# 14.6

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## 14.6 Directional Derivatives and

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### Directional Derivatives and the Gradient Vector

In this section we introduce a type of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more variables in any direction.

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### Directional Derivatives

Recall that if z = f(x, y), then the partial derivatives fx and fy are defined as

and represent the rates of change of z in the

x- and y-directions, that is, in the directions of the unit vectors i and j.

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### Directional Derivatives

Suppose that we now wish to find the rate of change of z at (x0, y0) in the direction of an arbitrary unit vector u =

a, b

### 〉

.

(See Figure 2.)

To do this we consider the surface S with the equation

z = f(x, y) (the graph of f) and we let z0 = f(x0, y0). Then the point P(x0, y0, z0) lies on S.

Figure 2

A unit vector u = 〈a, b〉 = 〈cos u, sin u 〉

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### Directional Derivatives

The vertical plane that passes through P in the direction of u intersects S in a curve C. (See Figure 3.)

Figure 3

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### Directional Derivatives

The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u.

If Q(x, y, z) is another point on C and P′, Q′ are the

projections of P, Q onto the xy-plane, then the vector is parallel to u and so

= hu =

ha, hb

### 〉

for some scalar h. Therefore x – x0 = ha, y – y0 = hb, so x = x0 + ha, y = y0 + hb, and

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### Directional Derivatives

If we take the limit as h → 0, we obtain the rate of change of z (with respect to distance) in the direction of u, which is

called the directional derivative of f in the direction of u.

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### Directional Derivatives

By comparing Definition 2 with Equations 1, we see that if u = i =

1, 0

### 〉

, then Dif = fx and if u = j =

0, 1

### 〉

, then Djf = fy.

In other words, the partial derivatives of f with respect to x and y are just special cases of the directional derivative.

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### Example 1

Use the weather map in Figure 1 to estimate the value of the directional derivative of the temperature function at Reno in the southeasterly direction.

Figure 1

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### Example 1 – Solution

The unit vector directed toward the southeast is

but we won’t need to use this expression.

We start by drawing a line through Reno toward the southeast (see Figure 4).

Figure 4

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### Example 1 – Solution

We approximate the directional derivative DuT by the average rate of change of the temperature between the points where this line intersects the isothermals

T = 50 and T = 60.

The temperature at the point southeast of Reno is T = 60°F and the temperature at the point northwest of Reno is

T = 50°F.

The distance between these points looks to be about

75 miles. So the rate of change of the temperature in the southeasterly direction is

cont’d

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### Directional Derivatives

When we compute the directional derivative of a function defined by a formula, we generally use the following

theorem.

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### Directional Derivatives

If the unit vector u makes an angle θ with the positive x-axis (as in Figure 2), then we can write u =

cos θ, sin θ

### 〉

and the formula in Theorem 3 becomes

Duf(x, y) = fx(x, y) cos θ + fy(x, y) sin θ

Figure 2

A unit vector u = 〈a, b〉 = 〈cos u, sin u 〉

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### The Gradient Vector

Notice from Theorem 3 that the directional derivative of a differentiable function can be written as the dot product of two vectors:

Duf(x, y) = fx(x, y)a + fy(x, y)b

=

### 〈

fx(x, y), fy(x, y)

a, b

=

### 〈

fx(x, y), fy(x, y)

### 〉

u

The first vector in this dot product occurs not only in computing directional derivatives but in many other contexts as well.

So we give it a special name (the gradient of f ) and a special notation (grad f or ∇f, which is read “del f ”).

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### Example 3

If f(x, y) = sin x + exy, then

∇f(x, y) =

fx, fy

=

### 〈

cos x + yexy, xexy

and ∇f(0, 1) =

2, 0

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### The Gradient Vector

With the notation for the gradient vector, we can rewrite Equation 7 for the directional derivative of a differentiable function as

This expresses the directional derivative in the direction of a unit vector u as the scalar projection of the gradient

vector onto u.

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### Functions of Three Variables

For functions of three variables we can define directional derivatives in a similar manner.

