8derivadas-parciales.pdf

8/11/2019 8Derivadas-parciales.pdf

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44. Let

(a) Show that as along any paththrough of the form with .

(b) Despite part (a), show that is discontinuous at .

(c) Show that is discontinuous on two entire curves.

45. Show that the function given by is continuous

on . [Hint: Consider .]

46. If , show that the function f given by is

continuous on . nfx c xc Vn

x a 2 x a x a nfx x f

f

0, 0fa 4y mx a0, 0

x,y l 0, 0fx,y l0

fx,y 0 ify 0 or y x 41 if 0 y x 4

; 42. At the beginning of this section we considered the function

and guessed that as on the basis

of numerical evidence. Use polar coordinates to confirm the

value of the limit. Then graph the function.

; 43. Graph and discuss the continuity of the function

fx,y 1

sinxy

xyif

if

xy 0

xy 0

x,y l 0, 0fx,y l 1

fx,y sinx2 y2

x2 y2

878 | | | | CHAPTER 14 PARTIAL DERIVATIVES

PARTIAL DERIVATIVES

On a hot day, extreme humidity makes us think the temperature is higher than it really

is, whereas in very dry air we perceive the temperature to be lower than the thermom-

eter indicates. The National Weather Service has devised the heat index (also called the

temperature-humidity index, or humidex, in some countries) to describe the combined

effects of temperature and humidity. The heat indexIis the perceived air temperature when

the actual temperature is Tand the relative humidity isH. SoIis a function of TandHand

we can write The following table of values of I is an excerpt from a table

compiled by the National Weather Service.

If we concentrate on the highlighted column of the table, which corresponds to a rela-

tive humidity ofH 70%, we are considering the heat index as a function of the single

variable Tfor a fixed value ofH. Lets write . Then describes how the

heat indexIincreases as the actual temperature Tincreases when the relative humidity is

70%. The derivative of t when is the rate of change ofIwith respect to Twhen:

t96 limhl0

t96 h t96

h lim

hl0f96 h, 70 f96, 70

h

T 96F

T 96F

tTtT fT, 70

T H

Relative humidity (%)

Actual

temperature

(F)

90

92

94

96

98

100

50 55 60 65 70 75 80 85 90

96

100

104

109

114

119

98

103

107

113

118

124

100

105

111

116

123

129

103

108

114

121

127

135

106

112

118

125

133

141

109

115

122

130

138

147

112

119

127

135

144

154

115

123

132

141

150

161

119

128

137

146

157

168

IfT,H.

14.3

TABLE 1

Heat index as a function of

temperature and humidity

I


2/14

We can approximate using the values in Table 1 by taking and :

Averaging these values, we can say that the derivative is approximately 3.75. This

means that, when the actual temperature is and the relative humidity is 70%, the

apparent temperature (heat index) rises by about for every degree that the actual

temperature rises!

Now lets look at the highlighted row in Table 1, which corresponds to a fixed temper-

ature of . The numbers in this row are values of the function ,which describes how the heat index increases as the relative humidity Hincreases when

the actual temperature is . The derivative of this function whenH 70% is the

rate of change ofIwith respect toHwhenH 70%:

By taking h 5 and 5, we approximate using the tabular values:

By averaging these values we get the estimate . This says that, when the tem-

perature is and the relative humidity is 70%, the heat index rises about for

every percent that the relative humidity rises.

