We begin with a nonconstant function that is continuously differentiable. (This means is differentiable and its gradient is continuous.) We have seen that at each point of the domain the gradient vector, if not , points in the direction of the most rapid increase of Here we show that

**Theorem.** At each point of the domain, the gradient vector, if
not
is perpendicular to the level curve that passes through that point.

Proof. We choose a point
in the domain and assume that
.
The level curve through this point has equation

Under our assumptions on
,
this curve can be parametrized in a neighborhood of
by a continuously differentiable vector function

with nonzero tangent vector
.

Now take
such that

We
will show that

Since
is constantly
on the curve, we have

For
such

In particular

and
thus

**Example 1.** For the function

the
level curves are concentric circles:

At
each point
the gradient vector

points away from the origin along the line of the radius vector and is thus
perpendicular to the circle in question. At the origin the level curve is
reduced to a point and the gradient is simply
.

Consider now a curve in the
-plane

As
before we assume that
is nonconstant and continuously differentiable. Suppose that
lies on the curve and
.
We can view
as the
-level
curve of
and conclude that the gradient

is
perpendicular to
at
.
We call it a **normal vector.**

The vector

is
perpendicular to the gradient:

It
is therefore a **tangent vector**. The line through
perpendicular to the gradient is the tangent line. A point
will lie on the tangent line if and only if

that is, if and only if

This is an equation for the** tangent line.**

The line through
perpendicular to the tangent vector
is the normal line. A point
will lie on the normal line if and only if

This is an equation for the **normal line.**

**Example 2. **Choose a point
on the hyperbola

To
avoid denominators, we write

This equation is of the form

Partial differentiation gives

At
the gradient

is
normal to the curve and the vector

is
tangent to the curve. The equation of the tangent line can be written

Dividing by
and noting that
we can simplify the equation to

The
equation of the normal line takes the form

This simplifies to

Here, instead of level curves, we have level surfaces, but the results are
similar. *If
is nonconstant and continuously differentiable, then at each point of the
domain, the gradient vector, if not
is perpendicular to the level surface that passes through that point.*

Proof. We choose a point
in the domain and assume that
.
The level surface through this point has equation

We
suppose now that

is
a differentiable curve that lies on this surface and passes through the point
.
We choose
so that

and
suppose that

Since the curve lies on the given surface, we have

For
such

In particular,

The
gradient vector

is
thus perpendicular to the curve in question.

This same argument applies to *every* differentiable curve that lies on
this level surface and passes through the point
with nonzero tangent vector. Consequently, we view
as perpendicular to the surface itself.

**Example 3.** For the function

the
level surfaces are concentric spheres:

At
each point
the gradient vector

points away from the origin along the line of the radius vector and is thus
perpendicular to the sphere in question. At the origin the level surface and
the gradient is

The **tangent plane** to a surface

at
a point
is the plane through
with normal
The tangent plane at a point
is the plane through
that best approximates the surface in a neighborhood of
.

A point
lies on the tangent plane through
if and only if

In
Cartesian coordinates the equation takes the form

Problem 4. Find an equation in
for the plane tangent to the surface

*Solution*. The surface is of the form

Observe that

At
the point

The equation for the tangent plane can therefore be written

This simplifies to

**Problem 5.** The curve

intersects the surface
at the point
What is the angle of intersection?

*Solution.* We want the angle between the tangent vector of the curve
and the tangent plane of the surface at the point of intersection. A simple
calculation shows that the curve passes through the point
at
.
Since

we
have

Now set

This function has gradient
At the point

Now let
be the angle between
and this gradient.

A surface of the form

can be written in the form

by
setting

If
is differentiable, so is
.
Moreover,

The tangent plane at
thus has equation

which we can rewrite as

If
,
then both partials of
are zero at
and the equation reduces to

In
this case the tangent plane is horizontal.

**Problem 6.** Find an equation for the plane tangent to the
surface

*Solution*. Set

Partial differentiation gives

When
and

At the point
the tangent plane has equation

This simplifies to

**Problem 7.** At what points of the surface

is
the tangent plane horizontal?

*Solution.* The function

has first partials

These partials are both zero only at
and
The surface has a horizontal tangent plane only at
and
.

**DEFINITION** LOCAL MAXIMUM AND LOCAL MINIMUM

Let be a function of several variables and let be an interior point of the domain:

is said to have a **local maximum** at
if
for all
in some neighborhood of
;

is said to have a **local minimum** at
if
for all
in some neighborhood of
.

As in the one-variable case, the local maxima and local minima together
comprise the *local extreme values*. In the one-variable case we know
that, if
has a local extreme value at
,
then

either or does not exist.

We have a similar result for functions of several variables.

**THEOREM **If
has a local extreme value at
,
then

either or does not exist.

Proof. We assume that has a local extreme value at and that is differentiable at [that exists]. We need to show that . The three-variable case is similar.

Since
has a local extreme value at
,
the function
has a local extreme value at
.
Since
is differentiable at
,
is differentiable at
and therefore

Similarly, the function
has a local extreme value at
and, being differentiable there, satisfies the relation

The gradient is
since both partials are
.

