Chap14-et-student-solutions

DIFFERENTIATION IN 15 S E VERAL VA RIABLES 15.1 Functions of Two or More Variables

(ET Section 14.1)

Preliminary Questions 1. What is the difference between a horizontal trace and a level curve? How are they related? SOLUTION A horizontal trace at height c consists of all points (x, y, c) such that f (x, y) = c. A level curve is the curve f (x, y) = c in the x y-plane. The horizontal trace is in the z = c plane. The two curves are related in the sense that the level curve is the projection of the horizontal trace on the x y-plane. The two curves have the same shape but they are located in parallel planes.

2. Describe the trace of f (x, y) = x 2 − sin(x 3 y) in the x z-plane. The intersection of the graph of f (x, y) = x 2 − sin(x 3 y) with the x z-plane is obtained by setting y = 0 in the equation z = x 2 − sin(x 3 y). We get the equation z = x 2 − sin 0 = x 2 . This is the parabola z = x 2 in the x z-plane. SOLUTION

3. Is it possible for two different level curves of a function to intersect? Explain. Two different level curves of f (x, y) are the curves in the x y-plane defined by equations f (x, y) = c1 and f (x, y) = c2 for c1  = c2 . If the curves intersect at a point (x 0 , y0 ), then f (x 0 , y0 ) = c1 and f (x 0 , y0 ) = c2 , which implies that c1 = c2 . Therefore, two different level curves of a function do not intersect. SOLUTION

4. Describe the contour map of f (x, y) = x with contour interval 1. SOLUTION The level curves of the function f (x, y) = x are the vertical lines x = c. Therefore, the contour map of f with contour interval 1 consists of vertical lines so that every two adjacent lines are distanced one unit from another.

5. How will the contour maps of f (x, y) = x and g(x, y) = 2x with contour interval 1 look different? SOLUTION The level curves of f (x, y) = x are the vertical lines x = c, and the level curves of g(x, y) = 2x are the vertical lines 2x = c or x = 2c . Therefore, the contour map of f (x, y) = x with contour interval 1 consists of vertical lines with distance one unit between adjacent lines, whereas in the contour map of g(x, y) = 2x (with contour interval 1) the distance between two adjacent vertical lines is 12 .

Exercises In Exercises 1–4, evaluate the function at the specified points. 1. f (x, y) = x + yx 3 , (2, 2), (−1, 4), (6, 12 ) SOLUTION

We substitute the values for x and y in f (x, y) and compute the values of f at the given points. This gives f (2, 2) = 2 + 2 · 23 = 18 f (−1, 4) = −1 + 4 · (−1)3 = −5   f 6, 12 = 6 + 12 · 63 = 114

3. h(x, y, z) = x yz −2 y , (3, 8, 2), (3, −2, −6) , (1, 3), (3, −2) g(x, y) = 2 SOLUTION Substituting x + y 2(x, y, z) = (3, 8, 2) and (x, y, z) = (3, −2, −6) in the function, we obtain h(3, 8, 2) = 3 · 8 · 2−2 = 3 · 8 ·

1 =6 4

h(3, −2, −6) = 3 · (−2) · (−6)−2 = −6 ·

1 1 =− 36 6

In Exercises 5–16, sketch the domain of the function. Q(y, z) = y 2 + y sin z, (y, z) = (2, π2 ), (−2, π6 ) 5. f (x, y) = 4x − 7y SOLUTION

The function is defined for all x and y, hence the domain is the entire x y-plane.

7. f (x, y) = ln(y− 2x) f (x, y) = 9 − x 2

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y = 2x.

(ET CHAPTER 14)

The function is defined if y − 2x > 0 or y > 2x. This is the region in the x y-plane that is above the line D = {(x, y) : y > 2x} y y = 2x y > 2x x

9. G(x, t) = e1/(x+t) 1 h(x, t) = SOLUTION The xfunction is defined if x + t  = 0, that is, x  = −t. The domain is the same as the domain of h(x, t) in +t the previous exercise. y

x

x  = −t y 11. f (x, y) = sin 1 x g(y, z) = z + y2 SOLUTION The function is defined for all x  = 0. The domain is the x y-plane with the y-axis excluded. D = {(x, y) : x  = 0} y

x

x = 0 √ 13. F(I, R) = I R2 4 H (r, s) = 3r s SOLUTION The function is defined if I R ≥ 0. Therefore the domain is the first and the third quadrants of the I R-plane including both axes. y

x

IR ≥ 0 f (x, y) = cos−1 (x + y)

Functions of Two or More Variables

S E C T I O N 15.1

(ET Section 14.1)

325

1 15. g(r, t) = 2 r −t SOLUTION The function is defined if r 2 − t  = 0, that is, t  = r 2 . The domain is the r t-plane with the parabola t = r 2 excluded.   D = (r, t) : t  = r 2 y

t = r2 x

t = r 2 In Exercises 17–19, describe the domain and range of the function. f (x, y) = ln(4x 2 − y) 17. f (x, y, z) = x z + e y SOLUTION

The domain of f is the entire (x, y, z)-space. Since f takes all the real values, the range is the entire real

line.

 19. f (x, y, z) 9 − x 2 − y2 − z2 P(r, s) = = er/s SOLUTION The function is defined if 9 − x 2 − y 2 − z 2 ≥ 0, i.e, x 2 + y 2 + z 2 ≤ 9. Therefore, the domain is the set {(x, y, z) : x 2 + y 2 + z 2 ≤ 9}, which is the ball of radius 3 centered at the origin, including the sphere x 2 + y 2 + z 2 = 9. To find the range, we examine the values ω taken by f . For ω in the range, the following equation has solutions x, y, z:  9 − x 2 − y2 − z2 = ω (1) Raising to the power of two and transfering sides gives 9 − x 2 − y2 − z2 = ω 2 x 2 + y2 + z2 = 9 − ω 2 The left-hand side is nonnegative, hence also 9 − ω 2 ≥ 0 or ω 2 ≤ 9. Therefore, −3 ≤ ω ≤ 3. By (1), ω ≥ 0, hence we must satisfy 0 ≤ ω ≤ 3. We obtain the following range: {ω ∈ R : 0 ≤ ω ≤ 3} . In Exercises 21–30, sketch the graph andtheir describe the(A)–(F) verticalinand horizontal traces. Match the functions (a)–(f) with graphs Figure 21. (x,=y)12=−|x| 21. (a) f (x,fy) 3x+−|y| 4y SOLUTION (x, y) (b) f (x, The y) =graph cos(xof− fy)

= 12 − 3x − 4y is shown in the figure:

−1 (c) f (x, y) = 1 + 9x 2 + y 2

z 12

2 2 (d) f (x, y) = cos(x 2 )e−0.1(x +y )

(e) f (x, y) =

−1 1 + 9x 2 + 9y 2

2 2 (f) f (x, y) = cos(x 2 + y 2 )e−0.1(x +y )

3 4

y

x

The horizontal trace at height c is the line 12 − 3x − 4y = c or 3x + 4y = 12 − c in the plane z = c. z

y x

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(ET CHAPTER 14)

The vertical traces obtained by setting x = a or y = a are the lines z = (12 − 3a) − 4y and z = −3x + (12 − 4a) in the planes x = a and y = a, respectively. z

z

y

y x

x

23. f (x, y) = x 2 +24y 2 f (x, y) = x + y SOLUTION The graph of the function is shown in the figure: z

y x

The horizontal trace at height c is the curve x 2 + 4y 2 = c, where c ≥ 0 (if c = 0, it is the origin). The horizontal traces are ellipses for c > 0. z

y x

The vertical trace in the plane x = a is the parabola z = a 2 + 4y 2 , and the vertical trace in the plane y = a is the parabola z = x 2 + 4a 2 . z z

y

x

1 25. f (x,f (x, y) = y) x=2 y+2 y 2 + 1 SOLUTION

The graph of the function is shown in the figure:

y x

Functions of Two or More Variables

S E C T I O N 15.1 z

y

The horizontal trace at height c is the following curve in the plane z = c: 1 1 1 ⇒ x 2 + y2 = − 1 = c ⇒ x 2 + y2 + 1 = c c x 2 + y2 + 1  For 0 < c < 1 it is a circle of radius 1c − 1 centered at (0, 0), and for c = 1 it is the origin. z

y x

The vertical trace in the plane x = a is the following curve in the plane x = a: 1 z= 2 a + y2 + 1

⇒

z=

1 (1 + a 2 ) + y 2

z

y x

1 The vertical trace in the plane y = a is the curve z = 2 in this plane. x + a2 + 1 z

y x

27. f (x, y) = x + |y| f (x, y) = 7

(ET Section 14.1)

327

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(ET CHAPTER 14)

The graph of f is shown in the figure: z

y x

The horizontal trace at height c is x + |y| = c, or in other words, x = c − |y|. z

y x

The vertical traces obtained by setting x = a or y = a are z = a + |y| or z = x + |a| in the planes x = a and y = a, respectively. z

z

y

y

x

x

 29. f (x, y) = 1 − x 2 2− y 2 2 f (x, y) = 9 − x − y SOLUTION The graph of f is shown in the figure below: z

y x

The horizontal trace at height c ≥ 0 is  1 − x 2 − y2 = c For 0 ≤ c < 1 it is the circle of radius

⇒

1 − x 2 − y 2 = c2

⇒

x 2 + y 2 = 1 − c2

 1 − c2 centered at the origin, and for c = 1 it is the point at the origin.

Functions of Two or More Variables

S E C T I O N 15.1

(ET Section 14.1)

329

z

y x

The vertical traces in the planes x = a and y = a (for |a| ≤ 1) are z = planes.

  1 − a 2 − y 2 and z = 1 − a 2 − x 2 in these

z

z

y

y

x

x

31. Sketch the contour  maps of f (x, y) = x + y with contour intervals m = 1 and 2. f (x, y) = 1 + x 2 + y 2 SOLUTION The level curves are x + y = c or y = c − x. Using contour interval m = 1, we plot y = c − x for various values of c. 4 2 0 −2 −4 −4 −2

0

2

4

Using contour interval m = 2, we plot y = c − x for various values of c. 4 2 0 −2 −4 −4 −2

0

2

4

The (x, t) t −1/2 e−xx 2/t+, whose shown in 0, Figure 22, 16. models the temperature along a 33. Sketch thefunction contour fmap of =f (x, y) = y 2 withgraph level is curves c= 4, 8, 12, metal bar after an intense burst of heat is applied at its center point. (a) Sketch the graphs of the vertical traces at times t = 0.5, 1, 1.5, 2. What do these traces tell us about the way heat diffuses through the bar? (b) Sketch the vertical trace at x = 0. Describe how temperature varies in time at the center point. (c) Sketch the vertical trace x = c for c = ±0.2, ±0.4. Describe how temperature varies in time at points near the center. 2

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(ET CHAPTER 14)

Temperature T

4 3 −0.4−0.2 0 0.2 0.4

2

Time t

1 x

Metal bar 2 FIGURE 22 Graph of f (x, t) = t −1/2 e−x /t beginning shortly after t = 0.

SOLUTION

(a) The vertical traces at times t = 0.5, 1, 1.5, 2 are √ 2 z = 2e−2x in the plane t = 0.5 2 z = e−x in the plane t = 1 2 z = √ 1 e−2x /3 in the plane t = 1.5 3/2

2 z = √1 e−x /2 in the plane t = 2

2

These vertical traces are shown in the following figure:

4

4

3

3

2

2

1

1

Trace at t = 0.5

Trace at t = 1

4

4

3

3

2

2

1

1

Trace at t = 1.5

Trace at t = 2

At each time the temperature decreases as we move away from the center point. Also, as t increases, the temperature at each point in the bar (except at the middle) increases and then decreases (as can be seen in Figure 22). (b) The vertical trace at x = 0 is z = t −1/2 = √1 . This trace is shown in the following figure: t

z

T

y t

x

As suggested by the graph, the temperature is decreasing with time at the center point, according to √1 . t

(c) The vertical traces x = c for the given values of c are: 0.04 z = √1 e− t in the planes x = 0.2 and x = −0.2

t

0.16 z = √1 e− t in the planes x = 0.4 and x = −0.4.

t

We see that for small values of t the temperature increases quickly and then slowly decreases as t increases.

Functions of Two or More Variables

S E C T I O N 15.1

(ET Section 14.1)

331

T

t

In Exercises 34–43, draw a contour map of f (x, y) with an appropriate contour interval, showing at least six level curves. 35. f (x, y) = e y/xy f (x, y) = x curves are SOLUTION The level y

ex = c

y = ln c x

⇒

⇒

y = x ln c

For c > 1, the level curves are the lines y = x ln c, and for c = 1 the level curve is the x-axis. We use the contour interval m = 3 and show a contour map with the level curves c = 1, 4, 7, 10, 13, 16: c = e3 c = e2

c=e

0.1

0.05

0 0

0.05

0.1

37. f (x, y) = x y y f (x, y) = 2 SOLUTION The level curves are x y = c or y = xc . These are hyperbolas in the x y-plane. We draw a contour map of x the function using contour interval m = 1 and c = 0, ±1, ±2, ±3: 4 2 0 −2 −4 −4 −2

0

2

4

39. f (x, y) = x + 2y − 1 f (x, y) = x − y x c+1 SOLUTION The level curves are the lines x + 2y − 1 = c or y = − 2 + 2 . We draw a contour map using the contour interval m = 4 and c = −9, −5, −1, 3, 7, 11. The corresponding level curves are: y=−

x − 4, 2

c=−9

y=−

x − 2, 2

x y=− , 2

c=−5

x y = − + 4, 2 c=7

c=−1

y=−

x + 2, 2

c=3

x y =− +6 2 c=11

4 2 0 −2 −4 −4 −2

0

2

4

41. f (x, y) = x 2 2 f (x, y) = x − y √ √ SOLUTION The level curves are x 2 = c. For c > 0 these are the two vertical lines x = c and x = − c and for c = 0 it is the y-axis. We draw a contour map using contour interval m = 4 and c = 0, 4, 8, 12, 16, 20:

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(ET CHAPTER 14)

4 2 0 −2 −4 −4 −2

0

2

4

1 + x 2 + y2 =

10 c

10 2 43. f (x,f (x, y) = y) 1=+3xx 22 − + yy 2 SOLUTION

The level curves are: 10 =c 1 + x 2 + y2

⇒

x 2 + y2 =

⇒

10 −1 c

 For 0 < c < 10 these are circles of radius 10 c − 1 centered at the origin. For c = 10 it is the point at the origin. We plot a contour map with contour interval m = 1.5 for c = 10, 8.5, 7, 5.5, 4, 2.5.

C B i P

470 450

ii iii iv

400

E

D A

Contour interval = 10 meters

0

1

2 km

Which linear maps function has (B) the in contour shown in Figure 24 f(with c =g(x, 0 as 45. Match the contour (A) and Figure map 23 with the two functions (x, y)level = x curve − 2y and y) indicated), = 2x − y. assuming that the contour interval is m = 6? What if m = 3? y

2 1 −6

−3

x 3

−1

6

−2 c=0

FIGURE 24 SOLUTION

We denote the linear function by f (x, y) = α x + β y + γ

(1)

αx + β y + γ = c

(2)

The level curves of f are

By the given information, the level curve for c = 0 is the line passing through the points (0, −1) and (−3, 0). We find the equation of this line: y=

0 − (−1) (x + 3) −3 − 0

⇒

1 y = − (x + 3) 3

⇒

x + 3y + 3 = 0

Setting c = 0 in (2) gives α x + β y + γ = 0. Hence,

α β γ = = 1 3 3

⇒

β = 3α ,

γ = 3α

Substituting into (2) gives

α x + 3α y + 3α = c or x + 3y + 3 =

c α

(3)

Functions of Two or More Variables

S E C T I O N 15.1

(ET Section 14.1)

333

Case 1: m = 6. In this case, the closest line above c = 0 corresponds to c = 6. This line is parallel to x + 3y + 3 = 0 and passes through the origin, hence its equation is x + 3y = 0. Setting c = 6 in (3) gives x + 3y + 3 =

6 α

or

x + 3y =

6 −3 α

This line coincides with the line x + 3y = 0 only if α6 − 3 = 0, that is, α = 2. Substituting α = 2, β = 3α = 6, and γ = 3α = 6 in (1) gives

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(ET CHAPTER 14)

53. Sketch the paths of steepest ascent beginning at D and E.

C B i P

470 450

ii iii iv

400

E

D A

Contour interval = 10 meters

0

1

2 km

FIGURE 26 Contour map of mountain. SOLUTION Starting at D or E, we draw a path that everywhere along the way points on the steepest direction, that is, moves as straight as possible from one level curve to the next. We obtain the following paths:

C B i P

470 450

ii iii iv

400

E

D A

Contour interval = 10 m

0

1

2 km

Refer to Figure 27 to answer the following questions. Further Insights and Challenges

(a) At which of (A)–(E) is temperature increasing in the easterly direction? x 55. Let f (x, y) =  for (x, y)  = 0. Write f as a function f (r, θ ) in polar coordinates and use this to find the (b) At which of (A)–(E) x 2 + y 2is temperature decreasing most rapidly in the northern direction? level(c) curves of f . direction at (B) is temperature increasing most rapidly? In which  SOLUTION In polar coordinates x = r cos θ and r = x 2 + y 2 . Hence, f (r, θ ) =

r cos θ = cos θ . r z

x y

The level curves are the curves cos θ = c in the r θ -plane, for |c| ≤ 1. For −1 < c < 1, c  = 0, the level curves cos θ = c are the two rays θ = cos−1 c and θ = − cos−1 c.

S E C T I O N 15.2

Limits and Continuity in Several Variables

(ET Section 14.2)

335

z

x y

For c = 0, the level curve cos θ = 0 is the y-axis; for c = 1 the level curve cos θ = 1 is the nonnegative x-axis. z

x y

For c = −1, the level curve cos θ = −1 is the negative x-axis. xy Use a computer algebra system to draw the graph of f (x, y) = 2 and its contour map. x + y2 + 1 15.2(a) Limits and Continuity iny)Several Variables (ET Section 14.2) Show that the level curve f (x, = c has equation 2  1 2− x− 1 y − 1 y =1 2c 4c2D ∗ (P, r )? 1. What is the difference between D(P, r ) and 

Preliminary Questions

1 and 1 . What SOLUTION r ) is the curve open disk of radius center b). It consists of all pointsisdistanced than is a(a, hyperbola if |c| < the level less curve forrcfrom = 12 ?P, (b) ShowD(P, that the level is empty if |c|r>and 2 2 hence D(P, r ) includes the point P. D ∗ (P, r ) consists of all points in D(P, r ) other than P itself.

2. Suppose that f (x, y) is continuous at (2, 3) and that f (2, y) = y 3 for y  = 3. What is the value f (2, 3)? SOLUTION

f (x, y) is continuous at (2, 3), hence the following holds: f (2, 3) =

lim

(x,y)→(2,3)

f (x, y)

Since the limit exists, we may compute it by approaching (2, 3) along the vertical line x = 2. This gives f (2, 3) =

lim

(x,y)→(2,3)

f (x, y) = lim f (2, y) = lim y 3 = 33 = 27 y→3

y→3

We conclude that f (2, 3) = 27. 3. Suppose that Q(x, y) is a function such that

1 is continuous for all (x, y). Which of the following statements Q(x, y)

are true? (a) Q(x, y) is continuous at all (x, y). (b) Q(x, y) is continuous for (x, y)  = (0, 0). (c) Q(x, y)  = 0 for all (x, y). 1 . Hence Q(x, y) = 1 All three statements are true. Let f (x, y) = Q(x,y) f (x,y) . (a) Since f is continuous, Q is continuous whenever f (x, y)  = 0. But by the definition of f it is never zero, therefore Q is continuous at all (x, y). (b) Q is continuous everywhere including at (0, 0). SOLUTION

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(ET CHAPTER 14)

1 (c) Since f (x, y) = Q(x,y) is continuous, the denominator is never zero, that is, Q(x, y)  = 0 for all (x, y). Moreover, there are no points where Q(x, y) = 0. (The equality Q(x, y) = (0, 0) is meaningless since the range of Q consists of real numbers.)

4. Suppose that f (x, 0) = 3 for all x  = 0 and f (0, y) = 5 for all y  = 0. What can you conclude about lim f (x, y)?

(x,y)→(0,0)

SOLUTION We show that the limit lim(x,y)→(0,0) f (x, y) does not exist. Indeed, if the limit exists, it may be computed by approaching (0, 0) along the x-axis or along the y-axis. We compute these two limits:

lim

f (x, y) = lim f (x, 0) = lim 3 = 3

lim

f (x, y) = lim f (0, y) = lim 5 = 5

(x,y)→(0,0) along y=0 (x,y)→(0,0) along x=0

x→0

y→0

x→0

y→0

Since the limits are different, f (x, y) does not approach one limit as (x, y) → (0, 0), hence the limit lim(x,y)→(0,0) f (x, y) does not exist.

Exercises In Exercises 1–10, use continuity to evaluate the limit. 1.

(x 2 + y)

lim

(x,y)→(1,2)

SOLUTION

Since the function x 2 + y is continuous, we evaluate the limit by substitution: lim

(x 2 + y) = 12 + 2 = 3

(x,y)→(1,2)

3.

lim (3xx2 y − 2x y 3 ) (x,y)→(3,−1) lim (x,y)→( 94 , 29 ) y 2

SOLUTION

The function 3x y − 2x y 3 is continuous everywhere because it is a polynomial. We evaluate the limit by

substitution: lim

(3x 2 y − 2x y 3 ) = 3 · 32 · (−1) − 2 · 3 · (−1)3 = −27 + 6 = −21

(x,y)→(3,−1)

x 2 + y2 x − 3y 2 lim x + y 2 (x,y)→(1,2) (x,y)→(−2,4) 4x + y x 2 +y SOLUTION The rational function is continuous at (1, 2) since the denominator is not zero at this point. We x+y 2 compute the limit using substitution: 5.

lim

x2 + y 12 + 2 3 = = 5 (x,y)→(1,2) x + y 2 1 + 22 lim

7.

y lim lim arctan tanxx cos y

(x,y)→(4,4) (x,y)→( π4 ,0)

SOLUTION

Since the function arctan xy is continuous at the point (4, 4), we compute the limit by substitution: lim

(x,y)→(4,4) −y ex − 2e 2 e x −y (x,y)→(1,1) (x,y)→(0,0) x + y 2

9.

arctan

π 4 y = arctan = arctan 1 = x 4 4

2

limlim

SOLUTION The function is the quotient of two continuous functions, and the denominator is not zero at the point (1, 1). Therefore, the function is continuous at this point, and we may compute the limit by substitution: 2 2 2 2 e − 1e e x − e−y e1 − e−1 1 = = = (e − e−1 ) lim x+y 1+1 2 2 (x,y)→(1,1)

In Exerciseslim 11–18,ln(x evaluate − y) the limit or determine that it does not exist. (x,y)→(1,0)

Limits and Continuity in Several Variables

S E C T I O N 15.2

11.

(ET Section 14.2)

337

x

lim

(x,y)→(0,1) y

The function xy is continuous at the point (0, 1), hence we compute the limit by substitution:

SOLUTION

0 x = =0 1 (x,y)→(0,1) y lim

13.

e x y xln(1 + x y) lim (x,y)→(1,1) lim (x,y)→(1,0) y

The function is continuous at (1, 1), hence we compute the limit using substitution:

SOLUTION

lim

(x,y)→(1,1)

e x y ln(1 + x y) = e1·1 ln(1 + 1 · 1) = e ln 2

lim x 22|y|3 2 (x,y)→(−1,−2) x + y lim (x,y)→(0,0) 1 + y 2 SOLUTION The function x 2 |y|3 is continuous everywhere. We compute the limit using substitution:

15.

lim

(x,y)→(−1,−2)

x 2 |y|3 = (−1)2 |−2|3 = 8

y−2  x y2 (x,y)→(4,2) lim x 2 − 4 (x,y)→(−1,−2) |x| SOLUTION The function is continuous at the point (4, 2), since it is the quotient of two continuous functions and the denominator is not zero at (4, 2). We compute the limit by substitution:

17.

lim

lim

(x,y)→(4,2)

In Exercises 19–22, sin assume that x lim sin y (x,y)→(π ,0)

lim

(x,y)→(2,5)

19.

lim



(x,y)→(2,5)

f (x, y) + 4g(x, y)

2−2 0 y−2  =  = √ =0 2 2 12 x −4 4 −4

f (x, y) = 3,

lim

(x,y)→(2,5)

g(x, y) = 7



Using the Sum Law and the Constant Multiples Law we get

SOLUTION

lim

( f (x, y) + 4g(x, y)) =

(x,y)→(2,5)

lim

(x,y)→(2,5)

f (x, y) + 4

lim

(x,y)→(2,5)

g(x, y) = 3 + 4 · 7 = 31

lim e f (x,y) lim f (x, y)g(x, y)2 (x,y)→(2,5) (x,y)→(2,5) SOLUTION e f (x,y) is the composition of G(u) = eu and u = f (x, y). Since eu is continuous, we may evaluate the

21.

limit as follows: lim

(x,y)→(2,5)

e f (x,y) = elim(x,y)→(2,5) f (x,y) = e3

 In Exercises 23–32, evaluate the limit or determine that the limit does not exist. You may evaluate the limit of a product lim ln g(x, y) − 2 f (x, y) function(x,y)→(2,5) as a product of limits as in Example 3. 23.

lim

(x,y)→(0,0)

SOLUTION

25.

(sin x)(sin y) xy

We evaluate the limit as a product of limits:   (sin x)(sin y) sin x sin y = lim lim lim = 1 · 1 = 1. xy (x,y)→(0,0) x→0 x y→0 y

lim (z 2 w2 − 29z) (z,w)→(−1,2) e x −y lim (x,y)→(2,1)

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(ET CHAPTER 14)

The function is continuous everywhere since it is a polynomial. Therefore we use substitution to evaluate

the limit: (z 2 ω − 9z) = (−1)2 · 2 − 9 · (−1) = 11.

lim

(z,ω )→(−1,2)

(2 + k)2 − 4 limlim h 4 sin x cos 1 k y (h,k)→(2,0) (x,y)→(0,0) SOLUTION We write the limit as a product of limits:

27.

lim

(h,k)→(2,0)

h4

(2 + k)2 − 4 (2 + k)2 − 4 (2 + k)2 − 4 = lim h 4 lim = 4 · lim k k k h→2 k→0 k→0

(1)

We show that this limit exists. Notice that the limit is the derivative of f (x) = x 2 at x = 2. That is, (2 + k)2 − 4 f (2 + k) − f (2) = lim = f  (2) = 2x =4 lim k k k→0 k→0 x=2 Combining with (1), we conclude that the given limit is 4 · 4 = 16. 29.

1 lim e1/xh(k tan2−1 + 4) y (x,y)→(0,0) lim k (h,k)→(0,0)

SOLUTION

We show that the limit along the line y = x does not exist. We have, lim

(x,y)→(0,0) along y=x

e1/x tan−1

1 1 = lim e1/x tan−1 y x x→0

(1)

We consider the two one-sided limits. As x → 0+, e1/x is increasing without bound and lim tan−1 1x = π2 . Therefore, x→0+ lim e1/x tan−1

x→0+

As x → 0−, we have lim e1/x tan−1

x→0−

1 = +∞ x

   −π 1 1 = lim e1/x lim tan−1 =0· =0 x x 2 x→0− x→0−

Since the one-sided limits are not equal, the limit in (1) does not exist. Therefore, the given limit does exist. x ln y  ek − 1 x2 lim k the limit as (x, y) approaches the origin along the positive y-axis (x = 0, y > 0): (x,k)→(1,0) SOLUTION We first compute

31.

lim

(x,y)→(0,0)

lim

(x,y)→(0,0) along the positive y -axis

x ln y = lim 0 · ln y = lim 0 = 0 y→0+

y→0

We now compute the limit as (x, y) approaches the origin along the curve y = e−1/x , x > 0 (notice that y → 0+ as x → 0+):  1 x ln y = lim x ln e−1/x = lim x · − lim = lim (−1) = −1 x x→0+ x→0+ x→0+ (x,y)→(0,0) along y=e−1/x , x>0

Since the limits along the two paths are not equal, the limit itself does not exist.  x 3 + y3 1 33. Let f (x, y) = 2 1 2 . lim x 2x+y y− x y(x + 2) (x,y)→(0,0) (a) Show that |x 3 | ≤ |x|(x 2 + y 2 ),

|y 3 | ≤ |y|(x 2 + y 2 )

(b) Show that | f (x, y)| ≤ |x| + |y|. (c) Use (b) and the formal definition of the limit to prove that

lim

(x,y)→(0,0)

f (x, y) = 0.

(d) Verify the conclusion of (c) again using polar coordinates as in Example 6. SOLUTION

Limits and Continuity in Several Variables

S E C T I O N 15.2

(ET Section 14.2)

339

(a) Since |x|y 2 ≥ 0, we have |x 3 | ≤ |x 3 | + |x|y 2 = |x|3 + |x|y 2 = |x|(x 2 + y 2 ) Similarly, since |y|x 2 ≥ 0, we have |y 3 | ≤ |y 3 | + |y|x 2 = |y|3 + |y|x 2 = |y|(x 2 + y 2 ) (b) We use the triangle inequality to write | f (x, y)| =

|x 3 | + |y 3 | |x 3 + y 3 | ≤ 2 2 x +y x 2 + y2

We continue using the inequality in part (a): | f (x, y)| ≤

(|x| + |y|)(x 2 + y 2 ) |x|(x 2 + y 2 ) + |y|(x 2 + y 2 ) = = |x| + |y| x 2 + y2 x 2 + y2

That is, | f (x, y)| ≤ |x| + |y| (c) In part (b) we showed that | f (x, y)| ≤ |x| + |y|

(1)

Let  > 0. Then if |x| < 2 and |y| < 2 , we have by (1) | f (x, y) − 0| ≤ |x| + |y| <

  + = 2 2

(2)

Notice that if x 2 + y 2 < 4 , then x 2 < 4 and y 2 < 4 . Hence |x| < 2 and |y| < 2 , so (1) holds. In other words, using

 D  2 to represent the punctured disc of radius /2 centered at the origin, we have    ⇒ |x| < (x, y) ∈ D  2 2 2

2

2

and |y| <

 ⇒ | f (x, y) − 0| <  2

We conclude by the limit definition that lim

(x,y)→(0,0)

f (x, y) = 0

(d) We verify the limit using polar coordinates: x = r cos θ ,

y = r sin θ

Then, (x, y) → (0, 0) if and only if x 2 + y 2 → 0, that is, if and only if r → 0+. We have x 2 + y2 = r 2 x 3 + y 3 = r 3 cos3 θ + r 3 sin3 θ = r 3 (cos3 θ + sin3 θ ) Hence, r 3 (cos3 θ + sin3 θ ) x 3 + y3 = = r (cos3 θ + sin3 θ ) f (x, y) = 2 x + y2 r2

(3)

Since | cos3 θ + sin3 θ | ≤ | cos3 θ | + | sin3 θ | ≤ 2, we have 0 ≤ |r (cos3 θ + sin3 θ )| ≤ 2r

(4)

As r approaches zero, 2r approaches zero, hence (4) and the Squeeze Theorem imply that lim r (cos3 θ + sin3 θ ) = 0

r →0+

By (3) and (5) we conclude that lim

(x,y)→(0,0)

f (x, y) = lim r (cos3 θ + sin3 θ ) = 0 r →0+

(5)

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(ET CHAPTER 14)

x does not exist. Hint: Consider the limits along the lines y = mx. 35. Show that lim 2 (x,y)→(0,0) x + y y Show that lim does not exist. 2 + y2 (x,y)→(0,0) SOLUTION We compute thex limit as (x, y) approaches the origin along the line y = mx, for a fixed value of m. x , we get Substituting y = mx in the function f (x, y) = x+y f (x, mx) =

x x = x + mx x(1 + m)

As (x, y) approaches (0, 0), (x, y)  = (0, 0). Therefore x  = 0 on the line y = mx. Thus, lim

(x,y)→(0,0) along y=mx

f (x, y) = lim

x

x→0 x(1 + m)

= lim

1

x→0 1 + m

=

1 1+m

We see that the limits along the lines y = mx are different, hence f (x, y) does not approach one limit as (x, y) → (0, 0). We conclude that the given limit does not exist. x a yb 37. LetShow a, b ≥that 0. Show that  limx = 0 if a + b > 2 and the limit does not exist if a + b ≤ 2. lim does not 2 2 exist. Hint: Use polar coordinates. (x,y)→(0,0) (x,y)→(0,0) x 2 + y2x + y SOLUTION We first show that the limit is zero if a + b > 2. We compute the limit using the polar coordinates x = r cos θ , y = r sin θ . Then (x, y) → (0, 0) if and only if x 2 + y 2 → 0, that is, if and only if r → 0+. Therefore, x a yb

lim

(x,y)→(0,0) x 2 + y 2

(r cos θ )a (r sin θ )b r a+b cosa θ sinb θ = lim 2 r →0+ r →0+ r r2

= lim

= lim (r a+b−2 cosa θ sinb θ )

(1)

r →0+

The following inequality holds: 0 ≤ |r a+b−2 cosa θ sinb θ | ≤ r a+b−2

(2)

Since a + b > 2, lim r a+b−2 = 0, therefore (2) and the Squeeze Theorem imply that r →0+

lim (r a+b−2 cosa θ sinb θ ) = 0

(3)

r →0

We combine (1) and (3) to conclude that if a + b > 2, then x a yb =0 (x,y)→(0,0) x 2 + y 2 lim

We now consider the case a + b < 2. We examine the limit as (x, y) approaches the origin along the line y = x. Along this line, θ = π4 , therefore (1) gives

    1 a 1 b x a yb π π r a+b−2 a+b−2 a b a+b−2 sin = lim r = lim r cos · √ · √ = lim √ a+b lim 2 2 4 4 r →0+ r →0+ r →0+ ( 2) (x,y)→(0,0) x + y 2 2 Since a + b < 2, we have a + b − 2 < 0 therefore lim r a+b−2 does not exist. It follows that if a + b < 2, the given r →0+

limit does not exist. Finally we examine the case a + b = 2. By (1) we get lim

x a yb

(x,y)→(0,0) x 2 + y 2

= lim (r 0 cosa θ sinb θ ) = lim cosa θ sinb θ = cosa θ sinb θ r →0+

r →0+

We see that the function does not approach one limit. For example, approaching the origin along the lines y = x (i.e.,  √ a+b θ = π4 ) and y = 0 (i.e., θ = 0) gives two different limits cosa π4 sinb π4 = 22 and cosa 0 sinb 0 = 0. We conclude that if a + b = 2, the limit does not exist. 39. Figure 8 shows the contour x 2 + y 2maps of two functions with contour interval m = 2. Explain why the limit  Does lim g(x, . y) appear to exist in (B)? If so, what is its limit? lim Evaluate f (x, y) doeslim not exist. 2+1−1 (x,y)→(0,0) x 2 + y(x,y)→P (x,y)→P 4

8

P

P

12 8 4

0

0 4 (A) Contour map of f(x, y)

(B) Contour map of g(x, y)

FIGURE 8

Limits and Continuity in Several Variables

S E C T I O N 15.2

(ET Section 14.2)

341

As (x, y) approaches arbitrarily close to P, the function f (x, y) takes the values 0, 4, and 8. Therefore f (x, y) does not approach one limit as (x, y) → P. Rather, the limit depends on the contour along which (x, y) is approaching P. This implies that the limit lim(x,y)→P f (x, y) does not exist. In (B) the limit lim(x,y)→P g(x, y) appears to exist. If it exists, it must be 6, which is the level curve of P. SOLUTION

Further Insights and Challenges sin(x y) is defined for x y  = 0. xy (x,y)→(0,2) (a) Is it possible to extend the domain of f (x, y) to all of R2 so that the result is a continuous function? (b) Use a computer algebra system to plot f (x, y). Does the result support your conclusion in (a)?

