Section 14.7 (3/24/08) Functions of three variables

(3/24/08)Section 14.7

Functions of three variables

Overview: In this section we modify definitions and results from Sections 14.1 through 14.6 to apply to

functions with three variables. We discuss level surfaces, partial derivatives, the Chain Rule, directional

derivatives, gradient vectors, and tangent planes to level surfaces.

Topics:

• Functions of three variables and their level surfaces

• Quadric surfaces

• Partial derivatives

• Gradient vectors and directional derivatives

• Tangent planes to level surfaces

Functions of three variables and their level surfacesA function w = f of three variables is a rule that assigns a single number denoted f(x, y, z) to each point

(x, y, z) in its domain. The domain is a subset or all of xyz-space.

Example 1 (a) What is the domain of G(x, y, z) =1

x+

1

y+

1

z? What is the value of G at (2, 4, 4)?

Solution (a) G(x, y, z) is defined unless x or y or z is zero, so the domain consists of allxyz-space except for the xy- xz-, and yz-coordinate planes.(b) G(2, 4, 4) = 1

2+ 1

4+ 1

4= 1. �

Example 2 The frustum of a right circular cone with base of radius R (meters), top of radius r

(meters), and height h (meters) (Figure 1) has volume V (R, r, h) = 13π(R2 +rR+r2)h.

(a) What does V (6, 3.2) represent and what is its value if lengths are measured inmeters? (b) What is the geometric interpretation of V (R,0, h)?

FIGURE 1 FIGURE 2

Solution (a) V (6, 3, 2) is the volume of a frustum of a right circular cone that has a base ofradius 6 meters, has a top of radius 3 meters, and is 2 meters high. The volume is

V (6, 3, 2) = 13π(62 + 6(3) + 32)(2) = 1

3π(36 + 18 + 9)(2) = 42π cubic meters.

(b) If the radius r of the top of the frustum is zero, then the frustum is a cone. Thus,

V (R, 0, h) = 13πR2h is the volume of a right circular cone whose base has radius R and

whose height is h (Figure 2). �

354

Section 14.7, Functions of three variables p. 355 (3/24/08)

We cannot visualize graphs of functions of three variables because they would be in four dimensionalspace. We can, however, study their level surfaces.

Definition 1 (Level surfaces with three variables) A level surface of a function w = f(x, y, z)with three variables is a surface f(x, y, z) = c where the function has the constant value c.

Example 3 Describe the level surfaces of f(x, y, z) = x2 + y2 + z2.

Solution Because x2 + y2 + z2 is the square of the distance from (x, y, z) to the origin (0, 0, 0),the level surface f = c for c > 0 consists of the points that are a distance

√c from

the origin (Figure 3). This sphere expands as c increases through positive values. Forc = 0, the level surface is the origin in xyz-space. For c < 0, it is empty �

A level surface of A level surface of

f(x, y, z) = x2 + y2 + z2 g(x, y, z) = x2 + y2

FIGURE 3 FIGURE 4

Example 4 Describe the level surfaces of g(x, y, z) = x2 + y2.

Solution Since z does not appear in the equations x2 + y2 = c of level surfaces of g, the surfacesconsist of vertical lines, parallel to the z-axis. If the constant c is positive, the surfaceis the cylinder in Figure 4 formed by the vertical lines through the circle x2 + y2 = cof radius

√c in the xy-plane. The level surface x2 + y2 = 0 with c = 0 is the z-axis,

and the surface x2 + y2 = c is empty if c < 0. �

p. 356 (3/24/08) Section 14.7, Functions of three variables

Example 5 Figure 5 is a cross-sectional view of level surfaces of the Van Allen belts of cosmicradiation that surround the earth. (To visualize the level surfaces imagine that thecurves as drawn are rotated around the north-south axis of the earth.) The radiation ismeasured in counts per second and the scale shows the distance in earth radii (≈ 4000miles). At approximately what distances from the equator is the radiation the greatest?

