GRADIENT AS A NORMAL; TANGENT LINES AND TANGENT PLANES

Functions of Two Variables

We begin with a nonconstant function $f=f(x,y)$ that is continuously differentiable. (This means $f$ is differentiable and its gradient $\nabla f$ is continuous.) We have seen that at each point of the domain the gradient vector, if not $0$, points in the direction of the most rapid increase of $f. $ Here we show that

Theorem. At each point of the domain, the gradient vector, if not $0,$ is perpendicular to the level curve that passes through that point.

Proof. We choose a point $(x_0,y_0)$ in the domain and assume that MATH. The level curve through this point has equation
MATH

Under our assumptions on $f$, this curve can be parametrized in a neighborhood of $(x_0,y_0)$ by a continuously differentiable vector function
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with nonzero tangent vector $r^{\prime }(t)$.

Now take $t_0$ such that
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We will show that
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Since $f$ is constantly $c$ on the curve, we have
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For such $t$
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In particular
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and thus
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Example 1. For the function
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the level curves are concentric circles:
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At each point $(x,y)\ne (0,0)$ the gradient vector
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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 $0$.

Consider now a curve in the $xy$-plane
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As before we assume that $f$ is nonconstant and continuously differentiable. Suppose that $(x_0,y_0)$ lies on the curve and MATH. We can view $C$ as the $c$-level curve of $f$ and conclude that the gradient
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is perpendicular to $C$ at $(x_0,y_0)$. We call it a normal vector.

The vector
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is perpendicular to the gradient:
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It is therefore a tangent vector. The line through $(x_0,y_0)$ perpendicular to the gradient is the tangent line. A point $(x,y)$ will lie on the tangent line if and only if
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that is, if and only if
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This is an equation for the tangent line.

The line through $(x_0,y_0)$ perpendicular to the tangent vector $t(x_0,y_0)$ is the normal line. A point $(x,y)$ will lie on the normal line if and only if
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This is an equation for the normal line.

Example 2. Choose a point $(x_0,y_0)$ on the hyperbola
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To avoid denominators, we write
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This equation is of the form


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Partial differentiation gives
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At $(x_0,y_0)$ the gradient
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is normal to the curve and the vector
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is tangent to the curve. The equation of the tangent line can be written
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Dividing by $2$ and noting that MATH we can simplify the equation to
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The equation of the normal line takes the form
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This simplifies to
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Functions of Three Variables

Here, instead of level curves, we have level surfaces, but the results are similar. If $f=f(x,y,z)$ is nonconstant and continuously differentiable, then at each point of the domain, the gradient vector, if not $0,$ is perpendicular to the level surface that passes through that point.

Proof. We choose a point $(x_0,y_0,z_0)$ in the domain and assume that MATH. The level surface through this point has equation
MATH
We suppose now that
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is a differentiable curve that lies on this surface and passes through the point $(x_0,y_0,z_0)$. We choose $t_0$ so that


MATH
and suppose that MATH

Since the curve lies on the given surface, we have
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For such $t$
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In particular,
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The gradient vector
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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 $(x_0,y_0,z_0)$ with nonzero tangent vector. Consequently, we view MATH as perpendicular to the surface itself.

Example 3. For the function
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the level surfaces are concentric spheres:
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At each point $(x,y,z)\ne (0,0,0)$ the gradient vector
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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 $0.$

The tangent plane to a surface
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at a point $(x_0,y_0,z_0)$ is the plane through $(x_0,y_0,z_0)$ with normal MATH The tangent plane at a point $(x_0,y_0,z_0)$ is the plane through $(x_0,y_0,z_0)$ that best approximates the surface in a neighborhood of $(x_0,y_0,z_0)$.

A point $(x,y,z)$ lies on the tangent plane through $(x_0,y_0,z_0)$ if and only if
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In Cartesian coordinates the equation takes the form
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Problem 4. Find an equation in $x,y,z$ for the plane tangent to the surface
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Solution. The surface is of the form


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Observe that
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At the point $(1,2,3)$
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The equation for the tangent plane can therefore be written
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This simplifies to
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Problem 5. The curve
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intersects the surface $x^2-4y^2-4z=0$ at the point $(2,2,-3).$ 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 $(2,2,-3)$ at $t=2$. Since
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we have
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Now set
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This function has gradient $2xi-8yj-4k.$ At the point $(2,2,-3)$
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Now let $\theta $ be the angle between $r^{\prime }(2)$ and this gradient.
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A surface of the form
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can be written in the form
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by setting
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If $g$ is differentiable, so is $f$. Moreover,
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The tangent plane at $(x_0,y_0,z_0)$ thus has equation
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which we can rewrite as
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If MATH, then both partials of $g$ are zero at $(x_0,y_0)$ and the equation reduces to
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In this case the tangent plane is horizontal.

Problem 6. Find an equation for the plane tangent to the surface
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Solution. Set
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Partial differentiation gives
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When $x=2p^2$ and $y=2q^2,$
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At the point MATH the tangent plane has equation
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This simplifies to
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Problem 7. At what points of the surface
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is the tangent plane horizontal?

Solution. The function
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has first partials
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These partials are both zero only at $(0,0)$ and $(1,1).$ The surface has a horizontal tangent plane only at $(0,0,0)$ and $(1,1,1)$.