Again Duf(x, y, z) can be interpreted as the rate of change of the function in the direction of a unit vector u.

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### Functions of Three Variables

If we use vector notation, then we can write both

definitions (2 and 10) of the directional derivative in the compact form

where x0 =

x0, y0

### 〉

if n = 2 and x0 =

x0, y0, z0

### 〉

if n = 3.

This is reasonable because the vector equation of the line through x0 in the direction of the vector u is given by

x = x0 + t u and so f(x0 + hu) represents the value of f at a point on this line.

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### Functions of Three Variables

If f(x, y, z) is differentiable and u =

a, b, c

### 〉

, then

Duf(x, y, z) = fx(x, y, z)a + fy(x, y, z)b + fz(x, y, z)c For a function f of three variables, the gradient vector, denoted by ∇f or grad f, is

∇f(x, y, z) =

### 〈

fx(x, y, z), fy(x, y, z), fz(x, y, z)

or, for short,

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### Functions of Three Variables

Then, just as with functions of two variables, Formula 12 for the directional derivative can be rewritten as

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### Example 5

If f(x, y, z) = x sin yz, (a) find the gradient of f and (b) find the directional derivative of f at (1, 3, 0) in the direction of v = i + 2 j – k.

Solution:

(a) The gradient of f is

∇f(x, y, z) =

### 〈

fx(x, y, z), fy(x, y, z), fz(x, y, z)

=

### 〈

sin yz, xz cos yz, xy cos yz

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### Example 5 – Solution

(b) At (1, 3, 0) we have ∇f(1, 3, 0) =

0, 0, 3

### 〉

.

The unit vector in the direction of v = i + 2 j – k is

Therefore Equation 14 gives Duf(1, 3, 0) = ∇f(1, 3, 0) u

cont’d

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### Maximizing the Directional Derivatives

Suppose we have a function f of two or three variables and we consider all possible directional derivatives of f at a

given point.

These give the rates of change of f in all possible directions.

We can then ask the questions: In which of these directions does f change fastest and what is the maximum rate of

change? The answers are provided by the following theorem.

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### Example 6

(a) If f(x, y) = xey, find the rate of change of f at the point P(2, 0) in the direction from P to .

(b) In what direction does f have the maximum rate of change? What is this maximum rate of change?

Solution:

(a) We first compute the gradient vector:

∇f(x, y) =

fx, fy

=

ey, xey

∇f(2, 0) =

1, 2

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### Example 6 – Solution

The unit vector in the direction of = is u = so the rate of change of f in the direction from P to Q is

Duf(2, 0) = ∇f(2, 0) u

(b) According to Theorem 15, f increases fastest in the direction of the gradient vector ∇f(2, 0) =

1, 2

### 〉

.

The maximum rate of change is

|∇f(2, 0)| = |

1, 2

| =

cont’d

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### Tangent Planes to Level Surfaces

Suppose S is a surface with equation F(x, y, z) = k, that is, it is a level surface of a function F of three variables, and let P(x0, y0, z0) be a point on S.

Let C be any curve that lies on the surface S and passes through the point P. Recall that the curve C is described by a continuous vector function r(t) =

x(t), y(t), z(t)

### 〉

.

Let t0 be the parameter value corresponding to P; that is,

r(t0) =

x0, y0, z0

### 〉

. Since C lies on S, any point (x(t), y(t), z(t)) must satisfy the equation of S, that is,

F(x(t), y(t), z(t)) = k

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### Tangent Planes to Level Surfaces

If x, y, and z are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to

differentiate both sides of Equation 16 as follows:

But, since ∇F =

Fx, Fy, Fz

and r′(t) =

### 〈

x′(t), y′(t), z′(t)

### 〉

,

Equation 17 can be written in terms of a dot product as

∇F r′(t) = 0

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### Tangent Planes to Level Surfaces

In particular, when t = t0 we have r(t0) =

x0, y0, z0

### 〉

, so

∇F(x0, y0, z0) r′(t0) = 0 Equation 18 says that the gradient vector at P,

∇F(x0, y0, z0), is perpendicular to the tangent vector r′(t0) to any curve C on S that passes through P. (See Figure 9.)

