11.3

PARTIAL DERIVATIVES

If is a function of two variables and , suppose we let only vary while keeping fixed, say , where is a constant. Then we are really considering a function of a single variable , namely, . If has a derivative at , then we call it the partial derivative of with respect to x at and denote it by . Thus

By the definition of a derivative, we have

and so Equation 1 becomes

Similarly, the partial derivative of with respect to y at , denoted by , is obtained by keeping fixed and finding the ordinary derivative at of the function :

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 the first variable) or fx &f&x. But here f1 &f&xD1fcan’t be interpreted as a ratio of differentials.

fyx, y  lim

hl 0

fx, y  h  f x, y

h fxx, y  lim

hl 0

fx  h, y  f x, y

h fy

fx

f 4

fy

fx

a, b

fya, b  lim

hl 0

fa, b  h  f a, b

3 h

Gy  f a, yx x  a b

fya, b

a, b

f fxa, b  lim

hl 0

fa  h, b  f a, b

2 h

ta  lim

hl 0

ta  h  ta

h

tx  f x, b

where fxa, b  ta

1

fxa, b

a, b

f

t a tx  f x, b

x

b y b

y

x y

x f

11.3

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 .

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

■

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 vertical 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 .

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.z y x

&z&y z x y

&z&x x a z  f x, y

y b fxa, bPa, b, cTfy2a, bP Gb  fC1 ^ya, bC2 S Gy  f a, yT1 P ta  f^xa, b C2

tx  f x, b

C1

P C2

C1

C2

S x a

y b S y by b C1 S C1

Pa, b, c S f fa, b  c

z  f x, y

fy2, 1  3  2² 1² 4  1  8 fyx, y  3x²y² 4y

y x

fx2, 1  3  2² 2  2  1³ 16 fxx, y  3x² 2xy³

x y

fy2, 1

fx2, 1

fx, y  x³ x²y³ 2y²

y fx, y

x fy

x fx, y

y fx

fx, y

y

t x

fyx, y  f^y &f

&y  &

&y fx, y  &z

&y  f² D²f D^yf fxx, y  f^x &f

&x  &

&x fx, y  &z

&x  f¹ D¹f D^xf z  f x, y

FIGURE 1

The partial derivatives of f at (a, b) are the slopes of the tangents to C¡ and C™.

0

(a, b, 0)

C™

C¡

T¡

P (a, b, c)

S T™

z

x y

EXAMPLE 2 If , find and and interpret these numbers as slopes.

SOLUTION We have

The graph of is the paraboloid and the vertical plane

intersects 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 paraboloid is the parabola , , and the slope of the tangent line

at is . (See Figure 3.) ■

EXAMPLE 3 If , calculate and . SOLUTION Using the Chain Rule for functions of one variable, we have

■

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

&z ■

&y  y² 2xz z² 2xy y

&z

&x  x² 2yz z² 2xy

&z&x 3x² 3z² &z

&x  6yz  6xy &z

&x  0 y

&z&x x

x³ y³ z³ 6xyz  1 y

z x

&z&y

&z&x

V

&f

&y  cos













&f

&x  cos













&f

&y

&f

&x fx, y  sin





V

fy1, 1  4

1, 1, 1 fx1, 1  2 z  3  2y² x 1C2 x 1

1, 1, 1C1

y 1

z  2  xz  4  x^{2 2} 2y² y 1 f

fy1, 1  4 fx1, 1  2

fyx, y  4y fxx, y  2x

fy1, 1

fx1, 1

fx, y  4  x² 2y²

FIGURE 2 (1, 1, 1)

z=4-≈-2¥

(1, 1) 2

y=1 C¡

z

y

x

(1, 1, 1) z=4-≈-2¥

(1, 1) 2

x=1 C™

FIGURE 3 z

y

x

FIGURE 4

■ Some computer algebra systems can plot surfaces defined by implicit equa- tions in three variables. Figure 4 shows such a plot of the surface defined by the equation in Example 4.

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 to x when y and are held fixed. But we can’t interpret it geometri- cally because the graph of f lies in four-dimensional space.

In general, if is a function of variables, , its partial deriva- tive with respect to the ith variable is

and we also write

EXAMPLE 5 Find , , and if .

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

Similarly, ■

HIGHER DERIVATIVES

If is a function of two variables, then its partial derivatives and are also func- tions 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 y fyxthe order is reversed.

&²f&y &x x fx y

 f^y^y f^yy f²² &

&y





 f^y^x f^yx f²¹ &

&x





&x &y

 f^x^y f^xy f¹² &

&y





&y &x

 f^x^x f^xx f¹¹ &

&x





z  f x, y

 f^y^y ffx  ff^xy^x  f^x^y  f^y^x

f

fz e^xy and z

fy xe^{x y}lnz

fx ye^{x y}lnz z x

y

fx, y, z  e^{x y}lnz f_z

fy

fx

&u

&xⁱ  &f

&xⁱ  f^xi fⁱ Dⁱf

&u

&xⁱ  lim

hl 0

fx¹, . . . , xi1, xi h, xⁱ1, . . . , xn  f x¹, . . . , xi, . . . , xn h

xi

u f x¹, x2, . . . , xn n

u

z w

fx &^w&x w f x, y, z

x

fx, y, z

z y fxx, y, z  lim

hl 0

fx  h, y, z  f x, y, z

h x

z y x f

EXAMPLE 6 Find the second partial derivatives of

SOLUTION In Example 1 we found that

Therefore

■ 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 (1713–1765), gives conditions under which we can assert that

The proof is given in Appendix B.

CLAIRAUT’S 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,

and using Clairaut’s Theorem it can be shown that if these functions are continuous.

EXAMPLE 7 Calculate if .

SOLUTION

■

PARTIAL DIFFERENTIAL EQUATIONS

Partial derivatives occur in partial differential equations that express certain physical laws. For instance, the partial differential equation

&²u

&x²  &²u

&y²  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  sin3x  yz

fxx yz

V

fx yy f^{yx y} f^yyx fx yy  f^{x y}^y &

&y





fx ya, b  f^yxa, b

D fyx

fx y

a, b f D

fx y f^yx. fyx

fx y

fx y f^yx

fyy &

&y 3x²y² 4y  6x²y 4 fyx &

&x 3x²y² 4y  6xy²

fxy &

&y 3x² 2xy³  6xy² fxx &

&x 3x² 2xy³  6x  2y³

fyx, y  3x²y² 4y fxx, y  3x² 2xy³

fx, y  x³ x²y³ 2y²

■ Alexis Clairaut was a child prodigy in mathematics, having read l’Hospital’s 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 à double courbure, which was the first systematic treatise on three-dimensional analytic geometry and included the calculus of space curves.

is called Laplace’s equation after Pierre Laplace (1749–1827). Solutions of this equa- tion are called harmonic functions and play a role in problems of heat conduction, fluid flow, and electric potential.

EXAMPLE 8 Show that the function is a solution of Laplace’s equation.

SOLUTION

Therefore, satisfies Laplace’s equation. ■

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 repre- sents the displacement of a vibrating violin string at time and at a distance from one end of the string (as in Figure 5), 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.u _■

utt a²sinx  at  a²uxx

ut a cosx  at

uxx sinx  at

ux cosx  at

ux, t  sinx  at

a

ux, t t x

ux, t

&²u

&t²  a²&²u

&x² u

uxx u^yy e^x sin y e^x sin y 0 uyy e^xsin y uxx e^xsin y

uy e^xcos y ux e^xsin y

ux, y  e^xsin y

FIGURE 5

u(x, t) x

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

3 3

_2

0 6 8

10 14 16 12

18 2

4 _4

1 x

y fy2, 1

fx2, 1

The temperature at a location in the Northern Hemisphere f depends on the longitude , latitude , and time , so we can write . Let’s 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 . Sup- pose that at 9:00 AMon 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.

ft158, 21, 9

fx158, 21, 9, f^y158, 21, 9

21 N 158 W

&T&t

&T&y &T&x,

T f x, y, t

t y

x 1. T

EXERCISES

11.3

32. ;

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

33–34 ^■ Use the definition of partial derivatives as limits (4) to find and .

33. 34.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

;^{35–36 ■ Find and} and graph , , and with domains and viewpoints that enable you to see the relationships between them.

35. 36.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

37– 40 ^■ Use implicit differentiation to find and .

37. 38.

39. 40.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

41– 42 ^■ Find and .

41. (a) (b)

(a) (b)

(c)

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

43– 48 ^■ Find all the second partial derivatives.

43. 44.

45. 46.

47. 48.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

49–50 ^■ Verify that the conclusion of Clairaut’s Theorem holds,

that is, .

49. 50.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

51–56 ^■ Find the indicated partial derivative.

51. ; ,

52. ; ,

53. ; ,

54. ; ,

55. ;

56. ;

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

&⁶u

&x &y²&z³ u x^ay^bz^c

&³u

&r²&

u e^r sin 

fr s t

fr s s

fr, s, t  r lnrs²t³

fyzz

fx yz

fx, y, z  cos4x  3y  2z

ft x x

ft t t

fx, t  x²e^ct

fy y y

fx x y

fx, y  3xy⁴ x³y²

u x⁴y² 2xy⁵ u x sinx  2y

ux y u^{y x}

v sx  y² u e^s sin t

z  y tan 2x z  xx  y

fx, y  ln3x  5y

fx, y  x⁴ 3x²y³

z  f xy z  f xy

z  f xty

42.

z  f x  y

z  f x  ty

&z&y

&z&x

sinxyz  x  2y  3z x z  arctanyz

yz  lnx  z

x² y² z² 3xyz

&z&y

&z&x fx, y  xe^x2y2 fx, y  x² y² x²y

fy

fx

f fy

fx

fx, y  x x y² fx, y  xy² x³y

fyx, y

fxx, y

fv2, 0, 3

fu,^v,w ^w tanu^v 3– 4 ^■ Determine the signs of the partial derivatives for the

function whose graph is shown.

(a) (b)

4. (a) (b)

(c) (d)

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

5. If , find and and

interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots.

6. If , find and and

interpret these numbers as slopes. Illustrate with either hand-drawn sketches or computer plots.

7–28 ^■ Find the first partial derivatives of the function.

7.

8.

9. 10.

12.

13. 14.

15. 16.

17. 18.

19. 20.

22.

23. 24.

25. 26.

27.

28.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

29–32 ^■ Find the indicated partial derivatives.

29. ;

30. ;

31. fx, y, z  xy  z; f_z3, 2, 1

fy6, 4

fx, y  sin2x  3y

fx3, 4

fx, y  sx² y²

u sinx¹ 2x²  nxⁿ u sx1² x2²  xn²

fx, y, z, t  x y² t 2z fx, y, z, t  xyz² tanyt

u x^yz u xe^t sin 

w sr² s² t² w lnx  2y  3z

21.

fx, y, z  x²e^yz fx, y, z  xy²z³ 3yz

fx, y 

y

fx, t  arctan

(

)

fr, s  r lnr² s²

fs, t  st²s² t² w sin cos 

fx, y  x^y fx, y  x y

x y 11.

z  y ln x z  xe^{3 y}

fx, y  x⁵ 3x³y² 3xy⁴ fx, y  3x  2y⁴

fy1, 0

fx1, 0

fx, y  s4  x² 4y²

fy1, 2

fx1, 2

fx, y  16  4x² y²

fy y1, 2

fx x1, 2

fy1, 2

fx1, 2

fy1, 2

fx1, 2

3.

1 2

x

y z

f

67. For the ideal gas of Exercise 66, show that

68. The wind-chill index is a measure of how cold it feels in windy weather. It 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 the actual temperature decreases by ? What if the wind speed increases by ?

69. 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 of the Law of Cosines.

You are told that there is a function whose partial deriva-

tives are and . Should

you believe it?

;^72. The paraboloid intersects the plane in a parabola. Find parametric equations for the tan- gent line to this parabola at the point . Use a com- puter to graph the paraboloid, the parabola, and the tangent line on the same screen.

73. The ellipsoid intersects the plane in an ellipse. Find parametric equations for the tangent line to this ellipse at the point .

74. In a study of frost penetration it was found that the tempera- ture 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 certain constant .

; ^{(d) If , , and} , use a computer to

graph .

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

75. Use Clairaut’s Theorem to show that if the third-order par- tial derivatives of are continuous, then

fx y y f^{y x y} f^{y y x} f

sint  x x

Tx, t

T1 10 T0 0

  0.2 k

Tt kT^{x x} T

&T&t

&T&x 

  2365

Tx, t  T⁰ T¹e^xsint  x

x t

T

1, 2, 2

y 2

4x² 2y² z² 16

1, 2, 4

x 1 z  6  x  x² 2y²

fyx, y  3x  y fxx, y  x  4y

71. f

&A&c

&A&b

&A&a

C B A c

b a 70.

&K

&m

&²K

&^v2  K K¹2mv²

m v 1 kmh

1C v 30 kmh T 15C

kmh

C v T

W 13.12  0.6215T  11.37^v0.16 0.3965T^v0.16 T&P

&T

&V

&T  mR 57. Verify that the function is a solution of the

heat conduction equation .

58. Determine whether each of the following functions is a solution of Laplace’s equation .

(a) (b) (c) (d) (e) (f )

59. Verify that the function is a

solution of the three-dimensional Laplace equation .

60. Show that each of the following functions is a solution of the wave equation .

(a) (b) (c) (d)

61. 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 60.

62. If , where ,

show that

63. Show that the function is a solution of the equation

64. 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 temper- ature with respect to distance at the point in (a) the -direction and (b) the -direction.

The total resistance produced by three conductors with resistances , , connected in a parallel electrical cir- cuit is given by the formula

Find .

66. The gas law for a fixed mass of an ideal gas at absolute temperature , pressure , and volume is , where is the gas constant. Show that

&P

&V

&V

&T

&T

&P  1 R

PV mRT V

P T

m

&R&R¹ 1 R  1

R1  1 R2  1

R3

R3

R2

R1

65. R

y x

2, 1

 x, y

T Tx, y  601  xx, y² y²

&³z

&x³  &³z

&y³  x &³z

&x &y²  y &³z

&x²&y z  xe^y ye^x

&²u

&x²¹  &²u

&x²²   &²u

&x²ⁿ  u

a²1 a²²  a²ⁿ 1 u e^{a1x1a2x2 anxn}

ux, t  f x  at  tx  at

t f

u sinx  at  lnx  at

u x  at⁶ x  at⁶ u ta²t² x² u sinkx sinakt

ut t a²ux x

ux x u^{y y} uzz 0

u 1sx² y² z² u e^x cos y e^y cos x

u sin x cosh y  cos x sinh y u ln sx² y²

u x³ 3xy² u x² y² u x² y²

ux x u^{y y} 0 ut²ux x

u e^2k2tsin kx

79. 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 Clairaut’s Theorem? Use graphs of and to illustrate your answer.

fy x

fx y CAS

fy x0, 0  1 fx y0, 0  1

fy0, 0

fx0, 0 fyx, y x, y  0, 0

fxx, y

f

fx, y 



x, y  0, 0

76. (a) How many th-order partial derivatives does a function of two variables have?

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

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

77. If , find .

[Hint: Instead of finding first, note that it’s easier to use Equation 1 or Equation 2.]

78. If , fx, y  s³x³ y³ find fx0, 0. fxx, y

fx1, 0

fx, y  xx² y²^32e^sinx2y n

TANGENT PLANES AND LINEAR APPROXIMATIONS

One of the most important ideas in single-variable calculus is that as we zoom in toward a point on the graph of a differentiable function, the graph becomes indistin- guishable from its tangent line and we can approximate the function by a linear func- tion. (See Section 2.8.) Here we develop similar ideas in three dimensions. As we zoom in toward a point on a surface that is the graph of a differentiable function of two variables, the surface looks more and more like a plane (its tangent plane) and we can approximate the function by a linear function of two variables. We also extend the idea of a differential to functions of two or more variables.

TANGENT PLANES

Suppose a surface has equation , where has continuous first partial derivatives, and let be a point on . As in the preceding section, let and be the curves obtained by intersecting the vertical planes and with the surface . Then the point lies on both and . Let and be the tangent lines to the curves and at the point . Then the tangent plane to the surface at the point is defined to be the plane that contains both tangent lines and . (See Figure 1.)

We will see in Section 11.6 that if is any other curve that lies on the surface and passes through , then its tangent line at also lies in the tangent plane. There- fore, you can think of the tangent plane to at as consisting of all possible tangent lines at to curves that lie on and pass through . The tangent plane at is the plane that most closely approximates the surface near the point .

We know from Equation 10.5.7 that any plane passing through the point has an equation of the form

By dividing this equation by and letting and , we can write it in the form

If Equation 1 represents the tangent plane at , then its intersection with the plane P z  z⁰ ax  x⁰  by  y⁰

1

b BC a AC

C

Ax  x⁰  By  y⁰  Cz  z⁰  0

Px⁰, y0,z⁰ P

S

P P

S P

P S

P P

S C

T2

T1

P

S P

C2

C1

T2

T1

C2

C1

P S

x x⁰ y y⁰

C2

C1

S

PSx⁰, y0,z⁰ z  f x, y f

11.4

z

FIGURE 1

The tangent plane contains the tangent lines T¡TT and T™TT .

y

x