to a limit as
Our object here is to extend the notion of differentiability from functions of
one variable to functions of several variables. Partial derivatives alone do
not fulfill this role because they reflect behavior only in particular
directions. In the one-variable case we formed the difference quotient
and called
differentiable at
provided that this quotient had a limit as
tended to zero. In the multivariable case we can still form the difference
but the ``quotient''
makes no sense because to divide by a vector makes no sense.
We can get around this difficulty this way. If
is an expression in
,
we say that
is
and write
if
For a function of one variable the following statements are equivalent:
Thus, for a function of one variable, the derivative of
at
is the unique number
such that
It is this view of the derivative that inspires the notion of
differentiability in the multivariable case.
Let us agree to call
any expression
for which
Let
We say that
is differentiable at
if there exists a vector
such that
It is not hard to show that, if such a vector
exists, it is unique. We call this unique vector the gradient
of
at
and denote it by
:
if
is differentiable at
,
the gradient of
at
is the unique vector
such that
The similarities between the one-variable case,
and the multivariable case,
are obvious. We point to the differences. There are essentially two of them:
(1) While the derivative is a number, the gradient is a vector.
(2) While is the ordinary product of two real numbers, is the dot product of two vectors.
We write
in the two-variable case and
for the three-variable case.
Example 1. For the function
we have
The remainder
is
Thus
Example 2. For the function
we have
Here the remainder is identically zero and is thus certainly
Therefore
Theorem. If
has continuous first partials in a neighborhood of
,
then
is differentiable at
and
In two variables,
Proof. We prove the theorem in the two-variable case. A similar argument yields a proof in the three-variable case.
Adding and subtracting
we have
By the mean-value theorem for functions of one variable, we know that there
are numbers
such that
and
By the continuity of
,
where
By the continuity of
,
where
Substituting these expressions, we find that
To complete the proof of the theorem we need only show that
From Schwarz's inequality,
we know that
It follows that
by the triangle inequality. The last term tends to
as
This completes the proof of the theorem.
Example 3. For
we have
and therefore
When there is no reason to emphasize the point of evaluation, we don't write
or
but simply
.
Thus for the function
we write
and
Example 4. For
we have
and
Example 5. We take the function
and evaluate
at
Here
At
and thus
Of special interest for later work are the powers of where, as usual,
We begin by showing that
Proof.
The formulas we just derived can be generalized: for each integer
As in the one-variable case, differentiability implies continuity; namely,
if is differentiable at , then is continuous at .
To see this, write
and note that
As
,
It follows that
In many respects gradients behave just as derivatives do in the one-variable
case. In particular, if
and
exist, then
,
and
all exist, and
To derive the third formula, let's assume that and both exist. Our task is to show that
Now
Here we take up an idea that generalizes the notion of partial derivative. Its
connection to gradients will be made clear as we go on. The partial
derivatives
expressed in vector notation take the form
Each partial is thus the limit of a quotient
where
is one of the unit coordinate vectors
or
.
There is no reason to be so restrictive on
.
If
is defined in a neighborhood of
,
then, for small
,
the difference quotient
makes sense for any unit vector
.
DEFINITION For each unit vector
,
the limit
if it exists, is called the directional derivative of
at
in the direction
.
The partials are of course themselves directional derivatives:
As the partials of give the rates of change of in the directions, the directional derivative gives the rate of change of in the direction .
There is an important connection between the gradient at and the directional derivatives at .
THEOREM If
is differentiable at
,
then
has a directional derivative at
in every direction
and
Proof. We take
as a unit vector and assume that
is differentiable at
.
The differentiability at
tells us that
exists and
Division by
gives
Since
we have
Earlier we saw that, if
has continuous first partials in a neighborhood of
,
then
is differentiable at
and
The next theorem shows that this formula for
holds wherever
is differentiable.
THEOREM If
is differentiable at
,
then all the first partials exist at
and
Proof. Assume that
is differentiable at
.
Then
exists and we can write
The result follows from observing that
Problem 1. Find the directional derivative of the function
at the point
in the direction of the vector
.
Solution. In the first place,
is not a unit vector; its norm is
.
The unit vector in the direction of
is the vector
Next
and therefore
Hence
Problem 2. Find the directional derivative of the function
at the point
in the direction of the vector
Solution. The unit vector in the direction of
is the vector
Here
so that
Therefore
Note that the directional derivative in a direction is the component of the gradient vector in that direction.
If
,
then
where
is the angle between
and
.
It follows from the Cauchy-Schwarz's inequality that
for all directions
.
If
points in the direction of
,
then
and, if
points in the direction of
Since the directional derivative gives the rate of change of the function in
that direction, it is clear that
a differentiable function increases most rapidly in the direction of the gradient (the rate of change is then and it decreases most rapidly in the opposite direction (the rate of change is then .
Example 3. The graph of the function
is the upper half of the unit sphere
The function is defined on the closed unit disc but differentiable only on the
open unit disc.
The gradient
is a negative multiple of
Since
points from the origin to
,
the gradient points from
to the origin. This means that
increases most rapidly toward the origin. This is borne out by the observation
that along the hemispherical surface the path of steepest ascent from the
point
is the ``great circle route to the north pole.''
Problem 4. The temperature at each point of a metal plate is
given by the function
(a) In what direction does the temperature increase most rapidly at the point ? What is this rate of increase?
(b) In what direction does the temperature decrease most rapidly at ?
Solution
(a) At
the temperature increases most rapidly in the direction of the gradient
This rate of increase is
(b) The temperature decreases most rapidly in the direction of
Problem 5. The mass density (mass per unit volume) of a metal
ball centered at the origin is given by the function
a positive constant.
(a) In what direction does the density increase most rapidly at the point ? What is this rate of density increase?
(b) In what direction does the density decrease most rapidly?
(c) What are the rates of density change at in the directions?
Solution. The gradient
points from
in the direction opposite to that of the radius vector.
(a) The density increases most rapidly toward the origin. The rate of increase
is
(b) The density decreases most rapidly directly away from the origin.
(c) The rates of density change in the
directions are given by the directional derivatives
These are just the first partials of
.
Problem 6. The temperature at each point of a metal plate is
given by the function
Find the path followed by a heat-seeking particle that originates at
Solution. The particle moves in the direction of the gradient vector
We want the curve
which begins at
and at each point has its tangent vector in the direction of
.
We can satisfy the first condition by setting
We can satisfy the second condition by setting
These differential equations, together with initial conditions at
,
imply that
We can eliminate the parameter
by noting that
In terms of just
and
we have
The particle moves from the point
along the left branch of the hyperbola
in the direction of decreasing
.
Remark. The pair of differential equations
can be set as a single differential equation in
and
:
the relation
gives
This equation is readily solved directly:
Thus
is constant:
Since the curve passes through the point
and once again we have the curve