41.

Evaluate

The function (x,. y) = lim (1 + x)fy/x

SOLUTION

(a) We define f (x, y) on the x- and y-axes by f (x, y) = 1 if x y = 0. We now show that f is continuous. f is continuous at the points where x y  = 0. We next show continuity at (x 0 , 0) (including x 0 = 0). For the points (0, y0 ), the proof is similar and hence will be omitted. To prove continuity at P = (x 0 , 0) we have to show that lim

(x,y)→P

f (x, y) =

lim

(x,y)→P

sin x y =1 xy

(1)

Let us denote u = x y. As (x, y) → (x 0 , 0), u = x · y → x 0 · 0 = 0. Thus, lim

(x,y)→P

f (x, y) =

lim

(x,y)→(x0 ,0)

sin x y sin u = lim = 1 = f (x 0 , 0). xy u→0 u

(b) The following figure shows the graph of f (x, y) = sinx yx y : z

y

x

The graph shows that, near the axes, the values of f (x, y) are approaching 1, as shown in part (a). x2 y xy xy The function f (x, provides andexample g(x, y) where = 2 the 2limit . as (x, y) → (0, 0) does Repeat Exercise 41y) for=the4functions f (x, y)an=interesting x + y2 x+y x +y not exist, even though the limit along every line y = mx exists and is zero (Figure 9). (a) Show that the limit along any line y = mx exists and is equal to 0. (b) Calculate f (x, y) at the points (10−1 , 10−2 ), (10−5 , 10−10 ), (10−20 , 10−40 ). Do not use a calculator. (c) Show that lim f (x, y) does not exist. Hint: Compute the limit along the parabola y = x 2 . 43.

(x,y)→(0,0)

SOLUTION 2 (a) Substituting y = mx in f (x, y) = 4x y 2 , we get

x +y

f (x, mx) =

x 2 · mx

mx 3 mx = = 2 2 2 2 2 4 x (x + m ) x + m2 x + (mx)

We compute the limit as x → 0 by substitution: lim f (x, mx) = lim

x→0

mx

x→0 x 2 + m 2

m·0 = 2 =0 0 + m2

(b) We compute f (x, y) at the given points: f (10−1 , 10−2 ) =

10−2 · 10−2 10−4 + 10−4

=

10−4 2 · 10−4

=

1 2

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f (10−5 , 10−10 ) = f (10−20 , 10−40 ) =

(ET CHAPTER 14)

10−10 · 10−10

10−20 1 = = −20 −20 −20 2 10 + 10 2 · 10 10−40 · 10−40 10−80 + 10−80

=

10−80 2 · 10−80

=

1 2

(c) We compute the limit as (x, y) approaches the origin along the parabola y = x 2 (by part (b), the limit appears to be 1 ). We substitute y = x 2 in the function and compute the limit as x → 0. This gives 2 lim

(x,y)→0 along y=x 2

f (x, y) = lim f (x, x 2 ) = lim

x2 · x2

x→0 x 4 + (x 2 )2

x→0

1 x4 1 = lim = 2 x→0 2x 4 x→0 2

= lim

However, in part (a), we showed that the limit along the lines y = mx is zero. Therefore f (x, y) does not approach one f (x, y) does not exist. limit as (x, y) → (0, 0), so the limit lim (x,y)→(0,0)

Is the following function continuous?

15.3 Partial Derivatives



(ET Section 14.3) x 2 + y2 f (x, y) = 1

Preliminary Questions

if x 2 + y 2 < 1 if x 2 + y 2 ≥ 1

1. Patricia derived the following incorrect formula by misapplying the Product Rule: ∂ 2 2 (x y ) = x 2 (2y) + y 2 (2x) ∂x What was her mistake and what is the correct calculation? SOLUTION To compute the partial derivative with respect to x, we treat y as a constant. Therefore the Constant Multiple Rule must be used rather than the Product Rule. The correct calculation is:

∂ 2 2 ∂ (x y ) = y 2 (x 2 ) = y 2 · 2x = 2x y 2 . ∂x ∂x ∂ 2. Explain why it is not necessary to use the Quotient Rule to compute ∂ x  ∂ x+y ? to compute ∂y y + 1



x+y . Should the Quotient Rule be used y+1

SOLUTION In differentiating with respect to x, y is considered a constant. Therefore in this case the Constant Multiple Rule can be used to obtain  ∂ x+y 1 1 1 ∂ (x + y) = ·1= . = ∂x y + 1 y + 1 ∂x y+1 y+1

As for the second part, since y appears in both the numerator and the denominator, the Quotient Rule is indeed needed. 3. Which of the following partial derivatives should be evaluated without using the Quotient Rule? y2 ∂ xy ∂ ∂ xy (b) (c) (a) 2 2 2 ∂x y + 1 ∂y y + 1 ∂x y + 1 SOLUTION

(a) This partial derivative does not require use of the Quotient Rule, since the Constant Multiple Rule gives  xy y y ∂ y ∂ (x) = 2 = 2 ·1= 2 . ∂ x y2 + 1 y + 1 ∂x y +1 y +1 (b) This partial derivative requires use of the Quotient Rule. (c) Since y is considered a constant in differentiating with respect to x, we do not need the Quotient Rule to state that y2 ∂ = 0. 2 ∂x y + 1 √

3 −1 y ? 4. What is f x , where f (x, y, z) = (sin yz)e z −z

In differentiating with respect to x, we treat y and z as constants. Therefore, the whole expression for f (x, y, z) is treated as constant, so the derivative is zero:

SOLUTION

3 −1 √ y  ∂

sin yze z −z = 0. ∂x

Partial Derivatives

S E C T I O N 15.3

5. Which of the following partial derivatives are equal to f x x y ? (b) f yyx (c) f x yy (a) f x yx

(ET Section 14.3)

343

(d) f yx x

f x x y involves two differentiations with respect to x and one differentiation with respect to y. Therefore, if f satisfies the assumptions of Clairaut’s Theorem, then

SOLUTION

f x x y = f x yx = f yx x

Exercises 1. Use the limit definition of the partial derivative to verify the formulas ∂ x y2 = y2, ∂x SOLUTION

∂ x y 2 = 2x y ∂y

Using the limit definition of the partial derivative, we have

∂ (x + h)y 2 − x y 2 x y 2 + hy 2 − x y 2 hy 2 x y 2 = lim = lim = lim = lim y 2 = y 2 ∂x h h h→0 h→0 h→0 h h→0 ∂ x(y + k)2 − x y 2 x(y 2 + 2yk + k 2 ) − x y 2 x y 2 + 2x yk + xk 2 − x y 2 x y 2 = lim = lim = lim ∂y k k k k→0 k→0 k→0 = lim

k→0

k(2x y + xk) = lim (2x y + k) = 2x y + 0 = 2x y k k→0

y ∂ 3. Use the Quotient Rule to compute ∂ 2. ∂y x + (x y + y)(x + y 4 ). Use the Product Rule to compute ∂y SOLUTION Using the Quotient Rule we obtain (x + y) ∂∂y (y) − y ∂∂y (x + y) y (x + y) · 1 − y · 1 x ∂ = = = 2 2 ∂y x + y (x + y) (x + y) (x + y)2 5. Calculate f z (2, 3, 1), where f (x, y, z) = x yz. ∂ ln(u 2f z+(x, Use theWe Chain Rulethe to compute uv). SOLUTION first find partial derivative y, z): ∂u f z (x, y, z) =

∂ (x yz) = x y ∂z

Substituting the given point we get f z (2, 3, 1) = 2 · 3 = 6 7. The plane y = 1 intersects the surface z = x 4 + 6x y − y 4 in a certain curve (Figure 7). Find the slope of the tangent Explain the relation between the two formulas (c is a constant) line to this curve at the point P = (1, 1, 6). d sin(cx) = c cos(cx), dx

z

∂ sin(x y) = y cos(x y) ∂x

x y

FIGURE 7 Graph of f (x, y) = x 4 + 6x y − y 4 . SOLUTION

The slope of the tangent line to the curve z = z(x, 1) = x 4 + 6x − 1, obtained by intersecting the surface

z = x 4 + 6x y − y 4 with the plane y = 1, is the partial derivative ∂∂zx (1, 1).

∂z ∂ 4 = (x + 6x y − y 4 ) = 4x 3 + 6y ∂x ∂x ∂z (1, 1) = 4 · 13 + 6 · 1 = 10 m= ∂x

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(ET CHAPTER 14)

In Exercises 9–11, refer to Figure 8. ∂f ∂f and are positive or negative at the point P in Figure 7. Determine if the partial derivatives ∂x ∂y 90 50 y

10

4

−50

2 C

B

−10

0 −2

−10

10 50 90 130

A

−30

−4 −4

x

−2

0

2

4

FIGURE 8 Contour map of f (x, y).

9. Estimate f x and f y at point A. SOLUTION To estimate f x we move horizontally to the next level curve in the direction of growing x, to a point A . The change in f from A to A is the contour interval,  f = 70 − 50 = 20. 50 70

A 1.25

A′

The distance between A and A is approximately x ≈ 1.25. Hence, f x ( A) ≈

20 f = = 16 x 1.25

To estimate f y we move vertically from A to a point A on the next level curve in the direction of growing y. The change in f from A to A is  f = 30 − 50 = −20. 30 A′′ 0.6

50

A

The distance between A and A is y ≈ 0.6. Hence, f y ( A) ≈

−20 f = ≈ −33.3. y 0.6

11. At which of A, B, or C is f y smallest? Starting at point B, in which direction does f increase most rapidly? SOLUTION We consider vertical lines through A, B, and C. The distance between each point A, B, C and the intersection of the vertical line with the adjacent level curves is the largest at C. It means that f y is smallest at C. In Exercises 12–39, compute the partial derivatives. 13. z = x 4 y 32 z = x + y2 SOLUTION Treating y as a constant (to find z x ) and x as a constant (to find z y ) and using Rules for Differentiation, we get, ∂ 4 3 (x y ) = ∂x ∂ 4 3 (x y ) = ∂y 15. V = π r 2 4h z = x y + x y −2

∂ 4 (x ) = ∂x ∂ x 4 (y 3 ) = ∂y y3

y 3 · 4x 3 = 4x 3 y 3 x 4 · 3y 2 = 3x 4 y 2

Partial Derivatives

S E C T I O N 15.3 SOLUTION

(ET Section 14.3)

V: We find ∂∂rV and ∂∂h

∂ ∂V ∂ = (π r 2 h) = π h (r 2 ) = π h · 2r = 2π hr ∂r ∂r ∂r ∂V ∂ = (π r 2 h) = π r 2 ∂h ∂h x 17. z = x z x=− y y SOLUTION We differentiate with respect to x, using the Quotient Rule. We get ∂ ∂x



x x−y

=

(x − y) ∂∂x (x) − x ∂∂x (x − y) (x − y)2

=

(x − y) · 1 − x · 1 (x − y)2

=

−y (x − y)2

We now differentiate with respect to y, using the Chain Rule:   x 1 −1 ∂ ∂ x ∂ −1 (x − y) = x · · (−1) = =x =x· ∂y x − y ∂y x − y (x − y)2 ∂y (x − y)2 (x − y)2 x 19. z =   z =x 2 9+−y 2x 2 − y 2 SOLUTION

We compute ∂∂zx using the Quotient Rule and the Chain Rule:

1· ∂z = ∂x



  x 2 + y 2 − x · √ 2x2 2 x 2 + y 2 − x ∂∂x x 2 + y 2 x 2 + y2 − x 2 y2 2 x +y = = =  2 3/2 3/2 x 2 + y2 (x 2 + y 2 ) (x 2 + y 2 ) x 2 + y2

We compute ∂∂zy using the Chain Rule:  ∂z ∂ 2 1 −x y −1/2 −3/2 2 = x (x + y ) =x· − · 2y = (x 2 + y 2 ) 3/2 ∂y ∂y 2 2 (x + y 2 ) 2 21. z =z sin(u = (sinv)x)(sin y) SOLUTION By the Chain Rule,

dω d sin ω = cos ω du du

and

d dω sin ω = cos ω . dv dv

Applying this with ω = u 2 v gives ∂ ∂ sin(u 2 v) = cos(u 2 v) (u 2 v) = cos(u 2 v) · 2uv = 2uv cos(u 2 v) ∂u ∂u ∂ ∂ sin(u 2 v) = cos(u 2 v) (u 2 v) = cos(u 2 v) · u 2 = u 2 cos(u 2 v) ∂v ∂v 23. S = tan−1 (wz) x z = tan SOLUTION Byy the Chain Rule, d du 1 tan−1 u = 2 dw dw 1+u

and

d 1 du tan−1 u = dz 1 + u 2 dz

Using this rule with u = wz gives ∂ ∂ z 1 dS = tan−1 (wz) = (wz) = 2 2 z2 dw ∂w ∂w 1 + w 1 + (wz) ∂ w dS ∂ 1 = tan−1 (wz) = (wz) = 2 dz ∂z ∂z 1 + w2 z 2 1 + (wz) 2 + y2) 25. z =z ln(x = ln(x + y)

345

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(ET CHAPTER 14)

Using the Chain Rule we have

SOLUTION

1 ∂z ∂ 2 1 2x = 2 (x + y 2 ) = 2 · 2x = 2 2 2 ∂x ∂ x x +y x +y x + y2 1 ∂z ∂ 2 1 2y = 2 (x + y 2 ) = 2 · 2y = 2 2 2 ∂y ∂y x +y x +y x + y2 27. z = e x y r +s W =e d u u du SOLUTION We use the Chain Rule, ddx eu = eu du d x ; d y e = e d y with u = x y to obtain ∂ xy ∂ e = e x y (x y) = e x y y = ye x y ∂x ∂x ∂ xy ∂ e = e x y (x y) = e x y x = xe x y ∂y ∂y 29. z = e−x −y 2 R = e−v /k ∂z ∂z SOLUTION We use the Chain Rule to find ∂ x and ∂ y : 2

2

2 2 ∂ 2 2 2 2 ∂z = e−x −y (−x 2 − y 2 ) = e−x −y · (−2x) = −2xe−x −y ∂x ∂x 2 2 ∂ 2 2 2 2 ∂z = e−x −y (−x 2 − y 2 ) = e−x −y · (−2y) = −2ye−x −y ∂y ∂y

31. z = y x √ 2 2 P = e y +z ∂z SOLUTION To find ∂ y , we use the Power Rule for differentiation: ∂z = x y x−1 ∂y To find ∂∂zx , we use the derivative of the exponent function: ∂z = y x ln y ∂x 2 y) 33. z =z sinh(x = cosh(3x + 2y) d du d du SOLUTION By the Chain Rule, d x sinh u = cosh u d x and d y sinh u = cosh u d y . We use the Chain Rule with u = x 2 y to obtain

∂ ∂ sinh(x 2 y) = cosh(x 2 y) (x 2 y) = 2x y cosh(x 2 y) ∂x ∂x ∂ ∂ sinh(x 2 y) = cosh(x 2 y) (x 2 y) = x 2 cosh(x 2 y) ∂y ∂y x 35. w = w y=+x yz 2 z 3 SOLUTION

We have ∂w ∂ = ∂x ∂x



x y+z

=

1 1 ∂ (x) = y + z ∂x y+z

∂w To find ∂w ∂ y and ∂z , we use the Chain Rule:  ∂w ∂ 1 −1 ∂ −x −1 =x (y + z) = x · ·1= =x· 2 2 ∂y ∂y y + z ∂y (y + z) (y + z) (y + z)2  ∂ 1 −1 ∂w ∂ −x −1 =x (y + z) = x · ·1= =x· ∂z ∂z y + z (y + z)2 ∂z (y + z)2 (y + z)2

x 37. w = 2 θ 2 Q (x = r+ e y + z 2 )3/2

S E C T I O N 15.3 SOLUTION

(ET Section 14.3)

To find ∂w ∂ x , we use the Quotient Rule and the Chain Rule: 3/2

1 · (x 2 + y 2 + z 2 ) − x · 32 (x 2 + y 2 + z 2 ) ∂w = 3 ∂x (x 2 + y 2 + z 2 ) =

Partial Derivatives

x 2 + y 2 + z 2 − 3x 2 5/2 (x 2 + y 2 + z 2 )

=

1/2

· 2x

= (x 2 + y 2 + z 2 )1/2

(x 2 + y 2 + z 2 ) − x · 3x (x 2 + y 2 + z 2 )3

y 2 + z 2 − 2x 2 (x 2 + y 2 + z 2 )

5/2

∂w We now use the Chain Rule to compute ∂w ∂ y and ∂z :

∂ 1 ∂ ∂w −3/2 =x = x (x 2 + y 2 + z 2 ) ∂y ∂y (x 2 + y 2 + z 2 )3/2 ∂y  3 3x y −5/2 =x· − · 2y = −

(x 2 + y 2 + z 2 ) 5/2 2 x 2 + y2 + z2 ∂ 1 ∂w ∂ −3/2 =x

= x (x 2 + y 2 + z 2 ) ∂z ∂z x 2 + y 2 + z 2 3/2 ∂z  3 3x z −5/2 =x· − · 2z = − (x 2 + y 2 + z 2 ) 5/2 2 (x 2 + y 2 + z 2 ) e−r t 39. U =A = sin(4θ − 9t) r SOLUTION

We have −te−r t · r − e−r t · 1 ∂U −(1 + r t)e−r t = = 2 ∂r r r2

and also ∂U −r e−r t = = −e−r t ∂t r In Exercises 40–44, compute the given partial derivatives. 41. f (x, y) = sin(x 22− y), 3 f y2(0, π ) 5 f (x, y) = 3x y + 4x y − 7x y , f x (1, 4) SOLUTION We differentiate with respect to y, using the Chain Rule. This gives f y (x, y) = cos(x 2 − y)

∂ 2 (x − y) = cos(x 2 − y) · (−1) = − cos(x 2 − y) ∂y

Evaluating at (0, π ) we obtain f y (0, π ) = − cos(02 − π ) = − cos(−π ) = − cos π = 1. 2 3

x z−x z 43. h(x,g(u, z) = 1)u (1, 2) z (2, g v)e= u ln(u, + hv), SOLUTION We first find the partial derivative h z (x, z) using the Chain Rule: 2 3 ∂ 2 3 2 3 (x z − x 2 z 3 ) = e x z−x z (x − x 2 · 3z 2 ) = (x − 3x 2 z 2 )e x z−x z h z (x, z) = e x z−x z ∂z

At the point (2, 1) we get 2 3 h z (2, 1) = (2 − 3 · 22 · 12 )e2·1−2 ·1 = −10e−2 .

∂W ∂W W = e−E/kT . 45. Calculate z−x 2 z 3 , , where h(x, z) ∂=Ee xand ∂ T h z (1, 0) SOLUTION

We use the Chain Rule d u du e = eu dE dE

and

du d u e = eu dT dT

347

348

C H A P T E R 15

D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S E , to obtain with u = − kT

 1 −E/kT 1 =− e − kT kT      ∂ 1 ∂W ∂ E E E −E/kT E 1 = e−E/kT e − = e−E/kT · − = e−E/kT − − 2 = ∂T ∂T kT k ∂T T k T kT 2 ∂ ∂W = e−E/kT ∂E ∂E



−

E kT



(ET CHAPTER 14)

= e−E/kT

∂V ∂V 47. TheAccording volume ofto a right-circular cone PofVradius r and height and and . temperature (R r 2 h.the Calculate the ideal gas law, = n RT , where P,hVis, V and=Tπ3 are pressure, volume, ∂r ∂h ∂P ∂P and . and n are We constants). View P as a function of V and T , and compute SOLUTION obtain the following derivatives: ∂T ∂V ∂  π 2  πh ∂ 2 2π hr πh ∂V = r h = r = · 2r = ∂r ∂r 3 3 ∂r 3 3   ∂ π 2 π ∂V = r h = r2 ∂h ∂h 3 3 Use the contour Figure 9 toleads explain the increase followinginare 49. A right-circular cone map has rof=f (x, h =y)12incm. What to awhy greater V ,true: a 1-cm increase in r or 1-cm h? both Arguelarger usingatpartial P thanderivatives. at Q. (a) increase f x and f in y are (b) f x (x, y) is decreasing as a function of y, that is, for any x, f x (x, b1 ) > f x (x, b2 ) if b1 < b2 . y 20 Q

16 14 10 6 2

P

–2

x

FIGURE 9 Contour interval 2. SOLUTION

(a) Since the vertical and horizontal lines through P meet more level curves than vertical and horizontal lines through Q, f is increasing more rapidly in the y and x direction at P than at Q. Therefore, f x and f y are both larger at P than at Q. (b) For any fixed value of x, a horizontal line through (x, b2 ) meets fewer level curves than a horizontal line through (x, b1 ), for b1 < b2 . Thus, f x is decreasing as a function of y. 51. Seawater Density to Example 6 and 4. to true (geographic) north; instead, it points at some Over most of the earth, aRefer magnetic compass doesFigure not point ∂ ρtrue north. The angle D between magnetic north and true north is called the magnetic declinaρ west of angle east∂or and at points B and C. (a) Estimate tion. Use ∂Figure 10 T ∂ Sto determine which of the following statements is true. Note that because of the way longitude at B,toa left. 1◦ increase in temperature or a 1-ppt increase in salinity? (b) Which has a greater effect on density is measured, the x-axis increases fromρright (c) True∂or The density of warm seawater∂ is D false: D more sensitive to a change in temperature ∂ D than the density of cold > ∂D (a) (b) > 0 (c) >0 seawater?∂y Explain. ∂y ∂x ∂y A

B

C

C

28◦ larger at A than at C. ∂ρ (a) We estimate ∂ T at B. Since T varies in the vertical direction, we move vertically from B to a point B  on the next level curve, in the direction of increasing T (upward). The change in ρ from B to B  is ρ = 1.0235 − 1.0240 = −0.0005. The change in T from B to B  is T = 2.5◦ . Hence, −0.0005 ∂ ρ ρ ≈ = −2 · 10−4 ≈ ∂T B T 2.5 SOLUTION (d) D is

B′

1.0235 1.0240

B

∂ρ

We now compute ∂ T |C . Following the same procedure as before, we find that ∂ ρ −0.0005 ρ ≈ = −0.0002 ≈ ∂ T C T 2.5 ∂ρ

We now estimate ∂ S at C. Since S varies in the horizontal direction, we move horizontally from C to a point C  on the next level curve, in the direction of increasing S (to the right). The change in ρ from C to C  is ρ = 1.0255 − 1.0250 = 0.0005. The change in S from C to C  is S = 0.5. Hence, 0.0005 ∂ ρ ρ = = 0.001. ≈ ∂ S C S 0.5

S E C T I O N 15.3

Partial Derivatives

(ET Section 14.3)

349

∂ρ

Finally, we now estimate ∂ S at B, using the same technique. We get ∂ ρ 0.0005 ρ ≈ = 0.0008 ≈ ∂ S B S 0.6 (b) The change ρ in the density as a result of a change T in temperature (where S remains unchanged), and as a result of a change S in salinity (where T remains unchanged), can be estimated by ∂ ρ ∂ ρ ρ ≈ T ;  ρ ≈ S (1) ∂ T C ∂ S C ∂ρ ∂ρ We substitute T = 1◦ , S = 1 ppt, ∂ T C ≈ −0.0002, and ∂ S C ≈ 0.001 (obtained in part (a)) in (1), to obtain ρ ≈ −0.0002 for the change in temperature ρ ≈ 0.001 for the change in salinity. We conclude that the effect caused by the change in salinity is greater than the effect caused by the change in temperature, at the point C. (c) At low temperatures the vertical distances between adjacent level curves are greater than at higher temperatures. This means that the density of cold water is less sensitive to a change in the temperature than the density of warm water. Thus, the statement is true. In Exercises 53–58, the derivative Estimate the compute partial derivatives at P indicated. of the function whose contour map is shown in Figure 11. ∂2 f ∂2 f and ∂x2 ∂y 2

53. f (x, y) = 3x 2 y − 6x y 4 , SOLUTION

We first compute the partial derivatives ∂∂ xf and ∂∂ yf : ∂f = 6x y − 6y 4 ; ∂x

∂f = 3x 2 − 6x · 4y 3 = 3x 2 − 24x y 3 ∂y

We now differentiate ∂∂ xf with respect to x and ∂∂ yf with respect to y. We get ∂2 f ∂ f x = 6y; = ∂x ∂x2

∂2 f ∂ f y = −24x · 3y 2 = −72x y 2 . = ∂y ∂y 2

xy ∂2g 55. g(x,f (x, y) =y) = x ln(y , 2 ), f yy (2, 3) x−y ∂ x ∂y SOLUTION

By definition we have ∂2g ∂ = g yx = ∂ x∂y ∂x



∂g ∂y



Thus, we must find ∂g ∂y : ∂ ∂g =x ∂y ∂y



y x−y

=x

1 · (x − y) − y · (−1) (x − y)2

=

x2 (x − y)2

Differentiating ∂g ∂ y with respect to x, using the Quotient Rule, we obtain ∂ ∂2g = ∂ x∂y ∂x



∂g ∂y



u 57. h(u,g(x, v) = y) u=+x vye, −yh,uv (u, g yyv) (1, 0) SOLUTION

∂ By definition h uv = ∂v

=



x2 2x(x − y)2 − x 2 · 2(x − y) 2x y ∂ = =− 2 ∂ x (x − y) (x − y)4 (x − y)3



∂h . We compute ∂h using the Quotient Rule: ∂u ∂u

∂h ∂ u 1 · (u + v) − u · 1 v = = = 2 ∂u ∂u u + v (u + v) (u + v)2 We now use again the Quotient Rule to differentiate ∂h ∂u with respect to υ . We obtain h uv (u, v) =

v ∂ 1 · (u + v)2 − v · 2(u + v) u−v = = 2 ∂v (u + v) (u + v)4 (u + v)3

350

C H A P T E R 15

S E C T I O N 15.3 SOLUTION

Partial Derivatives

(ET Section 14.3)

351

Using the Chain Rule, we have ∂ ∂ cos(u + v 2 ) = − sin(u + v 2 ) · (u + v 2 ) = − sin(u + v 2 ) ∂u ∂u  ∂

− sin(u + v 2 ) = − cos(u + v 2 ) f uu = ∂u  ∂

∂ − cos(u + v 2 ) = sin(u + v 2 ) · (u + v 2 ) = 2v sin(u + v 2 ) f uuv = ∂v ∂v fu =

65. F(r, s, t) = r (s 2 + t 2 ), Fr st g(x, y, z) = x 4 y 5 z 6 , gx x yz SOLUTION For F(r, s, t) = r (s 2 + t 2 ), we have Fr = s 2 + t 2 Fr s = 2s Fr st = 0 /4t) , u 67. u(x, t) = t −1/2 e−(x x g(x, y, z) = x 4 y 5 z 3 , gx xxyz SOLUTION Using the Chain Rule we obtain 2

2 2 ∂ ∂ u x = t −1/2 (e−x /4t ) = t −1/2 · e−x /4t

∂x

∂x

x2 − 4t



2 2 1 −2x = − xt −3/2 e−x /4t = t −1/2 · e−x /4t · 4t 2

We now differentiate u x with respect to x, using the Product Rule and the Chain Rule:  2 2 ∂ −x 2 /4t  1 −2x 1 xe u x x = − t −3/2 = − t −3/2 1 · e−x /4t + x · e−x /4t · 2 ∂x 2 4t



 2 2 1 x 2 −x 2 /4t 1 x2 e = − t −3/2 e−x /4t − = − t −3/2 e−x /4t 1 − 2 2t 2 2t  69. g(x, y, z) = x 2 +uy 2 + z 2 , gx yz , Ruvw R(u, v, w) = v+w SOLUTION Differentiating with respect to x, using the Chain Rule, we get  ∂ 2 ∂ 1 1 x (x + y 2 + z 2 ) =  gx = x 2 + y2 + z2 =  · 2x =  ∂x 2 x 2 + y2 + z2 ∂ x 2 x 2 + y2 + z2 x 2 + y2 + z2 We now differentiate gx with respect to y, using the Chain Rule. This gives  ∂ 1 −x y −1/2 −3/2 gx y = x (x 2 + y 2 + z 2 ) =x· − · 2y = (x 2 + y 2 + z 2 ) 3/2 ∂y 2 2 (x + y 2 + z 2 ) Finally, we differentiate gx y with respect to z, obtaining  ∂ 2 3 3x yz −3/2 −5/2 2 2 gx yz = −x y (x + y + z ) = −x y · − · 2z = (x 2 + y 2 + z 2 ) 5/2 ∂z 2 2 (x + y 2 + z 2 ) ∂f ∂f = 2x y and = x 2. 71. Findu(x, (byt)guessing) a function that = sech2 (x − t), usuch xxx ∂x ∂y SOLUTION

The function f (x, y) = x 2 y satisfies ∂∂ yf = x 2 and ∂∂ xf = 2x y.