FIGURE 5

Solution Based on the drawing, the radiation is greatest at approximately 0.5 earth radii (≈ 2000miles) and at 2.5 earth radii (≈ 10, 000 miles) from the equator. �

Quadric surfacesWe saw in Section 12.2 that level curves in the xy-plane of second-degree polynomials in two variables,

p(x, y) = Ax2 + Bxy + Cy2 + Dx + Ey + F

are conic sections or degenerate conic sections. Level surfaces in xyz-space of second-degreepolynomials with three variables,

p(x, y, z) = Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J

are called quadric surfaces. The sphere in Figure 3 and the cylinder in Figure 4 are quadric surfaces,as is the level surface in the next example.

Example 6 The level surface 19x2 + 1

16y2 + z2 = 1 of h(x, y, z) = 1

9x2 + 1

16y2 + z2 in Figure 6

is called an ellipsoid because all of its cross sections are ellipses. Find equations forand draw its cross sections (a) in the xz-plane, (b) in the yz-plane, and (c) in thexy-plane.

A level surface of

h(x, y, z) = 19x2 + 1

16y2 + z2

FIGURE 6


Solution (a) The xz-plane has the equation y = 0. Its intersection with 19x2 + 1

16y2 + z2 = 1 is

the ellipse 19x2 + z2 = 1 with x-intercepts x = ±3 and z-intercepts z = ±1 in Figure 7.

(b) The yz-plane has the equation x = 0. Its intersection with 19x2 + 1

16y2 + z2 = 1 is

the ellipse 116

y2 +z2 = 1 with y-intercepts y = ±4 and z-intercepts z = ±1 in Figure 8.

(c) The xy-plane has the equation z = 0. Its intersection with 19x2 + 1

16y2 + z2 = 1

is the ellipse 19x2 + 1

16y2 = 1 with x-intercepts x = ±3 and y-intercepts y = ±4 in

Figure 9. �

x3

z

2

19x2 + z2 = 1

y

z

2

116

y2 + z2 = 1

4 x2

y

2

19x2 + 1

16y2 = 1

FIGURE 7 FIGURE 8 FIGURE 9

Example 7 Determine the shape of the level surface k = 0 of k(x, y, z) = x2 + y2 − z2 by findingits cross sections in the vertical xz-plane y = 0 and in the horizontal planes z = c.

Solution The level surface k = 0 of k(x, y, z) = x2 + y2 − z2 has the equation x2 + y2 − z2 = 0,

which can be written z2 = x2 + y2. Its intersection with the vertical xz-plane y = 0has the equation z2 = x2 and consists of the perpendicular lines z = ±x in Figure 10.The intersection of z2 = x2 + y2 with the horizontal plane z = c has the equationx2 + y2 = c2. It is a circle of radius |c| with its center at the origin that is shownin Figure 11. Consequently, the level surface consists of horizontal circles with centerson the z-axis positioned between the lines in Figure 10. It is a double cone, as inFigure 12. �

x2

z

2

x2 − z2 = 0

x

yx2 + y2 = c2

|c|

The intersection of The intersection ofx2 + y2 − z2 = 0 x2 + y2 − z2 = 0 The level surfacewith the xz-plane with the plane z = c x2 + y2 − z2 = 0

FIGURE 10 FIGURE 11 FIGURE 12


Example 8 Determine the shape of the level surface k = 1 of k(x, y, z) = x2 + y2 − z2 by findingits cross sections in the vertical xz-plane y = 0 and in the horizontal planes z = c.

Solution (a) The level surface k = 1 of k(x, y, z) = x2 + y2 − z2 is x2 + y2 − z2 = 1. Its

intersection with the plane y = 0 has the equation x2 − z2 = 1. It is the hyperbola inFigure 13 with x-intercepts x = ±1 and no z-intercepts.

The horizontal cross section of k = 1 in the plane z = c has the equation

x2 + y2 = c2 + 1. It is a circle of radius√

c2 + 1.