Maximum and Minimum Values

DEFINITION LOCAL MAXIMUM AND LOCAL MINIMUM

Let $f$ be a function of several variables and let $x_0$ be an interior point of the domain:

$f$ is said to have a local maximum at $x_0$ if $f(x_0)\ge f(x)$ for all $x$ in some neighborhood of $x_0$;

$f$ is said to have a local minimum at $x_0$ if $f(x_0)\leq f(x)$ for all $x$ in some neighborhood of $x_0$.

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 $f$ has a local extreme value at $x_0$, then

either $f^{\prime }(x_0)=0$ or $f^{\prime }(x_0)$ does not exist.

We have a similar result for functions of several variables.

THEOREM If $f$ has a local extreme value at $x_0$, then

either $\nabla f(x_0)=0$ or $\nabla f(x_0)$ does not exist.

Proof. We assume that $f$ has a local extreme value at $(x_0,y_0)$ and that $f$ is differentiable at $(x_0,y_0)$ [that $\nabla f(x_0,y_0)$ exists]. We need to show that MATH. The three-variable case is similar.

Since $f$ has a local extreme value at $(x_0,y_0)$, the function $g(x)=f(x,y_0)$ has a local extreme value at $x_0$. Since $f$ is differentiable at $(x_0,y_0)$, $g$ is differentiable at $x_0$ and therefore
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Similarly, the function $h(y)=f(x_0,y)$ has a local extreme value at $y_0$ and, being differentiable there, satisfies the relation
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The gradient is $0$ since both partials are $0$.

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 $f=f(x,y)$ is defined on an open set and is continuously differentiable there. The graph of $f$ is a surface
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Where $f$ has a local maximum, the surface has a local high point. Where $f$ has a local minimum, the surface has a local low point. Where $f$ has either a local maximum or a local minimum, the gradient is $0$ 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
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we have
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To find the stationary points, we set $\nabla f(x,y)=0$. This gives

$4x-y=0$ and $2y-x-7=0.$

The only simultaneous solution to these equations is $x=1,y=4$. The point $(1,4)$ is therefore the only stationary point. We now compare the value of $f $ at $(1,4)$ with the values of $f$ at nearby points $(1+h,4+k):$


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MATH

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The difference


MATH


MATH

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is nonnegative for all small $h$ and $k$ (in fact for all real $h$ and $k$). It follows that $f$ has a local minimum at $(1,4)$. This local minimum is $-14.$

Example 2. In the case of
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we have
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The gradient is $0$ where

$2-y=0$ and $2y-x+1=0$.

The only simultaneous solution to these equations is $x=5,y=2$. The point $(5,2)$ is the only stationary point.

We now compare the value of $f$ at $(5,2)$ with the values of $f$ at nearby points $(5+h,2+k):$
MATH


MATH


MATH

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The difference
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does not keep a constant sign for small $h$ and $k$. It follows that $(5,2)$ is a saddle point.

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

Example 3. For all points $(x,y)$ with $x+y\ne 0$, set
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Then
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The gradient is $0$ where
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Solving these two equations simultaneously, we get
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The stationary points are all the points of the line $x=2y$ other than the point $(0,0),$ which is not in the domain of $f.$ At each stationary point $(x_0,y_0)$ the function takes on the value $-\frac 23$:
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In a moment we will show that for all $(x,y)$ in the domain,
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It will then follow that the number $-\frac 23$ is a local maximum value.

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

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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
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is everywhere defined and everywhere continuous. The graph is the upper nappe of a cone. The number $f(0,0)=1$ is obviously a local minimum.

Since the partials
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are not defined at $(0,0)$, the gradient is not defined at $(0,0).$ The point $(0,0)$ is thus a critical point, but not a stationary point. At $(0,0,1)$ 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 $f$ 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 $f$ 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 $r=r(t)$ and then work with $f(r(t))$. This is the approach we take in the example below.

Example 5. The triangular region
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is a closed, bounded set. The function
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being continuous everywhere, is continuous on $\Omega $. Therefore, we know that $f$ takes on an absolute maximum on MATH and an absolute minimum.

To see whether either of these values is taken on in the interior of the set, we form the gradient
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The gradient, defined everywhere, is $0$ only at the point $(3,2)$. We check,
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Since this difference does not keep a constant sign for small $h$ and $k$, the point $(3,2)$ 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 $r=r(t)$ and then analyzing $f(r(t)).$ We have
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MATH

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The values of $f$ on these line segments are given by the functions
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The maximum value of $f$ is the maximum of the maxima of $f_1,f_2,f_3$. And, the minimum value of $f$ is the minimum of the minima of $f_1,f_2,f_3$. In this case, it happens that each of the $f_i$ has a maximum of $0$ and a minimum of $-8$. It follows that the maximum value of $f$ is $0$ and the minimum value of $f$ is $-8.$

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
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is a bounded closed subset of the plane. The function


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being everywhere continuous, is continuous on this rectangle. Thus we can be sure that $f$ takes on both an absolute maximum and an absolute minimum on this set. The absolute maximum is taken on at the points $(a,-b)$ and $(a,b)$, the points of the rectangle furthest away from the origin. The value at these points is $1+\sqrt{a^2+b^2}$. The absolute minimum is taken on at the origin $(0,0)$. The value there is $1$.

Now let's continue with the same function but apply it instead to the rectangle
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This rectangle is bounded but not closed. On this set $f$ 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), $f$ takes on an absolute minimum ($1$ at the origin) but no maximum.

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