Vector Calculus by Michael Corral - HTML preview

∂ f

f ( a + hv , b

)

, b)

( a

, b

) .

1

+ hv 2 − f ( a + hv 1

= hv 2

+ hv

+ αhv

∂y

1

2

By a similar argument, there exists a number 0 < β < 1 such that

∂ f

f ( a + hv , b)

( a

, b) .

1

− f ( a, b) = hv 1

+ βhv

∂x

1

Thus, by equation (2.11), we have

∂ f

∂ f

f ( a + hv , b

)

hv

( a

, b

)

( a

, b)

2

+ hv 1

+ αhv 2 + hv 1

+ βhv 1

1

+ hv 2 − f ( a, b)

∂ y

∂x

=

h

h

∂ f

∂ f

= v

( a

, b

)

( a

, b)

2

+ hv

+ αhv + v

+ βhv

∂y

1

2

1 ∂x

1

so by formula (2.9) we have

f ( a + hv , b + hv ) − f ( a, b)

D f ( a, b)

1

2

v

= lim

h→0

h

∂ f

∂ f

= lim v

( a

, b

)

( a

, b)

2

+ hv 1

+ αhv 2 + v 1

+ βhv 1

h→0

∂y

∂x

∂ f

∂ f

∂ f

∂ f

= v

( a, b)

( a, b) by the continuity of

and

, so

2

+ v

∂y

1 ∂x

∂x

∂y

∂ f

∂ f

D f ( a, b)

( a, b)

( a, b)

v

= v 1

+ v

∂x

2 ∂y

after reversing the order of summation.

QED

Note that D f ( a, b)

( a, b), ∂f ( a, b) . The second vector has a special name:

v

= v · ∂f

∂x

∂ y

80

CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES

Definition 2.6. For a real-valued function f ( x, y), the gradient of f , denoted by ∇ f , is the

vector

∂ f ∂ f

∇ f =

,

(2.12)

∂x ∂y

in R2. For a real-valued function f ( x, y, z), the gradient is the vector

∂ f ∂ f ∂ f

∇ f =

,

,

(2.13)

∂x ∂y ∂z

in R3. The symbol ∇ is pronounced “del” .5

Corollary 2.3. D f

v

= v · ∇ f

Example 2.15. Find the directional derivative of f ( x, y) = xy 2 + x 3 y at the point (1,2) in the direction of v = 1 , 1 .

2

2

Solution: We see that ∇ f = ( y 2 + 3 x 2 y,2 xy + x 3), so

D f (1, 2)

, 1

v

= v · ∇ f (1,2) =

1

· (22 + 3(1)2(2),2(1)(2) + 13) = 15

2

2

2

A real-valued function z = f ( x, y) whose partial derivatives ∂f and ∂f exist and are con-

∂x

∂ y

tinuous is called continuously differentiable. Assume that f ( x, y) is such a function and that

∇ f = 0. Let c be a real number in the range of f and let v be a unit vector in R2 which is

tangent to the level curve f ( x, y) = c (see Figure 2.4.1).

y

v

∇ f

f ( x, y) = c

x

0

Figure 2.4.1

5Sometimes the notation grad( f ) is used instead of ∇ f .

2.4 Directional Derivatives and the Gradient

81

The value of f ( x, y) is constant along a level curve, so since v is a tangent vector to this

curve, then the rate of change of f in the direction of v is 0, i.e. D f

v

= 0. But we know that

D f

v

= v · ∇ f = v

∇ f cos θ, where θ is the angle between v and ∇ f . So since v = 1 then

D f

f

v

= ∇ f cos θ. So since ∇ f = 0 then Dv = 0 ⇒ cos θ = 0 ⇒ θ = 90◦. In other words, ∇ f ⊥ v, which means that ∇ f is normal to the level curve.

In general, for any unit vector v in R2, we still have D f

v

= ∇ f cos θ, where θ is the angle

between v and ∇ f . At a fixed point ( x, y) the length ∇ f is fixed, and the value of D f then

v

varies as θ varies. The largest value that D f can take is when cos θ

v

= 1 ( θ = 0◦), while the

smallest value occurs when cos θ = −1 ( θ = 180◦). In other words, the value of the function

f increases the fastest in the direction of ∇ f (since θ = 0◦ in that case), and the value of

f decreases the fastest in the direction of −∇ f (since θ = 180◦ in that case). We have thus

proved the following theorem:

Theorem 2.4. Let f ( x, y) be a continuously differentiable real-valued function, with ∇ f = 0.

Then:

(a) The gradient ∇ f is normal to any level curve f ( x, y) = c.

(b) The value of f ( x, y) increases the fastest in the direction of ∇ f .

(c) The value of f ( x, y) decreases the fastest in the direction of −∇ f .

Example 2.16. In which direction does the function f ( x, y) = xy 2 + x 3 y increase the fastest from the point (1, 2)? In which direction does it decrease the fastest?