The level surface surface in Figure 14 can be formed by placing horizontal circlesbetween the two branches of the hyperbola in Figure 13. The surface is a hyperboloid

of one sheet.

x2

z

2

x2 − z2 = 1

The intersection of The level surface k = 1 ofx2 + y2 − z2 = 1 k(x, y, z) = x2 + y2 − z2

with the xz-plane

FIGURE 13 FIGURE 14

Example 9 Determine the shape of the level surface k = −1 of k(x, y, z) = x2 + y2 − z2 by findingits cross sections in the vertical xz-plane y = 0 and in the horizontal planes z = c.

Solution The level surface k = −1 of k(x, y, z) = x2+y2−z2 is x2+y2−z2 = −1. Its intersection

with the plane y = 0 has the equation x2 − z2 = −1. It is the hyperbola of Figure 15with z-intercepts z = ±1 and no x-intercepts.

The horizontal cross section of k = −1 in the plane z = c has the equationx2 +y2 = c2−1. It is a circle of radius

√c2 − 1 for |c| > 1, consists of one point (0, 0, c)

if c = ±1, and is empty if |c| < 1.

The level surface in Figure 16 can be formed by placing horizontal circles betweenthe two sides of the hyperbola in Figure 15. This surface is a hyperboloid of two

sheets. �


x2

z

2

x2 − z2 = −1

The intersection of The level surfacex2 + y2 − z2 = −1 x2 + y2 − z2 = −1with the xz-plane

FIGURE 15 FIGURE 16

Example 10 Describe all the level surfaces of k(x, y, z) = x2 + y2 − z2.

Solution The double cone k = 0 of Example 8, the hyperboloid of one sheet k = 1 of Example 9,and the hyperboloid of two sheets k = −1 of Example 10 are shown together inFigure 17. The other level surfaces k = c with c > 0 are hyperboloids of one sheetthat surround the double cone and expand as c increases through positive values. Theother level surfaces k = c with c < 0 are hyperboloids of two sheets inside the doublecone that move away from the xy-plane as c decreases through negative values. �

FIGURE 17


Partial derivativesAs we saw in earlier sections, the partial derivatives, fx(x, y) and fy(x, y) of a function of two variablesare found by differentiating with respect to one variable while holding the other variable constant.Similarly, the partial derivatives, fx(x, y, z), fy(x, y, z), and fz(x, y, z) with three variables are obtainedby differentiating with respect to one variable while holding the other two constant.

Example 11 What are the first-order partial derivatives of f = x2y3z4?

Solution The derivatives are fx =∂

∂x(x2y3z4) = 2xy3z4, fy =

∂

∂y(x2y3z4) = 3x2y2z4, and

fz =∂

∂z(x2y3z4) = 4x2y3z3. �

Example 12 A rectangular box has width x, depth y, and height z feet and consequently volumeV = xyz cubic feet (Figure 18). (a) Find formulas for Vx, Vy, and Vz . (b) Givegeometric explanations of the formulas from part (a).

FIGURE 18

Solution (a) Vx =∂

∂x(xyz) = yz, Vy =

∂

∂y(xyz) = xz, and Vz =

∂

∂z(xyz) = xy.

(b) Imagine that the left, lower, front corner of the box is fixed, as suggested by thedot in Figure 18. Then as x increases, the right side moves to the right, and the rate ofchange ∂V/∂x of the volume with respect to x is equal to the area yz of the right side.As y increases, the back moves back, and the rate of change ∂V/∂y of the volume withrespect to y is equal to the area xz of the back. And, as z increases, the top moves up,and the rate of change ∂V/∂z of the volume with respect to z equals the area xy of thetop. �

As with functions of two variables, mixed partial derivatives with three variables can be calculated

in any order, provided that the resulting derivatives are continuous.†

†The limit of f (x, y, z) as (x, y, z) tends to (x0, y0, z0) is L if f it is defined in a sphere of positive radius containing

(x0, y0, z0) and f (x, y, z) approaches L as (x, y, z) approaches (x0, y0, z0) along all paths in the sphere. The function f is

continuous at (x0, y0, z0) if the limit of f (x, y, z) as (x, y, z) tends to (x0, y0, z0) is f (x0, y0, z0).