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### Tangent Planes to Level Surfaces

If ∇F(x0, y0, z0) ≠ 0, it is therefore natural to define the tangent plane to the level surface F(x, y, z) = k at

P(x0, y0, z0) as the plane that passes through P and has normal vector ∇F(x0, y0, z0).

Using the standard equation of a plane, we can write the equation of this tangent plane as

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### Tangent Planes to Level Surfaces

The normal line to S at P is the line passing through P and perpendicular to the tangent plane.

The direction of the normal line is therefore given by the gradient vector ∇F(x0, y0, z0) and so, its symmetric

equations are

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### Tangent Planes to Level Surfaces

In the special case in which the equation of a surface S is of the form z = f(x, y) (that is, S is the graph of a function f of two variables), we can rewrite the equation as

F(x, y, z) = f(x, y) – z = 0

and regard S as a level surface (with k = 0) of F. Then Fx(x0, y0, z0) = fx(x0, y0)

Fy(x0, y0, z0) = fy(x0, y0) Fz(x0, y0, z0) = –1

so Equation 19 becomes

fx(x0, y0)(x – x0) + fy(x0, y0)(y – y0) – (z – z0) = 0

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### Example 8

Find the equations of the tangent plane and normal line at the point (–2, 1, –3) to the ellipsoid

Solution:

The ellipsoid is the level surface (with k = 3) of the function

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### Example 8 – Solution

Therefore we have

Fx(x, y, z) = Fy(x, y, z) = 2y Fz(x, y, z) = Fx(–2, 1, –3) = –1 Fy(–2, 1, –3) = 2 Fz(–2, 1, –3) = Then Equation 19 gives the equation of the tangent plane at (–2, 1, –3) as

–1(x + 2) + 2(y – 1) – (z + 3) = 0 which simplifies to 3x – 6y + 2z + 18 = 0.

By Equation 20, symmetric equations of the normal line are

cont’d

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### Significance of the Gradient Vector

We now summarize the ways in which the gradient vector is significant.

We first consider a function f of three variables and a point P(x0, y0, z0) in its domain.

On the one hand, we know from Theorem 15 that the

gradient vector ∇f(x0, y0, z0) gives the direction of fastest increase of f.

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### Significance of the Gradient Vector

On the other hand, we know that ∇f(x0, y0, z0) is orthogonal to the level surface S of f through P. (Refer to Figure 9.)

These two properties are quite compatible intuitively

because as we move away from P on the level surface S, the value of f does not change at all.

Figure 9

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### Significance of the Gradient Vector

So it seems reasonable that if we move in the

perpendicular direction, we get the maximum increase.

In like manner we consider a function f of two variables and a point P(x0, y0) in its domain.

Again the gradient vector ∇f(x0, y0) gives the direction of fastest increase of f. Also, by considerations similar to our discussion of tangent planes, it can be shown that ∇f(x0, y0) is perpendicular to the level curve f(x, y) = k that passes

through P.

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### Significance of the Gradient Vector

Again this is intuitively plausible because the values of f remain constant as we move along the curve.

(See Figure 11.)

Figure 11

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### Significance of the Gradient Vector

If we consider a topographical map of a hill and let f(x, y) represent the height above sea level at a point with

coordinates (x, y), then a curve of steepest ascent can be drawn as in Figure 12 by making it perpendicular to all of the contour lines.

Figure 12

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### Significance of the Gradient Vector

Computer algebra systems have commands that plot sample gradient vectors.

Each gradient vector ∇f(a, b) is plotted starting at the point (a, b). Figure 13 shows such a plot (called a gradient vector field ) for the function f(x, y) = x2 – y2 superimposed on a

contour map of f.

As expected, the gradient vectors point “uphill” and are perpendicular to the level curves.

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