# GRADIENT AS A NORMAL; TANGENT LINES AND TANGENT PLANES

## Functions of Two Variables

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

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

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

## Functions of Three Variables

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,

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 .

# Maximum and Minimum Values

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.

## Two Variables

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

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

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.

# Absolute Extreme Values

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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