Example 13 Find a formula for hyz in two ways where h(x, y, z, w) = ex sin y cos z.

Solution If we take the y-derivative first and then the z-derivative, we obtain

hy =∂

∂y(ex sin y cos z) = ex cos y cos z

hyz =∂

∂z(ex cos y cos z) = −ex cos y sin z.

Differentiating first with respect to z and then with respect to y yields the same result:

hz =∂

∂z(ex sin y cos z) = −ex sin y sin z

hzy =∂

∂y(−ex sin y sin z) = −ex cos y sin z �

Gradient vectors and directional derivativesThe definitions and results in Section 14.5 concerning gradient vectors and directional derivatives withtwo variables can be converted to the three-variable case by just allowing for the extra variable in theformulas.

Definition 2 (Gradient vectors and directional derivatives) (a) The gradient vector ofw = f(x, y, z) at (x0, y0, z0) is

∇f(x, y, z) = 〈fx(x0, y0, z0), fy(x0, y0, z0), fz(x0, y0, z0)〉 . (1)

(b) The directional derivative Duf(x0, y0, z0) of z = f(x, y, z) at (x0, y, z) in the direction of the unitvector u = 〈u1, u2, u3〉 is the t-derivative of the cross section w = f(x0 + tu1, y0 + tu2, z0 + tu3) at t = 0.

The directional derivative Duf(x0, y0, z0) is the derivative of f with respect to distance in thedirection of the vector u. It can be calculated with part (a) of the following theorem. Here, as in allresults of this section, we assume that the functions involved have continuous first-order derivatives.

Theorem 1 (a) The directional derivative Duf(x0, y0, z0) of w = f(x, y, z) is equal to the dot productof the gradient of f at (x0, y0, z0) with the unit vector u:

Duf(x0, y0, z0) = ∇f(x0, y0, z0) · u= fx(x0, y0, z0)u1 + fy(x0, y0, z0)u2 + fz(x0, y0, z0)u3.

(2)

(b) The greatest directional derivative of f at (x0, y0, z0) is the length |∇f | of the gradient vector atthat point, and the least directional derivative is the negative of the length of the gradient vector.

(c) If the gradient vector is not zero at (x0, y0, z0), then the greatest directional derivative at the pointis in the direction of the gradient, the least directional derivative is in the direction opposite to that ofthe gradient, and the directional derivative is zero in the directions perpendicular to the gradient.

This theorem can be established by modifying the proofs of Theorems 1 through 3 in Section 14.5to deal with three variables instead of two.


Example 14 (a) What is the gradient of of f(x, y, z) = xyz at (1, 2, 3)? (b) What is the directionalderivative of f(x, y, z) = xyz at (1, 2, 3) in the direction toward (2, 3, 4)? (c) What isthe greatest directional derivative of f at (1, 2, 3)?

Solution (a) The derivatives fx =∂

∂x(xyz) = yz, fy =

∂

∂y(xyz) = xz, and fz =

∂

∂z(xyz) = xy

have the values fx(1, 2, 3) = 2(3) = 6, fy(1, 2, 3) = 1(3) = 3, and fz(1, 2, 3) = 1(2) = 2,definition (1) gives ∇f(x, y, z) = 〈6, 3, 2〉.(b) Set P = (1, 1, 1) and Q = (2, 3, 4). The displacement vector from P to Q is−→PQ = 〈2 − 1, 3 − 1, 3 − 1〉 = 〈1, 2, 3〉 and its length is

√12 + 22 + 32 =

√14. The unit

vector in the direction from P to Q is

−→PQ

|−→PQ|=

〈1, 2, 3〉√14

. By formula (2), the directional

derivative of f at (1, 2, 3) in the direction toward (2, 3, 4) is

∇f(1, 2, 3) · u = 〈6, 3, 2〉 ·(

〈1, 2, 3〉√14

)

=6(1) + 3(2) + 2(3)√

14=

18√14

.

