Chapter 5 Partial Derivatives

Chapter 5

Partial Derivatives

5.1 Functions of Several Variables

The functions discussed in previous chapter involved only one independent variable. Such functions have many applications; however, in some problems several independent variables occur. For instance, the ideal gas law P = RT states that the pressure P is a function of both its density and its temperature T . (The gas constant R is a material property and not a variable).

In this section, we expand our concept of function to include functions that depend on more than one variable, that is, functions whose domain is multi-dimensional.

The terminology and notation for functions of several variables is similar to that for functions of one variable. For example, the expression

z = f (x, y) means that z is a function of x and y in the sense that a unique value of the dependent variable z is determined by specifying values for the independent variables x and y. Similarly,

w = f (x, y, z) expresses w as a function x, y, and z, and

u = f (x1, x2, . . . , xn) expresses u as a function of x1, x2, . . . , xn.

As with functions of one variable, the independent variables of a function of two or more variables may be restricted to lie in some set D, which we call the domain of f .

Definition 5.1 A function f of two variables, x and y, is a rule that assigns a unique real number f (x, y) to each point (x, y) in some set D in the xy-plane.

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Definition 5.2

A function f of three variables, x, y, and z, is a rule that assigns a unique real number f (x, y, z) to each point (x, y, z) in some set D in three dimensional space.

Example 5.1 Let f (x, y) = 3x2y - 1. Find f (1, 4), f (t2, t), f (ab, 9b), and the domain of f .

Solution

Example 5.2

Find

and

sketch

the

domain

for

(a)

f (x, y)

=

ln(x2 -y )

and

(b)

g(x, y)

=

y

2x - x2

.

Solution

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Example 5.3

Let f (x, y, z) =

1 - x2 - y2 - z2. Find f

0,

1 2

,

-

1 2

and the domain of f .

Solution

Graphs of Functions of Two Variables

Recall that for a function f of one variable, the graph of f (x) in the xy-plane was defined to be the graph of the equation y = f (x). Similarly, if f is a function of two variables, we define the graph of f (x, y) in xyz-space to be the graph of the equation z = f (x, y). In general, such a graph will be a surface in 3-space.

Example 5.4

In each part, describe the graph of the function in an xyz-coordinate system.

(a)

f (x,

y)

=

1

-

x

-

1 2

y

(c) f (x, y) = - x2 + y2

(b) f (x, y) = 1 - x2 - y2

Solution

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Exercise 5.1

1 - 8 These exercises are concerned with functions of two variables.

1. Let f (x, y) = x2y + 1. Find

(a) f (2, 1)

(b) f (1, 2)

(c) f (0, 0)

(d) f (1, -3)

(e) f (3a, a)

2. Let f (x, y) = x + 3 xy. Find

(a) f (t, t2)

(b) f (x, x2)

(f) f (ab, a - b) (c) f (2y2, 4y)

3. Let f (x, y) = xy + 3. Find

(a) f (x + y, x - y)

(b) f (xy, 3x2y3)

4. Let g(x, y) = x sin y. Find

(a) g(x/y)

(b) g(xy)

(c) g(x - y)

5. Find F (g(x), h(x)) if F (x, y) = xexy, g(x) = x3, and h(y) = 3y + 1.

6. Find g(u(x, y), v(x, y)) if g(x, y) = y sin(x2y), u(x, y) = x2y3, and v(x, y) = xy.

7. Let f (x, y) = x + 3x2y2, x(t) = t2, and y(t) = t3. Find

(a) f (x(t), y(t))

(b) f (x(0), y(0))

(c) f (x(2), y(2))

8.

Let

g(x, y) = ye-3x,

x(t) = ln(t2 + 1),

and

y(t) = t.

Find

g(x(t), y(t)).

9 - 12 These exercises involve functions of three variables.

9. Let f (x, y, z) = xy2z3 + 3. Find (a) f (2, 1, 2)

(b) f (-3, 2, 1)

(c) f (0, 0, 0)

(d) f (a, a, a)

10. Let f (x, y, z) = zxy + x. Find (a) f (x + y, x - y, x2)

(b) f (xy, y/x, xz)

11. Find F (f (x), g(y), h(z)) if F (x, y, z) = yexyz, f (x) = x2, g(y) = y + 1, and h(z) = z2.

12. Find g(u(x, y, z), v(x, y, z), w(x, y, z)) if g(x, y, z) = z sin xy, u(x, y, z) = x2z3, v(x, y, z) = xyz, and w(x, y, z) = xy/z.

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13 - 14 These exercises are concerned with functions of four or more variables.

13.

(a)

Let

f (x, y, z, t) = x2y3z + t.

Find

f ( 5, 2, , 3).

n

(b) Let f (x1, x2, . . . , xn) = kxk. Find f (1, 1, . . . , 1).

k=1

14. (a) Let f (u, v, , ) = eu+v cos tan . Find f (-2, 2, 0, /4).

(b) Let f (x1, x2, . . . , xn) = x21 + x22 + ? ? ? + x2n. Find f (1, 2, . . . , n).

15 - 18 Sketch the domain of f . Use solid lines for portions of the boundary included in the domain and dashed lines for portions not included.

