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**14.6** Directional Derivatives and

### the Gradient Vector

### 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.

3 3

### Directional Derivatives

### Directional Derivatives

*Recall that if z = f(x, y), then the partial derivatives f*_{x}*and f** _{y}*
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.**

5 5

### Directional Derivatives

*Suppose that we now wish to find the rate of change of z at *
*(x*_{0}*, y*_{0}**) 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 z*_{0} *= f(x*_{0}*, y*_{0}). Then the
*point P(x*_{0}*, y*_{0}*, z*_{0}*) lies on S.*

**Figure 2**

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

### 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**

7 7

### 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 – x*_{0} *= ha, y – y*_{0} *= hb, *
*so x = x*_{0} *+ ha, y = y*_{0} *+ hb, and*

### 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.**

9 9

### Directional Derivatives

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

### 〈

^{1, 0}

### 〉

^{, then D}

_{i}

^{f = f}

_{x}**and if u = j =**

### 〈

^{0, 1}

### 〉

^{, then D}

_{j}

^{f = f}

_{y}^{.}

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

### 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**

11 11

*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**

*Example 1 – Solution*

*We approximate the directional derivative D*_{u}*T 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

13 13

### Directional Derivatives

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

theorem.

### 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

*D*_{u}*f(x, y) = f*_{x}*(x, y) cos *θ *+ f*_{y}*(x, y) sin *θ

**Figure 2**

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

15 15

### The Gradient Vector

### 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:

*D*_{u}*f(x, y) = f*_{x}*(x, y)a + f*_{y}*(x, y)b*

=

### 〈

^{f}

_{x}

^{(x, y), f}

_{y}

^{(x, y)}### 〉

^{}

### 〈

^{a, b}### 〉

=

### 〈

^{f}

_{x}

^{(x, y), f}

_{y}

^{(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 ”).**

17 17

### The Gradient Vector

### Example 3

*If f(x, y) = sin x + e** ^{xy}*, then

*∇f(x, y) = *

### 〈

^{f}

_{x}

^{, f}

_{y}### 〉

=

### 〈

^{cos x + ye}

^{xy}

^{, xe}

^{xy}### 〉

and *∇f(0, 1) = *

### 〈

^{2, 0}

### 〉

19 19

### 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.**

### Functions of Three Variables

21 21

### Functions of Three Variables

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

*Again D*_{u}*f(x, y, z) can be interpreted as the rate of change *
**of the function in the direction of a unit vector u.**

### 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 x**_{0} =

### 〈

*x*

_{0}

*, y*

_{0}

### 〉

**if n = 2 and x**_{0}=

### 〈

*x*

_{0}

*, y*

_{0}

*, z*

_{0}

### 〉

*if n = 3.*

This is reasonable because the vector equation of the line
**through x**_{0} **in the direction of the vector u is given by **

**x = x**_{0} **+ t u and so f(x**_{0} * + hu) represents the value of f at a *
point on this line.

23 23

### Functions of Three Variables

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

### 〈

*a, b, c*

### 〉

, then*D*_{u}*f(x, y, z) = f*_{x}*(x, y, z)a + f*_{y}*(x, y, z)b + f*_{z}*(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) = *

### 〈

^{f}

_{x}*(x, y, z), f*

_{y}*(x, y, z), f*

_{z}*(x, y, z)*

### 〉

or, for short,

### Functions of Three Variables

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

25 25

### 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) = *

### 〈

^{f}

_{x}*(x, y, z), f*

_{y}*(x, y, z), f*

_{z}*(x, y, z)*

### 〉

=

### 〈

*sin yz, xz cos yz, xy cos yz*

### 〉

*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
*D*_{u}*f(1, 3, 0) = ∇f(1, 3, 0) *^{} **u**

cont’d

27 27

### Maximizing the Directional

### Derivatives

### 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.

29 29

### Example 6

**(a) If f(x, y) = xe**^{y}*, 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) = *

### 〈

^{f}

_{x}

^{, f}

_{y}### 〉

^{=}

### 〈

^{e}

^{y}

^{, xe}

^{y}### 〉

*∇f(2, 0) = *

### 〈

1, 2### 〉

*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*

*D*_{u}*f(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

31 31

### Tangent Planes to Level Surfaces

### 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(x*_{0}*, y*_{0}*, z*_{0}*) 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 t*_{0} *be the parameter value corresponding to P; that is, *