Interior points of the domain at which the gradient is zero or the gradient
does not exist are called **critical points**. By the above
theorem these are the only points at which a local extreme value can occur.

Although the ideas introduced so far are completely general, their application to functions of more than two variables is generally laborious. We restrict ourselves mostly to functions of two variables. Not only are the computations less formidable, but also we can make use of our geometric intuition.

We suppose for the moment that
is defined on an open set and is continuously differentiable there. The graph
of
is a surface

Where
has a local maximum, the surface has a local high point. Where
has a local minimum, the surface has a local low point. Where
has either a local maximum or a local minimum, the gradient is
and therefore the tangent plane is horizontal.

A zero gradient signals the possibility of a local extreme value; it does not guarantee it. For example, in the case of the saddle-shaped surface, there is a horizontal tangent plane at the origin and therefore the gradient is zero there, yet the origin gives neither a local maximum nor a local minimum.

Critical points at which the gradient is zero are called **stationary
points**. The stationary points that do not give rise to extreme values
are called **saddle points.**

Below we test some differentiable functions for extreme values. In each case, our first step is to seek out the stationary points.

**Example 1.** For the function

we
have

To
find the stationary points, we set
.
This gives

and

The only simultaneous solution to these equations is . The point is therefore the only stationary point. We now compare the value of at with the values of at nearby points

The difference

is
nonnegative for all small
and
(in fact *for all real*
and
).
It follows that
has a local minimum at
.
This local minimum is

**Example 2**. In the case of

we
have

The gradient is
where

and .

The only simultaneous solution to these equations is . The point is the only stationary point.

We now compare the value of
at
with the values of
at nearby points

The difference

does not keep a constant sign for small
and
.
It follows that
is a saddle point.

The function in the next example has an infinite number of stationary points.

**Example 3.** For all points
with
,
set

Then

The gradient is
where

Solving these two equations simultaneously, we get

The stationary points are all the points of the line
other than the point
which is not in the domain of
At each stationary point
the function takes on the value
:

In a moment we will show that for all
in the domain,

It
will then follow that the number
is a local maximum value.

The inequality

can be justified by the following sequence of equivalent inequalities, the
last of which is obvious:

Each function we have considered so far was differentiable on its entire domain. The only critical points were thus the stationary points. Our next example exhibits a function that is everywhere defined but is not differentiable at the origin. The origin is therefore a critical point, though not a stationary point.

**Example 4.** The function

is
everywhere defined and everywhere continuous. The graph is the upper nappe of
a cone. The number
is obviously a local minimum.

Since the partials

are not defined at
,
the gradient is not defined at
The point
is thus a critical point, but not a stationary point. At
the surface comes to a sharp point and there is no tangent plane.

Recall that a function of one variable that is continuous on a bounded closed
interval must take on both an absolute maximum and an absolute minimum on that
interval. More generally, it can be proved that, if
now a function of one or several variables, is *continuous on a bounded
closed set* (a closed set that can be contained in a ball of finite
radius), then
takes on both an absolute maximum and an absolute minimum on that set.

In the search for local extreme values the critical points are interior points of the domain: the stationary points and the interior points at which the gradient does not exist. In the search for absolute extreme values we must also test the boundary points. This usually requires special methods. One approach is to try to parametrize the boundary by some vector function and then work with . This is the approach we take in the example below.

**Example 5.** The triangular region

is
a closed, bounded set. The function

being continuous everywhere, is continuous on
.
Therefore, we know that
takes on an absolute maximum on
and an absolute minimum.

To see whether either of these values is taken on in the interior of the set,
we form the gradient

The gradient, defined everywhere, is
only at the point
.
We check,

Since this difference does not keep a constant sign for small
and
,
the point
does not give rise to an extreme value. It is a saddle point.

Now we will look for extreme values on the boundary by writing each side of
the triangle in the form
and then analyzing
We have

The values of
on these line segments are given by the functions

The maximum value of
is the maximum of the maxima of
.
And, the minimum value of
is the minimum of the minima of
.
In this case, it happens that each of the
has a maximum of
and a minimum of
.
It follows that the maximum value of
is
and the minimum value of
is

In some cases, physical or geometric considerations may allow us to conclude that an absolute maximum or an absolute minimum exists even if the function is not continuous, or the domain is not bounded or not closed.

**Example 6.** The rectangle

is
a bounded closed subset of the plane. The function

being everywhere continuous, is continuous on this rectangle. Thus we can be
sure that
takes on both an absolute maximum and an absolute minimum on this set. The
absolute maximum is taken on at the points
and
,
the points of the rectangle furthest away from the origin. The value at these
points is
.
The absolute minimum is taken on at the origin
.
The value there is
.

Now let's continue with the same function but apply it instead to the
rectangle

This rectangle is bounded but not closed. On this set
takes on an absolute maximum (the same maximum as before and at the same
points), but it takes on no absolute minimum (the origin is not in the set).

Finally, on the entire plane (which is closed but not bounded), takes on an absolute minimum ( at the origin) but no maximum.

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