73. Assume that f x y and f yx are continuous and that f yx x exists. Show that f x yx also exists and that f yx x = f x yx . ∂f ∂f = x y and = x 2 . Hint: Show that f Prove that fthere does not exist any function f (x, y) such that that SOLUTION Since x y and f yx are continuous, Clairaut’s Theorem implies ∂x ∂y cannot satisfy Clairaut’s Theorem. f x y = f yx (1) We are given that f yx x exists. Using (1) we get f yx x =

∂ ∂ ∂ ∂ fy = f yx = f x y = f x yx ∂x ∂x ∂x ∂x

Therefore, f x yx also exists and f yx x = f x yx . 2 Show that u(x, t) = sin(nx) e−n t satisfies the heat equation (Example 10) for any constant n:

352

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D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S

(ET CHAPTER 14)

75. Find all values of A and B such that f (x, t) = e Ax+Bt satisfies Eq. (3). SOLUTION

We compute the following partials, using the Chain Rule: ∂f ∂ ∂ = (e Ax+Bt ) = e Ax+Bt ( Ax + Bt) = Be Ax+Bt ∂t ∂t ∂t ∂ Ax+Bt ∂f ∂ = (e ) = e Ax+Bt ( Ax + Bt) = Ae Ax+Bt ∂x ∂x ∂x ∂2 f ∂ ∂ ∂ = ( Ae Ax+Bt ) = A (e Ax+Bt ) = Ae Ax+Bt ( Ax + Bt) = A2 e Ax+Bt ∂x ∂x ∂x ∂x2

Substituting these partials in the differential equation (3), we get Be Ax+Bt = A2 e Ax+Bt We divide by the nonzero e Ax+Bt to obtain B = A2 We conclude that f (x, t) = e Ax+Bt satisfies equation (5) if and only if B = A2 , where A is arbitrary. In Exercises 76–79, the Laplace operator  is defined by  f = f x x + f yy . A function u(x, y) satisfying the Laplace equation u = 0 is called harmonic. 77. FindShow all harmonic polynomials u(x, y)are of harmonic: degree three, that is, u(x, y) = ax 3 + bx 2 y + cx y 2 + d y 3 . that the following functions SOLUTION We=compute the first-order partials u x and u y and second-order u x x and u yy of the given (a) u(x, y) x (b) the u(x, y) = e x cos partials y polynomial u(x, y). This gives y (d) u(x, y) = ln(x 2 + y 2 ) (c) u(x, y) = tan−1 x 2 u x = 3ax + 2bx y + cy 2 u y = bx 2 + 2cx y + 3d y 2 u x x = 6ax + 2by u yy = 2cx + 6d y The polynomial is harmonic if u x x + u yy = 0, that is, if for all x and y 6ax + 2by + 2cx + 6d y = 0 This equality holds for all x and y if and only if the coefficients of x and y are both zero. That is, 6a + 2c = 0 (so c = −3a) and 2b + 6d = 0 (so b = −3d). We conclude that the harmonic polynomials in the given form are u(x, y) = ax 3 − 3d x 2 y − 3ax y 2 + d y 3 79. Find all constants a, b such that u(x, y) = cos(ax)eby is harmonic. ∂u ∂u and are harmonic. Show that if u(x, y) is harmonic, then the partial derivatives SOLUTION To determine if the functions cos(ax)eby are harmonic,∂ we x compute ∂y the following derivatives: (cos ax) = −a sin ax ⇒ (cos ax) = −a 2 cos ax 



(eby ) = beby ⇒ (eby ) = b2 eby = a 2 eby Thus, we can conclude ux x =

∂2 cos(ax)eby = −a 2 cos(ax)eby = −a 2 u ∂x2

u yy =

∂2 cos(ax)eby = b2 cos(ax)eby = b2 u ∂y 2

Thus, u x x + u yy = (b2 − a 2 )u, which equals 0 if and only if a 2 = b2 . 2

Show that u(x,and t) = Challenges sech (x − t) satisfies the Korteweg–deVries equation (which arises in the study of water Further Insights 81.

waves)

Assumptions Matter This exercise shows that the hypotheses of Clairaut’s Theorem are needed. Let x 2 − y2 4u0) t +=u x0.x x + 12uu x = 0 f (x, y) = x y 2 for (x, y)  = (0, 0) and f (0, x + y2

Partial Derivatives

S E C T I O N 15.3

(ET Section 14.3)

353

(a) Use a computer algebra system to verify the following formulas for (x, y)  = (0, 0): f x (x, y) =

y(x 4 + 4x 2 y 2 − y 4 ) (x 2 + y 2 )2

f y (x, y) =

x(x 4 − 4x 2 y 2 − y 4 ) (x 2 + y 2 )2

(b) Use the limit definition of the partial derivative to show that f x (0, 0) = f y (0, 0) = 0 and that f yx (0, 0) and f x y (0, 0) both exist but are not equal. (c) Use a computer algebra system to show that for (x, y)  = (0, 0): f x y (x, y) = f yx (x, y) =

x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6 (x 2 + y 2 )3

Show that f x y is not continuous at (0, 0). Hint: Show that lim f x y (h, 0)  = lim f x y (0, h). h→0

h→0

(d) Explain why the result of (b) does not contradict Clairaut’s Theorem. SOLUTION

(a) We use a CAS to verify the following partials for (x, y)  = (0, 0): f x (x, y) = f y (x, y) =

y(x 4 + 4x 2 y 2 − y 4 ) (x 2 + y 2 )

2

x(x 4 − 4x 2 y 2 − y 4 ) (x 2 + y 2 )

2

(b) Using the limit definition of the partial derivatives at the point (0, 0) we have h · 0 h 2 −02 − 0 f (h, 0) − f (0, 0) 0 h +0 = lim = lim =0 h h h→0 h→0 h→0 h 2

2

f x (0, 0) = lim

0 · k 02 −k 2 − 0 f (0, k) − f (0, 0) 0 0 +k = lim = lim = 0 f y (0, 0) = lim k k k→0 k→0 k→0 k 2

2

We now use the derivatives in part (a) and the limit definition of the partial derivatives to compute f yx (0, 0) and f x y (0, 0). By the formulas in part (a), we have

Thus,

f y (0, 0) = 0,

f y (h, 0) =

f x (0, 0) = 0,

f x (0, k) =

h(h 4 − 0 − 0) (h 2 + 0)

=h

2

k(0 + 0 − k 4 ) (02 + k 2 )

2

= −k

f y (h, 0) − f y (0, 0) h−0 ∂ f y = lim = lim 1 = 1 = lim ∂x h h h→0 h→0 h→0 (0,0) f x (0, k) − f x (0, 0) −k − 0 ∂ f x = lim = lim (−1) = −1 = lim f x y (0, 0) = ∂y k k k→0 k→0 k→0 (0,0)

f yx (0, 0) =

We see that the mixed partials at the point (0, 0) exist but are not equal. (c) We verify, using a CAS, that for (x, y)  = (0, 0) the following derivatives hold: f x y (x, y) = f yx (x, y) =

x 6 + 9x 4 y 2 − 9x 2 y 4 − y 6 (x 2 + y 2 )

3

To show that f x y is not continuous at (0, 0), we show that the limit lim(x,y)→(0,0) f x y (x, y) does not exist. We compute the limit as (x, y) approaches the origin along the x-axis. Along this axis, y = 0; hence, lim

(x,y)→(0,0)

f x y (x, y) = lim f x y (h, 0) = lim h→0

h 6 + 9h 4 · 0 − 9h 2 · 0 − 0

h→0

along the x -axis

(0 + h 2 )

3

= lim 1 = 1 h→0

We compute the limit as (x, y) approaches the origin along the y-axis. Along this axis, x = 0, hence, lim

(x,y)→(0,0)

along the y -axis

f x y (x, y) = lim f x y (0, h) = lim h→0

h→0

0 + 0 + 0 − h6 (0 + h 2 )

3

= lim (−1) = −1 h→0

354

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(ET CHAPTER 14)

Since the limits are not equal f (x, y) does not approach one value as (x, y) → (0, 0), hence the limit lim f x y (x, y) does not exist, and f x y (x, y) is not continuous at the origin. (x,y)→(0,0)

(d) The result of part (b) does not contradict Clairaut’s Theorem since f x y is not continuous at the origin. The continuity of the mixed derivative is essential in Clairaut’s Theorem.

15.4 Differentiability, Linear Approximation, and Tangent Planes

(ET Section 14.4)

Preliminary Questions 1. How is the linearization of f (x, y) at (a, b) defined? SOLUTION

The linearization of f (x, y) at (a, b) is the linear function

L(x, y) = f (a, b) + f x (a, b)(x − a) + f y (a, b)(y − b)

 This function is the equation of the tangent plane to the surface z = f (x, y) at a, b, f (a, b) . 2. Define local linearity for functions of two variables. SOLUTION f (x, y) is locally linear at (a, b) if the linear approximation L(x, y) at (a, b) approximates f (x, y) at (a, b) to first order. That is, if there exists a function (x, y) satisfying lim (x, y) = 0 such that

(x,y)→(a,b)

 f (x, y) − L(x, y) = (x, y) (x − a)2 + (y − b)2 for all (x, y) in an open disk D containing (a, b). In Questions 3–5, assume that f (2, 3) = 8,

f x (2, 3) = 5,

f y (2, 3) = 7

3. Which of (a)–(b) is the linearization of f at (2, 3)? (a) L(x, y) = 8 + 5x + 7y (b) L(x, y) = 8 + 5(x − 2) + 7(y − 3) SOLUTION

The linearization of f at (2, 3) is the following linear function: L(x, y) = f (2, 3) + f x (2, 3)(x − 2) + f y (2, 3)(y − 3)

That is, L(x, y) = 8 + 5(x − 2) + 7(y − 3) = −23 + 5x + 7y The function in (b) is the correct answer. 4. Estimate f (2, 3.1). SOLUTION

We use the linear approximation f (a + h, b + k) ≈ f (a, b) + f x (a, b)h + f y (a, b)k

We let (a, b) = (2, 3), h = 0, k = 3.1 − 3 = 0.1. Then, f (2, 3.1) ≈ f (2, 3) + f x (2, 3) · 0 + f y (2, 3) · 0.1 = 8 + 0 + 7 · 0.1 = 8.7 We get the estimation f (2, 3.1) ≈ 8.7. 5. Estimate  f at (2, 3) if x = −0.3 and y = 0.2. SOLUTION

The change in f can be estimated by the linear approximation as follows:  f ≈ f x (a, b)x + f y (a, b)y  f ≈ f x (2, 3) · (−0.3) + f y (2, 3) · 0.2

or  f ≈ 5 · (−0.3) + 7 · 0.2 = −0.1 The estimated change is  f ≈ −0.1. 6. Which theorem allows us to conclude that f (x, y) = x 3 y 8 is differentiable? The function f (x, y) = x 3 y 8 is a polynomial, hence f x (x, y) and f y (x, y) exist and are continuous. Therefore the Criterion for Differentiability implies that f is differentiable everywhere. SOLUTION

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Exercises 1. Let f (x, y) = x 2 y 3 . (a) Find the linearization of f at (a, b) = (2, 1). (b) Use the linear approximation to estimate f (2.01, 1.02) and f (1.97, 1.01), and compare your estimates with the values obtained using a calculator. SOLUTION

(a) We compute the value of the function and its partial derivatives at (a, b) = (2, 1): f (x, y) = x 2 y 3 f x (x, y) = 2x y 3

f (2, 1) = 4 ⇒

f y (x, y) = 3x 2 y 2

f x (2, 1) = 4 f y (2, 1) = 12

The linear approximation is therefore f (x, y) ≈ f (2, 1) + f x (2, 1)(x − 2) + f y (2, 1)(y − 1) f (x, y) ≈ 4 + 4(x − 2) + 12(y − 1) = −16 + 4x + 12y (b) For h = x − 2 and k = y − 1 we have the following form of the linear approximation at (a, b) = (2, 1): f (x, y) ≈ f (2, 1) + f x (2, 1)h + f y (2, 1)k = 4 + 4h + 12k To compute f (2.01, 1.02) we set h = 2.01 − 2 = 0.01, k = 1.02 − 1 = 0.02 to obtain f (2.01, 1.02) ≈ 4 + 4 · 0.01 + 12 · 0.02 = 4.28 The actual value is f (2.01, 1.02) = 2.012 · 1.023 = 4.2874 To compute f (1.97, 1.01) we set h = 1.97 − 2 = −0.03, k = 1.01 − 1 = 0.01 to obtain f (1.97, 1.01) ≈ 4 + 4 · (−0.03) + 12 · 0.01 = 4. The actual value is f (1.97, 1.01) = 1.972 · 1.013 = 3.998. 3. Let f (x, y) = x 3 y −4 . Use Eq. (5) to estimate the change−1  f = f (2.03, 0.9) − f (2, 1). Write the linear approximation to f (x, y) = x(1 + y) at (a, b) = (8, 1) in the form SOLUTION We compute the function and its partial derivatives at (a, b) = (2, 1): f (a + h, b + k) ≈ f (a, b) + f x (a, b)h + f y (a, b)k f (2, 1) = 8 f (x, y) = x 3 y −4 Use it to estimate 7.98/2.02 and compare the estimate with the value obtained using a calculator. ⇒ f x (2, 1) = 12 f x (x, y) = 3x 2 y −4 f y (x, y) = −4x 3 y −5

f y (2, 1) = −32

Also, x = 2.03 − 2 = 0.03 and y = 0.9 − 1 = −0.1. Therefore,  f = f (2.03, 0.9) − f (2, 1) ≈ f x (2, 1)x + f y y = 12 · 0.03 + (−32) · (−0.1) = 3.56  f ≈ 3.56 x +y at (0, 0) to estimate f (0.01, −0.02). Compare with the value 5. UseAssume the linear y) = (3, 2) = e−1, and f y (3, 2) = 3. Use the linear approximation to estimate thatapproximation f (3, 2) = 2,of ffx (x, obtained using f (3.1, 1.8).a calculator. 2

SOLUTION

The linear approximation of f at the point (0, 0) is f (h, k) ≈ f (0, 0) + f x (0, 0)h + f y (0, 0)k

We first must compute f and its partial derivative at the point (0, 0). Using the Chain Rule we obtain 2 f (x, y) = e x +y 2 f x (x, y) = 2xe x +y 2 f y (x, y) = e x +y

f (0, 0) = e0 = 1 ⇒

f x (0, 0) = 2 · 0 · e0 = 0 f y (0, 0) = e0 = 1

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We substitute these values and h = 0.01, k = −0.02 in (1) to obtain f (0.01, −0.02) ≈ 1 + 0 · 0.01 + 1 · (−0.02) = 0.98 2 The actual value is f (0.01, −0.02) = e0.01 −0.02 ≈ 0.9803.

7. Use Eq. (3) to find an equation of the tangent plane to the graph of f (x, y) = 2x 2 − 4x y 2 at the point (−1, 2). x2 Let f (x, = 2 of .the Usetangent the linear approximation at an2,appropriate point (a, b) to estimate f (4.01, 0.98). SOLUTION They)equation plane at the point (−1, 18) is y +1 z = f (−1, 2) + f x (−1, 2)(x + 1) + f y (−1, 2)(y − 2) (1) We compute the function and its partial derivatives at the point (−1, 2): f (x, y) = 2x 2 − 4x y 2

f (−1, 2) = 18

f x (x, y) = 4x − 4y 2

⇒

f x (−1, 2) = −20

f y (x, y) = −8x y

f y (−1, 2) = 16

Substituting in (1) we obtain the following equation of the tangent plane: z = 18 − 20(x + 1) + 16(y − 2) = −34 − 20x + 16y That is, z = −34 − 20x + 16y xy at the point (2, 1, 2). 9. Find the linear approximation to f (x, y, z) = Let f (x, y) = e x/y and P = (2, 1, e2 ). z (a) Find an of the tangent to plane thepoint graph (x,is:y) at P. SOLUTION Theequation linear approximation f attothe (2,of1,f2) (b) Plot f (x, y) and the tangent plane at P on the same screen to illustrate local linearity. f (x, y, z) ≈ f (2, 1, 2) + f x (2, 1, 2)(x − 2) + f y (2, 1, 2)(y − 1) + f z (2, 1, 2)(z − 2)

(1)

We compute the values of f and its partial derivatives at (2, 1, 2): xy z y f x (x, y, z) = z x f y (x, y, z) = z xy f z (x, y, z) = − 2 z

f (2, 1, 2) = 1 1 f x (2, 1, 2) = 2 f y (2, 1, 2) = 1

f (x, y, z) =

⇒

f z (2, 1, 2) = −

1 2

We substitute these values in (1) to obtain the following linear approximation: 1 1 xy ≈ 1 + (x − 2) + 1 · (y − 1) − (z − 2) z 2 2 xy 1 1 ≈ x+y− z z 2 2 In Exercises 11–18, the given point. Assume that find f (1,an 0,equation 0) = −3,off xthe (1,tangent 0, 0) =plane −2, fat y (1, 0, 0) = 4, and f z (1, 0, 0) = 2. Use the linear approximation to estimate f (1.02, 0.01, −0.03). 11. f (x, y) = x 2 y + x y 3 , (2, 1) SOLUTION

The equation of the tangent plane at (2, 1) is z = f (2, 1) + f x (2, 1)(x − 2) + f y (2, 1)(y − 1)

We compute the values of f and its partial derivatives at (2, 1): f (x, y) = x 2 y + x y 3 f x (x, y) = 2x y + y 3 f y (x, y) = x 2 + 3x y 2

f (2, 1) = 6 ⇒

f x (2, 1) = 5 f y (2, 1) = 10

We now substitute these values in (1) to obtain the following equation of the tangent plane: z = 6 + 5(x − 2) + 10(y − 1) = 5x + 10y − 14 That is, z = 5x + 10y − 14.

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13. f (x, y) = x 2 +xy −2 , (4, 1) f (x, y) = √ , (4, 4) y SOLUTION The equation of the tangent plane at (4, 1) is z = f (4, 1) + f x (4, 1)(x − 4) + f y (4, 1)(y − 1)

(1)

We compute the values of f and its partial derivatives at (4, 1): f (x, y) = x 2 + y −2 f x (x, y) = 2x

f (4, 1) = 17 ⇒

f y (x, y) = −2y −3

f x (4, 1) = 8 f y (4, 1) = −2

Substituting in (1) we obtain the following equation of the tangent plane: z = 17 + 8(x − 4) − 2(y − 1) = 8x − 2y − 13. 15. F(r, s) = r 2 s −1/2 + s −3 , π (2, 1) G(u, w) = sin(uw), ( 6 , 1) SOLUTION The equation of the tangent plane at (2, 1) is z = f (2, 1) + fr (2, 1)(r − 2) + f s (2, 1)(s − 1)

(1)

We compute f and its partial derivatives at (2, 1): f (r, s) = r 2 s −1/2 + s −3

f (2, 1) = 5

fr (r, s) = 2r s −1/2

⇒

1 f s (r, s) = − r 2 s −3/2 − 3s −4 2

fr (2, 1) = 4 f s (2, 1) = −5

We substitute these values in (1) to obtain the following equation of the tangent plane: z = 5 + 4(r − 2) − 5(s − 1) = 4r − 5s + 2. y, (0, 1) 17. f (x, y) = e x ln st H (s, t) = se , (0, 0) SOLUTION The equation of the tangent plane at (0, 1) is z = f (0, 1) + f x (0, 1)x + f y (0, 1)(y − 1) We compute the values of f and its partial derivatives at (0, 1): f (x, y) = e x ln y f x (x, y) = e x ln y f y (x, y) =

ex

f (0, 1) = 0 ⇒

f x (0, 1) = 0 f y (0, 1) = 1

y

Substituting in (1) gives the following equation of the tangent plane: z = 0 + 0x + 1(y − 1) = y − 1 That is, z = y − 1. 19. Find the points on 2the graph of z = 3x 2 − 4y 2 at which the vector n = 3, 2, 2 is normal to the tangent plane. f (x, y) = ln(x + y 2 ), (1, 1)

 SOLUTION The equation of the tangent plane at the point a, b, f (a, b) on the graph of z = f (x, y) is z = f (a, b) + f x (a, b)(x − a) + f y (a, b)(y − b) or f x (a, b)(x − a) + f y (a, b)(y − b) − z + f (a, b) = 0 Therefore, the following vector is normal to the plane:   v = f x (a, b), f y (a, b), −1 We compute the partial derivatives of the function f (x, y) = 3x 2 − 4y 2 : f x (x, y) = 6x

⇒

f x (a, b) = 6a

f y (x, y) = −8y

⇒

f y (a, b) = −8b

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Therefore, the vector v = 6a, −8b, −1 is normal to the tangent plane at (a, b). Since we want n = 3, 2, 2 to be normal to the plane, the vectors v and n must be parallel. That is, the following must hold: −8b 1 6a = =− 3 2 2 which implies that a = − 14 and b = 18 . We compute the z-coordinate of the point:   2 1 1 2 1 z =3· − −4 = 4 8 8   The point on the graph at which the vector n = 3, 2, 2 is normal to the tangent plane is − 14 , 18 , 18 . Use a computer algebra system to graph of f (x, y) = x 2 + 3x y + y together with the tangent plane 21. 2 eplot y at the which the tangent plane is parallel to 5x − 2y + 12 z = 0. Find the points on the graph of z = x at (x, y) = (1, 1) on the same screen. SOLUTION

The equation of the tangent plane at the point (1, 1) is z = f (1, 1) + f x (1, 1)(x − 1) + f y (1, 1)(y − 1) z = 5 + 5(x − 1) + 4(1, 1)(y − 1) z = 5x + 4y − 4

A sketch is shown here: 20 15 10 5 0

−1

1 3

3

23. The following values are given: Suppose that the plane tangent to the surface z = f (x, y) at (−2, 3, 4) has equation z + 4x + 2y = 2. Estimate f (−2.1, 3.1). f (1, 2) = 10, f (1.1, 2.01) = 10.3, f (1.04, 2.1) = 9.7 Find an approximation to the equation of the tangent plane to the graph of f at (1, 2, 10). SOLUTION

The equation of the tangent plane at the point (1, 2) is z = f (1, 2) + f x (1, 2)(x − 1) + f y (1, 2)(y − 2) z = 10 + f x (1, 2)(x − 1) + f y (1, 2)(y − 2)

Since the values of the partial derivatives at (1, 2) are not given, we approximate them as follows: f (1.1, 2.01) − f (1, 2) f (1.1, 2) − f (1, 2) ≈ =3 0.1 0.1 f (1, 2.1) − f (1, 2) f y (1, 2) ≈ 0.1

f x (1, 2) ≈

(1)

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359

SOLUTION The number (2.01)3 (1.02)2 is a value of the function f (x, y) = x 3 y 2 . We use the linear approximation at (2, 1), which is

f (2 + h, 1 + k) ≈ f (2, 1) + f x (2, 1)h + f y (2, 1)k

(1)

We compute the value of the function and its partial derivatives at (2, 1): f (x, y) = x 3 y 2

f (2, 1) = 8

f x (x, y) = 3x 2 y 2 f y (x, y) = 2x 3 y

⇒

f x (2, 1) = 12 f y (2, 1) = 16

Substituting these values and h = 0.01, k = 0.02 in (1) gives the approximation (2.01)3 (1.02)2 ≈ 8 + 12 · 0.01 + 16 · 0.02 = 8.44 The value given by a calculator is 8.4487. The error is 0.0087 and the percentage error is Percentage error ≈

0.0087 · 100 = 0.103% 8.4487

 2 + 3.992 3.01 4.1/7.9  SOLUTION This is a value of the function f (x, y) = x 2 + y 2 . We use the linear approximation at the point (3, 4), which is 27.

f (3 + h, 4 + k) ≈ f (3, 4) + f x (3, 4)h + f y (3, 4)k Using the chain Rule gives the following partial derivatives:  f (x, y) = x 2 + y 2 2x x =  f x (x, y) =  2 x 2 + y2 x 2 + y2 2y y =  f y (x, y) =  2 2 2 2 x +y x + y2

(1)

f (3, 4) = 5 ⇒

f x (3, 4) =

3 5

f y (3, 4) =

4 5

Substituting these values and h = 0.01, k = −0.01 in (1) gives the following approximation:  4 3 3.012 + 3.992 ≈ 5 + · 0.01 + · (−0.01) = 4.998 5 5  The value given by a calculator is 3.012 + 3.992 ≈ 4.99802. The error is 0.00002 and the percentage error is at most Percentage error ≈

0.00002 · 100 = 0.0004002% 4.99802

√

(1.9)(2.02)(4.05) 0.982 √ SOLUTION 3 We use the linear approximation of the function f (x, y, z) = x yz at the point (2, 2, 4), which is 2.01 + 1

29.

f (2 + h, 2 + k, 4 + l) ≈ f (2, 2, 4) + f x (2, 2, 4)h + f y (2, 2, 4)k + f z (2, 2, 4)l We compute the values of the function and its partial derivatives at (2, 2, 4): √ f (2, 2, 4) = 4 f (x, y, z) = x yz  1 yz yz = ⇒ f x (2, 2, 4) = 1 f x (x, y, z) = √ 2 x yz 2 x  1 xz xz = f y (2, 2, 4) = 1 f y (x, y, z) = √ 2 x yz 2 y  1 xy xy 1 = f z (2, 2, 4) = f z (x, y, z) = √ 2 x yz 2 z 2 Substituting these values and h = −0.1, k = 0.02, l = 0.05 in (1) gives the following approximation:  1 (1.9)(2.02)(4.05) = 4 + 1 · (−0.1) + 1 · 0.02 + (0.05) = 3.945 2

(1)

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The value given by a calculator is:  (1.9)(2.02)(4.05) ≈ 3.9426 31. Estimate f (2.1, 3.8) given that f (2, 4) = 5, f x (2, 4) = 0.3, and f y (2, 4) = −0.2. 8.01 √ SOLUTION We use the linear approximation of f at the point (2, 4), which is (1.99)(2.01) f (2 + h, 4 + k) ≈ f (2, 4) + f x (2, 4)h + f y (2, 4)k Substituting the given values and h = 0.1, k = −0.2 we obtain the following approximation: f (2.1, 3.8) ≈ 5 + 0.3 · 0.1 + 0.2 · 0.2 = 5.07. In Exercises 32–34, let I = W/H 2 denote the BMI described in Example 6. 33. Suppose that (W, H ) = (34, 1.3) and W increases to 35. Use the linear approximation to estimate the increase in H A boy has weight W = 34 kg and height H = 1.3. Use the linear approximation to estimate the change in I if required to keep I constant. (W, H ) changes to (36, 1.32). W SOLUTION The linear approximation of I = 2 at the point (34, 1.3) is: H

I = I (34 + h, 1.3 + k) − I (34, 1.3) ≈

∂I ∂I (34, 1.3)h + (34, 1.3)k ∂W ∂H

(1)

In the earlier exercise, we found that ∂I (34, 1.3) = 0.5917, ∂W

∂I (34, 1.3) = −30.9513 ∂H

We substitute these derivatives and h = 1 in (1), equate the resulting expression to zero and solve for k. This gives: I ≈ 0.5917 · 1 − 30.9513 · k = 0 0.5917 = 30.9513k

⇒

k = 0.0191

That is, for an increase in weight of 1 kg, the increase in height must be approximately 0.0191 meters (or 1.91 centimeters) in order to keep I constant. 35. The volume of a cylinder of radius r and height h is V = π r 2 h. (a) Show that I ≈ 0 if H/W ≈ H/2W . (a) Use the linear approximation to show that (b) Suppose that (W, H ) = (25, 1.1). What increase in H will leave I (approximately) constant if W is increased 2r h V by 1 kg? ≈ + V r h (b) Calculate the percentage increase in V if r and h are each increased by 2%. (c) The volume of a certain cylinder V is determined by measuring r and h. Which will lead to a greater error in V : a 1% error in r or a 1% error in h? SOLUTION

(a) The linear approximation is V ≈ Vr r + Vh h We compute the partial derivatives of V = π r 2 h: ∂ 2 r = 2π hr ∂r ∂ Vh = π r 2 h = π r 2 ∂h Vr = π h

Substituting in (1) gives V ≈ 2π hr r + π r 2 h We divide by V = π r 2 h to obtain V 2π hr r 2π hr r h π r 2 h π r 2 h 2r ≈ + = + + = V V V r h πr 2h πr 2h That is, 2r h V ≈ + V r h

(1)

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361

(b) The percentage increase in V is, by part (a), V r h · 100 ≈ 2 · 100 + · 100 V r h h We are given that r r · 100 = 2 and h · 100 = 2, hence the percentage increase in V is

V · 100 = 2 · 2 + 2 = 6% V (c) The percentage error in V is V r h · 100 = 2 · 100 + · 100 V r h A 1% error in r implies that r r · 100 = 1. Assuming that there is no error in h, we get V · 100 = 2 · 1 + 0 = 2% V A 1% in h implies that h h · 100 = 1. Assuming that there is no error in r , we get V · 100 = 0 + 1 = 1% V We conclude that a 1% error in r leads to a greater error in V than a 1% error in h. 37. The monthly payment for a home loan is given by a function f (P, r, N ), where P is the principal (the initial size b , then Use the linear approximation show thatof if the I =loan x a y in of the loan), r the interest rate, and Ntothe length months. Interest rates are expressed as a decimal: A 6% interest rate is denoted by r = 0.06. If P = $100,000, r = 0.06, and N = 240 (a 20-year loan), then the monthly x y I ≈ awith these + b values, we have payment is f (100,000, 0.06, 240) = 716.43. Furthermore, I x y ∂f ∂f ∂f = 0.0071, = 5,769, = −1.5467 ∂P ∂r ∂N Estimate: (a) The change in monthly payment per $1,000 increase in loan principal. (b) The change in monthly payment if the interest rate increases to r = 6.5% and r = 7%. (c) The change in monthly payment if the length of the loan increases to 24 years. SOLUTION

(a) The linear approximation to f (P, r, N ) is f ≈

∂f ∂f ∂f P + r + N ∂P ∂r ∂N

We are given that ∂∂ Pf = 0.0071, ∂∂rf = 5769, ∂∂Nf = −1.5467, and P = 1000. Assuming that r = 0 and N = 0, we get  f ≈ 0.0071 · 1000 = 7.1 The change in monthly payment per thousand dollar increase in loan principal is $7.1. (b) By the given data, we have  f ≈ 0.0071P + 5769r − 1.5467N

(1)

The interest rate 6.5% corresponds to r = 0.065, and the interest rate 7% corresponds to r = 0.07. In the first case r = 0.065 − 0.06 = 0.005 and in the second case r = 0.07 − 0.06 = 0.01. Substituting in (1), assuming that P = 0 and N = 0, gives  f = 5769 · 0.005 = $28.845  f = 5769 · 0.01 = $57.69 (c) We substitute N = (24 − 20) · 12 = 48 months and r = N = 0 in (1) to obtain  f ≈ −1.5467 · 48 = −74.2416 The monthly payment will be reduced by $74.2416. 39. The volume V of a cylinder is computed using the values 3.5 m for diameter and 6.2 m for height. Use the linear Automobile traffic passes a point P on a road of width w ft at an average rate of R vehicles per second. The approximation to estimate the maximum error in V if each of these values has a possible error of at mostRt 5%. arrival of automobiles is irregular, however, and traffic engineers have found that the formula T = te gives a good model for the average waiting time T until there is a gap in traffic of at least t seconds. A pedestrian walking at a speed of 3.5 ft/s (5.1 mph) requires t = w/3.5 s to cross the road. Therefore, the average time the pedestrian will

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SOLUTION We denote by d and h the diameter and height of the cylinder, respectively. By the Formula for the Volume of a Cylinder we have

V =π

 2 π d h = d2h 2 4

V ≈

∂V ∂V d + h ∂d ∂h

The linear approximation is (1)

We compute the partial derivatives at (d, h) = (3.5, 6.2): ∂V (d, h) = ∂d ∂V (d, h) = ∂h

π π h · 2d = hd 4 2 π 2 d 4

⇒

∂V (3.5, 6.2) ≈ 34.086 ∂d ∂V (3.5, 6.2) = 9.621 ∂h

Substituting these derivatives in (1) gives V ≈ 34.086d + 9.621h

(2)

We are given that the errors in the measurements of d and h are at most 5%. Hence, d = 0.05 3.5 h = 0.05 6.2

⇒

d = 0.175

⇒

h = 0.31

Substituting in (2) we obtain V ≈ 34.086 · 0.175 + 9.621 · 0.31 ≈ 8.948 The error in V is approximately 8.948 meters. The percentage error is at most V · 100 8.948 · 100 = 15% = π 2 V 4 · 3.5 · 6.2

Further Insights and Challenges 41. Assume that f (0) = 0. By the discussion in this section, f (x) is differentiable at x = 0 if there is a constant M such Show that if f (x, y) is differentiable at (a, b), then the function of one variable f (x, b) is differentiable at x = a.  that Use this to prove that f (x, y) = x 2 + y 2 is not differentiable at (0, 0). f (h) = Mh + h(h) 7 where lim (h) = 0 [in this case, M = f  (0)]. h→0

(a) (b) (c) √ h

Use this definition to verify that f (x) = 1/(x + 1) is differentiable with M = −1 and E(h) = h/(h + 1). Use this definition to verify that f (x) = x 3/2 is differentiable with M = 0. Show that f (x) = x 1/2 is not differentiable at x = 0 by showing that if M is any constant, and if we write = Mh + h(h), then the function (h) does not approach zero as h → 0.

SOLUTION

(a) We define the function F(x) = f (x) − 1 =

−x 1 −1= x +1 x +1

Then F(0) = 0, and F is differentiable at x = 0 if and only if f is differentiable at x = 0. We show that the definition (7) is satisfied by F. Differentiating F gives  −1 d 1 −1 = ⇒ M = F  (0) = −1 F  (x) = dx x + 1 (x + 1)2 h satisfies lim E(h) = 0, and the following equality holds: The function E(h) = h+1 h→0

Mh + h E(h) = −1 · h + h ·

−h(h + 1) + h 2 −h h = = = F(h) h+1 h+1 h+1

That is, F(x) is differentiable at x = 0, hence also f (x) is differentiable at x = 0.