15. f (x, y) = ln(1 - x2 - y2)

17.

f (x, y)

=

1 x - y2

16. f (x, y) = x2 + y2 - 4 18. f (x, y) = ln xy

19 - 20 Describe the domain of f in words.

19. (a) f (x, y) = xe-y+2

(c) f (x, y, z) = exyz

20.

(a) f (x, y) =

4 - x2 y2 + 3

(c)

f (x, y, z)

=

x

xyz +y+

z

21 - 30 Sketch the graph of f .

21. f (x, y) = 3 23. f (x, y) = x2 + y2 25. f (x, y) = x2 - y2 27. f (x, y) = x2 + y2 + 1 29. f (x, y) = y + 1

(b) f (x, y, z) = 25 - x2 - y2 - z2

(b) f (x, y) = ln(y - 2x)

22. f (x, y) = 9 - x2 - y2 24. f (x, y) = x2 + y2 26. f (x, y) = 4 - x2 - y2 28. f (x, y) = x2 + y2 - 1 30. f (x, y) = x2

Answers to Exercise 5.1

1. (a) 5 (b) 3 (c) 1 (d) -2 (e) 9a3 + 1 (f) a3b2 - a2b3 + 1 3. (a) x2 - y2 + 3 (b) 3x3y4 + 3 5. x3ex3(3y+1) 7. (a) t2 + 3t10 (b) 0 (c) 3076

9. (a) 19 (b) -9 (c) 4 (d) a6 + 3 (e) -t8 + 3 (f) (a + b)(a - b)2b3 + 3 11. (y + 1)ex2(y+1)z2 13. (a) 80 (b) n(n + 1)/2

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15.

y

1x

17.

y

x

19. (a) all points above or on the line y = -2 (b) all points on or within the sphere x2 + y2 + z2 = 25 (c) all points in 3-space

5.2 Limits and Continuity

For a function of a single variable, if we write

lim f (x) = L,

xa

we mean that x gets closer and closer to a, f (x) gets closer and closer to the number L. Recall that when we say that x gets closer and closer to a, we mean that x gets arbitrarily close to a and can approach a from either side of a (x < a or x > a). Further, the limit must be the same as x approaches a from either side.

For functions of several variables, the idea is very similar. When we write

lim f (x, y) = L,

(x,y)(a,b)

we mean that as (x, y) gets closer and closer to (a, b), f (x, y) is getting closer and closer to the number L. In this case, (x, y) may approach (a, b) along any path through. Note that unlike the case for functions of a single variable, there are many (in fact, infinitely many) different paths passing through any given point (a, b).

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y

(a, b)

(x, y) x

For instance,

lim (x2 - xy + y2 - 2)

(x,y)(2,3)

asks us to identify what happens to the function x2 - xy + y2 - 2 as x approaches 2 and y approaches 3. Clearly, x2 - xy + y2 - 2 approaches 22 - 2(3) + 32 - 2 = 5 and we write

lim (x2 - xy + y2 - 2) = 5.

(x,y)(2,3)

Similarly, we can reason that

lim (sin xy + x2y) = sin(-) + (-) = -

(x,y)(1,-)

In other words, for many functions, we can compute limits simply by substituting in to the function.

Unfortunately, as with functions of a single variables, the limits we're most interested in cannot computed by simply substituting values for x and y. For instance, for

lim

(x,y)(3,1)

x

-

y 2y

-

1

substituting

in

x

=

3

and

y

=

1

gives

the

indeterminate

form

0 0

.

To

evaluate

this

limit,

we must investigate further.

Definition 5.3

Let f be defined on the interior of a circle center at the point (a, b), except possibly at (a, b) itself. We say that

lim f (x, y) = L,

(x,y)(a,b)

if for every > 0 there exists a > 0 such that |f (x, y) - L| < whenever

0 < (x - a)2 + (y - b)2 < .

The basic properties of limits of functions of one variable discussed in Calculus I are extended to the case of functions of several variables.

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Theorem 5.1 (Properties of the Limit)

Let c be a number. Suppose that

lim f (x, y) = L1 and lim g(x, y) = L2.

(x,y)(a,b)

(x,y)(a,b)

Then the following properties hold:

1. lim c = c (x,y)(a,b)

2. lim x = a and lim y = b

(x,y)(a,b)

(x,y)(a,b)

3. lim cf (x, y) = c lim f (x, y) = cL1

(x,y)(a,b)

(x,y)(a,b)

4. lim [f (x, y) ? g(x, y)] = lim f (x, y) ? lim g(x, y) = L1 ? L2

(x,y)(a,b)

(x,y)(a,b)

(x,y)(a,b)

5. lim f (x, y) ? g(x, y) = lim f (x, y) ? lim g(x, y) = L1 ? L2

(x,y)(a,b)

(x,y)(a,b)

(x,y)(a,b)

6.

lim

(x,y)(a,b)

f (x, g(x,

y) y)

=

lim f (x, y)

(x,y)(a,b)

lim g(x, y)

=

L1 L2

,

if L2

=0

(x,y)(a,b)

Remark: As for single-variable functions, limits of polynomial and rational functions in two variables may be found by substituting for x and y.

Example 5.5

Evaluate

lim

(x,y)(2,1)

5xy2 + 3y 2x2y + 3xy

.

Solution

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