In general, if is a function of two variables and , suppose we let only vary whilekeeping fixed, say , where is a constant. Then we are really considering a func-

tion of a single variable , namely, . If has a derivative at , then we call it

the partial derivative of with respect tox at and denote it by . Thus

By the definition of a derivative, we have

and so Equation 1 becomes

fxa, b limhl0

fa h, b fa, b

h2

ta limhl0

ta h ta

h

tx fx, bwherefxa, b ta1

fxa, ba, bfattx fx, bx

by byxyxf

0.9F96F

G70 0.9

G70 G65 G70

5

f96, 65 f96, 70

5

121 125

5 0.8

G70 G75 G705

f96, 75 f96, 70

5

130 125

5 1

G70

G70 limhl0

G70 h G70

h lim

hl0f96, 70 h f96, 70

h

T 96F

GH f96,HT 96F

3.75F

96F

t96

t96 t94 t96

2

f94, 70 f96, 70

2

118 125

2 3.5

t96 t98 t96

2

f98, 70 f96, 70

2

133 125

2 4

2h 2t96

SECTION 14.3 PARTIAL DERIVATIVES | | | | 879


3/14

Similarly, the partial derivative of with respect toy at , denoted by , is

obtained by keeping fixed and finding the ordinary derivative at of the func-

tion :

With this notation for partial derivatives, we can write the rates of change of the heat

indexIwith respect to the actual temperature Tand relative humidityHwhen

andH 70% as follows:

If we now let the point vary in Equations 2 and 3, and become functions of

two variables.

If is a function of two variables, its partial derivatives are the functions

and defined by

There are many alternative notations for partial derivatives. For instance, instead of

we can write or (to indicate differentiation with respect to thefirstvariable) or

. But here cant be interpreted as a ratio of differentials.

NOTATIONS FOR PARTIAL DERIVATIVES If , we write

To compute partial derivatives, all we have to do is remember from Equation 1 that

the partial derivative with respect to is just the ordinary derivative of the function of a

single variable that we get by keeping fixed. Thus we have the following rule.

RULE FOR FINDING PARTIAL DERIVATIVES OF z

1. To find , regard as a constant and differentiate with respect to .

2. To find , regard as a constant and differentiate with respect to .yfx,yxfy

xfx,yyfx

fx, y

y

tx

fyx,y fy f

y

yfx,y

z

yf2 D2fDyf

fxx,y fx f

x

xfx,y

z

xf1 D1fDxf

z fx,y

fxfxD1ff1fx

fyx,y limhl0

fx,y h fx,y

h

fxx,y

limhl0

fx h,y fx,y

h

fy

fxf4

fyfxa, b

fH96, 70 0.9fT96, 70 3.75

T 96F

fya, b limhl0

fa, b h fa, b

h3

Gy fa,ybx ax

fya, ba, bf



4/14

EXAMPLE 1 If , find and .

SOLUTION Holding constant and differentiating with respect to , we get

and so

Holding constant and differentiating with respect to , we get

M

INTERPRETATIONS OF PARTIAL DERIVATIVES

To give a geometric interpretation of partial derivatives, we recall that the equation

represents a surface (the graph of ). If , then the point

lies on . By fixing , we are restricting our attention to the curve in which the ver-

tical plane intersects S. (In other words, is the trace of in the plane .)

Likewise, the vertical plane intersects in a curve . Both of the curves and

pass through the point . (See Figure 1.)

Notice that the curve is the graph of the function , so the slope of its

tangent at is . The curve is the graph of the function ,

so the slope of its tangent at is .

Thus the partial derivatives and can be interpreted geometrically as the

slopes of the tangent lines at to the traces and of in the planes

and .

As we have seen in the case of the heat index function, partial derivatives can also be

interpreted as rates of change. If , then represents the rate of change of

with respect to when is fixed. Similarly, represents the rate of change of with

respect to when is fixed.

EXAMPLE 2 If , find and and interpret these num-

bers as slopes.

SOLUTION We have

fy1, 1

4fx1, 1

2

fyx,y 4yfxx,y 2x

fy1, 1fx1, 1fx,y 4 x 2 2y 2

xy

zzyyxzzxz fx,y

x a

y bSC2C1Pa, b, cfy a, bfxa, b

Gb fya, bPT2

Gy fa,yC2ta fxa, bPT1tx fx, bC1

FIGURE 1

The partial derivatives of fat (a, b)arethe slopes of the tangents to C and C .

0

(a, b, 0)

C

C

T

P(a, b, c)

S T

z

yx

P

C2C1C2Sx a

y bSC1y b

C1y bS

Pa, b, cfa, b cfSz fx,y

fy2, 1 3 22 12 4 1 8

fyx,y 3x2y 2 4y

yx

fx2, 1 3 22 2 2 13 16

fxx,y

3x2

2xy3

xy

fy2, 1fx2, 1fx,y x3x 2y 3 2y 2



5/14

The graph of is the paraboloid and the vertical plane inter-

sects it in the parabola , . (As in the preceding discussion, we label

it in Figure 2.) The slope of the tangent line to this parabola at the point is

. Similarly, the curve in which the plane intersects the parabo-

loid is the parabola , , and the slope of the tangent line at is

. (See Figure 3.)

M

Figure 4 is a computer-drawn counterpart to Figure 2. Part (a) shows the plane

intersecting the surface to form the curve and part (b) shows and . [We have used

the vector equations for and for .]

Similarly, Figure 5 corresponds to Figure 3.

FIGURE 4

FIGURE 5

1y0

4

3

2z

1

0

21

x

0

(a)

1y0

4

3

2z

1

0

21

x

0

(b)

1y0

4

3

2z

1

0

21 x

0

1y0

4

3

2z

1

0

21 x

0

T1rt 1 t, 1, 1 2tC1rt t, 1, 2 t2 T1C1C1

y 1

FIGURE 2

(1, 1, 1)

z=4--2

(1, 1)2

y=1

C

(1, 1, 1)

z=4--2

(1, 1)2

x=1

C

FIGURE 3

z

y

x

z

y

x

fy1, 1 41, 1, 1x 1z 3 2y 2

x 1C2fx1, 1 21, 1, 1C1

y 1z 2 x 2y 1z 4 x 2 2y 2f



6/14

EXAMPLE 3 If , calculate and .

SOLUTION Using the Chain Rule for functions of one variable, we have

M

EXAMPLE 4 Find and if is defined implicitly as a function of and

by the equation

SOLUTION To find , we differentiate implicitly with respect to , being careful to treat

as a constant:

Solving this equation for , we obtain

Similarly, implicit differentiation with respect to gives

M

FUNCTIONS OF MORE THAN TWO VARIABLES

Partial derivatives can also be defined for functions of three or more variables. For example,

if is a function of three variables , , and , then its partial derivative with respect to

is defined as

and it is found by regarding and as constants and differentiating with respect

to . If , then can be interpreted as the rate of change of with

respect tox wheny and are held fixed. But we cant interpret it geometrically because the

graph off lies in four-dimensional space.

In general, if is a function of variables, , its partial derivative

with respect to the ith variable is

uxi lim

hl0 fx1, . . . ,xi

1,xi

h,xi

1, . . . ,xn

fx1, . . . , xi , . . . ,xn

h

xi

u fx1,x2, . . . ,xn nu

z

wfx wxw fx,y, zxfx,y, zzy

fxx,y, z limhl0fx h,y, z fx,y, z

h

xzyxf

z

y

y 2 2xz

z2 2xy

y

z

x

x 2 2yz

z2 2xy

zx

3x 2 3z2z

x 6yz 6xy

z

x 0

y

xzx

x3 y 3 z3 6xyz 1

yxzzyzxV

f

y cos x

1 y

y x

1 y cos x

1 y x

1 y2

f

x cos x

1 y

x x

1 y cos x

1 y 1

1 y

f

y

f

xfx,y sin x

1 yV


FIGURE 6

N Some computer algebra systems can plot

surfaces defined by implicit equations in three

variables. Figure 6 shows such a plot of the

surface defined by the equation in Example 4.


7/14

and we also write

EXAMPLE 5 Find , , and if .