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(b) We have f (0) = 0, hence f is differentiable at x = 0 if (7) holds. We compute M: f  (x) =

3 1/2 x 2

⇒

M = f  (0) = 0

We define the function E(h) by E(h) =

h 3/2 f (h) = = h 1/2 h h

Then lim E(h) = 0 and equality (7) holds, since h→0

f (h) = h E(h) = 0h + h E(h) = Mh + h E(h) (c) We show that f (x) = x 1/2 is not differentiable at x = 0, by showing that for any constant M, the function E(h) defined by Eq. (7) does not approach zero as h → 0. For f (h) = h 1/2 , Eq. (7) becomes √ h = Mh + h E(h) We solve for E(h) to obtain

√

h − Mh = h E(h) √ 1 h − Mh = √ −M E(h) = h h

As h →√0+ , E(h) is increasing without bound, therefore E(h) does not approach zero as h → 0. We conclude that f (x) = x is not differentiable at x = 0. Assumptions Matter Recall that if f (x, y) is differentiable at (a, b), then the partial derivatives f x (a, b) and f y (a, b) exist. This exercise shows that the mere existence of f x (a, b) and f y (a, b) does not imply f (x, y) is 15.5 The Gradient and Directional (ET Section 14.5) 2x y(x + y)Derivatives for (x, y)  = 0 and g(0, 0) = 0. differentiable. Define g(x, y) = 2 2 x +y Preliminary Questions (a) Prove that |g(x, y)| ≤ |x + y|. Hint: Use the inequality 1. Which of the following is a possible value of the gradient ∇ f of a function f (x, y) of two variables? (x ± y)2 ≥ 0 (a) 5 (b) 3, 4 (c) 3, 4, 5 SOLUTION The gradient 2x y of f (x, y) is a vector with two components, hence the possible value of the gradient ∇ f =  to show  ≤ 1. that 2 ∂f ∂f x + y2 ∂ x , ∂ y is (b). (b) Use (a) to show that g(x, y) is continuous at (0, 0). 2. True or false: A differentiable function increases at the rate ∇ f P  in the direction of ∇ f P ? (c) Use the limit definitions to show that gx (0, 0) and g y (0, 0) exist and both are equal to zero. SOLUTION The statement is true. The value ∇ f P  is the rate of increase of f in the direction ∇ f P . (d) Show that if a function f (x, y) is locally linear at (0, 0) and f (0, 0) = 0, then f (h, h)/ h approaches 0 as 3. Describe the two the gradient ∇ f . is not differentiable at (0, 0). h → 0. Show thatmain g(x,geometric y) does notproperties have thisof property and hence SOLUTION Thewhy gradient points in the direction of be maximum rateatof(0, increase of f since and isg(x, normal the level curve (e) Explain gx (x,ofy)fand g y (x, y) cannot both continuous 0). In fact, y) istosymmetric in x (or surface) of f partial . and y, both derivatives are discontinuous at (0, 0). (f) Show lim∂ f asgxa(x, y) does not exist byDverifying that gx (x,vector 0) = u. 0 for x  = 0 and gx (0, y) = 2 directional derivative 4. Express thedirectly partialthat derivative u f for some unit (x,y)→(0,0) ∂x for y  = 0. ∂f SOLUTION The partial derivative ∂ x is the following limit: f (a + t, b) − f (a, b) ∂f (a, b) = lim ∂x t t→0 Considering the unit vector i = 1, 0, we find that f (a + t · 1, b + t · 0) − f (a, b) ∂f (a, b) = lim = Di f (a, b) ∂x t t→0 We see that the partial derivative ∂∂ xf can be viewed as a directional derivative in the direction of i. 5. You are standing at point where the temperature gradient vector is pointing in the northeast (NE) direction. In which direction(s) should you walk to avoid a change in temperature? (a) NE (b) NW (c) SE (d) SW SOLUTION The rate of change of the temperature T at a point P in the direction of a unit vector u, is the directional derivative Du T (P), which is given by the formula

Du T (P) = ∇ f P  cos θ To avoid a change in temperature, we must choose the direction u so that Du T (P) = 0, that is, cos θ = 0, so θ = π2 or θ = 32π . Since the gradient at P is pointing NE, we should walk NW or SE to avoid a change in temperature. Thus, the answer is (b) and (c).

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NW

NE ∇T(P)

W

E

P

SE S

6. What is the rate of change of f (x, y) at (0, 0) in the direction making an angle of 45◦ with the x-axis if ∇ f (0, 0) = 2, 4? √   √ SOLUTION By the formula for directional derivatives, and using the unit vector 1/ 2, 1/ 2 , we get 2, 4 · √  √ √  √ 1/ 2, 1/ 2 = 6/ 2 = 3 2.

Exercises 1. Let f (x, y) = x y 2 and c(t) = ( 12 t 2 , t 3 ). (a) Calculate ∇ f · c (t). (b) Use the Chain Rule for Paths to evaluate

d f (c(t)) at t = 1 and t = −1. dt

SOLUTION

(a) We compute the partial derivatives of f (x, y) = x y 2 : ∂f = y2, ∂x

∂f = 2x y ∂y

The gradient vector is thus   ∇ f = y 2 , 2x y . Also, 

   1 2   3  t , t = t, 3t 2 2     ∇ f · c (t) = y 2 , 2x y · t, 3t 2 = y 2 t + 6x yt 2 . c (t) =

(b) Using the Chain Rule gives d d f (c(t)) = dt dt



1 2 6 t ·t 2

=

d dt



1 8 t 2

= 4t 7

Substituting x = 12 t 2 and y = t 3 , we obtain d 1 f (c(t)) = t 6 · t + 2 · t 2 · 3 · t 3 · t 2 = 4t 7 dt 2 At the point t = 1 and t = −1, we get d = 4 · 17 = 4, ( f (c(t))) dt t=1

d = 4 · (−1)7 = −4. ( f (c(t))) dt t=−1

3. Let f (x, y) = x 2 +x yy 2 and c(t) = (cos t, sin t). Letd f (x, y) = e and c(t) = (t 3 , 1 + t). f (c(t)) any calculations. Explain. (a) Find (a) Calculate ∇ without f and cmaking (t). dt (b) Verify your to (a)for using thetoChain Rule.d f (c(t)) as ∇ f · c (t). (b) Use the answer Chain Rule Paths calculate dt SOLUTION (c) Write out the composite f (c(t)) as a function of t and differentiate. Check that the result agrees with (b). (a) The level curves of f (x, y) are the circles x 2 + y 2 = c2 . Since c(t) is a parametrization of the unit circle, f has d f (c(t)) = 0. constant value 1 on c. That is, f (c(t)) = 1, which implies that dt

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d f (c(t)) using the Chain Rule: (b) We now find dt

∂ f dx ∂ f dy d f (c(t)) = + dt ∂ x dt ∂y dt We compute the derivatives involved in (1):  ∂f ∂  2 = x + y 2 = 2x, ∂x ∂x

(1)

 ∂  2 ∂f = x + y 2 = 2y ∂y ∂y

d dx = (cos t) = − sin t, dt dt

dy d = (sin t) = cos t dt dt

Substituting the derivatives in (1) gives d f (c(t)) = 2x(− sin t) + 2y cos t dt Finally, we substitute x = cos t and y = sin t to obtain d f (c(t)) = −2 cos t sin t + 2 sin t cos t = 0. dt In Exercises 5–8, gradient. Figure 16 calculate shows thethe level curves of a function f (x, y) and a path c(t), traversed in the direction indicated. State d 2 thecos(x derivative 5. whether f (x, y) = + y) f (c(t)) is positive, negative, or zero at points A–D. dt SOLUTION We find the partial derivatives using the Chain Rule:   ∂ ∂f = − sin x 2 + y ∂x ∂x   ∂f ∂ = − sin x 2 + y ∂y ∂y

 

   x 2 + y = −2x sin x 2 + y

   x 2 + y = − sin x 2 + y

The gradient vector is thus          ∂f ∂f , ∇f = = −2x sin x 2 + y , − sin x 2 + y = − sin x 2 + y 2x, 1 ∂ x ∂y 7. h(x, y, z) = x yz −3 x g(x, y) = 2 SOLUTION We compute x + y 2 the partial derivatives of h(x, y, z) = x yz −3 , obtaining ∂h = x z −3 , ∂y

∂h = yz −3 , ∂x

  ∂h = x y · −3z −4 = −3x yz −4 ∂z

The gradient vector is thus  ∇h =

   ∂h ∂h ∂h , , = yz −3 , x z −3 , −3x yz −4 . ∂ x ∂y ∂z

yw

r (x, y, z, w) =the x zeChain Rule to calculate In Exercises 9–20, use 9. f (x, y) = 3x − 7y, c(t) = (cos t, sin t), SOLUTION

d f (c(t)). dt

t =0

By the Chain Rule for paths, we have d f (c(t)) = ∇ f c(t) · c (t) dt

We compute the gradient and the derivative c (t):   ∂f ∂f , ∇f = = 3, −7 , ∂ x ∂y

c (t) = − sin t, cos t

We determine these vectors at t = 0: c (0) = − sin 0, cos 0 = 0, 1 and since the gradient is a constant vector, we have ∇ f c(0) = ∇ f (1,0) = 3, −7

(1)

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Substituting these vectors in (1) gives d f (c(t)) = 3, −7 · 0, 1 = 0 − 7 = −7 dt t=0 11. f (x, y) = x 2 − 3x y, c(t) = (cos2t, sin t), t = 0 f (x, y) = 3x − 7y, c(t) = (t , t 3 ), t = 2 SOLUTION By the Chain Rule For Paths we have d f (c(t)) = ∇ f c(t) · c (t) dt We compute the gradient and c (t):

 ∇f =

(1)

 ∂f ∂f , = 2x − 3y, −3x ∂ x ∂y

c (t) = − sin t, cos t At the point t = 0 we have c(0) = (cos 0, sin 0) = (1, 0) c (0) = − sin 0, cos 0 = 0, 1 = ∇ f (1,0) = 2 · 1 − 3 · 0, −3 · 1 = 2, −3 ∇ f c(0)

Substituting in (1) we obtain d f (c(t)) = 2, −3 · 0, 1 = −3 dt t=0 13. f (x, y) = sin(x2y), c(t) = (e2t , e3t ), t = 0 f (x, y) = x − 3x y, c(t) = (cos t, sin t), SOLUTION By the Chain Rule for Paths we have

t = π2

d f (c(t)) = ∇ f c(t) · c (t) dt We compute the gradient and c (t):



 ∂f ∂f , = y cos(x y), x cos(x y) ∂ x ∂y   c (t) = 2e2t , 3e3t ∇f =

At the point t = 0 we have

  c(0) = e0 , e0 = (1, 1)   c (0) = 2e0 , 3e0 = 2, 3 ∇ f c(0) = ∇ f (1,1) = cos 1, cos 1

Substituting the vectors in (1) we get d f (c(t)) = cos 1, cos 1 · 2, 3 = 5 cos 1 dt t=0 15. f (x, y) = x − x y, c(t) = (t 2 , t 2 − t4t),2t t = 4 f (x, y) = cos(y − x), c(t) = (e , e ), t = ln 3 SOLUTION We compute the gradient and c (t):   ∂f ∂f ∇f = , = 1 − y, −x ∂ x ∂y c (t) = (2t, 2t − 4) At the point t = 4 we have

  c(4) = 42 , 42 − 4 · 4 = (16, 0)

(1)

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c (4) = 2 · 4, 2 · 4 − 4 = 8, 4 ∇ f c(4) = ∇ f (16,0) = 1 − 0, −16 = 1, −16 We now use the Chain Rule for Paths to compute the following derivative: d f (c(t)) = ∇ f c(4) · c (4) = 1, −16 · 8, 4 = 8 − 64 = −56 dt t=4 17. f (x, y) = ln x +yln y, c(t) =2 (cos t, t 2 ), t = π f (x, y) = xe , c(t) = (t , t 2 − 4t), t = 04 SOLUTION We compute the gradient and c (t):     ∂f ∂f 1 1 , , ∇f = = ∂ x ∂y x y c (t) = − sin t, 2t At the point t = π4 we have

  √ π  π 2 2 π2 = cos , , = 4 4 4 2 16   √  π   π 2π 2 π  = − sin , , c = − 4 4 4 2 2   √ 16 ∇ f c π  = ∇ f  √2 π 2  = 2, 2 4 π 2 , 16 c

π 

Using the Chain Rule for Paths we obtain the following derivative:  √   π  √ 16  2 π d 8 

 f (c(t)) = , = ∇ fc π · c 2, 2 · − = −1 + ≈ 1.546 4 dt 4 2 2 π π t= π 4

19. g(x, y, z) = x yz −1 ,z c(t) = (et ,2t, t32 ), t = 1 g(x, y, z) = x ye , c(t) = (t , t , t − 1), t = 1 SOLUTION By the Chain Rule for Paths we have d g (c(t)) = ∇g c(t) · c (t) dt

(1)

We compute the gradient and c (t): 

   ∂g ∂g ∂g , , = yz −1 , x z −1 , −x yz −2 ∂ x ∂y ∂z  t   c (t) = e , 1, 2t ∇g =

At the point t = 1 we have c(1) = (e, 1, 1) c (1) = e, 1, 2 ∇gc(1) = ∇g(e,1,1) = 1, e, −e Substituting the vectors in (1) gives the following derivative: d g (c(t)) = 1, e, −e · e, 1, 2 = e + e − 2e = 0 dt t=1 In Exercises 21–30, calculate the directional derivative2 in3the direction of v at the given point. Remember to normalize g(x, y, z, w) = x + 2y + 3z + 5w, c(t) = (t , t , t, t−2), t = 1 the direction vector or use Eq. (2). 21. f (x, y) = x 2 + y 3 , v = 4, 3,

P = (1, 2)

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We first normalize the direction vector v: u=

  4, 3 4 3 v =  = , v 5 5 42 + 32

We compute the gradient of f (x, y) = x 2 + y 3 at the given point:     ∂f ∂f , ⇒ ∇f = = 2x, 3y 2 ∂ x ∂y

∇ f (1,2) = 2, 12

Using the Theorem on Evaluating Directional Derivatives, we get   4 3 44 8 36 = = 8.8 = + Du f (1, 2) = ∇ f (1,2) · u = 2, 12 · , 5 5 5 5 5 f (x, y) = x 2 y 3

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29. g(x, y, z) = xe−yz , v = 1, 1, 1, P = (1, 2, 0) g(x, y, z) = z 2 − x y 2 , v = −1, 2, 2, P = (2, 1, 3) SOLUTION We first compute a unit vector u in the direction v: u=

1, 1, 1 v 1 =  = √ 1, 1, 1 2 2 2 v 3 1 +1 +1

We find the gradient of f (x, y, z) = xe−yz at the point P = (1, 2, 0):     ∂f ∂f ∂f ∇f = , , = e−yz , −x ze−yz , −x ye−yz = e−yz 1, −x z, −x y ∂ x ∂y ∂z ∇ f (1,2,0) = e0 1, 0, −2 = 1, 0, −2 The directional derivative in the direction v is thus 1 1 1 Du f (1, 2, 0) = ∇ f (1,2,0) · u = 1, 0, −2 · √ 1, 1, 1 = √ (1 + 0 − 2) = − √ 3 3 3 31. Findg(x, the y, directional derivative x 2 +P4y=2 (2, at Pe, = z) = x ln(y + z), of v =f (x, 2i −y)j = + k, e) (3, 2) in the direction pointing to the origin. →

SOLUTION

The direction vector is v = P O= −3, −2. A unit vector u in the direction v is obtained by normalizing

v. That is, u=

−3, −2 v −1 =  = √ 3, 2 2 2 v 13 3 +2

We compute the gradient of f (x, y) = x 2 + 4y 2 at the point P = (3, 2):   ∂f ∂f , ∇f = = 2x, 8y ⇒ ∇ f (3,2) = 6, 16 ∂ x ∂y The directional derivative is thus −50 −1 Du f (3, 2) = ∇ f (3,2) · u = 6, 16 · √ 3, 2 = √ 13 13 33. A bug located at (3, 9, 4) begins walking in a straight line toward (5, 7, 3). At what rate is the bug’s temperature y−z at meters P = (3,and −2, −1) inCelsius. the direction pointing to the origin. Find thetemperature directional derivative z)?=Units x y +are z 3 in changing if the is T (x, y,of z) f=(x, xey, degrees SOLUTION The bug is walking in a straight line from the point P = (3, 9, 4) towards Q = (5, 7, 3), hence the rate of −→ change in the temperature is the directional derivative in the direction of v = P Q. We first normalize v to obtain

→

v = P Q= 5 − 3, 7 − 9, 3 − 4 = 2, −2, −1 u=

2, −2, −1 1 v = √ = 2, −2, −1 v 3 4+4+1

We compute the gradient of T (x, y, z) = xe y−z at P = (3, 9, 4):     ∂T ∂T ∂T , , ∇T = = e y−z , xe y−z , −xe y−z = e y−z 1, x, −x ∂ x ∂y ∂z ∇T(3,9,4) = e9−4 1, 3, −3 = e5 1, 3, −3 The rate of change of the bug’s temperature at the starting point P is the directional derivative Du f (P) = ∇T (3,9,4) · u = e5 1, 3, −3 ·

e5 1 2, −2, −1 = − ≈ −49.47 3 3

The answer is −49.47 degrees Celsius per meter. 35. (a) (b) (c)

x −y LetSuppose f (x, y) = = (1, 2,P−4, 4. 1). Is f increasing or decreasing at P in the direction v = 2, 1, 3? thatxe∇ f P =and Calculate ∇ f P . Find the rate of change of f in the direction ∇ f P . Find the rate of change of f in the direction of a vector making an angle of 45◦ with ∇ f P .

SOLUTION

2

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2 (a) We compute the gradient of f (x, y) = xe x −y . The partial derivatives are

  2 2 2 ∂f = 1 · e x −y + xe x −y · 2x = e x −y 1 + 2x 2 ∂x 2 ∂f = −xe x −y ∂y The gradient vector is thus  ∇f =

       2 2 2 ∂f ∂f , = e x −y 1 + 2x 2 , −xe x −y = e x −y 1 + 2x 2 , −x ∂ x ∂y

At the point P = (1, 1) we have ∇ f P = e0 1 + 2, −1 = 3, −1

⇒

∇ f P  =

 √ 32 + (−1)2 = 10

√ (b) The rate of change of f in the direction of the gradient vector is the length of the gradient, that is, ∇ f P  = 10. (c) Let ev be the unit vector making an angle of 45◦ with ∇ f P . The rate of change of f in the direction of ev is the directional derivative of f in the direction ev , which is the following dot product: Dev f (P) = ∇ f P · ev = ∇ f P ev  cos 45◦ =

√

√ 1 10 · 1 · √ = 5 ≈ 2.236 2

y −where 37. LetLet T (x,f (x, y) be the=temperature location y). Assume that ∇T 4, x + Let vector c(t) = making (t 2 , t) be path y, z) sin(x y + z)atand P = (x, (0, −1, π ). Calculate Du= f (P) u 2y. is a unit anaangle in theθ plane. Find the values of t such that ◦ = 30 with ∇ f . P

d T (c(t)) = 0 dt SOLUTION

By the Chain Rule for Paths we have d T (c(t)) = ∇Tc(t) · c (t) dt

(1)

We compute the gradient vector ∇T for x = t 2 and y = t:   ∇T = t − 4, t 2 + 2t Also c (t) = 2t, 1. Substituting in (1) gives     d T (c(t)) = t − 4, t 2 + 2t · 2t, 1 = (t − 4) · 2t + t 2 + 2t · 1 = 3t 2 − 6t dt We are asked to find the values of t such that d T (c(t)) = 3t 2 − 6t = 0 dt We solve to obtain 3t 2 − 6t = 3t (t − 2) = 0

⇒

t1 = 0,

t2 = 2

39. Find a vector normal to the surface 3z 3 + x 2 y − y 2 x = 1 at P = (1, −1, 1). Find a vector normal to the surface x 2 + y 2 − z 2 = 6 at P = (3, 1, 2). SOLUTION The gradient is normal to the level surfaces, that is ∇ f P is normal to the level surface f (x, y, z) = 3z 3 + x 2 y − y 2 x = 1. We compute the gradient vector at P = (1, −1, 1):     ∂f ∂f ∂f , , ∇f = = 2x y − y 2 , x 2 − 2yx, 9z 2 ∂ x ∂y ∂z ∇ f P = −3, 3, 9 In Exercises 41–44, find an equation of the tangent plane to the surface at the given point. y2 x2 + + z 2 = 1 where the tangent plane is normal to v = 1, 1, −2. Find the two points on the ellipsoid 9 41. x 2 + 3y 2 + 4z 2 = 20, P = (2, 2, 1) 4

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371

The equation of the tangent plane is ∇ f P · x − 2, y − 2, z − 1 = 0

(1)

We compute the gradient of f (x, y, z) = x 2 + 3y 2 + 4z 2 at P = (2, 2, 1):   ∂f ∂f ∂f , , ∇f = = 2x, 6y, 8z ∂ x ∂y ∂z At the point P we have ∇ f P = 2 · 2, 6 · 2, 8 · 1 = 4, 12, 8 Substituting in (1) we obtain the following equation of the tangent plane: 4, 12, 8 · x − 2, y − 2, z − 1 = 0 4(x − 2) + 12(y − 2) + 8(z − 1) = 0 x − 2 + 3(y − 2) + 2(z − 1) = 0 or x + 3y + 2z = 10  3 2 e y−x 2 2 3 z = 13, P = 2, 3, √ 43. x 2 + x z + 2x y + y z = 11, P = (2, e 1, 1) SOLUTION

  We compute the gradient of f (x, y, z) = x 2 + z 2 e y−x at the point P = 2, 3, √3 : e

 ∇f =



  ∂f ∂f ∂f , , = 2x − z 2 e y−x , z 2 e y−x , 2ze y−x ∂ x ∂y ∂z

  At the point P = 2, 3, √3 we have e

   √  9 3 9 ∇ f P = 4 − · e, · e, 2 · √ · e = −5, 9, 6 e e e e The equation of the tangent plane at P is   3 ∇ f P · x − 2, y − 3, z − √ = 0 e That is,  √ 3 −5(x − 2) + 9(y − 3) + 6 e z − √ =0 e or √ −5x + 9y + 6 ez = 35 45. Verify what is2 clear 4from Figure 17: Every tangent plane to the cone x 2 + y 2 − z 2 = 0 passes through the origin. ln[1 + 4x + 9y ] − 0.1z 2 = 0, P = (3, 1, 6.1876)

y x

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We compute the gradient of f (x, y, z) = x 2 + y 2 − z 2 at P:   ∂f ∂f ∂f , , = 2x, 2y, −2z ∇f = ∂ x ∂y ∂z Hence, ∇ f P = 2x 0 , 2y0 , −2z 0  Substituting in (1) we obtain the following equation of the tangent plane: 2x 0 , 2y0 , −2z 0  · x − x 0 , y − y0 , z − z 0  = 0 x 0 (x − x 0 ) + y0 (y − y0 ) − z 0 (z − z 0 ) = 0 x 0 x + y0 y − z 0 z = x 02 + y02 − z 02 Since P = (x 0 , y0 , z 0 ) is on the surface, we have x 02 + y02 − z 02 = 0. The equation of the tangent plane is thus x 0 x + y0 y − z 0 z = 0 This plane passes through the origin. 47. Find a function f (x, y, z) such that ∇ f is the constant vector 1, 3, 1. Use a computer algebra system to produce a contour plot of f (x, y) = x 2 − 3x y + y − y 2 together with its SOLUTION The gradient y, z)[−4, must4]satisfy the4]. equality gradient vector field onofthef (x, domain × [−4,   ∂f ∂f ∂f , , ∇f = = 1, 3, 1 ∂ x ∂y ∂z Equating corresponding components gives ∂f =1 ∂x ∂f =3 ∂y ∂f =1 ∂z One of the functions that satisfies these equalities is f (x, y, z) = x + 3y + z   3. 49. FindFind a function f (x,f (x, y, z)y,such thatthat ∇ f ∇=f = x, 2x, y 2 , z1, a function z) such 2. SOLUTION The following equality must hold:     ∂f ∂f ∂f , , ∇f = = x, y 2 , z 3 ∂ x ∂y ∂z That is, ∂f =x ∂x ∂f = y2 ∂y ∂f = z3 ∂z One of the functions that satisfies these equalities is f (x, y, z) =

1 2 1 3 1 4 x + y + z 2 3 4

51. Find a function f (x, y) such that ∇ f = y, x. Find a function f (x, y, z) such that ∇ f = z, 2y, x. SOLUTION We must find a function f (x, y) such that   ∂f ∂f , ∇f = = y, x ∂ x ∂y

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That is, ∂f = y, ∂x

∂f =x ∂y

We integrate the first equation with respect to x. Since y is treated as a constant, the constant of integration is a function of y. We get  f (x, y) = y d x = yx + g(y) (1) We differentiate f with respect to y and substitute in the second equation. This gives ∂ ∂f = (yx + g(y)) = x + g  (y) ∂y ∂y Hence, x + g  (y) = x

⇒

g  (y) = 0

⇒

g(y) = C

Substituting in (1) gives f (x, y) = yx + C One of the solutions is f (x, y) = yx (obtained for C = 0). 53. Let  f = f (a + h, b + k) − f (a, b) be the change in f at P = (a,  b). Set v = h, k. Show that the linear Show that does not exist a function f (x, y) such that ∇ f = y 2 , x . Hint: Use Clairaut’s Theorem f x y = approximation canthere be written f yx .  f ≈ ∇ f P · v 6 SOLUTION

The linear approximation is    f ≈ f x (a, b)h + f y (a, b)k = f x (a, b), f y (a, b) · h, k = ∇ f P · v

55. FindUse theEq. unit(6) vector n normal to the surface z 2 − 2x 4 − y 4 = 16 at P = (2, 2, 8) that points in the direction of the to estimate x y-plane.  f =to fthe (3.53, 8.98)f − (3.5, 9) z 2 − 2x 4 − y 4 = 16 at P. We find this SOLUTION The gradient vector ∇ f is normal surface (x,f y, z) = P

vector: assuming that ∇ f (3.5,9) = 2, −1.     ∂f ∂f ∂f , , ∇f = = −8x 3 , −4y 3 , 2z ∂ x ∂y ∂z

⇒

  ∇ f (2,2,8) = −8 · 23 , −4 · 23 , 2 · 8 = −64, −32, 16

We normalize to obtain a unit vector normal to the surface: −64, −32, 16 −64, −32, 16 1 ∇ fP =  = √ −4, −2, 1 = √ ∇ f P  16 21 21 2 2 2 (−64) + 32 + 16 There are two unit normals to the surface at P, namely, 1 n = ± √ −4, −2, 1 21 We need to find the normal that points in the direction of the x y-plane. Since the point P = (2, 2, 8) is above the x y-plane, the normal we need has negative z-component. Hence, 1 n = √ 4, 2, −1 21

57. (a) (b) (c) (d)

√ √  x 2that 2a particle located at the point P = (2, 2, 8) travels toward the x y-plane in −1 Suppose, in the previous exercise, and u = , Let f (x, y) = tan . y 2 2 the direction normal to the surface. Calculate the which gradient of fQ. on the x y-plane will the particle pass? (a) Through point √ f (1,axes 1) and Calculate Du the (b) Suppose are D calibrated in centimeters. Determine the path c(t) of the particle if it travels at a constant u f ( 3, 1). speed that of 8 the cm/s. How it take the level particle to reach Show lines y =long mx will for m  = 0 are curves for fQ? . Verify that ∇ f P is orthogonal to the level curve through P for P = (x, y)  = (0, 0).

SOLUTION

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(a) We compute the partial derivatives of f (x, y) = tan−1 xy . Using the Chain Rule we get ∂f = ∂x ∂f = ∂y

1 y 1  2 · y = 2 x + y2 x

1+ y

 1 x x · − =− 2  2 2 y x + y2 x

1+ y

The gradient of f is thus  ∇f =

 y x 1 y, −x , − = 2 x 2 + y2 x 2 + y2 x + y2

(b) By the Theorem on Evaluating Directional Derivatives, Du f (a, b) = ∇ f (a,b) · u

(1)

We find the values of the gradient at the two points: 1 1 1, −1 = 1, −1 ∇ f (1,1) = 2 2 1 + 12  √  √  1 1 1, − 3 1, − 3 = ∇ f √  = √ 2 3,1 4 3 + 12 Substituting in (1) we obtain the following directional derivatives √ √  2 2 1 , Du f (1, 1) = ∇ f (1,1) · u = 1, −1 · =0 2 2 2 √ √  √ √  √  √  2 2 2 1 1, − 3 · , = 1, − 3 · 1, 1 3, 1 = ∇ f √  · u = Du f 3,1 4 2 2 8 √ √  √ √  2 2− 6 1− 3 = = 8 8 (c) Note that f is not defined for y = 0. For x = 0, the level curve of f is the y-axis, and the gradient vector is  1y , 0, which is perpendicular to the y-axis. For y  = 0 and x  = 0, the level curves of f are the curves where f (x, y) is constant. That is, tan−1

x =k y x = tan k y y=

(for k  = 0)

1 x tan k

We conclude that the lines y = mx, m  = 0, are level curves for f .

  (d) By part (c), the level curve through P = (x 0 , y0 ) is the line y = xy00 x. This line has a direction vector 1, xy00 . The gradient at P is, by part (a), ∇ f P = 2 1 2 y0 , −x 0 . We verify that the two vectors are orthogonal: 

y 1, 0 x0



x0 +y0

   1 y x 0 y0 1 y  = , −x − · ∇ f P = 1, 0 · 2 y =0 0 0 x0 x + y 2 0 x0 x 02 + y02 0 0

Since the dot products is zero, the two vectors are orthogonal as expected (Theorem 6). 59. Let C be the curve of intersection of the two surfaces x 2 + y 2 + z 2 = 3 and (x − 2)2 + (y − 2)2 + z 2 = 3. Use the Suppose that the intersection of two surfaces F(x, y, z) = 0 and G(x, y, z) = 0 is a curve C and let P be a result of Exercise 58 to find parametric equations of the tangent line to C at P = (1, 1, 1). point on C. Explain why the vector v = ∇ FP × ∇G P is a direction vector for the tangent line to C at P. SOLUTION The parametric equations of the tangent line to C at P = (1, 1, 1) are x = 1 + at,

y = 1 + bt,

z = 1 + ct

(1)

where v = a, b, c is a direction vector for the line. By Exercise 58 v may be chosen as the following cross product: v = ∇ FP × ∇G P

(2)

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where F(x, y, z) = x 2 + y 2 + z 2 and G(x, y, z) = (x − 2)2 + (y − 2)2 + z 2 . We compute ∇ FP and ∇G P : Fx (x, y, z) = 2x Fy (x, y, z) = 2y

⇒

∇ F P = 2 · 1, 2 · 1, 2 · 1 = 2, 2, 2

⇒

∇G P = 2(1 − 2), 2(1 − 2), 2 · 1 = −2, −2, 2

Fz (x, y, z) = 2z G x (x, y, z) = 2(x − 2) G y (x, y, z) = 2(y − 2) G z (x, y, z) = 2z Hence,

i v = 2, 2, 2 × −2, −2, 2 = 2 −2

j 2 −2

k 2 2

= (4 + 4)i − (4 + 4)j + (−4 + 4)k = 8i − 8j = 8, −8, 0

Therefore, v = a, b, c = 8, −8, 0, yielding a = 8, b = −8, c = 0. Substituting in (1) gives the following equations of the tangent line: x = 1 + 8t, y = 1 − 8t, z = 1. 61. (a) (b)

Verify the linearity relations for gradients: 3 2 ∇( fLet + Cg)be=the ∇ fcurve + ∇gof intersection of the two surfaces x + 2x y + yz = 7 and 3x − yz = 1. Find the parametric equations of the tangent line to C at P = (1, 2, 1). ∇(c f ) = c∇ f

SOLUTION

(a) We use the linearity relations for partial derivative to write     ∇( f + g) = ( f + g) x , ( f + g) y , ( f + g)z = f x + gx , f y + g y , f z + gz     = f x , f y , f z + gx , g y , gz = ∇ f + ∇g (b) We use the linearity properties of partial derivatives to write       ∇(c f ) = (c f )x , (c f ) y , (c f )z = c f x , c f y , c f z = c f x , f y , f z = c∇ f 63. Prove the Product Rule for gradients (Theorem 1). Prove the Chain Rule for gradients (Theorem 1). SOLUTION We must show that if f (x, y, z) and g(x, y, z) are differentiable, then ∇( f g) = f ∇g + g∇ f Using the Product Rule for partial derivatives we get     ∇( f g) = ( f g) x , ( f g) y , ( f g)z = f x g + f gx , f y g + f g y , f z g + f gz         = f x g, f y g, f z g + f gx , f g y , f gz = f x , f y , f z g + f gx , g y , gz = g∇ f + f ∇g

Further Insights and Challenges 1/3 . 65. LetLet f (x, y) a=unit (x y)vector. u be Show that the directional derivative Du f is equal to the component of ∇ f along u. (a) Use the limit definition to show that f x (0, 0) = f y (0, 0) = 0. (b) Use the limit definition to show that the directional derivative Du f (0, 0) does not exist for any unit vector u other than i and j. (c) Is f differentiable at (0, 0)? SOLUTION

(a) By the limit definition and since f (0, 0) = 0, we have f (h, 0) − f (0, 0) (h · 0)1/3 − 0 0 = lim = lim =0 h h h→0 h→0 h→0 h

f x (0, 0) = lim

f (0, h) − f (0, 0) (0 · h)1/3 − 0 0 = lim = lim =0 h h h→0 h→0 h→0 h

f y (0, 0) = lim

(b) By the limit definition of the directional derivative, and for u = u 1 , u 2  a unit vector, we have  1/3 2u u −0 t 1 2 f (tu 1 , tu 2 ) − f (0, 0) u u = lim = lim 11/32 Du f (0, 0) = lim t t t→0 t→0 t→0 t This limit does not exist unless u 1 = 0 or u 2 = 0. u 1 = 0 corresponds to the unit vector j, and u 2 = 0 corresponds to the unit vector i.

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(c) If f was differentiable at (0, 0), then Du f (0, 0) would exist for any vector u. Therefore, using the result obtained in part (b), f is not differentiable at (0, 0). 67. This exercise shows that there exists a function which is not differentiable at (0, 0) even though all directional Use the definition of differentiability to show that if f (x, y) is differentiable at (0, 0) and x2 y derivatives at (0, 0) exist. Define f (x, y) = 2 for (x, y)  = 0 and f (0, 0) = 0. xf (0, + 0) y 2= f x (0, 0) = f y (0, 0) = 0 (a) Use the limit definition to show that Dv f (0, 0) exists for all vectors v. Show that f x (0, 0) = f y (0, 0) = 0. then that f is not differentiable at (0, 0) by showing that Eq. (7) does not hold. (b) Prove f (x, y) SOLUTION  =0 lim (x,y)→(0,0) x2 + y2 derivative Dv f (0, 0), we have (a) Let v  = 0 be the vector v = v1 , v2 . By the definition of the (tv1 ) tv2 −0 f (tv1 , tv2 ) − f (0, 0) (tv1 )2 +(tv2 )2 = lim Dv f (0, 0) = lim t t t→0 t→0 2

v12 v2 t 3 v 2 v2 v12 v2  1  = lim = t→0 t 3 v 2 + v 2 t→0 v 2 + v 2 v12 + v22 1 2 1 2

= lim

(1)

Therefore Dv f (0, 0) exists for all vectors v. (b) In Exercise 66 we showed that if f (x, y) is differentiable at (0, 0) and f (0, 0) = 0, then f (x, y) − f x (0, 0)x − f y (0, 0)y  =0 (x,y)→(0,0) x 2 + y2 lim

We now show that f does not satisfy the above equation. We first compute the partial derivatives f x (0, 0) and f y (0, 0). The partial derivatives f x and f y are the directional derivatives in the directions of v = 1, 0 and v = 0, 1, respectively. Substituting v1 = 1, v2 = 0 in (1) gives 12 · 0 f x (0, 0) = 2 =0 1 + 02 Substituting v1 = 0, v2 = 1 in (1) gives 02 · 1 f y (0, 0) = 2 =0 0 + 12 Also f (0, 0) = 0, therefore for (x, y)  = (0, 0) we have x2 y

− 0x − 0y f (x, y) − f x (0, 0)x − f y (0, 0)y x2 y x 2 +y 2   lim = lim = lim 3 (x,y)→(0,0) (x,y)→(0,0) (x,y)→(0,0) 2 x 2 + y2 x 2 + y2 (x + y 2 ) 2 √ We compute the limit along the line y = 3x: √ √ √ 3 √ 3 x2 y x 2 3x 3x 3x 3 = 0 lim = lim  = 3/2 = lim

3/2 = lim 3/2 3   8 (x,y)→(0,0) x→0 x→0 4x 2 x→0 8x √ 2 (x 2 + y 2 ) √ 2 3x x + along y= 3x Since this limit is not zero, f does not satisfy Eq. (7), hence f is not differentiable at (0, 0). 69. Prove the following Quotient Rule where f, g are differentiable: Prove that if f (x, y) is differentiable and ∇ f (x,y) = 0 for all (x, y), then f is constant.  f g∇ f − f ∇g ∇ = g g2 SOLUTION

∇

The Quotient Rule is valid for partial derivatives, therefore ∂f ∂g        ∂ f ∂f ∂g  g ∂ x − f ∂∂gx g ∂ y − f ∂ y g ∂z − f ∂z f ∂ f f f ∂ ∂ , , = , , = g ∂x g ∂y g ∂z g g2 g2 g2

=

 ∂f ∂f     ∂ f   ∂g f ∂g ∂g  f ∂x g ∂ x g ∂ y g ∂z g ∂f ∂f ∂f f ∂g ∂g ∂g ∂ y f ∂z , , , , − = , , , , − g2 g2 g2 g2 g2 g2 g 2 ∂ x ∂y ∂z g 2 ∂ x ∂y ∂z

f g∇ f − f ∇g g = 2 ∇ f − 2 ∇g = g g g2

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In Exercises 70–72, a path c(t) = (x(t), y(t)) follows the gradient of a function f (x, y) if the tangent vector c (t) points in the direction of ∇ f for all t. In other words, c (t) = k(t)∇ f c(t) for some positive function k(t). Note that in this case, c(t) crosses each level curve of f (x, y) at a right angle. 71. Find a path of the form c(t) = (t, g(t)) passing through (1, 2) that follows the gradient y  (t) (Figure Show 18). Hint: Use of(x(t), Variables. = that if theSeparation path c(t) = y(t)) follows the gradient of f (x, y), then  x (t)

of f (x, y) = 2x 2 + 8y 2 fy . fx

y 2 1 −2 −1 −1

x 1

2

−2

FIGURE 18 The path c(t) is orthogonal to the level curves of f (x, y) = 2x 2 + 8y 2 . SOLUTION

By the previous exercise, if c(t) = (x(t), y(t)) follows the gradient of f , then fy y  (t) dy =  = dx x (t) fx

(1)

We find the partial derivatives of f : fy =

 ∂  2 2x + 8y 2 = 16y, ∂y

fx =

 ∂  2 2x + 8y 2 = 4x ∂x

Substituting in (1) we get 16y 4y dy = = dx 4x x We solve the differential equation using separation of variables. We obtain dy dx =4 y x   dx dy =4 y x ln y = 4 ln x + c = ln x 4 + c or 4 y = eln x +c = ec x 4

Denoting k = ec , we obtain the following solution: y = kx 4 The corresponding path may be parametrized using the parameter x = t as   c(t) = t, kt 4

(2)

Since we want the path to pass through (1, 2), there must be a solution t for the equation   t, kt 4 = (1, 2) or t =1 kt 4 = 2

⇒

k · 14 = 2

⇒

k=2

Substituting in (2) we obtain the following path:   c(t) = t, 2t 4 We now show that c follows the gradient of f (x, y) = 2x 2 + 8y 2 . We have     c (t) = 1, 8t 3 and ∇ f = f x , f y = 4x, 16y

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    Therefore, ∇ f c(t) = 4t, 16 · 2t 4 = 4t, 32t 4 , so we obtain    1  1 4t, 32t 4 = ∇ f c(t) , c (t) = 1, 8t 3 = 4t 4t

t = 0

For t = 0, ∇ f c(0) = ∇ f (0,0) = 0, 0 and c (0) = 1, 0. We conclude that c follows the gradient of f for t  = 0. Find the curve y = g(x) passing through (0, 1) that crosses each level curve of f (x, y) = y sin x at a right angle. If you have a computer algebra system, graph y = g(x) together with the level curves of f .

15.6 The Chain Rule

(ET Section 14.6)

Preliminary Questions 1. Consider a function f (x, y) where x = uv and y = u/v. (a) What are the primary derivatives of f ? (b) What are the independent variables? SOLUTION

(a) The primary derivatives of f are ∂∂ xf and ∂∂ yf . (b) The independent variables are u and v, on which x and y depend. In Questions 2–4, suppose that f (u, v) = uev , where u = r s and v = r + s. 2. The composite function f (u, v) is equal to: (a) r ser +s (b) r es

(c) r ser s

SOLUTION The composite function f (u, v) is obtained by replacing u and v in the formula for f (u, v) by the corresponding functions u = r s and v = r + s. This gives

 f u(r, s), v(r, s) = u(r, s)ev(r,s) = r ser +s

Answer (a) is the correct answer. 3. What is the value of f (u, v) at (r, s) = (1, 1)? SOLUTION

We compute u = r s and v = r + s at the point (r, s) = (1, 1): u(1, 1) = 1 · 1 = 1;

Substituting in f (u, v) = uev , we get

f (u, v)

v(1, 1) = 1 + 1 = 2

(r,s)=(1,1)

4. According to the Chain Rule, (a)

∂ f ∂x ∂ f ∂x + ∂ x ∂r ∂ x ∂s

(b)

∂ f ∂y ∂ f ∂x + ∂ x ∂r ∂y ∂r

(c)

∂ f ∂s ∂ f ∂r + ∂r ∂ x ∂s ∂ x

SOLUTION

= 1 · e2 = e2 .

∂f is equal to (choose correct answer): ∂r

For a function f (x, y) where x = x(r, s) and y = y(r, s), the Chain Rule states that the partial derivative

∂f ∂r is as given in (b). That is,

∂ f ∂x ∂ f ∂y + ∂ x ∂r ∂y ∂r 5. Suppose that x, y, z are functions of the independent variables u, v, w. Given a function f (x, y, z), which of the ∂f following terms appear in the Chain Rule expression for ? ∂w ∂ f ∂x ∂ f ∂w (a) (b) ∂v ∂v ∂w ∂ x (c)

∂ f ∂z ∂z ∂w

(d)

∂ f ∂v ∂v ∂w

S E C T I O N 15.6 SOLUTION

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(ET Section 14.6)

379

∂f By the Chain Rule, the derivative ∂w is

∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f = + + ∂w ∂ x ∂w ∂y ∂w ∂z ∂w Therefore (c) is the only correct answer. 6. With notation as in the previous question, does SOLUTION

∂x ∂f appear in the Chain Rule expression for ? ∂v ∂u

The Chain Rule expression for ∂∂uf is ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f = + + ∂u ∂ x ∂u ∂y ∂u ∂z ∂u

The derivative ∂∂vx does not appear in differentiating f with respect to the independent variable u.

Exercises 1. Let f (x, y, z) = x 2 y 3 + z 4 and x = s 2 , y = st 2 , and z = s 2 t. ∂f ∂f ∂f , , . (a) Calculate the primary derivatives ∂ x ∂y ∂z ∂ x ∂y ∂z , , . (b) Calculate ∂s ∂s ∂s ∂f using the Chain Rule: (c) Compute ∂s ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = + + ∂s ∂ x ∂s ∂y ∂s ∂z ∂s Express the answer in terms of the independent variables s, t. SOLUTION

(a) The primary derivatives of f (x, y, z) = x 2 y 3 + z 4 are ∂f = 2x y 3 , ∂x

∂f = 3x 2 y 2 , ∂y

∂f = 4z 3 ∂z

(b) The partial derivatives of x, y, and z with respect to s are ∂x = 2s, ∂s

∂y = t 2, ∂s

∂z = 2st ∂s

(c) We use the Chain Rule and the partial derivatives computed in parts (a) and (b) to find the following derivative: ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = + + = 2x y 3 · 2s + 3x 2 y 2 t 2 + 4z 3 · 2st = 4x y 3 s + 3x 2 y 2 t 2 + 8z 3 st ∂s ∂ x ∂s ∂y ∂s ∂z ∂s To express the answer in terms of the independent variables s, t we substitute x = s 2 , y = st 2 , z = s 2 t. This gives ∂f 3 2 2 3 = 4s 2 (st 2 ) s + 3(s 2 ) (st 2 ) t 2 + 8(s 2 t) st = 4s 6 t 6 + 3s 6 t 6 + 8s 7 t 4 = 7s 6 t 6 + 8s 7 t 4 . ∂s In Exercises 3–10, use the Chain Rule to calculate the partial derivatives. Express the answer in terms of the independent Let f (x, y) = x cos(y) and x = u 2 + v 2 and y = u − v. variables. ∂f ∂f , . (a) the primary derivatives ∂ f Calculate ∂f ∂ ∂y2r s, z = r 2 2 2 3. , : f (x, y, z) = x y + z , x = s ,xy = ∂s ∂r ∂f (b) Use the Chain Rule to calculate . Leave the answer in terms of both the dependent and independent variables. ∂v SOLUTION We perform the following steps: ∂f Step(c) 1. Compute derivatives. primary derivatives (x,(2, y, 1). z) = x y + z 2 are at (u,ofv)f = Determinethe (x,primary y) for (u, v) = (2, The 1) and evaluate ∂v ∂f ∂f ∂f = y, = x, = 2z ∂x ∂y ∂z Step 2. Apply the Chain Rule. By the Chain Rule, ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = · + · + · ∂s ∂ x ∂s ∂y ∂s ∂z ∂s

(1)

∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = · + · + · ∂r ∂ x ∂r ∂y ∂r ∂z ∂r

(2)

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We compute the partial derivatives of x, y, z with respect to s and r : ∂x ∂y ∂z = 2s, = 2r, = 0. ∂s ∂s ∂s ∂x ∂y ∂z = 0, = 2s, = 2r. ∂r ∂r ∂r Substituting these derivatives and the primary derivatives computed in step 1 in (1) and (2), we get ∂f = y · 2s + x · 2r + 2z · 0 = 2ys + 2xr ∂s ∂f = y · 0 + x · 2s + 2z · 2r = 2xs + 4zr ∂r Step 3. Express the answer in terms of r and s. We substitute x = s 2 , y = 2r s, and z = r 2 in ∂∂sf and ∂∂rf in step 2, to obtain ∂f = 2r s · 2s + s 2 · 2r = 4r s 2 + 2r s 2 = 6r s 2 . ∂s ∂f = 2s 2 · s + 4r 2 · r = 2s 3 + 4r 3 . ∂r ∂g ∂g ,∂ f :∂g(x, y) = cos(x 2 − y 2 ), x = 2u − 3v, y = −5u + 8v f ∂u ∂v, : f (x, y, z) = x y + z 2 , x = r + s − 2t, y = 3r t, z = s 2 ∂r ∂t SOLUTION We use the following steps: 5.

Step 1. Compute the primary derivatives. The primary derivatives of g(x, y) = cos(x 2 − y 2 ) are ∂g = −2x sin(x 2 − y 2 ), ∂x

∂g = 2y sin(x 2 − y 2 ) ∂y

Step 2. Apply the Chain Rule. By the Chain Rule, ∂g ∂ x ∂g ∂y ∂g ∂x ∂y = + = −2x sin(x 2 − y 2 ) + 2y sin(x 2 − y 2 ) ∂u ∂ x ∂u ∂y ∂u ∂u ∂u ∂g ∂ x ∂g ∂y ∂x ∂y ∂g = + = −2x sin(x 2 − y 2 ) + 2y sin(x 2 − y 2 ) ∂v ∂ x ∂v ∂y ∂v ∂v ∂v We find the partial derivatives of x and y: ∂x ∂y = 2, = −5 ∂u ∂u ∂y ∂x = −3, =8 ∂v ∂v We substitute these derivatives to obtain ∂g = −4x sin(x 2 − y 2 ) − 10y sin(x 2 − y 2 ) = −(4x + 10y) sin(x 2 − y 2 ) ∂u ∂g = 6x sin(x 2 − y 2 ) + 16y sin(x 2 − y 2 ) = (6x + 16y) sin(x 2 − y 2 ) ∂v Step 3. Express the answer in terms of u and v. We substitute x = 2u − 3v, y = −5u + 8v in (1) and (2), to obtain

 ∂g = −(8u − 12v − 50u + 80v) · sin (2u − 3v)2 − (−5u + 8v)2 ∂u 

= (42u − 68v) sin 68uv − 21u 2 − 55v 2

 ∂g = (12u − 18v − 80u + 128v) · sin (2u − 3v)2 − (−5u + 8v)2 ∂v 

= (110v − 68u) sin 68uv − 21u 2 − 55v 2 ∂F ∂F , : F(u, v) =x eu+v , u = x 2 , v = x y ∂h ∂ x ∂y: h(x, y) = , x = t1 t2 , y = t12 t2 ∂t2 y SOLUTION We use the following steps: 7.

(1) (2)

S E C T I O N 15.6

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(ET Section 14.6)

381

Step 1. Compute the primary derivatives. The primary derivatives of F(u, v) = eu+v are ∂f = eu+v ∂v

∂f = eu+v , ∂u Step 2. Apply the Chain Rule. By the Chain Rule,

∂F ∂ F ∂u ∂ F ∂v ∂u ∂v = + = eu+v + eu+v = eu+v ∂x ∂u ∂ x ∂v ∂ x ∂x ∂x ∂ F ∂u ∂ F ∂v ∂u ∂v ∂F = + = eu+v + eu+v = eu+v ∂y ∂u ∂y ∂v ∂y ∂y ∂y

 

∂v ∂u + ∂x ∂x ∂u ∂v + ∂y ∂y



We compute the partial derivatives of u and v with respect to x and y: ∂v ∂u = 2x, =y ∂x ∂x ∂u ∂v = 0, =x ∂y ∂y We substitute to obtain ∂F = (2x + y)eu+v ∂x ∂F = xeu+v ∂y

(1) (2)

Step 3. Express the answer in terms of x and y. We substitute u = x 2 , v = x y in (1) and (2), obtaining 2 ∂F = (2x + y)e x +x y , ∂x

2 ∂F = xe x +x y . ∂y

∂f ∂f ,∂ f : f (r, θ ) = r sin2 θ , x = r cos θ , y = r sin θ ∂ x ∂y: f (x, y) = x 2 + y 2 , x = eu+v , y = u + v ∂u SOLUTION We use the following steps: 9.

Step 1. Compute the primary derivatives. The primary derivatives of f (r, θ ) = r sin2 θ are ∂f = sin2 θ , ∂r

∂f = r · 2 sin θ cos θ = r sin 2θ ∂θ

Step 2. Apply the Chain Rule. By the Chain Rule, ∂ f ∂r ∂f ∂f = + ∂x ∂r ∂ x ∂θ ∂f ∂ f ∂r ∂f = + ∂y ∂r ∂y ∂θ

∂θ ∂r ∂θ = sin2 θ + r sin 2θ ∂x ∂x ∂x ∂θ ∂r ∂θ = sin2 θ + r sin 2θ ∂y ∂y ∂y

(1) (2)

We compute the partial derivatives of r and θ with respect to x and y. Since r 2 = x 2 + y 2 , we have ∂r = 2x ∂x ∂r = 2y 2r ∂y

2r

⇒ ⇒

∂r x = ∂x r ∂r y = ∂y r

By the relation tan θ = xy , we get y 1 ∂θ =− 2 cos2 θ ∂ x x

⇒

y cos2 θ ∂θ =− ∂x x2

1 1 ∂θ = x cos2 θ ∂y

⇒

∂θ cos2 θ = ∂y x

Substituting these derivatives in (1) and (2) gives

 ∂f yr sin 2θ cos2 θ y cos2 θ x sin2 θ x 2 = sin θ · + r sin 2θ − − = 2 ∂x r r x x2 ∂f y sin2 θ r sin 2θ cos2 θ cos2 θ y = sin2 θ · + r sin 2θ = + ∂y r x r x

(3)

(4)

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(ET CHAPTER 14)

Step 3. Express answer in terms of x and y. We express r , cos θ , and sin 2θ in terms of x and y. Since r = we have x x cos θ = =  r x 2 + y2 sin θ =



x 2 + y2,

y y =  2 r x + y2

2x y sin 2θ = 2 sin θ cos θ = 2 x + y2 We substitute in (3) and (4) to obtain  ∂f x y2 y x 2 + y2 2x y x2 =  · 2 − · 2 · 2 2 2 2 ∂x x x +y x + y2 x 2 + y2 x + y =

x y2 3/2 (x 2 + y 2 )

−

2y 2 x 3/2 (x 2 + y 2 )

y ∂f y2 =  · 2 + ∂y x 2 + y2 x + y2 =



=

−x y 2 (x 2 + y 2 )

3/2

2x y x 2 + y2 x2 · 2 · x x + y2 x 2 + y2

y3

2x 2 y y(y 2 + 2x 2 ) + = 3/2 3/2 3/2 (x 2 + y 2 ) (x 2 + y 2 ) (x 2 + y 2 )

In Exercises use the Chain Rule to evaluate the partial derivative at the point specified. ∂ f ∂11–16, f : f (x, y, z) = x y − z 2 , x = r cos θ , y = cos2 θ , z = r , ∂f ∂ f ∂ θ ∂r and at (u, v) = (−1, −1), where f (x, y, z) = x 3 + yz 2 , x = u 2 + v, y = u + v 2 , z = uv. 11. ∂u ∂v SOLUTION

The primary derivatives of f (x, y, z) = x 3 + yz 2 are ∂f = 3x 2 , ∂x

∂f = z2, ∂y

∂f = 2yz ∂z

By the Chain Rule we have ∂x ∂y ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂z ∂f = + + = 3x 2 + z2 + 2yz ∂u ∂ x ∂u ∂y ∂u ∂z ∂u ∂u ∂u ∂u

(1)

∂f ∂x ∂y ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂z = + + = 3x 2 + z2 + 2yz ∂v ∂ x ∂v ∂y ∂v ∂z ∂v ∂v ∂v ∂v

(2)

We compute the partial derivatives of x, y, and z with respect to u and v: ∂y ∂x = 2u, = 1, ∂u ∂u ∂x ∂y = 1, = 2v, ∂v ∂v

∂z =v ∂u ∂z =u ∂v

Substituting in (1) and (2) we get ∂f = 6x 2 u + z 2 + 2yzv ∂u ∂f = 3x 2 + 2vz 2 + 2yzu ∂v

(3) (4)

We determine (x, y, z) for (u, v) = (−1, −1): x = (−1)2 − 1 = 0,

y = −1 + (−1)2 = 0,

z = (−1) · (−1) = 1.

Finally, we substitute (x, y, z) = (0, 0, 1) and (u, v) = (−1, −1) in (3), (4) to obtain the following derivatives: ∂ f = 6 · 02 · (−1) + 12 + 2 · 0 · 1 · (−1) = 1 ∂u (u,v)=(−1,−1) ∂ f = 3 · 02 + 2 · (−1) · 12 + 2 · 0 · 1 · (−1) = −2 ∂v (u,v)=(−1,−1) ∂f at (r, s) = (1, 0), where f (x, y) = ln(x y), x = 3r + 2s, y = 5r + 3s. ∂s

The Chain Rule

S E C T I O N 15.6

13.

(ET Section 14.6)

383

√ ∂g 1 at (r, θ ) = (2 2, π4 ), where g(x, y) = , x = r sin θ , y = r cos θ . ∂θ x + y2

SOLUTION

We compute the primary derivatives of g(x, y) = 1 ∂g =− , 2 ∂x (x + y 2 )

1 : x+y 2

2y ∂g =− 2 ∂y (x + y 2 )

By the Chain Rule we have ∂x ∂y ∂g ∂g ∂ x ∂g ∂y 1 2y 1 = + =− − =− 2 ∂θ 2 ∂θ 2 ∂θ ∂ x ∂θ ∂y ∂ θ 2 2 (x + y ) (x + y ) (x + y 2 )



∂y ∂x + 2y ∂θ ∂θ



We find the partial derivatives ∂∂ θx , ∂∂ θy : ∂x = r cos θ , ∂θ

∂y = −r sin θ ∂θ

Hence, ∂g r =− (cos θ − 2y sin θ ) (1) 2 ∂θ (x + y 2 ) √ √  

√

√ At the point (r, θ ) = 2 2, π4 , we have x = 2 2 sin π4 = 2 and y = 2 2 cos π4 = 2. Substituting (r, θ ) = 2 2, π4 and (x, y) = (2, 2) in (1) gives the following derivative: √  √ ∂g 4 1 π π − 2 −2 2  1 − − 4 sin = = √ √ cos = .   2 ∂ θ (r,θ )= 2√2, π 4 4 18 6 2 2 2 (2 + 2 ) 4

∂g at (u, v) = (0, 1), where g(x, y)2 = x 22 − y 2 , x2 = eu cos v, y = eu sin v. ∂g ∂u at s = 4, where g(x, y) = x − y , x = s + 1, y = 1 − 2s. ∂s SOLUTION The primary derivatives of g(x, y) = x 2 − y 2 are

15.

∂g = 2x, ∂x

∂g = −2y ∂y

By the Chain Rule we have ∂g ∂g ∂ x ∂g ∂y ∂x ∂y = · + · = 2x − 2y ∂u ∂ x ∂u ∂y ∂u ∂u ∂u

(1)

∂y We find ∂∂ux and ∂u :

∂x = eu cos v, ∂u

∂y = eu sin v ∂u

Substituting in (1) gives ∂g = 2xeu cos v − 2yeu sin v = 2eu (x cos v − y sin v) ∂u

(2)

We determine (x, y) for (u, v) = (0, 1): x = e0 cos 1 = cos 1,

y = e0 sin 1 = sin 1

Finally, we substitute (u, v) = (0, 1) and (x, y) = (cos 1, sin 1) in (2) and use the identity cos2 α − sin2 α = cos 2α , to obtain the following derivative:   ∂g = 2e0 cos2 1 − sin2 1 = 2 · cos 2 · 1 = 2 cos 2 ∂u (u,v)=(0,1) 17. The∂h temperature at a point (x, y) is T (x, y) = 20 + 0.1(x 2 − x y) (degrees Celsius). A particle moves clockwise v , uis=the at (q, r ) at = unit (3, 2), where h(u, v)How = uefast q 3particle’s , v = qr 2temperature . along the unit circle speed (1 cm/s). changing at time t = π ? ∂q SOLUTION The particle moves along the unit circle at unit speed, hence its trajectory is parametrized by the arc length parametrization of the unit circle. That is, c(t) = (cos t, sin t)

⇒

x(t) = cos t,

y(t) = sin t

384

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(ET CHAPTER 14)

2 We need to find dT dt t=π . We first compute the primary derivatives of T (x, y) = 20 + 0.1(x − x y): ∂T = 0.2x − 0.1y, ∂x

∂T = −0.1x ∂y

By the Chain Rule we have dT ∂T dx ∂T dy dx dy = + = (0.2x − 0.1y) − 0.1x dt ∂ x dt ∂y dt dt dt

(1)

We compute ddtx and ddty : dx = − sin t, dt

dy = cos t dt

Substituting in (1) gives dT = −(0.2x − 0.1y) sin t − 0.1x cos t dt

(2)

We determine (x, y) for t = π : x = cos π = −1,

y = sin π = 0

Substituting (x, y) = (−1, 0) and t = π in (2), we obtain the following derivative: dT = −(−0.2 − 0) sin π − 0.1 · (−1) cos π = 0 + 0.1 · (−1) = −0.1 dt We conclude that at time t = π , the particle’s temperature is decreasing at a rate of 0.1 degrees per second. 2 θ . Use Eq. (7) to compute ∇u2 . Then compute ∇u2 directly by observing that u(x, y) = 19. LetLet u(r,uθ= )= r 2 cos u(x, y) and let (r, θ ) be polar coordinates. Express u x and u y in terms of u r and u θ . Then show that 2 x and compare. 1 SOLUTION By Eq. (7) we have ∇u2 = u r2 + 2 u 2θ r 1 ∇u2 = u r2 + 2 u 2θ r

We compute the partial derivatives of u(r, θ ) = r 2 cos2 θ : u r = 2r cos2 θ ,

u θ = r 2 · 2 cos θ (− sin θ ) = −2r 2 cos θ sin θ

Substituting in Eq. (7) we get 1 2 2 ∇u2 = (2r cos2 θ ) + 2 (−2r 2 cos θ sin θ ) = 4r 2 cos4 θ + 4r 2 cos2 θ sin2 θ r = 4r 2 cos2 θ (cos2 θ + sin2 θ ) = 4r 2 cos2 θ That is, ∇u2 = 4r 2 cos2 θ

(1)

We now compute ∇u2 directly. We first express u(r, θ ) as a function of x and y. Since x = r cos θ , we have u(x, y) = x 2 Hence u x = 2x, u y = 0, so we obtain ∇u2 = u 2x + u 2y = (2x)2 + 02 = 4x 2 = 4(r cos θ )2 = 4r 2 cos2 θ The answer agrees with the result in (1), as expected. ∂f ∂f ∂f ∂f ∂f ∂f , , , where (ρ, θ ,fφ(x, in that terms ) are 21. Express Let xthe = sderivatives + t and y = s, − t., Show forofany differentiable function y),spherical coordinates. ∂ρ ∂θ ∂φ ∂ x ∂y ∂z  2  2 ∂f ∂f ∂f ∂f SOLUTION The spherical coordinates are − = ∂x ∂y ∂s ∂t x = ρ sin φ cos θ , y = ρ sin φ sin θ , z = ρ cos φ We apply the Chain Rule to write ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f = + + ∂ρ ∂ x ∂ρ ∂y ∂ ρ ∂z ∂ ρ

(1)

The Chain Rule

S E C T I O N 15.6

(ET Section 14.6)

385

∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = + + ∂θ ∂ x ∂θ ∂y ∂ θ ∂z ∂ θ ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = + + ∂φ ∂ x ∂φ ∂y ∂ φ ∂z ∂ φ

(2)

We use (1) to compute the partial derivatives of x, y, and z with respect to ρ, θ , and φ . This gives ∂x = −ρ sin φ sin θ , ∂θ ∂x = ρ cos φ cos θ , ∂φ

∂y = ρ sin φ cos θ , ∂θ ∂y = ρ cos φ sin θ , ∂φ

∂z =0 ∂θ ∂z = −ρ sin φ ∂φ

∂x = sin φ cos θ , ∂ρ

∂y = sin φ sin θ , ∂ρ

∂z = cos φ ∂ρ

Substituting these derivatives in (2), we get ∂f ∂f ∂f ∂f + (sin φ sin θ ) + (cos φ ) = (sin φ cos θ ) ∂ρ ∂x ∂y ∂z ∂f ∂f ∂f ∂f + (ρ cos φ sin θ ) − (ρ sin φ ) = (ρ cos φ cos θ ) ∂φ ∂x ∂y ∂z ∂f ∂f ∂f + (ρ sin φ cos θ ) = (−ρ sin φ sin θ ) ∂θ ∂x ∂y ∂z ∂z 4 z2 x 2 − 23. Calculate and is defined at the points (3, 2,as1)a and (3, 2,of−1), where is defined implicitly by = thex equation Suppose implicitly function x and y byzthe equation F(x, y, z) z 2 + y 2 z z+ + xy − 1 = ∂ xthat z ∂y y − 0. 8 = 0. (a) Calculate Fx , Fy , Fz . 4 SOLUTION For F(x, y, z) = z + z 2 x 2 − y − 8 = 0, we use the following equalities, (Eq. (6)): ∂z ∂z and . (b) Use Eq. (6) to calculate Fy Fx ∂z ∂x ∂y ∂z =− , =− (1) ∂x Fz ∂y Fz The partial derivatives of F are Fx = 2z 2 x,

Fy = −1,

Fz = 4z 3 + 2zx 2

Substituting in (1) gives 2z 2 x ∂z zx =− 3 =− 2 2 ∂x 4z + 2zx 2z + x 2 ∂z 1 = 3 ∂y 4z + 2zx 2 At the point (3, 2, 1), we have ∂z 1·3 3 =− =− , 2 2 ∂x 11 2·1 +3 (3,2,1)

At the point (3, 2, −1), we have

∂z 1 1 = = 3 2 ∂y (3,2,1) 22 4·1 +2·1·3

∂z −3 3 =− = ∂ x (3,2,−1) 11 2 · (−1)2 + 32 1 1 ∂z = =− 3 2 ∂y (3,2,−1) 22 4 · (−1) + 2 · (−1) · 3

In Exercises 24–29, calculate the derivative using implicit differentiation. ∂w , x 2 w + w 3 + wz 2 + 3yz = 0 ∂z ∂z , x 2 y + y 2 z + x z 2 = 10 ∂x SOLUTION We find the partial derivatives Fw and Fz of

25.

F(x, w, z) = x 2 w + w 3 + wz 2 + 3yz Fw = x 2 + 3w 2 + z 2 ,

Fz = 2wz + 3y

386

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(ET CHAPTER 14)

Using Eq. (6) we get ∂w Fz 2wz + 3y =− =− 2 . ∂z Fw x + 3w 2 + z 2 ∂t ∂r and x, y r 2 = te s/r ∂z ∂t , ∂re + sin(x z) + y = 0 ∂y SOLUTION We use the formulas obtained by implicit differentiation of F(r, s, t) = r 2 − tes/r (Eq. (6)):

27.

∂r Ft =− , ∂t Fr

Fr ∂t =− ∂r Ft

(1)

The partial derivatives of F are  Fr = 2r − tes/r

s − 2 r



st = 2r + 2 es/r r

Ft = −es/r Substituting in (1) gives ∂r es/r r 2 es/r = = 3 st s/r ∂t 2r + stes/r 2r + 2 e r

2r + st2 es/r ∂t 2r 3 + stes/r st r = = = 2r e−s/r + 2 s/r 2 s/r ∂r e r e r ∂T ∂U and , 1 (T U − V )21 ln(W − U V ) = 1 at (T, U, V , W ) = (1, 1, 2, 4) ∂w ∂T , ∂U 2 + 2 = 1 at (x, y, w) = (1, 1, 1) ∂y w + x2 w + y2 SOLUTION Using the formulas obtained by implicit differentiation (Eq. (6)) we have,

29.

∂U FT =− , ∂T FU

FU ∂T =− ∂U FT

(1)

We compute the partial derivatives of F(T, U, V , W ) = (T U − V )2 ln(W − U V ) − 1: FT = 2U (T U − V ) ln(W − U V ) FU = 2T (T U − V ) ln(W − U V ) + (T U − V )2 ·

 V (T U − V ) −V = (T U − V ) 2T ln(W − U V ) − W − UV W − UV

At the point (T, U, V , W ) = (1, 1, 2, 4) we have FT = 2(1 − 2) ln(4 − 2) = −2 ln 2  2(1 − 2) FU = (1 − 2) 2 ln(4 − 2) − = (−2 ln 2 − 1) = −1 − 2 ln 2 4−2 Substituting in (1) we obtain 2 ln 2 ∂U , =− ∂ T (1,1,2,4) 1 + 2 ln 2

∂ T 1 + 2 ln 2 . =− ∂U (1,1,2,4) 2 ln 2

 31. LetThe f (x,pressure y, z) = P, F(rvolume ), where = temperature x 2 + y 2 + zT2 .of Show V r, and a vanthat der Waals gas with n molecules (n constant) are related by the equation 8 ∇ f = F  (r )er

 an 2 r P + 2 (V − nb) = n RT and r = x, y, z. where er = V r The gradient of f is the following ∂ Pvector:∂ V and  . where a, b, and R are constant. Calculate ∂T ∂ P∂ f ∂ f ∂ f  ∇f = , , ∂ x ∂y ∂z

SOLUTION

We must express this vector in terms of r and r . Using the Chain Rule, we have ∂f ∂r 2x x = F  (r ) = F  (r ) ·  = F  (r ) · 2 2 2 ∂x ∂x r 2 x +y +z

S E C T I O N 15.6

The Chain Rule

(ET Section 14.6)

387

∂f ∂r 2y y = F  (r ) = F  (r ) ·  = F  (r ) · ∂y ∂y r 2 x 2 + y2 + z2 ∂f ∂r 2z z = F  (r ) = F  (r ) ·  = F  (r ) · 2 2 2 ∂z ∂z r 2 x +y +z Hence,  x y z r F  (r ) x, y, z = F  (r ) = F  (r )er ∇ f = F  (r ) , F  (r ) , F  (r ) = r r r r r  1 2 −y 2 −z 2and ∇(ln 2 ). 33. UseLet Eq.f(8) to compute ∇ −x (x, y, z) = e = e−r ,r with r as in Exercise 31. Compute ∇ f directly and using Eq. (8). r

1 1 SOLUTION To compute ∇ r using Eq. (8), we let F(r ) = r : 1 F  (r ) = − 2 r We obtain

 1 1 1 r = − 3r ∇ = F  (r )er = − 2 · r r r r

To compute ∇(ln r ) we let F(r ) = ln r , hence F  (r ) = r1 . Thus, ∇(ln r ) = F  (r )er =

r 1 r · = 2 r r r

35. Jessica and Matthew are running toward the point P along the straight paths that make a fixed angle of θ (Figure Show that if f (x) is differentiable and c  = 0 is a constant, then u(x, t) = f (x − ct) satisfies the so-called 3). Suppose that Matthew runs with velocity va m/s and Jessica with velocity vb m/s. Let f (x, y) be the distance from advection equation Matthew to Jessica when Matthew is x meters from P and Jessica is y meters from P.  ∂u ∂u (a) Show that f (x, y) = x 2 + y 2 − 2x y cos θ . +c =0 ∂t the ∂ xrate at which the distance between Matthew and Jessica (b) Assume that θ = π /3. Use the Chain Rule to determine is changing when x = 30, y = 20, va = 4 m/s, and vb = 3 m/s. P

va vb

x y

A B

FIGURE 3 SOLUTION

(a) This is a simple application of the Law of Cosines. Connect points A and B in the diagram to form a line segment 2 2 2 that we  will call f . Then, the Law of Cosines says that f = x + y − 2x y cos θ . By taking square roots, we find that f = x 2 + y 2 − 2x y cos θ . (b) Using the chain rule, df ∂ f dx ∂ f dy = + dt ∂ x dt ∂y dt so we get (x − y cos θ )d x/dt df (y − x cos θ )d y/dt =  + 2 2 dt x + y − 2x y cos θ x 2 + y 2 − 2x y cos θ and using x = 30, y = 20, and d x/dt = 4, d y/dt = 3, we get 180 − 170 cos θ df = √ dt 1300 − 1200 cos θ

388

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(ET CHAPTER 14)

Further Insights and Challenges In Exercises 37–40, a function f (x, y, z)2 is called homogeneous of degree n if f (λ x, λ y, λ z) = λ n f (x, y, z) for all The Law of Cosines states that c = a 2 + b2 − 2ab cos θ , where a, b, c are the sides of a triangle and θ is the λ ∈ R. angle opposite the side of length c. ∂θ ∂ θ ∂ θ their degree. 37. Show thatimplicit the following functions homogeneous and determine , , and . (a) Use differentiation to are compute the derivatives 2 ∂a ∂b (a) f (x, y, z) = x y + x yz (b) f (x, y, z) =∂c3x + 2y − 8z (b) Suppose that  a = 10, b = 16, c = 22. Estimate the change in θ if a and b are increased by 1 and c is increased xy 2. y, z) = ln (d) f (x, y, z) = z 4 (c) by f (x, z2 SOLUTION

(a) For f (x, y, z) = x 2 y + x yz we have f (λ x, λ y, λ z) = (λ x)2 (λ y) + (λ x)(λ y)(λ z) = λ 3 x 2 y + λ 3 x yz = λ 3 (x 2 y + x yz) = λ 3 f (x, y, z) Hence, f is homogeneous of degree 3. (b) For f (x, y, z) = 3x + 2y − 8z we have f (λ x, λ y, λ z) = 3(λ x) + 2(λ y) − 8(λ z) = λ (3x + 2y − 8z) = λ f (x, y, z) Hence, f is homogeneous of degree 1.   (c) For f (x, y, z) = ln x 2y we have, for λ  = 0, z

 f (λ x, λ y, λ z) = ln

(λ x)(λ y)



(λ z)2

λ 2x y = ln λ 2 z2



 = ln

xy z2

= f (x, y, z) = λ 0 f (x, y, z)

Thus, f is homogeneous of degree 0. (d) For f (z) = z 4 we have f (λ z) = (λ z)4 = λ 4 z 4 = λ 4 f (z) Hence, f is homogeneous of degree 4. 39. Prove that if f (x, y, z) is homogeneous of degree n, then Prove that if f (x, y, z) is homogeneous of degree n, then f x (x, y, z) is homogeneous of degree n − 1. Hint: Either use the limit definition or apply the Chain ∂ f Rule ∂ f to f (∂λ fx, λ y, λ z). x +y +z = nf 9 ∂x ∂y ∂z Hint: Let F(t) = f (t x, t y, t z) and calculate F  (1) using the Chain Rule. SOLUTION

We use the Chain Rule to differentiate the function F(t) = f (t x, t y, t z) with respect to t. This gives F  (t) =

∂f ∂f ∂f ∂ f ∂(t x) ∂ f ∂(t y) ∂ f ∂(t z) · + · + · =x +y +z ∂x ∂t ∂y ∂t ∂z ∂t ∂x ∂y ∂z

(1)

On the other hand, since f is homogeneous of degree n, we have F(t) = f (t x, t y, t z) = t n f (x, y, z) Differentiating with respect to t we get F  (t) = nt n−1 f (x, y, z)

(2)

By (1) and (2) we obtain x

∂f ∂f ∂f +y +z = nt n−1 f (x, y, z) ∂x ∂y ∂z

Substituting t = 1 gives x

∂f ∂f ∂f +y +z = nf ∂x ∂y ∂z

41. Suppose that x = g(t, s), y = h(t, s). Show that f tt is equal to Verify Eq. (9) for the functions in Exercise 37.  2    2 ∂x ∂x ∂y ∂y ∂2x ∂2 y fx x + 2 fx y + fx 2 + f y 2 + f yy ∂t ∂t ∂t ∂t ∂t ∂t

10

S E C T I O N 15.6

The Chain Rule

(ET Section 14.6)

389

We are given that x = g(t, s), y = h(t, s). We must compute f tt for a function f (x, y). We first compute f t using the Chain Rule:

SOLUTION

ft = f x

∂x ∂y + fy ∂t ∂t

To find f tt we differentiate the two sides with respect to t using the Product Rule. This gives f tt =

∂2x ∂2 y ∂ ∂x ∂ ∂y ( fx ) + fx 2 + ( f y ) + fy 2 ∂t ∂t ∂t ∂t ∂t ∂t

(1)

By the Chain Rule, ∂ ∂x ∂y ( fx ) = fx x + fx y ∂t ∂t ∂t ∂x ∂y ∂ ( f y ) = f yx + f yy ∂t ∂t ∂t Substituting in (1) we obtain 

 ∂x ∂y ∂ x ∂2x ∂x ∂y ∂y ∂2 y + fx y + f x 2 + f yx + f yy + fy 2 ∂t ∂t ∂t ∂t ∂t ∂t ∂t ∂t  2     2  ∂x ∂y ∂x ∂y ∂2x ∂2 y ∂x ∂y = fx x + fx y + fy 2 + f x 2 + f yx + f yy ∂t ∂t ∂t ∂t ∂t ∂t ∂t ∂t

f tt =

fx x

If f x y and f yx are continuous, Clairaut’s Theorem implies that f x y = f yx . Hence,  f tt = f x x

   2 ∂x 2 ∂x ∂y ∂y ∂2x ∂2 y + 2 fx y + fx 2 + f y 2 + f yy ∂t ∂t ∂t ∂t ∂t ∂t

43. Prove that ifg(r ) is a function of r as in Exercise 42, then Let r = x 12 + · · · + x n2 and let g(r ) be a function of r . Prove the formulas ∂2g ∂2g n−1 gr + · · · + 2 = grr + 2 2 r r2 − x2 ∂ x∂n2 g xi ∂g ∂ x 1 xi i = gr , = 2 grr + gr ∂ xi r r r3 ∂ xi2 SOLUTION In Exercise 42 we showed that ∂2g ∂ xi 2

x2 r 2 − xi2 = i2 grr + gr r r3

Hence, ∂2g ∂ xi 2

+ ··· +

∂2g ∂ xn 2

=

x 12 r2

grr +

r 2 − x 12 r3

 gr

+ ··· +

x n2 r 2 − x n2 grr + gr 2 r r3



 x 2 + · · · + x n2 1  2 2 ) + · · · + (r 2 − x 2 ) g + g − x (r = 1 rr r n 1 r2 r3   2 r 1 = 2 grr + 3 gr nr 2 − (x 12 + · · · + x n2 ) r r 1 r2 n−1 gr = grr + 3 gr (nr 2 − r 2 ) = grr + 3 gr (n − 1) = grr + r r r In Exercises 44–48, the Laplace operator is defined by  f = f x x + f yy . A function f (x, y) satisfying the Laplace  equation  f = 0 is called harmonic. A function f (x, y) is called radial if f (x, y) = g(r ), where r = x 2 + y 2 . 45. Use Eq. (11) to show that f (x, y) = ln r is harmonic. Use Eq. (10) to prove that in polar coordinates (r, θ ), SOLUTION We must show that f (r, θ ) = ln r satisfies

1 1  f = frr + 2 f θθ + fr 1 r 1 r  f = frr + 2 f θθ + fr = 0 r r

We compute the derivatives of f (r, θ ) = ln r : fr =

1 , r

1 frr = − 2 , r

f θ = 0,

f θθ = 0

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Hence, 1 1 1 1 1 1 1 1  f = frr + 2 f θθ + fr = − 2 + 2 · 0 + · = − 2 + 2 = 0 r r r r r r r r Since  f = 0, f is harmonic. −1 y is harmonic using both the rectangular and polar expressions for  f . 47. Verify thatthat f (x,f (x, y) = Verify y)tan = x and x f (x, y) = y are harmonic using both the rectangular and polar expressions for  f . SOLUTION

(a) Using the rectangular expression for  f :  f = f x x + f yy

 We compute the partial derivatives of f (x, y) = tan−1 xy . Using the Chain Rule we get  1 y y fx = · − =− 2

y 2 2 x x + y2 1+ x x 1 1

y 2 · x = 2 x + y2 1+

fy =

x

fx x = −

f yy =

−y x 2 + y2 −x

x 2 + y2

2 · 2x =

2 · 2y =

2x y x 2 + y2

−2x y x 2 + y2

2

2

Hence, f x x + f yy =

2x y 2 (x 2 + y 2 )

−

2x y (x 2 + y 2 )

2

=0

(b) Using the polar expression for  f , 1 1  f = frr + 2 f θθ + fr r r

y y r sin θ −1 −1 Since x = r cos θ = tan θ , we have f (x, y) = tan x = tan (θ ) = θ . We compute the partial derivatives: fr = 0,

f θ = 1,

frr = 0,

(1)

f θθ = 0.

Substituting in (1), we get 1 1 f = 0 + 2 · 0 + · 0 = 0 r r 49. Figure 4 shows the graph of the equation Use the Product Rule to show that F(x, y, z) = x 2 + y 2 − z 2 − 12x− 8z − 4 = 0 ∂ 1 ∂f frr + fr = r −1 r r of x and∂ry. This ∂r gives two formulas, depending on the choice (a) Use the quadratic formula to solve for z as a function of a sign. Use this formula to show that if f is a radial harmonic function, then r fr = C for some constant C. Conclude that (b) Which formula defines the portion of the surface satisfying z ≥ −4? Which formula defines the portion satisfying f (x, y) = C ln r + b for some constant b. z ≤ −4? ∂z using the formula z = f (x, y) (for both choices of sign) and again via implicit differentiation. Verify (c) Calculate ∂x that the two answers agree. z

y z = −4 x

FIGURE 4 Graph of x 2 + y 2 − z 2 − 12x − 8z − 4 = 0.

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

391

SOLUTION

(a) We rewrite F(x, y, z) = 0 as a quadratic equation in the variable z:   z 2 + 8z + 4 + 12x − x 2 − y 2 = 0 We solve for z. The discriminant is     82 − 4 4 + 12x − x 2 − y 2 = 4x 2 + 4y 2 − 48x + 48 = 4 x 2 + y 2 − 12x + 12 Hence, z 1,2 =

−8 ±



 4 x 2 + y 2 − 12x + 12 2

We obtain two functions: z = −4 +



x 2 + y 2 − 12x + 12,

= −4 ±

z = −4 −





x 2 + y 2 − 12x + 12

x 2 + y 2 − 12x + 12

(b) The formula with the positive root defines the portion of the surface satisfying z ≥ −4, and the formula with the negative root defines the portion satisfying z ≤ −4.  (c) Differentiating z = −4 + x 2 + y 2 − 12x + 12 with respect to x, using the Chain Rule, gives 2x − 12 x −6 ∂z =  =  2 2 2 ∂x 2 x + y − 12x + 12 x + y 2 − 12x + 12

(1)

Alternatively, using the formula for ∂∂zx obtained by implicit differentiation gives ∂z Fx =− ∂x Fz

(2)

We find the partial derivatives of F(x, y, z) = x 2 + y 2 − z 2 − 12x − 8z − 4: Fx = 2x − 12,

Fz = −2z − 8

Substituting in (2) gives 2x − 12 x −6 ∂z =− = ∂x −2z − 8 z+4  This result is the same as the result in (1), since z = −4 + x 2 + y 2 − 12x + 12 implies that  x 2 + y 2 − 12x + 12 = z + 4  For z = −4 − x 2 + y 2 − 12x + 12, differentiating with respect to x gives 2x − 12 ∂z x −6 x −6 =−  =  = 2 2 2 2 ∂x z+4 2 x + y − 12x + 12 − x + y − 12x + 12 which is equal to − FFx computed above. z



∂z ∂z to in place of ∂x y ∂x 15.7indicate Optimization in Severalz Variables (ET Section that in the differentiation, is treated as a function of x with14.7) y held constant (and similarly for the other variables). Preliminary (a) Use Eq.Questions (6) to prove the cyclic relation 2 2 1. The functions f (x, y) = x 2 + y 2 and g(x,  have  y) = x − y both a critical point at (0, 0). How is the behavior of ∂ x ∂y ∂z the two functions at the critical point different? = −1 ∂x ∂y ∂z x SOLUTION Let f (x, y) = x 2 + y 2 and g(x, y) = xy 2 − y 2 . zIn the domain R2 , the partial derivatives of f and g are When x, y, and z are related by an equation F(x, y, z) = 0, we sometimes write

(b) Verify Eq. (12) for F(x,f xy,=z)2x, = x +f xyx + = z2,= 0.f y = 2y, f yy = 2, f x y = 0 (c) Verify the cyclic relation for the variables P, V , T in the ideal gas law P V − n RT = 0 (n and R are constants). gx = 2x, gx x = 2, g y = −2y, g yy = −2, gx y = 0 Therefore, f x = f y = 0 at (0, 0) and gx = g y = 0 at (0, 0). That is, the two functions have one critical point, which is the origin. Since the discriminant of f is D = 4 > 0, f x x > 0, and the discriminant of g is D = −4 < 0, f has a local minimum (which is also a global minimum) at the origin, whereas g has a saddle point there. Moreover, since lim g(0, y) = −∞ and lim g(x, 0) = ∞, g does not have global extrema on the plane. Similarly, f does not have a y→∞

x→∞

global maximum but does have a global minimum, which is f (0, 0) = 0.

392

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(ET CHAPTER 14)

2. Identify the points indicated in the contour maps as local minima, maxima, saddle points, or neither (Figure 14). 3 −3

1 −1

0

−1

−3 −1

−3

1 3

0

−10

10

−6

1 3

6 −2

1

2

0

FIGURE 14 SOLUTION If f (P) is a local minimum or maximum, then the nearby level curves are closed curves encircling P. In Figure (C), f increases in all directions emanating from P and decreases in all directions emanating from Q. Hence, f has a local minimum at P and local maximum at Q.

P −10

Q 10

−6

6 −2

2 0

In Figure (A), the level curves through the point R consist of two intersecting lines that divide the neighborhood near R into four regions. f is decreasing in some directions and increasing in other directions. Therefore, R is a saddle point.

3 −3

1 −1 R

0

−1

−3

1 3 1

Figure (A) Point S in Figure (B) is neither a local extremum nor a saddle point of f .

−3 −1

S

0

1

3

Figure (B) 3. Let f (x, y) be a continuous function on a domain D in R2 . Determine which of the following statements are true: (a) If D is closed and bounded, then f takes on a maximum value on D. (b) If D is neither closed nor bounded, then f does not take on a maximum value of D. (c) f (x, y) need not have a maximum value on D = {(x, y) : 0 ≤ x, y ≤ 1}. (d) A continuous function takes on neither a minimum nor a maximum value on the open quadrant {(x, y) : x > 0, y > 0}. SOLUTION

(a) This statement is true. It follows by the Theorem on Existence of Global Extrema. (b) The statement is false. Consider the constant function f (x, y) = 2 in the following domain:

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

393

y

x 1

D = {(x, y) : 0 < x ≤ 1, 0 ≤ y < ∞} Obviously f is continuous and D is neither closed nor bounded. However, f takes on a maximum value (which is 2) on D. (c) The domain D = {(x, y) : 0 ≤ x, y ≤ 1} is the following rectangle: y

1

x 1

D = {(x, y) : 0 ≤ x, y ≤ 1} D is closed and bounded, hence f takes on a maximum value on D. Thus the statement is false. (d) The statement is false. The constant function f (x, y) = c takes on minimum and maximum values on the open quadrant.

Exercises 1. Let P = (a, b) be a critical point of f (x, y) = x 2 + y 4 − 4x y. √ √ (a) First use f x (, x, y) = 0 to show that a = 2b. Then use f y (x, y) = 0 to show that P = (0, 0), (2 2, 2), or √ √ (−2 2, − 2). (b) Referring to Figure 15, determine the local minima and saddle points of f (x, y) and find the absolute minimum value of f (x, y). z

y x

FIGURE 15 SOLUTION

(a) We find the partial derivatives: ∂ ∂x ∂ f y (x, y) = ∂y f x (x, y) =

 

 x 2 + y 4 − 4x y = 2x − 4y

 x 2 + y 4 − 4x y = 4y 3 − 4x

Since P = (a, b) is a critical point, f x (a, b) = 0. That is, 2a − 4b = 0

⇒

a = 2b

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(ET CHAPTER 14)

Also f y (a, b) = 0, hence, 4b3 − 4a = 0

a = b3

⇒

We obtain the following equations for the critical points (a, b):  a = 2b a = b3 Equating the two equations, we get 2b = b3 b3 − 2b = b(b2 − 2) = 0

⇒

⎧ ⎨ b1 = 0 √ b2 = √ 2 ⎩ b3 = − 2

√ √ Since a = 2b, we have a1 = 0, a2 = 2 2, a3 = −2 2. The critical points are thus  √ √   √ √  P1 = (0, 0), P2 = 2 2, 2 , P3 = −2 2, − 2  √ √   √ √  (b) Referring to Figure 14, we see that P1 = (0, 0) is a saddle point and P2 = 2 2, 2 , P3 = −2 2, − 2 are local minima. The absolute minimum value of f is −4. 3. Find the critical points of Find the critical points of the functions f (x, y) = 8y 4 + x 2 + x y − 3y 2 − y 3 f (x, y) = x 2 + 2y 2 − 4y + 6x, g(x, y) = x 2 − 12x y + y Use the contour map in Figure 17 to determine their nature (minimum, maximum, saddle point). Use the Second Derivative Test to determine the local minimum, maximum, and saddle points. Match f (x, y) and 1 g(x, y) with their graphs in Figure 16. −0.1 −0.2 −0.3

0.5 0 −0.5 −1

0

0.1 0.2 0.3

−1

−0.5

0

0.5

1

FIGURE 17 Contour map of f (x, y) = 8y 4 + x 2 + x y − 3y 2 − y 3 . SOLUTION

The critical points are the solutions of f x = 0 and f y = 0. That is, f x (x, y) = 2x + y = 0 f y (x, y) = 32y 3 + x − 6y − 3y 2 = 0

The first equation gives y = −2x. We substitute in the second equation and solve for x. This gives 32(−2x)3 + x − 6(−2x) − 3(−2x)2 = 0 −256x 3 + 13x − 12x 2 = 0 −x(256x 2 + 12x − 13) = 0 Hence x = 0 or 256x 2 + 12x − 13 = 0. Solving the quadratic,  −12 ± 122 − 4 · 256 · (−13) −12 ± 116 = x 1,2 = 512 512

⇒

x=

13 64

or

−

1 4

Substituting in y = −2x gives the y-coordinates of the critical points. The critical points are thus   13 13 1 1 ,− (0, 0), , − , 64 32 4 2 We now use the contour map to determine the type of each critical point. The level curves through (0, 0) consist of two intersecting lines that divide the neighborhood near (0, 0) into four regions. The function is decreasing in they direction  , − 13 and increasing in the x-direction. Therefore, (0, 0) is a saddle point. The level curves near the critical points 13 64 32   and − 14 , 12 are closed curves encircling the points, hence these are local minima or maxima. The graph shows that     13 , − 13 is a local maximum and − 1 , 1 is a local minimum. 64 32 4 2

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

395

In Exercises 5–20, find the critical points of the function. Then use the Second Derivative Test to determine whether they Let f (x, y) = y 2 x − yx 2 + x y. are local minima or maxima (or state that the test fails). (a) Show that the critical points (x, y) satisfy the equations 5. f (x, y) = x 2 + y 2 − x y + x y(y − 2x + 1) = 0, x(2y − x + 1) = 0 SOLUTION

Showthe thatcritical there are threeWe critical points with either x= 0 or y = 0of(or both) x, to y = 0. 2 +critical Step(b) 1. Find points. set the first-order partial derivatives f (x, y) and = xone y 2 − xpoint y + xwith equal zero (c) Use the second derivative to determine the nature of the critical points. and solve: f x (x, y) = 2x − y + 1 = 0

(1)

f y (x, y) = 2y − x = 0

(2)

Equation (2) implies that x = 2y. Substituting in (1) and solving for y gives 1 3     The corresponding value of x is x = 2 · − 13 = − 23 . The critical point is − 23 , − 13 . 2 · 2y − y + 1 = 0

⇒

3y = −1

⇒

y=−

Step 2. Compute the Discriminant. We find the second-order partials: f x x (x, y) = 2,

f yy (x, y) = 2,

f x y (x, y) = −1

The discriminant is D(x, y) = f x x f yy − f x2y = 2 · 2 − (−1)2 = 3 Step 3. Applying the Second Derivative Test. We have   2 1 1 2 D − ,− = 3 > 0 and f x x − , − =2>0 3 3 3 3   The Second Derivative Test implies that f − 23 , − 13 is a local minimum. 7. f (x, y) = x 3 y + 12x 2 − 8y f (x, y) = x 3 − x y + y 3 SOLUTION

Step 1. Find the critical points. We set the first-order partial derivatives of f (x, y) = x 3 y + 12x 2 − 8y equal to zero and solve: f x (x, y) = 3x 2 y + 24x = 3x(x y + 8) = 0

(1)

f y (x, y) = x 3 − 8 = 0

(2)

Equation (2) implies that x = 2. We substitute in equation (1) and solve for y to obtain 6(2y + 8) = 0 or

y = −4

The critical point is (2, −4). Step 2. Compute the Discriminant. We find the second-order partials: f x x (x, y) = 6x y + 24,

f yy = 0,

f x y = 3x 2

The discriminant is thus D(x, y) = f x x f yy − f x2y = −9x 4 Step 3. Apply the Second Derivative Test. We have D(2, −4) = −9 · 24 < 0 Hence (2, −4) is a saddle point. 9. f (x, y) = 4x −33x 3 − 2x y 2 2 f (x, y) = x + 2x y − 2y − 10x SOLUTION

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(ET CHAPTER 14)

Step 1. Find the critical points. We set the first-order derivatives of f (x, y) = 4x − 3x 3 − 2x y 2 equal to zero and solve: f x (x, y) = 4 − 9x 2 − 2y 2 = 0

(1)

f y (x, y) = −4x y = 0

(2)

Equation (2) implies that x = 0 or y = 0. If x = 0, then equation (1) gives √ 4 − 2y 2 = 0 ⇒ y 2 = 2 ⇒ y = 2,

√ y=− 2

If y = 0, then equation (1) gives 4 − 9x 2 = 0

9x 2 = 4

⇒

⇒

x=

2 , 3

x =−

2 3

The critical points are therefore  √  0, 2 ,

 √  0, − 2 ,



2 ,0 , 3

 2 − ,0 3

Step 2. Compute the discriminant. The second-order partials are f x x (x, y) = −18x,

f yy (x, y) = −4x,

f x y = −4y

The discriminant is thus D(x, y) = f x x f yy − f x2y = −18x · (−4x) − (−4y)2 = 72x 2 − 16y 2 Step 3. Apply the Second Derivative Test. We have  √  D 0, 2 = −32 < 0  √  D 0, − 2 = −32 < 0  2 4 , 0 = 72 · = 32 > 0, D 3 9  2 2 , 0 = −18 · = −12 < 0 fx x 3 3  4 2 D − , 0 = 72 · = 32 > 0, 3 9   2 2 f x x − , 0 = −18 · − = 12 > 0 3 3    √  The Second Derivative Test implies that the points 0, ± 2 are the saddle points, f 23 , 0 is a local maximum, and   f − 23 , 0 is a local minimum. 11. f (x, y) = x 4 +3y 4 −44x y f (x, y) = x + y − 6x − 2y 2 SOLUTION

Step 1. Find the critical points. We set the first-order derivatives of f (x, y) = x 4 + y 4 − 4x y equal to zero and solve: f x (x, y) = 4x 3 − 4y = 0,

f y (x, y) = 4y 3 − 4x = 0

(1)

Equation (1) implies that y = x 3 . Substituting in (2) and solving for x, we obtain 3 (x 3 ) − x = x 9 − x = x(x 8 − 1) = 0

⇒

x = 0,

x = 1,

The corresponding y coordinates are y = 03 = 0,

y = 13 = 1,

y = (−1)3 = −1

The critical points are therefore (0, 0),

(1, 1),

(−1, −1)

x = −1

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

397

Step 2. Compute the discriminant. We find the second-order partials: f x x (x, y) = 12x 2 ,

f yy (x, y) = 12y 2 ,

f x y (x, y) = −4

The discriminant is thus D(x, y) = f x x f yy − f x2y = 12x 2 · 12y 2 − (−4)2 = 144x 2 y 2 − 16 Step 3. Apply the Second Derivative Test. We have D(0, 0) = −16 < 0 D(1, 1) = 144 − 16 = 128 > 0, D(−1, −1) = 144 − 16 = 128 > 0,

f x x (1, 1) = 12 > 0 f x x (−1, −1) = 12 > 0

We conclude that (0, 0) is a saddle point, whereas f (1, 1) and f (−1, −1) are local minima. −y 13. f (x, y) = x ye−x 2 2 f (x, y) = e x −y +4y 2

2

SOLUTION

Step 1. Find the critical points. We compute the partial derivatives of f (x, y) = x ye−x −y , using the Product Rule and the Chain Rule:     2 2 2 2 2 2 f x (x, y, z) = y 1 · e−x −y + xe−x −y · (−2x) = ye−x −y 1 − 2x 2     2 2 2 2 2 2 f y (x, y, z) = x 1 · e−x −y + ye−x −y · (−2y) = xe−x −y 1 − 2y 2 2

2

We set the partial derivatives equal to zero and solve to find the critical points. This gives   2 2 ye−x −y 1 − 2x 2 = 0   2 2 xe−x −y 1 − 2y 2 = 0 2 2 Since e−x −y  = 0, the first equation gives y = 0 or 1 − 2x 2 = 0, that is, y = 0, x = √1 , x = − √1 . We substitute 2 2 each of these values in the second equation and solve to obtain

y = 0: 1 x= √ : 2 1 x = −√ : 2

xe−x = 0 2

1 √ 2

1 2 e− 2 −y

⇒ x =0   1 − 2y 2 = 0

⇒

  1 2 1 − √ e− 2 −y 1 − 2y 2 = 0 2

We obtain the following critical points: (0, 0),   1 1 1 1 √ ,√ , √ ,−√ , 2 2 2 2



⇒

1 − 2y 2 = 0 1 − 2y 2 = 0

1 1 −√ , √ , 2 2



⇒ ⇒

1 y = ±√ 2 1 y = ±√ 2

1 1 −√ ,−√ 2 2



Step 2. Compute the second-order partials.      2 2 2 2 ∂  −x 2 −y 2  e 1 − 2x 2 = y e−x −y (−2x) 1 − 2x 2 + e−x −y (−4x) ∂x   2 2 = −2x ye−x −y 3 − 2x 2

f x x (x, y) = y

     2 2 2 2 ∂  −x 2 −y 2  e 1 − 2y 2 = x e−x −y (−2y) 1 − 2y 2 + e−x −y (−4y) ∂y   2 2 = −2yxe−x −y 3 − 2y 2

f yy (x, y) = x

   ∂     2 2 2 2 2 2 ∂ f x = 1 − 2x 2 ye−x −y = 1 − 2x 2 1 · e−x −y + ye−x −y (−2y) ∂y ∂y     2 2 = e−x −y 1 − 2x 2 1 − 2y 2

f x y (x, y) =

The discriminant is D(x, y) = f x x f yy − f x2y

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Step 3. Apply the Second Derivative Test. We construct the following table: Critical Point  (0, 0) 

√1 , √1 2 2  √1 , − √1 2  2 − √1 , √1 2  2  − √1 , − √1 2 2



fx x 0

f yy 0

fx y 1

D −1

Type D < 0, saddle point

− 2e

− 2e

0

D > 0, f x x < 0 local maximum

2 e 2 e − 2e

2 e 2 e − 2e

0

4 e2 4 e2 4 e2 4 e2

0 0

D > 0, f x x > 0 local minimum D > 0, f x x > 0 local minimum D > 0, f x x < 0 local maximum

15. f (x, y) = e x − xe y f (x, y) = x ln(x + y) SOLUTION

Step 1. Find the critical points. We set the first-order derivatives of f (x, y) = e x − xe y equal to zero and solve: f x (x, y) = e x − e y = 0 f y (x, y) = −xe y = 0 Since e y  = 0, the second equation gives x = 0. Substituting in the first equation, we get e0 − e y = 1 − e y = 0

⇒

ey = 1

⇒

y=0

The critical point is (0, 0). Step 2. Compute the discriminant. We find the second-order partial derivatives: ∂ ∂x ∂ f yy (x, y) = ∂y

f x x (x, y) =

f x y (x, y) =

x  e − e y = ex

 −xe y = −xe y

 ∂ x e − e y = −e y ∂y

The discriminant is D(x, y) = f x x f yy − f x2y = −xe x+y − e2y Step 3. Apply the Second Derivative Test. We have D(0, 0) = 0 − e0 = −1 < 0 The point (0, 0) is a saddle point. 17. f (x, y) = ln x + 2 ln y − x − 4y 2 f (x, y) = (x + 3y)e y−x SOLUTION

Step 1. Find the critical points. We set the first-order partials of f (x, y) = ln x + 2 ln y − x − 4y equal to zero and solve: f x (x, y) =

1 − 1 = 0, x

f y (x, y) =

2 −4=0 y

  The first equation gives x = 1, and the second equation gives y = 12 . We obtain the critical point 1, 12 . Notice that f x and f y do not exist if x = 0 ory = 0, respectively, but these are not critical points since they are not in the domain of f . The critical point is thus 1, 12 . Step 2. Compute the discriminant. We find the second-order partials: 1 f x x (x, y) = − 2 , x

2 f yy (x, y) = − 2 , y

f x y (x, y) = 0

The discriminant is 2 D(x, y) = f x x f yy − f x2y = 2 2 x y

Optimization in Several Variables

S E C T I O N 15.7

Step 3. Apply the Second Derivative Test. We have  2 1 = D 1,  2 = 8 > 0, 2 12 · 12



1 f x x 1, 2



(ET Section 14.7)

399

1 = − 2 = −1 < 0 1

  We conclude that f 1, 12 is a local maximum. 19. f (x, y) = x − y 2 − ln(x + 2y) 2 f (x, y) = (x 2 + y 2 )e−x −y SOLUTION

Step 1. Find the critical points. We set the partial derivatives of f (x, y) = x − y 2 − ln(x + y) equal to zero and solve. f x (x, y) = 1 −

1 = 0, x+y

f y (x, y) = −2y −

1 =0 x+y

1 = 1. Substituting in the second equation gives The first equation implies that x+y

−2y − 1 = 0

⇒

2y = −1

⇒

y=−

1 2

We substitute y = − 12 in the first equation and solve for x: 1−

1 x − 12

=0

⇒

x−

1 =1 2

⇒

x=

3 2

  We obtain the critical point 32 , − 12 . Notice that although f x and f y do not exist where x + y = 0, these are not critical points since f is not defined at these points. Step 2. Compute the discriminant. We compute the second-order partial derivatives:  ∂ 1 1 f x x (x, y) = 1− = ∂x x+y (x + y)2  ∂ 1 1 f yy (x, y) = −2y − = −2 + ∂y x+y (x + y)2  ∂ 1 1 f x y (x, y) = 1− = ∂y x+y (x + y)2 The discriminant is D(x, y) = f x x f yy − f x2y =

1 (x + y)2

 −2 +

1 (x + y)2

−

1 (x + y)4

=

−2 (x + y)2

Step 3. Apply the Second Derivative Test. We have  3 1 −2 ,− =  D 2 = −2 < 0 2 2 3 − 1 2 2   We conclude that 32 , − 12 is a saddle point.  21. Show that f (x, y) = xx22−y+2 y 2 has one critical point P and that f is nondifferentiable at P. Show that f (P) is an (x, y) = (x − y)e absolutefminimum value.  SOLUTION Since f (x, y) = x 2 + y 2 ≥ 0 and f (0, 0) = 0, f (0, 0) is an absolute minimum value. To find the critical point of f we first find the first derivatives:  ∂ x 2x f x (x, y) = x 2 + y2 =  =  ∂x 2 x 2 + y2 x 2 + y2  ∂ y 2y f y (x, y) = x 2 + y2 =  =  2 2 2 ∂y 2 x +y x + y2 Since f x and f y do not exist at (0, 0) and the equations f x (x, y) = 0 and f y (x, y) = 0 have no solutions, the only critical point is P = (0, 0), a point where f is non-differentiable. Let f (x, y) = (x 2 + y 2 − 1)2 . (a) Find the critical points of f (x, y). (b) Use a computer algebra system to graph f (x, y) and identify the critical points on the graph.

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23.

(ET CHAPTER 14)

Use a computer algebra system to find numerical approximations to the critical points of 2

f (x, y) = (1 − x + x 2 )e y + (1 − y + y 2 )e x

2

Use Figure 18 to determine whether they correspond to local minima or maxima. z

x

y

2 2 FIGURE 18 Plot of the function f (x, y) = (1 − x + x 2 )e y + (1 − y + y 2 )e x .

SOLUTION

The critical points are the solutions of f x (x, y) = 0 and f y (x, y) = 0. We compute the partial derivatives:   2 2 f x (x, y) = (−1 + 2x)e y + 1 − y + y 2 e x · 2x   2 2 f y (x, y) = 1 − x + x 2 e y · 2y + (−1 + 2y)e x

Hence, the critical points are the solutions of the following equations:   2 2 (2x − 1)e y + 2x 1 − y + y 2 e x = 0   2 2 (2y − 1)e x + 2y 1 − x + x 2 e y = 0 Using a CAS we obtain the following solution: x = y = 0.27788, which from the figure is a local minimum. 25. Which of the following domains are closed and which are bounded? Use the contour in Figure 19 to determine whether the critical points A, B, C, D are local minima, maxima, 2 : x 2 +map (a) {(x, y) ∈ R y 2 ≤ 1} or saddle points. (b) {(x, y) ∈ R2 : x 2 + y 2 < 1} (c) {(x, y) ∈ R2 : x ≥ 0} (d) {(x, y) ∈ R2 : x > 0, y > 0} (e) {(x, y) ∈ R2 : 1 ≤ x ≤ 4, 5 ≤ y ≤ 10} (f) {(x, y) ∈ R2 : x > 0, x 2 + y 2 ≤ 10} SOLUTION

(a) {(x, y) ∈ R2 : x 2 + y 2 ≤ 1}: This domain is bounded since it is contained, for instance, in the disk x 2 + y 2 < 2. The domain is also closed since it contains all of its boundary points, which are the points on the unit circle x 2 + y 2 = 1. (b) {(x, y) ∈ R2 : x 2 + y 2 < 1}: The domain is contained in the disk x 2 + y 2 < 1, hence it is bounded. It is not closed since its boundary x 2 + y 2 = 1 is not contained in the domain. (c) {(x, y) ∈ R2 : x ≥ 0}: y

x

This domain is not contained in any disk, hence it is not bounded. However, the domain contains its boundary x = 0 (the y-axis), hence it is closed. (d) {(x, y) ∈ R2 : x > 0, y > 0}:

S E C T I O N 15.7

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401

y

x

The domain is not contained in any disk, hence it is not bounded. The boundary is the positive x and y axes, and it is not contained in the domain, therefore the domain is not closed. (e) {(x, y) ∈ R2 : 1 ≤ x ≤ 4, 5 ≤ y ≤ 10}: y 10

7.5

D (1, 10) C (4, 10)

A (1, 5) B (4, 5)

x 1

4

This domain is contained in the disk x 2 + y 2 ≤ 112 , hence it is bounded. Moreover, the domain contains its boundary, which consists of the segments AB, BC, C D, AD shown in the figure, therefore the domain is closed. (f) {(x, y) ∈ R2 : x > 0, x 2 + y 2 ≤ 10}: y

x

This domain is bounded since it is contained in the disk x 2 + y 2 ≤ 10. It is not closed since the part {(0, y) ∈ R2 : |y| ≤ √ 10} of its boundary is not contained in the domain. In Exercises 26–29, determine the global extreme values of the function on the given set without using calculus. 27. f (x, y) = 2x − y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 f (x, y) = x + y, 0 ≤ x, y ≤ 1 SOLUTION f is maximum when x is maximum and y is minimum, that is x = 1 and y = 0. f is minimum when x is minimum and y is maximum, that is, x = 0, y = 3. Therefore, the global maximum of f in the set is f (1, 0) = 2 · 1 − 0 = 2 and the global minimum is f (0, 3) = 2 · 0 − 3 = −3. , x 2 + y2 ≤ 1 29. f (x, y) = e−x −y f (x, y) = (x 2 + y 2 + 1)−1 , 0 ≤ x ≤ 3, 0 ≤ y ≤ 5 2 2 2 2 1 SOLUTION The function f (x, y) = e−(x +y ) = 2 2 is maximum when e x +y is minimum, that is, when x 2 + y 2 x +y 2

2

e

is minimum. The minimum value of x 2 + y 2 on the given set is zero, obtained at x = 0 and y = 0. We conclude that the maximum value of f on the given set is 2 2 f (0, 0) = e−0 −0 = e0 = 1

f is minimum when x 2 + y 2 is maximum, that is, when x 2 + y 2 = 1. Thus, the minimum value of f on the given disk is obtained on the boundary of the disk, and it is e−1 = 1e . 31. Let D = {(x, y) : x > 0, y > 0}. Show that D is not closed. Find a continuous function that does not have a global Assumptions minimum value on D. Matter Show that f (x, y) = x + y has no global minimum or maximum on the domain 0 < x, y < 1. Does this contradict Theorem 3? SOLUTION The boundary of D consists of two rays: the positive x and y axes.

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y

x

x > 0, y > 0

1 is continuous on D (since These rays are not contained in D, therefore D is not closed. The function f (x, y) = − x+y   1 = −∞. Thus, when x + y  = 0 in D). However, f does not have a global minimum value on D, since lim − x+x x→0+

x approaches zero along the ray {(x, x) : x > 0}, the values of f are decreasing without bound. 33. Let f (x, y) = (x + y) ln(x 2 + y 2 ), defined for (x, y)  =2(0, 0). The goal is to find the extreme values of f (x, y) = x − 2x y + 2y on the square 0 ≤ x, y ≤ 2 (Figure 20). (a) Show that if (x, y) is a critical point of f (x, y), then either x = y or x = −y. Determine the critical points (four in (a) Evaluate f at the critical points that lie in the square. all). (b) On the they)square, and f (x, 0) = x 2 .these Find critical the extreme of f on minima, the bottom edge. Plotbottom a graphedge of fof(x, and useyit=to0determine whether pointsvalues are maxima, or neither. (b) (c) Find the extreme values of f on the remaining edges of the square. SOLUTION (d) Find the done largest smallest to values f among theWe values computed (a),(r(b), (a) This is best byand converting polarofcoordinates. see that f (r, θin ) = cosand θ +(c). r sin θ ) ln r 2 , or in other words, f (r, θ ) = (cos θ + sin θ )2r ln r . We take derivatives and find that

∂f = (− sin θ + cos θ )(2r ln r ) ∂θ

∂f = (cos θ + sin θ )(2 ln r + 2), ∂r

Since a critical point makes both equations equal to zero, and since 2 ln r + 2 and 2r ln r can’t both be zero at the same x = y or√x = −y. These solutions r , we get that either cos θ + sin θ = 0 or − sin θ + cos θ = 0, which means either √ √ √ lead√to r = √ 1 and r = 1/e, giving us our four critical points of (1/e 2, 1/e 2), (−1/e 2, −1/e 2), √ respectively, √ (1/ 2, −1/ 2), and (−1/ 2, 1/ 2). √ √ √ √ √ √ (b) Looking at a graph, that (1/ 2, −1/ 2) and (−1/ 2, 1/ 2) are saddle points, while (1/e 2, 1/e 2) is a √ we see√ minimum and (−1/e 2, −1/e 2) is a maximum. z

1 0.5 −1.5 −1.5

0

1

y

x

Let f (x, y) =35–41, x 4 − 4x y + 2y 2 .the global extreme values of the function on the given domain. In Exercises determine (a) Find the critical points of f (x, y). 35. f (x, y) = x 3 − 2y, 0 ≤ x, y ≤ 1 (b) Use a plot to determine whether the critical points are maxima, minima, or saddle points. Then confirm this SOLUTION We use the following steps. result using the Second Derivative Test. Step 1. Find the critical points. We set the first derivative equal to zero and solve: f x (x, y) = 3x 2 = 0,

f y (x, y) = −2

The two equations have no solutions, hence there are no critical points. Step 2. Check the boundary. The extreme values occur either at the critical points or at a point on the boundary of the domain. Since there are no critical points, the extreme values occur at boundary points. We consider each edge of the square 0 ≤ x, y ≤ 1 separately. The segment O A: On this segment y = 0, 0 ≤ x ≤ 1, and f takes the values f (x, 0) = x 3 . The minimum value is f (0, 0) = 0 and the maximum value is f (1, 0) = 1. y C (0, 1)

B (1, 1)

D (0, 0)

A (1, 0)

x

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

403

The segment AB: On this segment x = 1, 0 ≤ y ≤ 1, and f takes the values f (1, y) = 1 − 2y. The minimum value is f (1, 1) = 1 − 2 · 1 = −1 and the maximum value is f (1, 0) = 1 − 2 · 0 = 1. The segment BC: On this segment y = 1, 0 ≤ x ≤ 1, and f takes the values f (x, 1) = x 3 − 2. The minimum value is f (0, 1) = 03 − 2 = −2 and the maximum value is f (1, 1) = 13 − 2 = −1. The segment OC: On this segment x = 0, 0 ≤ y ≤ 1, and f takes the values f (0, y) = −2y. The minimum value is f (0, 1) = −2 · 1 = −2 and the maximum value is f (0, 0) = −2 · 0 = 0. Step 3. Conclusions. The values obtained in the previous steps are f (0, 0) = 0,

f (1, 0) = 1,

f (1, 1) = −1,

f (0, 1) = −2

The smallest value is f (0, 1) = −2 and it is the global minimum of f on the square. The global maximum is the largest value f (1, 0) = 1. 2 2 , 0 ≤ x, y ≤ 1 37. f (x,f (x, y) = y)x= + 4x2y + 3y, 0 ≤ x, y ≤ 1 SOLUTION The sum x 2 + 2y 2 is maximum at the point (1, 1), where x 2 and y 2 are maximum. It is minimum if x = y = 0, that is, at the point (0, 0). Hence,

Global maximum = f (1, 1) = 12 + 2 · 12 = 3 Global minimum = f (0, 0) = 02 + 2 · 02 = 0 39. f (x, y) = x 3 + x 2 y + 2y 2 , 2 x, y2 ≥ 0, x + y ≤ 1 f (x, y) = (4y 2 − x 2 )e−x −y , x 2 + y 2 ≤ 2 SOLUTION We use the following steps. Step 1. Examine the critical points. We find the critical points of f (x, y) = x 3 + x 2 y + 2y 2 in the interior of the domain (the standard region in the figure). y

B (0, 1)

0

A (1, 0)

x

We set the partial derivatives of f equal to zero and solve: f x (x, y) = 3x 2 + 2x y = x(3x + 2y) = 0 f y (x, y) = x 2 + 4y = 0 The first equation gives x = 0 or y = − 32 x. Substituting x = 0 in the second equation gives 4y = 0 or y = 0. We obtain the critical point (0, 0). We now substitute y = − 32 x in the second equation and solve for x:  3 x 2 + 4 · − x = x 2 − 6x = x(x − 6) = 0 ⇒ x = 0, x = 6 2 We get the critical points (0, 0) and (6, −9). None of the critical points (0, 0) and (6, −9) is in the interior of the domain. Step 2. Check the boundary. The boundary consists of the three segments O A, O B, and AB shown in the figure. We consider each part of the boundary separately. The segment O A: On this segment y = 0, 0 ≤ x ≤ 1, and f (x, y) = f (x, 0) = x 3 . The minimum value is f (0, 0) = 03 = 0 and the maximum value is f (1, 0) = 13 = 1. The segment O B: On this segment x = 0, 0 ≤ y ≤ 1, and f (x, y) = f (0, y) = 2y 2 . The minimum value is f (0, 0) = 2 · 02 = 0 and the maximum value is f (0, 1) = 2 · 12 = 2. The segment AB: On this segment y = 1 − x, 0 ≤ x ≤ 1, and   f (x, y) = x 3 + x 2 (1 − x) + 2(1 − x)2 = x 3 + x 2 − x 3 + 2 1 − 2x + x 2 = 3x 2 − 4x + 2 We find the extreme values of g(x) = 3x 2 − 4x + 2 in the interval 0 ≤ x ≤ 1. With the aid of the graph of g(x), and with setting the derivative g  equal to 0, we find that the minimum value is g

   2 2 2 1 2 2 2 , −4· +2= = f =3· 3 3 3 3 3 3

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and the maximum value is g(0) = f (0, 1) = 3 · 02 − 4 · 0 + 2 = 2 y 2 g(x) = 3x 2 − 4x + 2

0

2 3

x

1

Step 3. Conclusions. We compare the values of f (x, y) at the points obtained in step (2), and determine the global extrema of f (x, y). This gives  2 2 1 , = f (0, 0) = 0, f (1, 0) = 1, f (0, 1) = 2, f 3 3 3 We conclude that the global minimum of f in the given domain is f (0, 0) = 0 and the global maximum is f (0, 1) = 2. √ 41. f (x, y) = x 3 y 5 , the set bounded by x = 0, y = 0, and y = 1 − x. Hint: Use a computer algebra system √ √ f (x, y) = x 3 + y 3 − 3x y, 0 ≤ x, y ≤ 1 to find the minimum along the boundary curve y = 1 − x, which is parametrized by (t, 1 − t) for 0 ≤ t ≤ 1. SOLUTION y

B (0, 1) y=1− x

0

A (1, 0)

x

We use the following steps. Step 1. Examine the critical points. We find the critical points of f (x, y) = x 3 y 5 in the interior of the domain. Setting the partial derivatives equal to zero and solving gives f x (x, y) = 3x 2 y 5 = 0,

f y (x, y) = 5y 4 x 3 = 0

The solutions are all the points with at least one zero coordinate. These points are not in the interior of the region, hence there are no critical points in the interior of the region. Step 2. Check the boundary. We consider each part of the boundary separately. The segment O A: On this segment y = 0 and f (x, 0) = 0, hence f has the constant value 0 on this part of the boundary. The segment O B: On this segment x = 0, hence f (0, y) = 03 · y 5 = 0. f attains the constant value 0 on this part of the boundary.

√ 5 √ The curve y = 1 − x, 0 ≤ x ≤ 1: On this curve we have f (x, y) = g(x) = x 3 1 − x .

√ 5 We find the points on the interval 0 ≤ x ≤ 1 where g(x) = x 3 1 − x has maximum and minimum values. Differentiating g gives 





√ 4 √ 5 √ 5 5 √ 4 −1 g  (x) = 3x 2 1 − x + x 3 · 5 1 − x · = 3x 2 1 − x − x 2.5 1 − x √ 2 2 x  



  √ √ √ √ √ 5 11 4 4 = x2 1 − x x = x2 1 − x x 3−3 x − 3− 2 2 We solve g  (x) = 0 in the interval 0 < x < 1: 

√ 4 11 √ x =0 3− g  (x) = x 2 1 − x 2

⇒

x = 0,

1−

√

x = 0,

3−

11 √ x=0 2

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

405

36 . The critical point in the interval 0 < x < 1 is x = 36 . We compute The solutions are x = 0, x = 1, and x = 121 121

√ 5 3 g(x) = x 1 − x at the critical point and at the endpoints x = 0, x = 1:

 g(0) = g(1) = 0,

g

36 121



 =

 36 3 6 5 ≈ 0.0005 1− 121 11

36 . We find the The points where g has extreme values in √ the interval 0 ≤ x ≤ 1 are x = 0, x = 1, and x = 121 y-coordinates of these points from y = 1 − x: √ x = 0: y = 1 − 0 = 1 √ x = 1: y = 1 − 1 = 0  5 36 36 : y =1− = x= 121 121 11 √ We conclude that the global extrema of f on the curve y = 1 − x, 0 ≤ x ≤ 1 are obtained at the points (0, 1), (1, 0),  36 , 5 . and 121 11

Step 3. Conclusions. We examine the values of f (x, y) = x 3 y 5 at the points obtained in the previous parts. The candidates for global extrema are f = 0, the values of f on the segments O A and O B:  f

f (0, 1) = f (1, 0) = 0   36 5 36 3 5 5 , = 0.0005 = 121 11 121 11

We conclude that the minimum value of f (x, y) in the given domain is 0 and the maximum value is f 0.0005.



36 5 121 , 11



≈

rectangularbox box(including B with a abottom and sides but top such thatVBwith has minimal surface area 43. ShowConsider that the arectangular top and bottom) withnofixed volume the smallest possible among all boxes with fixed(Figure volume21). V. surface area is a cube (a) Do you think B is a cube as in the solution to Exercise 42? If not, how would you make its shape differ from a cube? (b) Find the dimensions of B and compare with your response to (a). SOLUTION

(a) Each of the variables x and y is the length of a side of three faces (for example, x is the length of the front, back, and bottom sides), whereas z is the length of a side of four faces.

y

z x

Therefore, the variables x, y, and z do √ not have equal influence on the surface area. We expect that in the box B with minimal surface area, z is smaller than 3 V , which is the side of a cube with volume V . (b) We must find the dimensions of the box B, with fixed volume V and with smallest possible surface area, when the top is not included. Step 1. Find a function to be minimized. The surface area of the box with sides lengths x, y, z when the top is not included is S = 2x z + 2yz + x y

z

(1)

y

x

To express the surface in terms of x and y only, we use the formula for the volume of the box, V = x yz, giving z = xVy . We substitute in (1) to obtain S = 2x ·

V 2V 2V V + 2y · + xy = + + xy xy xy y x

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That is, S=

2V 2V + + x y. y x

Step 2. Determine the domain. The variables x, y denote lengths, hence they must be nonnegative. Moreover, S is not defined for x = 0 or y = 0. Since there are no other limitations on the variables, the domain is D = {(x, y) : x > 0, y > 0} We must find the minimum value of S on D. Because this domain is neither closed nor bounded, we are not sure that a minimum value exists. However, it can be proved (in like manner as in Exercise 42) that S does have a minimum value on D. This value occurs at a critical point in D, hence we set the partial derivatives equal to zero and solve. This gives 2V Sx (x, y) = − 2 + y = 0 x 2V S y (x, y) = − 2 + x = 0 y The first equation gives y = 2V2 . Substituting in the second equation yields x

x−

 x3 x4 = x 1− =x− =0 2V 2V

2V 4V 2 x4

The solutions are x = 0 and x = (2V )1/3 . The solution x = 0 is not included in D, so the only solution is x = (2V )1/3 . We find the value of y using y = 2V2 : x

y=

2V (2V )2/3

= (2V )1/3

  We conclude that the critical point, which is the point where the minimum value of S in D occurs, is (2V )1/3 , (2V )1/3 . We find the corresponding value of z using z = xVy . We get z=

V V 1/3 = = = 22/3 V 2/3 22/3 (2V )1/3 (2V )1/3 V



V 1/3 4

We conclude that the sizes of the box with minimum surface area are width: x = (2V )1/3 ; length: y = (2V )1/3 ;  1/3 height: z = V4 . We see that z is smaller than x and y as predicted.

Further Insights and Challenges 45. The power (in microwatts) of a laser is measured as a function of current (in milliamps). Find the linear least-squares Given n data points (x 1 , y1 ), . . . , (x n , yn ), we may seek a linear function f (x) = mx + b that best fits the data. fit (Exercise 44) for the data points. The linear least-squares fit is the linear function f (x) = mx + b that minimizes the sum of the squares (Figure 22) n 1.0 1.1 1.2 1.3  (y j − f (x j ))2 E(m, b) = Laser power (μ W) 0.52 0.56 0.82 0.78 j=1

Current (mA)

1.4

1.5

1.23

1.50

Show thatBy E Exercise is minimized forcoefficients m and b satisfying SOLUTION 44, the of the linear least-square fit f (x) = mx + b are determined by the following equations: n n   x j + bn  =n yj mn  j=1 j=1 x j + bn = yj m j=1

j=1

n n n    mn x 2j + xj  =n xj yj nb  j=1 j=1 j=1 m x 2j + b xj = xj · yj j=1

j=1

j=1

In our case there are n = 6 data points: (x 1 , y1 ) = (1, 0.52), (x 2 , y2 ) = (1.1, 0.56),

(1)

S E C T I O N 15.7

Optimization in Several Variables

(ET Section 14.7)

407

(x 3 , y3 ) = (1.2, 0.82), (x 4 , y4 ) = (1.3, 0.78), (x 5 , y5 ) = (1.4, 1.23), (x 6 , y6 ) = (1.5, 1.50). We compute the sums in (1): 6 

x j = 1 + 1.1 + 1.2 + 1.3 + 1.4 + 1.5 = 7.5

j=1 6 

y j = 0.52 + 0.56 + 0.82 + 0.78 + 1.23 + 1.50 = 5.41

j=1 6 

x 2j = 12 + 1.12 + 1.22 + 1.32 + 1.42 + 1.52 = 9.55

j=1 6 

x j · y j = 1 · 0.52 + 1.1 · 0.56 + 1.2 · 0.82 + 1.3 · 0.78 + 1.4 · 1.23 + 1.5 · 1.50 = 7.106

j=1

Substituting in (1) gives the following equations: 7.5m + 6b = 5.41 9.55m + 7.5b = 7.106

(2)

We multiply the first equation by 9.55 and the second by (−7.5), then add the resulting equations. This gives 71.625m + 57.3b = 51.6655 + −71.625m − 56.25b = −53.295

⇒

b = −1.5519

1.05b = −1.6295 We now substitute b = −1.5519 in the first equation in (2) and solve for m: 7.5m + 6 · (−1.5519)=5.41 7.5m=14.7214

⇒

m = 1.9629

The linear least squares fit f (x) = mx + b is thus f (x) = 1.9629x − 1.5519. 47. problem proposed Fermat thethe Italian scientist Evangelista Torricelli −1 = 1, to ProveThe thatfollowing if α , β ≥ 1 are was numbers suchby that α −1 as + aβchallenge then following inequality holds for all (1608–1647), x, y ≥ 0: a student of Galileo and inventor of the barometer. Given three points A = (a1 , a2 ), B = (b1 , b2 ), and C = (c1 , c2 ) in the plane, find the point P = (x, y) that minimizes the sum of the distances 1 α 1 β x ≥ xy f (x, y)α=x A+ P+ β BP + CP (a) Write out f (x, y) as a function of x and y, and show that f (x, y) is differentiable except at the points A, B, C. For example, (b) Define the unit vectors 2 5/2 3 5/3 1 3 2 3/2 x + −→ y x ≥ x y, + y ≥ x y, etc. −5→ −→ 3 3A P 5 BP CP e = −→ , f = −→ , g = −→ Pα −1 x α + β−1 B P C P Hint: Find the critical points of f (x, y)A= x β − x y. Show that the condition ∇ f = 0 is equivalent to e+f+g=0

3

Prove that Eq. (3) holds if and only if the mutual angles between the unit vectors are all 120◦ . (c) Define the Fermat point to be the point P such that angles between the segments A P, B P, C P are all 120◦ . Conclude that the Fermat point solves the minimization problem (Figure 23). (d) Show that the Fermat point does not exist if one of the angles in ABC is ≥ 120◦ . Where does the minimum occur in this case? Hint: The minimum must occur at a point where f (x, y) is not differentiable.

408

C H A P T E R 15

D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S

(ET CHAPTER 14) B

A

f

e

P

B

g

140˚ A

C (A) P is the Fermat point (the angles between e, f, and g are all 120˚).

C

(B) Fermat point does not exist.

FIGURE 23 SOLUTION

(a) P(x, y) A(a 1 , a 2 )

B(b 1 , b 2 )

C(c 1 , c 2 )

Using the formula for the length of a segment we obtain    f (x, y) = (x − a1 )2 + (y − a2 )2 + (x − b1 )2 + (y − b2 )2 + (x − c1 )2 + (y − c2 )2 We compute the partial derivatives of f : x − a1 x − b1 x − c1 f x (x, y) =  + + 2 2 2 2 (x − a1 ) + (y − a2 ) (x − b1 ) + (y − b2 ) (x − c1 )2 + (y − c2 )2

(1)

y − a2 y − b2 y − c2 + + f y (x, y) =  (x − a1 )2 + (y − a2 )2 (x − b1 )2 + (y − b2 )2 (x − c1 )2 + (y − c2 )2

(2)

For all (x, y) other then (a1 , a2 ), (b1 , b2 ), (c1 , c2 ) the partial derivatives are continuous, therefore the Criterion for Differentiability implies that f is differentiable at all points other than A, B, and C. (b) P

e

A

f B

g C

We compute the unit vectors e, f, and g: x − a1 , y − a2  e=  (x − a1 )2 + (y − a2 )2 x − b1 , y − b2  f=  (x − b1 )2 + (y − b2 )2 x − c1 , y − c2  g=  (x − c1 )2 + (y − c2 )2 We write the condition e + f + g = 0: x − a1 , y − a2  x − b1 , y − b2  x − c1 , y − c2  e+f+g=  + + (x − a1 )2 + (y − a2 )2 (x − b1 )2 + (y − b2 )2 (x − c1 )2 + (y − c2 )2

Optimization in Several Variables

S E C T I O N 15.7

(ET Section 14.7)

409



x − a1 x − b1 x − c1 =  + + , (x − a1 )2 + (y − a2 )2 (x − b1 )2 + (y − b2 )2 (x − c1 )2 + (y − c2 )2  y − a2 y − b2 y − c2  + + (x − a1 )2 + (y − a2 )2 (x − b1 )2 + (y − b2 )2 (x − c1 )2 + (y − c2 )2 Combining with (1) and (2) we get   e + f + g = f x (x, y), f y (x, y) = ∇ f Therefore, the condition ∇ f = 0 is equivalent to e + f + g = 0. We now show that Eq. (3) holds if and only if the mutual angles between the unit vectors are all 120◦ . We place the axes so that the positive x-axis is in the direction of e. y

g f

a q

x

e

Let θ and α be the angles that f and g make with e, respectively. Hence, e = 1, 0 , f = cos θ , sin θ  , g = cos α , sin α  Substituting in e + f + g = 0 we have cos θ + cos α + 1, sin θ + sin α  = 0, 0 or cos θ + cos α + 1 = 0 sin θ + sin α = 0 The second equation implies that sin θ = − sin α = sin(180 + α ) The solutions for 0 ≤ α , θ ≤ 360 are

θ = 180 + α ,

θ = 360 − α

We substitute each solution in the first equation and solve for α . This gives

θ = 180 + α

θ = 360◦ − α

cos(180 + α ) + cos α + 1 = 0

cos(360◦ − α ) + cos α + 1 = 0

− cos α + cos α + 1 = 0

cos α + cos α + 1 = 0

1=0

2 cos α = −1 cos α = − 12

⇒

α = 120◦ θ

= 360◦ − α = 240◦

α = 240◦ θ = 360◦ − α = 120◦

We obtain the following vectors: e = 1, 0 ,

  f = cos 240◦ , sin 240◦ ,

  g = cos 120◦ , sin 120◦

e = 1, 0 ,

  f = cos 120◦ , sin 120◦ ,

  g = cos 240◦ , sin 240◦

or

410

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(ET CHAPTER 14) y

f 120° x

e

120° 120° g

or y

B A

C 120°

e

g 120°

g

P e

120°

f

120°

x

A

120°

120° P

e

120°

120°

120° f

f

g

B

C

In either case the angles between the vectors are 120◦ . (c) f (x, y) has the minimum value at a critical point. The critical points are the points where f x and f y are 0 or do not exist, that is, the points A, B, C and the point where ∇ f = 0, which according to part (b) is the Fermat point. We now show that if the Fermat point P exists, then f (P) ≤ f ( A), f (B), f (C). A

120° 120° P 120° B

C

Suppose that the Fermat point P exists. The values of f at the critical points are f ( A) = AB + AC f (B) = AB + BC f (C) = AC + BC f (P) = A P + B P + PC We show that f (P) < f ( A). Similarly it can be shown that also f (P) < f (B) and f (P) < f (C). By the Cosine Theorem for the triangles AB P and AC P we have   2 2 2 2 ◦ AB = A P + B P − 2 A P · B P cos 120 = A P + B P + A P · B P   2 2 2 2 ◦ AC = A P + C P − 2 A P · PC cos 120 = A P + C P + A P · PC Hence

 f ( A) = AB + AC =

2

2

AP + B P + AP · B P +



2

2

A P + C P + A P · PC

≥ A P + B P + PC = f (P) The last inequality can be verified by squaring and transfering sides. It’s best to use a computer to help with the algebra; it’s a daunting task to do by hand. (d) We show that if one of the angles of  ABC is ≥ 120◦ , then the Fermat point does not exist. Notice that the Fermat point (if it exists) must fall inside the triangle ABC.

Lagrange Multipliers: Optimizing with a Constraint

S E C T I O N 15.8

(ET Section 14.8)

411

A

C B

120°

120°

120° P

P cannot lie outside  ABC Suppose the Fermat point P exists. A 1 2

120° 120° P 1

120°

1

B

C

We sum the angles in the triangles AB P and AC P, obtaining A1 + B1 + 120◦ = 180◦

⇒

 A1 = 60◦ − B1

 A2 + C1 + 120◦ = 180◦

⇒

 A2 = 60◦ − C1

Therefore,



 A = A1 +  A2 = 60◦ − B1 + 60◦ − C1 = 120◦ − (B1 + C1 ) < 120◦ We thus showed that if the Fermat point exists, then A < 120◦ . Similarly, one shows also that B and C must be smaller than 120◦ . We conclude that if one of the angles in  ABC is equal or greater than 120◦ , then the Fermat point does not exist. In that case, the minimum value of f (x, y) occurs at a point where f x or f y do not exist, that is, at one of the points A, B, or C.

15.8 Lagrange Multipliers: Optimizing with a Constraint

(ET Section 14.8)

Preliminary Questions 1. Suppose that the maximum of f (x, y) subject to the constraint g(x, y) = 0 occurs at a point P = (a, b) such that ∇ f P  = 0. Which of the following are true? (a) ∇ f P is tangent to g(x, y) = 0 at P

412

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(ET CHAPTER 14)

SOLUTION The level curve f (x, y) = 2 is tangent to the constraint curve at the point A. A close level curve that intersects the constraint curve is f (x, y) = 1, hence we may assume that f has a local maximum 2 under the constraint at A. The level curve f (x, y) = 3 is tangent to the constraint curve. However, in approaching B under the constraint, from one side f is increasing and from the other side f is decreasing. Therefore, f (B) is neither local minimum nor local maximum of f under the constraint.

3. On the contour map in Figure 9: (a) Identify the points where ∇ f = λ ∇g for some scalar λ . (b) Identify the minimum and maximum values of f (x, y) subject to g(x, y) = 0. y −4 −2 2

6

Graph of g (x, y) = 0

x

6

2 −2 −6

Contour plot of f(x, y) (contour interval 2)

FIGURE 9 Contour map of f (x, y); contour interval 2. SOLUTION

(a) The gradient ∇g is orthogonal to the constraint curve g(x, y) = 0, and ∇ f is orthogonal to the level curves of f . These two vectors are parallel at the points where the level curve of f is tangent to the constraint curve. These are the points A, B, C, D, E in the figure: ∇fA, ∇gA

−6 −2 2

A

6

g(x, y) = 0

B C D

E

6

2 −2 −6

(b) The minimum and maximum occur where the level curve of f is tangent to the constraint curve. The level curves tangent to the constraint curve are f ( A) = −4,

f (C) = 2,

f (B) = 6,

f (D) = −4,

f (E) = 4

Therefore the global minimum of f under the constraint is −4 and the global maximum is 6.

Exercises 1. Use Lagrange multipliers to find the extreme values of the function f (x, y) = 2x + 4y subject to the constraint g(x, y) = x 2 + y 2 − 5 = 0. (a) Show that the Lagrange equation ∇ f = λ ∇g gives λ x = 1 and λ y = 2. (b) Show that these equations imply λ  = 0 and y = 2x. (c) Use the constraint equation to determine the possible critical points (x, y). (d) Evaluate f (x, y) at the critical points and determine the minimum and maximum values. SOLUTION

(a) The Lagrange equations are determined by the equality ∇ f = λ ∇g. We find them:     ∇ f = f x , f y = 2, 4 , ∇g = gx , g y = 2x, 2y Hence, 2, 4 = λ 2x, 2y or

λ (2x) = 2 λ (2y) = 4

⇒

λx = 1 λy = 2

Lagrange Multipliers: Optimizing with a Constraint

S E C T I O N 15.8

(ET Section 14.8)

413

(b) The Lagrange equations in part (a) imply that λ  = 0. The first equation implies that x = λ1 and the second equation

gives y = λ2 . Therefore y = 2x. (c) We substitute y = 2x in the constraint equation x 2 + y 2 − 5 = 0 and solve for x and y. This gives x 2 + (2x)2 − 5 = 0 5x 2 = 5 x2 = 1

⇒

x 1 = −1,

x2 = 1

Since y = 2x, we have y1 = 2x 1 = −2, y2 = 2x 2 = 2. The critical points are thus (−1, −2) and

(1, 2).

Extreme values can also occur at the points where ∇g = 2x, 2y = 0, 0. However, (0, 0) is not on the constraint. (d) We evaluate f (x, y) = 2x + 4y at the critical points, obtaining f (−1, −2) = 2 · (−1) + 4 · (−2) = −10 f (1, 2) = 2 · 1 + 4 · 2 = 10 Since f is continuous and the graph of g = 0 is closed and bounded, global minimum and maximum points exist. So according to Theorem 1, we conclude that the maximum of f (x, y) on the constraint is 10 and the minimum is −10. 3. Apply the method of Lagrange multipliers to2 the function f (x, y) = (x 2 + 1)y subject to the constraint x 2 + y 2 = 5. 2 subject + 2y the constraint g(x, y) = 4x − 6y = 25. Find the extreme values of f (x, y) = x Hint: First show that y  = 0; then treat the cases x = 0 and x  = to 0 separately.  (a) Show that the Lagrange equations yield 2x = 4λ , 4y = −6λ .  SOLUTION We first write out the Lagrange Equations. We have ∇ f = 2x y, x 2 + 1 and ∇g = 2x, 2y. Hence, the (b) Show that if x = 0 or y = 0, then λ = 0 and the Lagrange equations give x = y = 0. Since (0, 0) does not Lagrange for ∇gyou  = may 0 is assume that x and y are nonzero. satisfyCondition the constraint, (c) Use the Lagrange equations to show that y = −∇34 x. f = λ ∇g  is a unique critical point P. (d) Substitute in the constraint equation to show that there 2y to Figure 10 to justify your answer. Hint: Do 1 = of λ 2x, 2x y, x 2 +value (e) Does P correspond to a minimum or maximum f ? Refer the values of f (x, y) increase or decrease as (x, y) moves away from P along the line g(x, y) = 0? We obtain the following equations: 2x y = λ (2x) x 2 + 1 = λ (2y)

2x(y − λ ) = 0

⇒

x 2 + 1 = 2λ y

(1)

The second equation implies that y  = 0, since there is no real value of x such that x 2 + 1 = 0. Likewise, λ  = 0. The solutions of the first equation are x = 0 and y = λ . Case 1: x = 0. Substituting x = 0 in the second equation gives 2λ y = 1, or y = 21λ . We substitute x = 0, y = 21λ (recall that λ  = 0) in the constraint to obtain 02 +

1 =5 4λ 2

4λ 2 =

⇒

1 5

⇒

1 1 λ = ±√ = ± √ 20 2 5

The corresponding values of y are y=

1 1 2· √ 2 5

=

√

y=

5 and

√ 1   =− 5 1 2· − √ 2 5

We obtain the critical points:  √  0, 5

and

 √  0, − 5

Case 2: x  = 0. Then the first equation in (1) implies y = λ . Substituting in the second equation gives x 2 + 1 = 2λ 2

⇒

x 2 = 2λ 2 − 1

We now substitute y = λ and x 2 = 2λ 2 − 1 in the constraint x 2 + y 2 = 5 to obtain 2λ 2 − 1 + λ 2 = 5 3λ 2 = 6

λ2 = 2 The solution (x, y) are thus

λ=

√

2:

y=

√

2,

⇒

√ λ =± 2

√ √ x =± 2·2−1=± 3

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√ λ = − 2: We obtain the critical points: √ √  3, 2 , We conclude that the critical points are   √  √  0, − 5 , 0, 5 ,

(ET CHAPTER 14)

√ y = − 2,

 √ √  − 3, 2 , √ √  3, 2 ,

√ √ x =± 2·2−1=± 3 √

√  3, − 2 ,

 √ √  − 3, 2 ,

 √ √  − 3, − 2

√

√  3, − 2 ,

 √ √  − 3, − 2 .

  We now calculate f (x, y) = x 2 + 1 y at the critical points:  √  √ f 0, 5 = 5 ≈ 2.24  √ √  f 0, − 5 = − 5 ≈ −2.24 √ √   √ √  √ f 3, 2 = f − 3, 2 = 4 2 ≈ 5.66 √  √ √  √  √ f 3, − 2 = f − 3, − 2 = −4 2 ≈ −5.66 Since the constraint gives a closed and bounded curve, f achieves a minimum √and a maximum under it. We conclude √ that the maximum of f (x, y) on the constraint is 4 2 and the minimum is −4 2. In Exercises 4–13, find the minimum and maximum values of the function subject to the given constraint. 5. f (x, y) = x 2 + y 2 , 2x +23y =2 6 f (x, y) = 2x + 3y, x + y = 4 SOLUTION We find the extreme values of f (x, y) = x 2 + y 2 under the constraint g(x, y) = 2x + 3y − 6 = 0. Step 1. Write out the Lagrange Equations. The gradients of f and g are ∇ f = 2x, 2y and ∇g = 2, 3. The Lagrange Condition is ∇ f = λ ∇g 2x, 2y = λ 2, 3 We obtain the following equations: 2x = λ · 2 2y = λ · 3 Step 2. Solve for λ in terms of x and y. Notice that if x = 0, then the first equation gives λ = 0, therefore by the second equation also y = 0. The point (0, 0) does not satisfy the constraint. Similarly, if y = 0 also x = 0. We therefore may assume that x  = 0 and y  = 0 and obtain by the two equations:

λ =x

and

λ=

2 y. 3

Step 3. Solve for x and y using the constraint. Equating the two expressions for λ gives x=

2 y 3

⇒

y=

3 x 2

We substitute y = 32 x in the constraint 2x + 3y = 6 and solve for x and y: 2x + 3 ·

3 x =6 2

13x = 12

⇒

x=

12 , 13

y=

3 12 18 · = 2 13 13

  18 . We obtain the critical point 12 , 13 13 Step 4. Calculate f at the critical point. We evaluate f (x, y) = x 2 + y 2 at the critical point:   2  2 18 12 18 12 468 , ≈ 2.77 + = f = 13 13 13 13 169 Rewriting the constraint as y = − 23 x + 2, we see that as |x| → +∞ then so does |y|, and hence x 2 + y 2 is increasing without bound on the constraint as |x| → ∞. We conclude that the value 468/169 is the minimum value of f under the constraint, rather than the maximum value.

Lagrange Multipliers: Optimizing with a Constraint

S E C T I O N 15.8

(ET Section 14.8)

415

2 + 9y 2 = 32 7. f (x, y) = x y, 4x f (x, y) = 4x 2 + 9y 2 , x y = 4 SOLUTION We find the extreme values of f (x, y) = x y under the constraint g(x, y) = 4x 2 + 9y 2 − 32 = 0. Step 1. Write out the Lagrange Equation. The gradient vectors are ∇ f = y, x and ∇g = 8x, 18y, hence the Lagrange Condition is

∇ f = λ ∇g y, x = λ 8x, 18y We obtain the following equations: y = λ (8x) x = λ (18y) Step 2. Solve for λ in terms of x and y. If x = 0, then the Lagrange equations also imply that y = 0 and vice versa. Since the point (0, 0) does not satisfy the equation of the constraint, we may assume that x  = 0 and y  = 0. The two equations give

λ=

y 8x

λ=

and

x 18y

Step 3. Solve for x and y using the constraint. We equate the two expressions for λ to obtain x y = 8x 18y

⇒

18y 2 = 8x 2

⇒

2 y=± x 3

We now substitute y = ± 23 x in the equation of the constraint and solve for x and y:  2 2 4x 2 + 9 · ± x = 32 3 4x 2 + 9 ·

4x 2 = 32 9 8x 2 = 32

⇒

x = −2,

x =2

We find y by the relation y = ± 23 x: y=

4 2 · (−2) = − , 3 3

2 4 y = − · (−2) = , 3 3

We obtain the following critical points:  4 , −2, − 3

 −2,

4 , 3

y=  2,

2 4 ·2= , 3 3

4 , 3

 2, −

2 4 y =− ·2=− 3 3 4 3



Extreme values can also occur at the point where ∇g = 8x, 18y = 0, 0, that is, at the point (0, 0). However, the point does not lie on the constraint. Step 4. Calculate f at the critical points. We evaluate f (x, y) = x y at the critical points:   4 4 8 f −2, − = f 2, = 3 3 3   4 4 8 f −2, = f 2, − =− 3 3 3 Since f is continuous and the constraint is a closed and bounded set in R 2 (an ellipse), f attains global extrema on the constraint. We conclude that 83 is the maximum value and − 83 is the minimum value. 9. f (x, y) = x 2 +2y 2 , x 4 + y4 = 1 f (x, y) = x y + x + y, x y = 4 SOLUTION We find the extreme values of f (x, y) = x 2 + y 2 under the constraint g(x, y) = x 4 + y 4 − 1 = 0.   Step 1. Write out the Lagrange Equations. We have ∇ f = 2x, 2y and ∇g = 4x 3 , 4y 3 , hence the Lagrange Condition ∇ f = λ ∇g gives   2x, 2y = λ 4x 3 , 4y 3 or

  2x = λ 4x 3   2y = λ 4y 3

⇒

x = 2λ x 3 y = 2λ y 3

(1)

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(ET CHAPTER 14)

Step 2. Solve for λ in terms of x and y. We first assume that x  = 0 and y  = 0. Then the Lagrange equations give

λ=

1 2x 2

and λ =

1 2y 2

Step 3. Solve for x and y using the constraint. Equating the two expressions for λ gives 1 1 = 2 2x 2 2y

⇒

y2 = x 2

⇒

y = ±x

We now substitute y = ±x in the equation of the constraint x 4 + y 4 = 1 and solve for x and y: x 4 + (±x)4 = 1 2x 4 = 1 1 x4 = 2

1 x = 1/4 , 2

⇒

1 x = − 1/4 2

The corresponding values of y are obtained by the relation y = ±x. The critical points are thus     1 1 1 1 1 1 1 1 , , − , , − , , − , − 21/4 21/4 21/4 21/4 21/4 21/4 21/4 21/4

(2)

We examine the case x = 0 or y = 0. Notice that the point (0, 0) does not satisfy the equation of the constraint, hence either x = 0 or y = 0 can hold, but not both at the same time. Case 1: x = 0. Substituting x = 0 in the constraint x 4 + y 4 = 1 gives y = ±1. We thus obtain the critical points (0, −1),

(0, 1)

(3)

Case 2: y = 0. We may interchange x and y in the discussion in case 1, and obtain the critical points: (−1, 0),

(1, 0)

(4)

Combining (2), (3), and (4) we conclude that the critical points are    1 1 1 1 1 1 A1 = , = , − = − , , A , A , 2 3 21/4 21/4 21/4 21/4 21/4 21/4  1 1 A4 = − 1/4 , − 1/4 , A5 = (0, −1), A6 = (0, 1), A7 = (−1, 0), 2 2   The point where ∇g = 4x 3 , 4y 3 = 0, 0, that is, (0, 0), does not lie on the constraint.

A8 = (1, 0)

Step 4. Compute f at the critical points. We evaluate f (x, y) = x 2 + y 2 at the critical points:  f ( A1 ) = f ( A2 ) = f ( A3 ) = f ( A4 ) =

1 21/4

2

 +

1 21/4

2

√ 2 = 1/2 = 2 2

f ( A5 ) = f ( A6 ) = f ( A7 ) = f ( A8 ) = 1 2 The constraint x 4 + y 4 = 1 is a closed √ and bounded set in R and f is continuous on this set, hence f has global extrema on the constraint. We conclude that 2 is the maximum value and 1 is the minimum value.

11. f (x, y, z) = 3x2+42y +24z, x22 + 2y 2 + 6z 2 = 1 f (x, y) = x y , x + 2y = 6 SOLUTION We find the extreme values of f (x, y, z) = 3x + 2y + 4z under the constraint g(x, y, z) = x 2 + 2y 2 + 6z 2 − 1 = 0. Step 1. Write out the Lagrange Equations. The gradient vectors are ∇ f = 3, 2, 4 and ∇g = 2x, 4y, 12z, therefore the Lagrange Condition ∇ f = λ ∇g is: 3, 2, 4 = λ 2x, 4y, 12z The Lagrange equations are, thus: 3 = λ (2x) 2 = λ (4y) 4 = λ (12z)

⇒

3 = λx 2 1 = λy 2 1 = λz 3

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Step 2. Solve for λ in terms of x, y, and z. The Lagrange equations imply that x  = 0, y  = 0, and z  = 0. Solving for λ we get

λ=

3 , 2x

λ=

1 , 2y

λ=

1 3z

Step 3. Solve for x, y, and z using the constraint. Equating the expressions for λ gives 1 1 3 = = 2x 2y 3z

⇒

x=

9 z, 2

y=

3 z 2

Substituting x = 92 z and y = 32 z in the equation of the constraint x 2 + 2y 2 + 6z 2 = 1 and solving for z we get  2  2 3 9 z + 2 z + 6z 2 = 1 2 2 123 2 z =1 4

⇒

z1 = √

2 123

, z2 = − √

2 123

Using the relations x = 92 z, y = 32 z we get x1 =

9 2 9 = √ , ·√ 2 123 123

x2 =

−2 9 9 ·√ = −√ , 2 123 123

y1 =

We obtain the following critical points:  9 3 2 ,√ ,√ p1 = √ 123 123 123

3 2 3 = √ , ·√ 2 123 123

y2 =

z1 = √

−2 3 3 ·√ = −√ , 2 123 123

and

2 123

z2 = − √

2 123

 9 2 3 ,−√ p2 = − √ ,−√ 123 123 123

Critical points are also the points on the constraint where ∇g = 0. However, ∇g = 2x, 4y, 12z = 0, 0, 0 only at the origin, and this point does not lie on the constraint. Step 4. Computing f at the critical points. We evaluate f (x, y, z) = 3x + 2y + 4z at the critical points:  6 8 41 41 27 ≈ 3.7 +√ +√ = √ = f ( p1 ) = √ 3 123 123 123 123  6 8 41 41 27 f ( p2 ) = − √ −√ −√ = −√ =− ≈ −3.7 3 123 123 123 123 Since f is continuous and the constraint is closed and bounded in R 3 , f has global extrema under the constraint. We conclude that the minimum value of f under the constraint is about −3.7 and the maximum value is about 3.7. 13. f (x, y, z) = x y + 3x z + 2yz, 5x + 9y + z = 10 f (x, y, z) = x 2 − y − z, x 2 − y 2 + z = 0 SOLUTION We show that f (x, y, z) = x y + 3x z + 2yz does not have minimum and maximum values subject to the constraint g(x, y, z) = 5x + 9y + z − 10 = 0. First notice that the curve c1 : (x, x, 10 − 14x) lies on the surface of the constraint since it satisfies the equation of the constraint. On c1 we have, f (x, y, z) = f (x, x, 10 − 14x) = x 2 + 3x(10 − 14x) + 2x(10 − 14x) = −69x 2 + 50x   Since lim −69x 2 + 50x = −∞, f does not have minimum value on the constraint. Notice that the curve c2 : x→∞ (x, −x, 10 + 4x) also lies on the surface of the constraint. The values of f on c2 are f (x, y, z) = f (x, −x, 10 + 4x) = −x 2 + 3x(10 + 4x) − 2x(10 + 4x) = 3x 2 + 10x The limit lim (3x 2 + 10x) = ∞ implies that f does not have a maximum value subject to the constraint. x→∞

15. Use Lagrange multipliers to find the point (a, b) on the graph of y = e x , where the value ab is as small as possible. Let f (x, y) = x 3 + x y + y 3 , g(x, y) = x 3 − x y + y 3 . SOLUTION We must find the point where f (x, y) = x y has a minimum value subject to the constraint g(x, y) = e x −(a) y =Show 0. that there is a unique point P = (a, b) on g(x, y) = 1 where ∇ f P = λ ∇g P for some scalar λ .   x maximum (b) Refer to Figure 11 to determine whether f (P) a local minimum of f subject to the ∇ constraint. , −1 , the Lagrange Condition f = λ ∇g Step 1. Write out the Lagrange Equations. Since ∇ f =isy, x and ∇g = eor is (c) Does Figure 11 suggest that f (P) is a global extremum subject to the constraint?   y, x = λ e x , −1 The Lagrange equations are thus y = λ ex x = −λ

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Step 2. Solve for λ in terms of x and y. The Lagrange equations imply that

λ = ye−x

and

λ = −x

Step 3. Solve for x and y using the constraint. We equate the two expressions for λ to obtain ye−x = −x

⇒

y = −xe x

We now substitute y = −xe x in the equation of the constraint and solve for x: e x − (−xe x ) = 0 e x (1 + x) = 0 Since e x  = 0 for all x, we have x = −1. The corresponding value of y is determined by the relation y = −xe x . That is, y = −(−1)e−1 = e−1 We obtain the critical point (−1, e−1 ) Step 4. Calculate f at the critical point. We evaluate f (x, y) = x y at the critical point. f (−1, e−1 ) = (−1) · e−1 = −e−1 We conclude (see Remark) that the minimum value of x y on the graph of y = e x is −e−1 , and it is obtained for x = −1 and y = e−1 . Remark: Since the constraint is not bounded, we need to justify the existence of a minimum value. The values f (x, y) = x y on the constraint y = e x are f (x, e x ) = h(x) = xe x . Since h(x) > 0 for x > 0, the minimum value (if it exists) occurs at a point x < 0. Since lim xe x =

x→−∞

lim

x

x→−∞ e−x

=

lim

1

x→−∞ −e−x

=

lim −e x = 0,

x→−∞

then for x < some negative number −R, we have | f (x) − 0| < 0.1, say. Thus, on the bounded region −R ≤ x ≤ 0, f has a minimum value of −e−1 ≈ −0.37, and this is thus a global minimum (for all x).  2 + h 2 , and its volume is V = 1 π r 2 h. 17. TheFind surface area of a right-circular cone of radius r and height h is S = π r the rectangular box of maximum volume if the sum of the lengths ofr the edges is 300 cm. 3 (a) Determine the ratio h/r for the cone with given surface area S and maximal volume V . (b) What is the ratio h/r for a cone with given volume V and minimal surface area S? (c) Does a cone with given volume V and maximal surface area exist? SOLUTION

(a) Let S0 denote a given surface area. We must find the ratio hr for which the function V (r, h) = 13 π r 2 h has maximum   value under the constraint S(r, h) = π r r 2 + h 2 = π r 4 + h 2 r 2 = S0 . Step 1. Write out the Lagrange Equation. We have     2r h r 2 2r 3 + h 2 r hr 2 , ∇V = π , and ∇ S = π  3 3 r 4 + h2r 2 r 4 + h 2r 2 The Lagrange Condition ∇V = λ ∇ S gives the following equations: 2r 3 + h 2 r 2r h =  λ 3 r 4 + h 2r 2

⇒

2r 2 + h 2 2h =  λ 3 r 4 + h 2r 2

hr 2 r2 =  λ 3 r 4 + h 2r 2

⇒

h 1 =  λ 3 r 4 + h2r 2

Step 2. Solve for λ in terms of r and h. These equations yield two expressions for λ that must be equal:  2h r 4 + h 2 r 2 1  4 λ= = r + h2r 2 3 2r 2 + h 2 3h Step 3. Solve for r and h using the constraint. We have  1  4 2h r 4 + h 2 r 2 = r + h 2r 2 2 2 3 2r + h 3h

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1 1 = h 2r 2 + h 2 2h 2 = 2r 2 + h 2

 We substitute h 2 = 2r 2 in the constraint π r r 2 + h 2 =  π r r 2 + 2r 2 = S0  π r 3r 2 = S0 √ 3π r 2 = S0 ⇒

h 2 = 2r 2

⇒

⇒

√ h = 2 r

S0 and solve for r . This gives

S r2 = √ 0 , 3π

2S h 2 = 2r 2 = √ 0 3π   2 +h 2 r 2 2r hr √ √ Extreme values can occur also at points on the constraint where ∇ S = , = 0, 0, that is, at 4 2 2 4 2 2 r +h r

r +h r

(r, h) = (0, h), h  = 0. However, since the radius of the cone is positive (r > 0), these points are irrelevant. We conclude that for the cone with surface area S0 and maximum volume, the following holds:   √ 2S0 S h = 2, h = √ , r = √ 0 r 3π 3π For the surface area S0 = 1 we get 

 2 ≈ 0.6, √ 3π

1 = 0.43 √ 3π  (b) We now must find the ratio hr that minimizes the function S(r, h) = π r r 2 + h 2 under the constraint h=

V (r, h) =

r=

1 2 π r h = V0 3

Using the gradients computed in part (a), the Lagrange Condition ∇ S = λ ∇V gives the following equations: 2r h 2r 3 + h 2 r  =λ 4 2 2 3 r +h r

2r 2 + h 2 2h  =λ 4 2 2 3 r +h r

⇒

r2 hr 2  =λ 3 r 4 + h2r 2

λ h  = 3 r 4 + h2r 2

These equations give 1 2r 2 + h 2 λ h  = =  4 3 2h r 4 + h 2 r 2 r + h 2r 2 We simplify and solve for hr : 2r 2 + h 2 =h 2h 2r 2 + h 2 = 2h 2 2r 2 = h 2

⇒

√ h = 2 r

We conclude that the ratio hr for a cone with a given volume and minimal surface area is √ h = 2 r (c) The constant V = 1 gives 13 π r 2 h = 1 or h = 3 2 . As r → ∞, we have h → 0, therefore πr

 lim S(r, h) = lim π r r 2 + h 2 = ∞

r →∞ h→0

r →∞ h→0

That is, S does not have maximum value on the constraint, hence there is no cone of volume 1 and maximal surface area. problem In Example 1, we found the maximum of f (x, y) = x + 2y on the ellipse 3x 2 + 4y 2 = 3. Solve this √ 3 sin t. again without using Lagrange multipliers. First, show that the ellipse is parametrized by x = cos t, y = 2

 √

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19. Use Lagrange multipliers to find the maximum area of a rectangle inscribed in the ellipse (Figure 12): x2 y2 + 2 =1 2 a b y (−x, y)

(x, y) x

(−x, −y)

(x, −y)

x2 y2 + = 1. a2 b2

FIGURE 12 Rectangle inscribed in the ellipse

SOLUTION Since (x, y) is in the first quadrant, x > 0 and y > 0. The area of the rectangle is 2x · 2y = 4x y. The vertices lie on the ellipse, hence their coordinates (±x, ±y) must satisfy the equation of the ellipse. Therefore, we must find the maximum value of the function f (x, y) = 4x y under the constraint

x2 y2 g(x, y) = 2 + 2 = 1, a b

x > 0,

y > 0.

  Step 1. Write out the Lagrange Equations. The gradient vectors are ∇ f = 4y, 4x and ∇g = 2x2 , 2y2 , hence the a b Lagrange Condition ∇ f = λ ∇g gives   2x 2y 4y, 4x = λ 2 , 2 a b or



2x a2

4y = λ 

2y b2

4x = λ



x 2y = λ 2 a y 2x = λ 2 b

⇒



Step 2. Solve for λ in terms of x and y. The Lagrange equations give the following two expressions for λ :

λ=

2ya 2 , x

λ=

2xb2 y

Equating the two equations we get 2xb2 2ya 2 = x y Step 3. Solve for x and y using the constraint. We solve the equation in step 2 for y in terms of x: 2xb2 2ya 2 = x y 2y 2 a 2 = 2x 2 b2 y2 =

x 2 b2 a2

⇒

y=

b x a

2 2 We now substitute y = ab x in the equation of the constraint x 2 + y2 = 1 and solve for x: a b  2 b ax x2 + =1 a2 b2

x2 x2 + 2 =1 2 a a 2x 2 =1 a2 x2 =

a2 2

⇒

a x= √ 2

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The corresponding value of y is obtained by the relation y = ab x: y=

b b a ·√ = √ a 2 2

    We obtain the critical point √a , √b . Extreme values can also occur at points on the constraint where ∇g = 2x2 , 2y2 = a b 2 2 0, 0. However, the point (0, 0) is not on the constraint. We conclude  that if f (x, y) = 4x y has a maximum value on the

2 2 ellipse x 2 + y2 = 1 with x > 0, y > 0, then it occurs at the point √a , √b

a

b

2

 f

a b √ ,√ 2 2



and the maximum value is

2

a b = 4 · √ · √ = 2ab 2 2

We now justify why the maximum value exists. We consider the problem of finding the extreme values of f (x, y) = 4x y 2

2

on the quarter ellipse x 2 + y2 = 1 in the first quadrant. Since the constraint curve is bounded and f (x, y) is continuous, a b f has a minimum and maximum values on the ellipse. The minimum volume occurs at the end points: x = 0, y = b ⇒ 4x y = 0   So the critical point √a , √b must be a maximum. 2

x = a,

or

y=0

⇒

4x y = 0

2

21. FindShow the maximum value y) = xtoa the y b for x, yon≥the 0 on theaxunit circle, a, b > 0 are constants. ) closest origin line + by = cwhere has coordinates that the point (x 0of, yf0(x, SOLUTION We must find the maximum value of f (x, y) = x a y b (a, b > 0) subject to the constraint g(x, y) = ac bc x0 = 2 , y0 = 2 x 2 + y 2 = 1. a + b2  a + b2  Step 1. Write out the Lagrange Equations. We have ∇ f = ax a−1 y b , bx a y b−1 and ∇g = 2x, 2y. Therefore the Lagrange Condition ∇ f = λ ∇g is   ax a−1 y b , bx a y b−1 = λ 2x, 2y or ax a−1 y b = 2λ x

(1)

bx a y b−1 = 2λ y

Step 2. Solve for λ in terms of x and y. If x = 0 or y = 0, f has the minimum value 0. We thus may assume that x > 0 and y > 0. The equations (1) imply that

λ=

ax a−2 y b , 2

λ=

bx a y b−2 2

Step 3. Solve for x and y using the constraint. Equating the two expressions for λ and solving for y in terms of x gives bx a y b−2 ax a−2 y b = 2 2 ax a−2 y b = bx a y b−2 ay 2 = bx 2 y2 = We now substitute y =



b 2 x a

 ⇒

y=

b x a

b 2 2 a x in the constraint x + y = 1 and solve for x > 0. We obtain

x2 +

b 2 x =1 a

(a + b)x 2 = a x2 = We find y using the relation y =



a a+b

 ⇒

x=

a a+b

b a x:

    b a ab b = = y= a a+b a(a + b) a+b

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We obtain the critical point:



a , a+b



b a+b



Extreme points can also occur where ∇g = 0, that is, 2x, 2y = 0, 0 or (x, y) = (0, 0). However, this point is not on the constraint. Step 4. Conclusions. We compute f (x, y) = x a y b at the critical point: 

    a/2  b/2 b a a b a a bb a a/2 bb/2 , f = = = (a+b)/2 a+b a+b a+b a+b (a + b)a+b (a + b) The function f (x, y) = x a y b is continuous on the set x 2 + y 2 = 1, x ≥ 0, y ≥ 0, which is a closed and bounded set in R 2 , hence f has minimum and maximum values on the set. The minimum value is 0 (obtained at (0, 1) and (1, 0)), hence the critical point that we found corresponds to the maximum value. We conclude that the maximum value of x a y b on x 2 + y 2 = 1, x > 0, y > 0 is  a a bb . (a + b)a+b 23. Find the maximum value of f (x, y, z) = x aay bbz c for x, y, z ≥ 0 on the unit sphere, where a, b, c > 0 are constants. Find the maximum value of f (x, y) = x y for x, y ≥ 0 on the line x + y = 1, where a, b > 0 are constants. SOLUTION We must find the maximum value of f (x, y, z) = x a y b z c subject to the constraint g(x, y, z) = x 2 + y 2 + z 2 − 1 = 0, x ≥ 0, y ