# Exercise

1. 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.

2. Let if Show that but that

3. Let whenever Show that

but that does not tend to a limit as

## Differentiability

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

# Definition of Differentiability

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.

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

# Differentiability Implies Continuity

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

# Some Elementary Formulas

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

# Directional Derivatives

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.

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?

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

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