(c) The greatest directional derivative of f at (1, 2, 3) is equal to the length

|∇f(1, 2, 3)| = |〈6, 3, 2〉| =√

62 + 32 + 22 =√

49 = 7 of the gradient vector at thatpoint. �

Tangent planes to level surfacesJust as not all level curves of functions of two variables are what we would normally call “curves,” not alllevel surfaces of functions with three variables are what we would call “surfaces.” However, we get whatwe would normally call a surface through any point where the function f(x, y, z) has a nonzero gradient,as is shown by the following theorem, which is proved in advanced classes.

Theorem 2 (The Implicit Function Theorem) If ∇f(x0, y0, z0) is not zero, then the portion in asphere centered at (x0, y0, z0) of the level surface of f through (x0, y0, z0) is a surface with a tangentplane at (x0, y0, z0). Moreover, ∇f(x0, y0, z0) is a normal vector to the surface (Figure 19), so the tangentplane has the equation

a(x − x0) + b(y − y0) + c(z − z0) = 0 (3)

where 〈a, b, c〉 is the gradient vector ∇f(x0, y0, z0) or a nonzero constant multiple of it.

FIGURE 19


Example 15 Give an equation of the tangent plane at the point (2, 2, 3) to the hyperboloid of two

sheets x2 + y2 − z2 = −1 of Figure 16.

Solution We set f = x2 + y2 − z2 so that the hyperboloid is the level surface f = −1. Then∇f = 〈fx, fy, fz〉 = 〈2x, 2y,−2z〉 and ∇f(2, 2, 3) = 〈4, 4,−6〉 is a normal vector to thelevel surface at (2, 2, 3). The tangent plane at that point has the equation4(x − 2) + 4(y − 2) − 6(z − 3) = 0, which simplifies to 2x + 2y − 3z = −1. �

Interactive Examples 14.7Interactive solutions are on the web page http//www.math.ucsd.edu/ ashenk/.†

1. Describe the level surfaces of the function f(x, y, z) = x2 + y2 − z.

2. Find the first-order partial derivatives of f(x, y, z) = x1/2y1/4 + y1/5z1/6.

3. What are the second-order partial derivatives of N(x, y, z) = x2y − y3z2 + xz?

4 (a) What is the gradient of f(x, y, z) = x sin(yz2) at (3, 2, 1)? (b) What is the derivative of

f(x, y, z) = x sin(yz2) at (3, 2, 1) in the direction toward the origin?

5. Find (a) the maximum directional derivative of f(x, y, z) = exyz at (2, 3, 4) and (b) the unitvector in the direction of the maximum directional derivative.

6 Give an equation of the tangent plane to 3xy2 + 2yz3 + xz = 93 at (−1, 2, 3).

Exercises 14.7AAnswer provided. OOutline of solution provided. CGraphing calculator or computer required.

CONCEPTS:

1. What does V (R,R, h) represent for the function V of Example 2?

2. How are the level surfaces of w = g(x, y, z) related to the level surfaces of w = f(x, y, z) ifg(x, y, z) = f(x, y, z) + 10 for all (x, y, z)?

3. How are the level surfaces of w = g(x, y, z) related to the level surfaces of w = f(x, y, z) if

g(x, y, z) = [f(x, y, z)]3 for all (x, y, z)?

4. Suppose that the gradient vector of w = h(x, y, z) is horizontal (parallel to the xy-plane) for all(x, y, z). Show that h is independent of z.

5. What can you say about a function w = k(x, y, z) whose gradient is vertical (perpendicular tothe xy-plane) for all (x, y, z)?