SOLUTION Holding and constant and differentiating with respect to , we have

Similarly, M

HIGHER DERIVATIVES

If is a function of two variables, then its partial derivatives and are also functions of

two variables, so we can consider their partial derivatives , , , and ,

which are called the second partial derivatives of . If , we use the following

notation:

Thus the notation (or ) means that we first differentiate with respect to and

then with respect to , whereas in computing the order is reversed.

EXAMPLE 6 Find the second partial derivatives of

SOLUTION In Example 1 we found that

Therefore

Mfyy

y3x 2y 2 4y 6x 2y 4fyx

x3x 2y 2 4y 6xy 2

fxy

y3x 2 2xy3 6xy 2fxx

x3x 2 2xy 3 6x 2y 3

fyx,y 3x2y2 4yfxx,y 3x

2 2xy 3

fx,y x 3 x 2y 3 2y 2

fyxy

x2fy xfx y

fyy fyy f22

y fy 2f

y 2

2z

y 2

fyx fyx f21

x fy 2f

xy

2z

xy

fxy fxy f12

y fx 2f

yx

2z

yx

fxx fxx f11

x fx 2f

x 2

2z

x 2

z fx,yffyyfyxfx yfxx

fyfxf

fz exy

z

andfy xex y ln z

fx yex y ln z

xzy

fx,y, z ex y ln zfzfyfx

u

x i

f

xifx i fi Dif



8/14

Notice that in Example 6. This is not just a coincidence. It turns out that the

mixed partial derivatives and are equal for most functions that one meets in practice.

The following theorem, which was discovered by the French mathematician Alexis Clairaut

(17131765), gives conditions under which we can assert that The proof is given

in Appendix F.

CLAIRAUTS THEOREM Suppose is defined on a disk that contains the point. If the functions and are both continuous on , then

Partial derivatives of order 3 or higher can also be defined. For instance,

fx yy fx yy

y

2f

yx

3f

y 2 x

fx ya, b fyxa, b

Dfyxfx ya, bDf

fx y fyx .

fyxfx y

fx y fyx

_1_2

2 1

2_2

20

_20

_1 0 10

y

x

z 0

f

fxx

FIGURE 7

40

_20

0

20

_2 _1 0 1 2 2 1 0

_1_2

y

x

z

1 0_1

_2

22_2

40

20

_40

_20

0

_1 0 1y

x

z

fx

fxy

fyx

_2_1

22_2

40

0

20

_1 0 1 1 0

y

x

z

2 2 1 0

_1_2

_2

40

20

_40

_20

0

_1 0 1y

x

z

fy

fyy

z

_2

2_2

20

_40

_20

0

_1 0 1 2 1 0

_1

y

x


N Figure 7 shows the graph of the function

in Example 6 and the graphs of its first- and

second-order partial derivatives for ,

. Notice that these graphs are con-

sistent with our interpretations of and as

slopes of tangent lines to traces of the graph of .

For instance, the graph of decreases if we start

at and move in the positive -direction.

This is reflected in the negative values of . You

should compare the graphs of and with the

graph of to see the relationships.fy

fyyfyx

fx

x0, 2f

f

fyfx

2 y 2

2 x 2

f

N Alexis Clairaut was a child prodigy inmathematics: he read lHospitals textbook

on calculus when he was ten and presented a

paper on geometry to the French Academy of

Sciences when he was 13. At the age of 18,

Clairaut published Recherches sur les courbes

double courbure, which was the first systematic

treatise on three-dimensional analytic geometry

and included the calculus of space curves.


9/14

and using Clairauts Theorem it can be shown that if these functions are

continuous.

EXAMPLE 7 Calculate if .

SOLUTION

M

PARTIAL DIFFERENTIAL EQUATIONS

Partial derivatives occur inpartial differential equations that express certain physical laws.

For instance, the partial differential equation

is called Laplaces equation after Pierre Laplace (17491827). Solutions of this equation

are called harmonic functions; they play a role in problems of heat conduction, fluid flow,

and electric potential.