Solution: Since ∇ f = ( y 2 + 3 x 2 y,2 xy + x 3), then ∇ f (1,2) = (10,5) = 0. A unit vector in that direction is v = ∇ f

, 1 . Thus, f increases the fastest in the direction of

2 , 1 and

∇ f

= 25 5

5

5

decreases the fastest in the direction of −2 , −1 .

5

5

Though we proved Theorem 2.4 for functions of two variables, a similar argument can

be used to show that it also applies to functions of three or more variables. Likewise, the

directional derivative in the three-dimensional case can also be defined by the formula D f

v

=

v · ∇ f .

Example 2.17. The temperature T of a solid is given by the function T( x, y, z) = e− x + e−2 y +

e 4 z, where x, y, z are space coordinates relative to the center of the solid. In which direction

from the point (1, 1, 1) will the temperature decrease the fastest?

Solution: Since ∇ f = (− e− x,−2 e−2 y,4 e 4 z), then the temperature will decrease the fastest in the direction of −∇ f (1,1,1) = ( e−1,2 e−2,−4 e 4).

82

CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES

Exercises

A

For Exercises 1-10, compute the gradient ∇ f .

1

1. f ( x, y) = x 2 + y 2 − 1

2. f ( x, y) = x 2 + y 2

3. f ( x, y) =

x 2 + y 2 + 4

4. f ( x, y) = x 2 ey

5. f ( x, y) = ln( xy)

6. f ( x, y) = 2 x + 5 y

7. f ( x, y, z) = sin( xyz)

8. f ( x, y, z) = x 2 eyz

9. f ( x, y, z) = x 2 + y 2 + z 2

10. f ( x, y, z) =

x 2 + y 2 + z 2

For Exercises 11-14, find the directional derivative of f at the point P in the direction of

v = 1 , 1 .

2

2

1

11. f ( x, y) = x 2 + y 2 − 1, P = (1,1)

12. f ( x, y) =

, P = (1,1)

x 2 + y 2

13. f ( x, y) =

x 2 + y 2 + 4, P = (1,1)

14. f ( x, y) = x 2 ey, P = (1,1)

For Exercises 15-16, find the directional derivative of f at the point P in the direction of

v = 1 , 1 , 1 .

3

3

3

15. f ( x, y, z) = sin( xyz), P = (1,1,1)

16. f ( x, y, z) = x 2 eyz, P = (1,1,1)

17. Repeat Example 2.16 at the point (2, 3).

18. Repeat Example 2.17 at the point (3, 1, 2).

B

For Exercises 19-26, let f ( x, y) and g( x, y) be continuously differentiable real-valued func-

tions, let c be a constant, and let v be a unit vector in R2. Show that:

19. ∇( c f ) = c ∇ f

20. ∇( f + g) = ∇ f + ∇ g

g ∇ f − f ∇ g

21. ∇( f g) = f ∇ g + g ∇ f

22. ∇( f / g) =

if g( x, y) = 0

g 2

23. D f

f

24. D ( c f )

f

−v

= − Dv

v

= c Dv

25. D ( f

f

g

26. D ( f g)

g

f

v

+ g) = Dv + Dv

v

= f Dv + g Dv

27. The function r( x, y) =

x 2 + y 2 is the length of the position vector r = x i + yj for each

1

point ( x, y) in R2. Show that ∇ r = r when ( x, y) = (0,0), and that ∇( r 2) = 2r.

r

2.5 Maxima and Minima

83

2.5 Maxima and Minima

The gradient can be used to find extreme points of real-valued functions of several variables,

that is, points where the function has a local maximum or local minimum. We will consider

only functions of two variables; functions of three or more variables require methods using

linear algebra.

Definition 2.7. Let f ( x, y) be a real-valued function, and let ( a, b) be a point in the domain

of f . We say that f has a local maximum at ( a, b) if f ( x, y) ≤ f ( a, b) for all ( x, y) inside some disk of positive radius centered at ( a, b), i.e. there is some sufficiently small r > 0 such that

f ( x, y) ≤ f ( a, b) for all ( x, y) for which ( x − a)2 + ( y − b)2 < r 2.

Likewise, we say that f has a local minimum at ( a, b) if f ( x, y) ≥ f ( a, b) for all ( x, y) inside some disk of positive radius centered at ( a, b).

If f ( x, y) ≤ f ( a, b) for all ( x, y) in the domain of f , then f has a global maximum at ( a, b). If f ( x, y) ≥ f ( a, b) for all ( x, y) in the domain of f , then f has a global minimum at ( a, b).

Suppose that ( a, b) is a local maximum point for f ( x, y), and that the first-order partial

derivatives of f exist at ( a, b). We know that f ( a, b) is the largest value of f ( x, y) as ( x, y) goes in all directions from the point ( a, b), in some sufficiently small disk centered at ( a, b).

In particular, f ( a, b) is the largest value of f in the x direction (around the point ( a, b)), that is, the single-variable function g( x) = f ( x, b) has a local maximum at x = a. So we know that

g ′( a) = 0. Since g ′( x) = ∂f ( x, b), then ∂f ( a, b)

∂x

∂x

= 0. Similarly, f ( a, b) is the largest value of f

near ( a, b) in the y direction and so