**r(t**_{0}) =

### 〈

^{x}_{0}

^{, y}_{0}

^{, z}_{0}

### 〉

*. 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*

33 33

### 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 =

### 〈

^{F}

_{x}

^{, F}

_{y}

^{, F}

_{z}### 〉

**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**

### Tangent Planes to Level Surfaces

*In particular, when t = t*_{0} **we have r(t**_{0}) =

### 〈

*x*

_{0}

*, y*

_{0}

*, z*

_{0}

### 〉

, so*∇F(x*_{0}*, y*_{0}*, z*_{0}) **r′(t**_{0}) = 0
*Equation 18 says that the gradient vector at P, *

*∇F(x*_{0}*, y*_{0}*, z*_{0}**), is perpendicular to the tangent vector r′(t**_{0}*) to *
*any curve C on S that passes through P. (See Figure 9.)*

35 35

### Tangent Planes to Level Surfaces

If ∇F(x_{0}*, y*_{0}*, z*_{0}) ≠ 0, it is therefore natural to define the
**tangent plane to the level surface F(x, y, z) = k at **

*P(x*_{0}*, y*_{0}*, z*_{0}*) as the plane that passes through P and has *
normal vector ∇F(x_{0}*, y*_{0}*, z*_{0}).

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

### 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(x_{0}*, y*_{0}*, z*_{0}) and so, its symmetric

equations are

37 37

### 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*
*F*_{x}*(x*_{0}*, y*_{0}*, z*_{0}*) = f*_{x}*(x*_{0}*, y*_{0})

*F*_{y}*(x*_{0}*, y*_{0}*, z*_{0}*) = f*_{y}*(x*_{0}*, y*_{0})
*F*_{z}*(x*_{0}*, y*_{0}*, z*_{0}) = –1

so Equation 19 becomes

*f*_{x}*(x*_{0}*, y*_{0}*)(x – x*_{0}*) + f*_{y}*(x*_{0}*, y*_{0}*)(y – y*_{0}*) – (z – z*_{0}) = 0

### 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*

39 39

*Example 8 – Solution*

Therefore we have

*F*_{x}*(x, y, z) =* *F*_{y}*(x, y, z) = 2y* *F*_{z}*(x, y, z) =*
*F** _{x}*(–2, 1, –3) = –1

*F*

*(–2, 1, –3) = 2*

_{y}*F*

*(–2, 1, –3) = Then Equation 19 gives the equation of the tangent plane at (–2, 1, –3) as*

_{z}*–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

### Significance of the Gradient Vector

41 41

### 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(x*_{0}*, y*_{0}*, z*_{0}) in its domain.

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

gradient vector ∇f(x_{0}*, y*_{0}*, z*_{0}) gives the direction of fastest
*increase of f. *

### Significance of the Gradient Vector

On the other hand, we know that ∇f(x_{0}*, y*_{0}*, z*_{0}) 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**

43 43

### 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(x*_{0}*, y*_{0}) in its domain.

Again the gradient vector ∇f(x_{0}*, y*_{0}) gives the direction of
*fastest increase of f. Also, by considerations similar to our *
discussion of tangent planes, it can be shown that ∇f(x_{0}*, y*_{0})
*is perpendicular to the level curve f(x, y) = k that passes *

*through P.*

### 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**

45 45

### 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**

### 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) = x*^{2} *– y*^{2} superimposed on a

*contour map of f. *

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