EXAMPLE 8 Show that the function is a solution of Laplaces

equation.

SOLUTION

Therefore satisfies Laplaces equation. M

The wave equation

describes the motion of a waveform, which could be an ocean wave, a sound wave, a light

wave, or a wave traveling along a vibrating string. For instance, if represents the dis-

placement of a vibrating violin string at time and at a distance from one end of thestring (as in Figure 8), then satisfies the wave equation. Here the constant depends

on the density of the string and on the tension in the string.

EXAMPLE 9 Verify that the function satisfies the wave equation.

SOLUTION

So satisfies the wave equation. Mu

utt a2 sinx at a 2uxxut a cosx at

uxx sinx atux cosx at

ux, t sinx at

aux, txt

ux, t

2u

t2 a2

2u

x 2

u

uxx uyy exsiny exsiny 0

uyy ex sinyuxx e

x siny

uy ex cosyux e

x siny

ux,y ex siny

2u

x 2

2u

y 2 0

fxx yz 9 cos3x yz 9yz sin3x yz

fxx y 9z cos3x yz

fxx 9 sin3x yz

fx 3 cos3x yz

fx,y, z

sin

3x yz

fxx yzV

fx yy fyx y fyy x


FIGURE 8

u(x, t)

x


10/14

THE COBB-DOUGLAS PRODUCTION FUNCTION

In Example 3 in Section 14.1 we described the work of Cobb and Douglas in modeling the

total production P of an economic system as a function of the amount of labor L and the

capital investment K. Here we use partial derivatives to show how the particular form oftheir model follows from certain assumptions they made about the economy.

If the production function is denoted by , then the partial derivative

is the rate at which production changes with respect to the amount of labor. Economists

call it the marginal production with respect to labor or the marginal productivity of labor.

Likewise, the partial derivative is the rate of change of production with respect to

capital and is called the marginal productivity of capital. In these terms, the assumptions

made by Cobb and Douglas can be stated as follows.

(i) If either labor or capital vanishes, then so will production.

(ii) The marginal productivity of labor is proportional to the amount of production

per unit of labor.

(iii) The marginal productivity of capital is proportional to the amount of production

per unit of capital.

Because the production per unit of labor is , assumption (ii) says that

for some constant . If we keep Kconstant , then this partial differential equa-

tion becomes an ordinary differential equation:

If we solve this separable differential equation by the methods of Section 9.3 (see also

Exercise 79), we get

Notice that we have written the constant as a function of because it could depend on

the value of .

Similarly, assumption (iii) says that

and we can solve this differential equation to get

Comparing Equations 6 and 7, we have

PL, K bLK8

PL0, K C2L0 K

7

P

K

P

K

K0

K0C1

PL, K0 C1K0 L

6

dP

dL

P

L5

K K0

P

L

P

L

PL

PK

PLP PL, K



11/14

where b is a constant that is independent of bothL and K. Assumption (i) shows that

and .

Notice from Equation 8 that if labor and capital are both increased by a factor m, then

If , then , which means that production is also increased

by a factor of m. That is why Cobb and Douglas assumed that and therefore

This is the Cobb-Douglas production function that we discussed in Section 14.1.

PL, K bLK1

1

PmL, mK mPL, K 1

PmL, mK bmLmK mbLK mPL, K

0

0


(b) In general, what can you say about the signs of

and ?

(c) What appears to be the value of the following limit?

4. The wave heights in the open sea depend on the speed

of the wind and the length of time that the wind has been

blowing at that speed. Values of the function are

recorded in feet in the following table.

(a) What are the meanings of the partial derivatives

and ?

(b) Estimate the values of and . What are

the practical interpretations of these values?

(c) What appears to be the value of the following limit?

limtl

ht

ft40, 15fv40, 15ht

hv

2

4

5

9

14

19

24

2

4

7

13

21

29

37

2

5

8

16

25

36

47

2

5

8

17

28

40

54

2

5

9

18

31

45

62

2

5

9

19

33

48

67

2

5

9

19

33

50

69

v

t

10

15

20

30

40

50

60

Duration (hours)

Windspeed(knots)

5 10 15 20 30 40 50

h fv, tt

vh

limv

l

Wv

WvWTThe temperature at a location in the Northern Hemisphere

depends on the longitude , latitude , and time , so we can

write . Lets measure time in hours from the

beginning of January.

(a) What are the meanings of the partial derivatives

, and ?

(b) Honolulu has longitude and latitude .

Suppose that at 9:00 AM on January 1 the wind is blowing

hot air to the northeast, so the air to the west and south is

warm and the air to the north and east is cooler. Would you

expect , and to be

positive or negative? Explain.

2. At the beginning of this section we discussed the function

, where is the heat index, is the temperature,and is the relative humidity. Use Table 1 to estimate

and . What are the practical interpretations

of these values?

3. The wind-chill index is the perceived temperature when the

actual temperature is and the wind speed is , so we can

write . The following table of values is an excerpt

from Table 1 in Section 14.1.

(a) Estimate the values of and . Whatare the practical interpretations of these values?

fv15, 30fT15, 30

18

24

30

37

20

26

33

39

21

27

34

41

22

29

35

42

23

30

36

43

Tv 20 30 40 50 60

10

15

20

25Actualtemperature(C)

70

23

30

37

44

Wind speed (km/h)

WfT, vvT

W

fH92, 60fT92, 60H

TIIfT,H

ft158, 21, 9fx158, 21, 9, fy158, 21, 9

21N158W

TtTyTx,

Tfx,y, ttyx

T1.

EXERCISES14.3


12/14


10. A contour map is given for a function . Use it to estimate

and .

11. If , find and and inter-pret these numbers as slopes. Illustrate with either hand-drawn

sketches or computer plots.

12. If , find and and

interpret these numbers as slopes. Illustrate with either hand-

drawn sketches or computer plots.

; 1314 Find and and graph , , and with domains andviewpoints that enable you to see the relationships between them.

13. 14.

1538 Find the first partial derivatives of the function.

15. 16.

17. 18.

19. 20.

22.

23. 24.

25. 26.

27. 28.

29. 30.

32.

33. 34.

35. 36.

37.

38.

3942 Find the indicated partial derivatives.

39. ;

40. ;

41. ; fy 2, 1, 1fx,y, z yx y z

fx 2, 3fx,y arctanyx

fx 3, 4fx,y ln(x sx 2 y 2 )

u sinx1 2x2 nx n

u sx 21 x 22 x 2nfx,y, z, t

xy 2

t 2zfx,y,z, t xyz 2tanyt

u xyzu xysin1yz

w zexyzw lnx 2y 3z31.

fx,y, z xsiny zfx,y, z xz 5x 2y 3z4

fx,y yxy

cost2 dtu te wt

fx, t arctan(xst)fr, s rlnr2 s 2

w ev

u v 2w sin cos

fx,y xyfx,y x y

x y21.

z tanxyz 2x 3y10

fx, t sx ln tfx, t etcos x

fx,y x 4y 3 8x 2yfx,y y 5 3xy

fx,y

xe

x 2y2

fx,y

x

2

y

2

x

2

y

fyfxffyfx

fy1, 0fx1, 0fx,y s4 x 2 4y 2

fy1, 2fx1, 2fx,y 16 4x 2 y 2

3 x

y

3

_2

06 8

10

14

16

12

18

2

4

_4

1

fy2, 1fx2, 1f5 8 Determine the signs of the partial derivatives for the function

whose graph is shown.

(a) (b)

6. (a) (b)

7. (a) (b)

8. (a) (b)

The following surfaces, labeled , , and , are graphs of a

function and its partial derivatives and . Identify each

surface and give reasons for your choices.

b_4

_3 _1 0 1 3

0 _2

y

x

z 0

2

4

2_2

a

8

_8

_4

_3 _1 0 1 3

0 _2

yx

z 0

2

4

2_2

c

8

_8

_3 _1 0 1 30

_2

yx

z 0

2

4

2_2

_4

fyfxf

cba9.

fxy 1, 2fxy 1, 2

fyy 1, 2fxx 1, 2

fy1, 2fx1, 2

fy1, 2fx1, 25.

1x

y

z

2

f


13/14


Use the table of values of to estimate the values of

, , and .

70. Level curves are shown for a function . Determine whether

the following partial derivatives are positive or negative at the

point .

(a) (b) (c)

(d) (e)

71. Verify that the function is a solution of theheat conduction equation .

72. Determine whether each of the following functions is a solution

of Laplaces equation .

(a) (b)

(c) (d)

(e)

(f )

73. Verify that the function is a solution of

the three-dimensional Laplace equation .

74. Show that each of the following functions is a solution of the

wave equation .

(a) (b)

(c)

(d)

75. If and are twice differentiable functions of a single vari-

able, show that the function

is a solution of the wave equation given in Exercise 74.

76. If , where ,

show that

77. Verify that the function is a solution of the

differential equations

z

x z

y 1

z lnex ey

2u

x 21

2u

x 22

2u

x 2n u

a 21 a22 a

2n 1u e

a1x1a2x2anxn

ux, t fx at tx at

tf

u sinx at lnx atu x at6 x at6

u ta 2t2 x 2u sinkx sinak tu t t a

2uxx

uxx uyy uzz 0

u 1sx 2 y 2 z 2u excosy eycosx

u sinx coshy cosx sinhy

u ln sx 2 y 2u x 3 3xy 2u x 2 y 2u x 2 y 2

uxx uyy 0

ut 2uxx

u e

2

k

2

tsin kx

10 8 6 4 2

y

x

P

fyyfxy

fxxfyfx

P

f

12.5

18.1

20.0

10.2

17.5

22.4

9.3

15.9

26.1

xy

2.5

3.0

3.5

1.8 2.0 2.2

fx y3, 2fx3, 2.2fx3, 2fx,y69.42. ;

4344 Use the definition of partial derivatives as limits (4) to find

and .

43. 44.

4548 Use implicit differentiation to find and .

45. 46.

47. 48.

4950 Find and .

49. (a) (b)

(a) (b)

(c)

5156 Find all the second partial derivatives.

51. 52.

53. 54.

55. 56.

5760 Verify that the conclusion of Clairauts Theorem holds, that

is, .

57. 58.

59. 60.

6168 Find the indicated partial derivative.

61. ; ,

62. ; ,

63. ; ,

64. ; ,

65. ;

66. ;

67. ; ,

68. ;

6u

x y 2 z 3u x ay bz c

3w

x 2 y

3w

z y xw

x

y 2z

3z

u v wz usv w

3u

r2 u e rsin

frs tfrs sfr, s, t rlnrs2t3

fyzzfxy zfx,y, z cos4x 3y 2z

ftx xftt tfx, t x2ect

fyy yfxx yfx,y 3xy4x 3y 2

u xyeyu ln sx 2 y 2u x 4y 2 2xy 5u xsinx 2y

ux y uyx

v exey

z arctanx y

1 xy

v xy

x yw su 2 v 2

fx,y sin2mx nyfx,y x 3y 5 2x 4y

z fxy

z fxyz fxty50.

z fx yz fx ty

zyzx

sinxyz x 2y 3zx z arctanyz

yz lnx zx 2 y 2 z2 3xyz

zyzx

fx,y x

x y 2fx,y xy 2 x 3y

fyx,yfxx,y

fz 0, 0, 4fx,y, z ssin2x sin2y sin2z


14/14


You are told that there is a function whose partial derivatives

are and . Should you

believe it?

;88. The paraboloid intersects the plane

in a parabola. Find parametric equations for the tangent

line to this parabola at the point . Use a computer to

graph the paraboloid, the parabola, and the tangent line on the

same screen.

89. The ellipsoid intersects the plane

in an ellipse. Find parametric equations for the tangent line to

this ellipse at the point .

90. In a study of frost penetration it was found that the temperature

at time (measured in days) at a depth (measured in feet)can be modeled by the function

where and is a positive constant.

(a) Find . What is its physical significance?

(b) Find . What is its physical significance?

(c) Show that satisfies the heat equation for a cer-

tain constant .

; (d) If , , and , use a computer tograph .

(e) What is the physical significance of the term in the

expression ?

91. Use Clairauts Theorem to show that if the third-order partial

derivatives of are continuous, then

92. (a) How many th-order partial derivatives does a function of

two variables have?

(b) If these partial derivatives are all continuous, how many ofthem can be distinct?

(c) Answer the question in part (a) for a function of three

variables.

93. If , find .

[Hint: Instead of finding first, note that its easier to

use Equation 1 or Equation 2.]

94. If , find .

95. Let

; (a) Use a computer to graph .(b) Find and when .

(c) Find and using Equations 2 and 3.

(d) Show that and .

(e) Does the result of part (d) contradict Clairauts Theorem?Use graphs of and to illustrate your answer.fyxfxy

CAS

fyx 0, 0 1fxy 0, 0 1fy0, 0fx0, 0

x,y 0, 0fyx,yfxx,yf

fx,y 0

x 3y xy 3

x 2 y 2if

if

x,y 0, 0

x,y 0, 0

fx0, 0fx,y s3x 3 y 3

fxx,yfx1, 0fx,y xx

2y 232e sinx

2y

n

fx yy fyx y fyy x

f

sint xx

Tx, t

T1 10T0 0 0.2

k

Tt kTxxT

TtTx

2365

Tx, t T0 T1ex sint x

xtT

1, 2, 2

y 24x 2 2y 2 z2 16

1, 2, 4x 1

z 6 x x 2 2y 2

fyx,y 3x yfxx,y x 4yf87.and

78. Show that the Cobb-Douglas production function

satisfies the equation

79. Show that the Cobb-Douglas production function satisfies

by solving the differential equation

(See Equation 5.)80. The temperature at a point on a flat metal plate is given

by , where is measured in C

and in meters. Find the rate of change of temperature with

respect to distance at the point in (a) the -direction and

(b) the -direction.

The total resistance produced by three conductors with resis-

tances , , connected in a parallel electrical circuit is

given by the formula

Find .

82. The gas law for a fixed mass of an ideal gas at absolute tem-

perature , pressure , and volume is , where is

the gas constant. Show that

83. For the ideal gas of Exercise 82, show that

84. The wind-chill index is modeled by the function

where is the temperature and is the wind speed

. When and , by how much

would you expect the apparent temperature to drop if theactual temperature decreases by ? What if the wind speed

increases by ?

85. The kinetic energy of a body with mass and velocity is

. Show that

If , , are the sides of a triangle and , , are the opposite

angles, find , , by implicit differentiation ofthe Law of Cosines.

AcAbAa

CBAcba86.

K

m

2K

v2 K

K1

2 mv2

vm

1 kmh1C

W

v 30 kmhT 15CkmhvCT

W 13.12 0.6215T 11.37v 0.16 0.3965Tv 0.16

TP

TV

T mR

P

VV

TT

P 1

RPV mRTVPT

m

RR1

1

R

1

R1

1

R2

1

R3

R3R2R1

R81.

y

x2, 1x,y

TTx,y 601 x 2 y 2x,y

dP

dL

P

L

PL, K0 C1K0L

LP

L K

P

K P

P bLK

2z

x 2

2z

y 2 2z

x y2 0

8derivadas-parciales.pdf

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