6. What are the maximum and minimum directional derivatives of w = f(x, y, z) at (x0, y0, z0) if∇f(x0, y0, z0) = 0?

BASICS:

7.O For a yacht to be in the R–5.5 class at the Olympic Games, its rating R(L, V,A)

= 0.9(

L√

A

123√

V+ 1

4(L +

√A )

)

must be less than 5.5, where L (meters) is the length of the yacht,

A (square meters) is the surface area of its sails, and V (cubic meters) is the volume of water itdisplaces.(1) Would your yacht qualify for the R–5.5 class if it is 10 meters long, has 36 squaremeters of sail, and displaces 8 cubic meters of water?

8.O (a) What is the domain of f(x, y, z) =√

25 − x2 − y2 − z2? (b) Describe the level surfacef = 4.

9. Describe the level surfaces of g(x, y, z) =√

x2 + y2 + z.

10.O Give a formula for Rxy where R(x, y, z) = (x + 2y + 3z)10.

†In the published text the interactive solutions of these examples will be on an accompanying CD disk which can be run by

any computer browser without using an internet connection.(1)Sailing Theory and Practice by C. Marchaj, New York, NY: Dodd, Mead, and Company, 1964, p. 75.


Describe the level surfaces of the functions in Exercises 11 through 17.

11.A g(x, y, z) = x + y + z

12. h(x, y, z) = xy − 2z

13.A k(x, y, z) = x2 + 14y2 + 1

9z2

14.O f(x, y, z) = x2 + 9y2

15.A g(x, y, z) =z

x2 + y2

16. h(x, y, z) = −y2 − z2

17.A k(x, y, z) =z2

x2 + y2

Find the first-order partial derivatives of the functions in Exercises 18 through 25.

18.O f(x, y, z) = x√

y − y√

x

19.A g(x, y, z) = ln(x + 3y − 4z)

20. T (u, v, w) = u1/3v1/4w1/5

21. h = x sin y − y cos z.

22.O f(x, y, z) = x(1 + yz)2

23.A g(x, y, z) = sin(xyz)

24. g(x, y, z) = xyz + x2y2z2

25. h(x, y, z) = ex2+y2

+x2

26.O Find (a) Mxy(x, y, z) and (b) Mxz(x, y, z) for M(x, y, z) = x2e−3y sin(4z).

27.A What are the second-order partial derivatives of P (x, y, z) = x2y3z4?

28. What is Hxxyyzz for H(x, y, z) = ez sin x cos y?

29.O (a) What is the gradient of g(x, y, z) = exyz at (1, 2, 3)? (b) What is the derivative of g at(1, 2, 3) in the direction toward the origin?

30.O What is the derivative of f = x4y−2z3 at (1, 2, 3) in the direction of the (nonunit) vector3i− 4j− 12k?

31.O (a) What is ∇H(x, y, z) for H(x, y, z) = x2+y3+z4? (b) What is the derivative of H at (3, 2, 1)in the direction toward (4, 4, 3)? (c) What is the maximum directional derivative of H at (3, 2, 1)and what is the unit vector in that direction?

32.O Find the derivative of g = e3x−2y+z at (0, 0, 0) in the direction toward (1,−1, 1).

33. Find the derivative of h = ln(xyz) at (5, 2, 6) in the direction toward (4, 3, 7).

34.O Find (a) the maximum directional derivative of f(x, y, z) = x2+y2+z2 at (2, 3, 4) and (b) Theunit vector in the direction of the maximum directional derivative.

35.A Find the minimum directional derivative of g(x, y, z) = x2yz at (10, 5,−1).

36. Find the minimum directional derivative of h(x, y, z) = y3z ln x at (2,−1, 4).

37.A Give the unit vector u such that Duh(1, 2, 2) is the minimum directional derivative of

h(x, y, z) = x2 + y2 + z2 at (1, 2, 2).

38. Give the unit vector u such that Duk(1, 1, 1) is the maximum directional derivative of

k(x, y, z) = 24x1/2y1/3z1/4 at (1, 1, 1).

In Exercises 39 through 44 give equations for tangent planes to the surfaces at the given points.

39.O The tangent plane to xyez = 2e3 at (1, 2, 3)

40.A The tangent plane to x2y3z4 = 1 at (−1, 1,−1)

41. The tangent plane to√

x +√

y +√

z = 6 at (1, 4, 9)

42. The tangent plane to x3 + y4 − z5 = 10 at (2,−1,−1)

43.A The tangent plane to sin x + cos y + tan z = 1 +√

2 at ( 14π, 1

4π, 1

4π)

44. The tangent plane to ln x + ln y + ln z = ln(24) at (2, 3, 4)

45. Find the sixth derivative∂6f

∂x2∂y2∂z2of f(x, y, z) = x2y2z2.


EXPLORATION:

Find the first-order derivatives of the functions in Exercises 46 through 49.

46.O f(x, y, z) = xy tan–1(yz)

47.A g(x, y, z) =xyz

x + y + z

48. h(x, y, z) = xyex2z

49. k(x, y, z) = sin–1(xyz)

50.O If two thin lenses with focal lengths f1 and f2 are separated a distance d < f1 +f2, then together

they have the same effect as one lens of focal length F =f1f2

f1 + f2 − d.(2) What are the derivatives

of F with respect to f1, f2, and d?

51.A The sides of lengths x and y in a triangle form an angle θ, so that by the Law of Cosines, the

third side has length z =√

x2 + y2 − 2xy cos θ. What are the rates of change of z with respectto x, y, and θ?

52. (aA) A silo has the shape of a right circular cylinder of radius r and height H with a rightcircular cone of radius r and height h on top of it. Give a formula for the volume V of the silo.(b) What are the rates of change of V with respect to r,H , and h? (c) The exterior surface

area of the silo is A = 2πrH + πr√

h2 + r2. What are the rates of change of A with respect tor,H , and h?

53. Find the first-order derivatives of Y (x, y, z) = Z(A(x, y, z)) at (0, 10, 20) where A(0, 10, 20) = 100,Ax(0, 10, 20) = 5, Ay(0, 10, 20) = 6, Az(0, 10, 20) = 7, and Z′(100) = 2.

54.A Find G(1, 2, 3), Gx(1, 2, 3), Gy(1, 2, 3) and Gz(1, 2, 3) where G(x, y, z) = f(x)[f(y)]2 + [f(z)]3,f(1) = 4, f(2) = 3, f(3) = −1, f ′(1) = 2, f ′(2) = −5, and f ′(3) = 10.

55. Find A(1, 10, 100), Au(1, 10, 100), Av(1, 10, 100) and Aw(1, 10, 100) where

A(u, v, w) = u2v2[B(w)]−1,B(100) = 100, and B′(100) = 300.

56. Express the first-order derivatives of K(x, y, z) = A(x2) + [A(y)]3 + A(z4) in terms of A and A′.

57. In a mathematical model of a vibrating drum head, it is assumed that the drum has the shapeof the surface z = f(x, y, t) at each fixed time t and that the function f(x, y, t) satisfies the wave

equation fxx +fyy = k−2ftt. Show that f(x, y, t) = sin x sin y cos(k√

2 t) satisfies this differentialequation.

58. Show that the tangent plane to the quadric surface Ax2 + By2 + Cz2 + D = 0 at the point(x0, y0, z0) on it has the equation Ax0x + By0y + Cz0z + D = 0.

59.A Give parametric equations of the normal line to the surface x2y3z4 + xyz = 2 at (2, 1,−1)

60. Give parametric equations of the normal line to the hyperboloid of two sheets z2−4x2−9y2 = 11at (2, 1, 6).

(End of Section 14.7)

(2)Adapted from CRC Handbook of Chemistry and Physics, edited by R. Weast, Boca Raton, FL: CRC Pres, Inc.,

1981, p. F-103.

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Section 14.7 (3/24/08) Functions of three variables