Exercise

  1. In each of the following examples a scalar field is defined by the given function for all points $(x,y)$ in the plane for which the expression on the right is defined. In each example determine the set of points $(x,y)$ at which $f$ is continuous.

    1. MATH

    2. MATH

    3. MATH

    4. MATH

    5. MATH

    6. MATH

    7. MATH

    8. MATH

    9. $f(x,y)=x^{(y^{2})}$

    10. MATH

  2. Let MATH if $x+y\ne 0.$ Show that MATH but that MATH

  3. Let MATH whenever MATH Show that
    MATH
    but that $f(x,y)$ does not tend to a limit as MATH

Differentiability and Gradient

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
MATH
and called $f$ differentiable at $x$ provided that this quotient had a limit as $h$ tended to zero. In the multivariable case we can still form the difference
MATH
but the ``quotient''
MATH
makes no sense because to divide by a vector makes no sense.

We can get around this difficulty this way. If $g(h)$ is an expression in $h$, we say that $g(h)$ is MATH and write $g(h)=o(h)$ if
MATH
For a function of one variable the following statements are equivalent:
MATH

MATH

MATH

MATH

MATH
Thus, for a function of one variable, the derivative of $f$ at $x$ is the unique number $f^{\prime }(x)$ such that
MATH
It is this view of the derivative that inspires the notion of differentiability in the multivariable case.

Let us agree to call $o(h)$ any expression $g(h)$ for which
MATH

Definition of Differentiability

Let MATH We say that $f$ is differentiable at $x$ if there exists a vector $y$ such that
MATH
It is not hard to show that, if such a vector $y$ exists, it is unique. We call this unique vector the gradient of $f$ at $x$ and denote it by $\nabla f(x)$:

if $f$ is differentiable at $x$, the gradient of $f$ at $x$ is the unique vector $\nabla f(x)$ such that
MATH
The similarities between the one-variable case,
MATH
and the multivariable case,
MATH
are obvious. We point to the differences. There are essentially two of them:

(1) While the derivative $f^{\prime }(x)$ is a number, the gradient $\nabla f(x)$ is a vector.

(2) While $f^{\prime }(x)h$ is the ordinary product of two real numbers, $\nabla f(x)\cdot h$ is the dot product of two vectors.

Calculating Gradients

We write
MATH
in the two-variable case and
MATH
for the three-variable case.

Example 1. For the function
MATH
we have
MATH

MATH

MATH
The remainder MATH is $o(h):$
MATH
Thus
MATH

Example 2. For the function
MATH
we have
MATH

MATH

MATH
Here the remainder is identically zero and is thus certainly $o(h).$ Therefore


MATH

Theorem. If $f$ has continuous first partials in a neighborhood of $x$, then $f$ is differentiable at $x$ and
MATH
In two variables,
MATH

Proof. We prove the theorem in the two-variable case. A similar argument yields a proof in the three-variable case.

Adding and subtracting $f(x,y+h_2),$ we have
MATH

MATH

By the mean-value theorem for functions of one variable, we know that there are numbers
MATH
such that
MATH
and
MATH
By the continuity of MATH,
MATH
where
MATH
By the continuity of MATH,
MATH
where
MATH

Substituting these expressions, we find that
MATH

MATH

MATH

To complete the proof of the theorem we need only show that
MATH
From Schwarz's inequality, MATH we know that
MATH
It follows that
MATH

MATH
by the triangle inequality. The last term tends to $0$ as $h\rightarrow 0.$ This completes the proof of the theorem.

Example 3. For
MATH
we have
MATH
and therefore
MATH

When there is no reason to emphasize the point of evaluation, we don't write $\nabla f(x,y)$ or $\nabla f(x,y,z)$ but simply $\nabla f$. Thus for the function
MATH
we write
MATH
and
MATH

Example 4. For
MATH
we have
MATH
and
MATH

Example 5. We take the function
MATH
and evaluate $\nabla f$ at $(0,1,2).$

Here
MATH
At $(0,1,2)$
MATH
and thus $\nabla f=(0,1,0).$

Of special interest for later work are the powers of $\left\| r\right\| $ where, as usual, $r=(x,y,z).$

We begin by showing that
MATH

Proof.
MATH

MATH

MATH

MATH

MATH

MATH

MATH

MATH

The formulas we just derived can be generalized: for each integer $n,$
MATH

Differentiability Implies Continuity

As in the one-variable case, differentiability implies continuity; namely,

if $f$ is differentiable at $x$, then $f$ is continuous at $x$.

To see this, write
MATH
and note that
MATH
As $h\rightarrow 0$,
MATH

It follows that
MATH

Some Elementary Formulas

In many respects gradients behave just as derivatives do in the one-variable case. In particular, if $\nabla f(x)$ and $\nabla g(x)$ exist, then MATH MATH, and $\nabla [f(x)g(x)]$ all exist, and
MATH

MATH

MATH

To derive the third formula, let's assume that $\nabla f(x)$ and $\nabla g(x) $ both exist. Our task is to show that


MATH

Now
MATH

MATH

MATH


MATH


MATH

MATH


MATH

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
MATH

MATH

MATH
expressed in vector notation take the form
MATH

MATH

MATH

Each partial is thus the limit of a quotient
MATH
where $u$ is one of the unit coordinate vectors $i,j,$ or $k$. There is no reason to be so restrictive on $u$. If $f$ is defined in a neighborhood of $x $, then, for small $h$, the difference quotient
MATH
makes sense for any unit vector $u$.

DEFINITION For each unit vector $u$, the limit
MATH
if it exists, is called the directional derivative of $f$ at $x$ in the direction $u$.

The partials are of course themselves directional derivatives:
MATH

As the partials of $f$ give the rates of change of $f$ in the $i,j,k$ directions, the directional derivative $f_u^{\prime }$ gives the rate of change of $f$ in the direction $u$.

There is an important connection between the gradient at $x$ and the directional derivatives at $x$.

THEOREM If $f$ is differentiable at $x$, then $f$ has a directional derivative at $x$ in every direction $u$ and
MATH

Proof. We take $u$ as a unit vector and assume that $f$ is differentiable at $x$. The differentiability at $x$ tells us that $\nabla f(x)$ exists and
MATH
Division by $h$ gives
MATH
Since
MATH
we have
MATH

Earlier we saw that, if $f$ has continuous first partials in a neighborhood of $x$, then $f$ is differentiable at $x$ and
MATH
The next theorem shows that this formula for $\nabla f(x)$ holds wherever $f$ is differentiable.

THEOREM If $f$ is differentiable at $x$, then all the first partials exist at $x$ and
MATH

Proof. Assume that $f$ is differentiable at $x$. Then $\nabla f(x)$ exists and we can write
MATH
The result follows from observing that
MATH

MATH

MATH

Problem 1. Find the directional derivative of the function
MATH
at the point $(1,2)$ in the direction of the vector $2i-3j$.

Solution. In the first place, $2i-3j$ is not a unit vector; its norm is $\sqrt{13}$. The unit vector in the direction of $2i-3j$ is the vector
MATH
Next
MATH
and therefore
MATH
Hence


MATH

Problem 2. Find the directional derivative of the function
MATH
at the point $(1,\pi ,\pi /4)$ in the direction of the vector $2i-j+4k.$

Solution. The unit vector in the direction of $2i-j+4k$ is the vector
MATH
Here
MATH
so that
MATH
Therefore
MATH

MATH

MATH

Note that the directional derivative in a direction $u$ is the component of the gradient vector in that direction.

If $\nabla f(x)\ne 0$, then
MATH
where $\theta $ is the angle between $\nabla f(x)$ and $u$. It follows from the Cauchy-Schwarz's inequality that
MATH
for all directions $u$. If $u$ points in the direction of $\nabla f(x)$, then
MATH
and, if $u$ points in the direction of $-\nabla f(x)$
MATH
Since the directional derivative gives the rate of change of the function in that direction, it is clear that

a differentiable function $f$ increases most rapidly in the direction of the gradient (the rate of change is then MATH and it decreases most rapidly in the opposite direction (the rate of change is then MATH.

Example 3. The graph of the function
MATH
is the upper half of the unit sphere $x^2+y^2+z^2=1.$ The function is defined on the closed unit disc but differentiable only on the open unit disc.

The gradient
MATH
is a negative multiple of $r:$
MATH
Since $r$ points from the origin to $(x,y)$, the gradient points from $(x,y)$ to the origin. This means that $f$ 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 $P(x,y,f(x,y))$ 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
MATH

(a) In what direction does the temperature increase most rapidly at the point $(0,0)$? What is this rate of increase?

(b) In what direction does the temperature decrease most rapidly at $(0,0)$?

Solution
MATH

(a) At $(0,0)$ the temperature increases most rapidly in the direction of the gradient
MATH
This rate of increase is
MATH

(b) The temperature decreases most rapidly in the direction of
MATH

Problem 5. The mass density (mass per unit volume) of a metal ball centered at the origin is given by the function
MATH
$k$ a positive constant.

(a) In what direction does the density increase most rapidly at the point $(x,y,z)$? 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 $(x,y,z)$ in the $i,j,k$ directions?

Solution. The gradient
MATH

MATH
points from $(x,y,z)$ in the direction opposite to that of the radius vector.

(a) The density increases most rapidly toward the origin. The rate of increase is
MATH

(b) The density decreases most rapidly directly away from the origin.

(c) The rates of density change in the $i,j,k$ directions are given by the directional derivatives
MATH

MATH

MATH
These are just the first partials of $\lambda $.

Problem 6. The temperature at each point of a metal plate is given by the function
MATH
Find the path followed by a heat-seeking particle that originates at $(-2,1). $

Solution. The particle moves in the direction of the gradient vector
MATH
We want the curve
MATH
which begins at $(-2,1)$ and at each point has its tangent vector in the direction of $\nabla T$. We can satisfy the first condition by setting
MATH
We can satisfy the second condition by setting
MATH
These differential equations, together with initial conditions at $t=0$, imply that
MATH
We can eliminate the parameter $t$ by noting that
MATH
In terms of just $x$ and $y$ we have
MATH
The particle moves from the point $(-2,1)$ along the left branch of the hyperbola $xy=-2$ in the direction of decreasing $x$.

Remark. The pair of differential equations


MATH
can be set as a single differential equation in $x$ and $y$: the relation
MATH
gives
MATH
This equation is readily solved directly:
MATH

MATH

MATH

MATH
Thus $xy$ is constant:
MATH
Since the curve passes through the point $(-2,1),k=-2$ and once again we have the curve
MATH

This document created by Scientific WorkPlace 4.0.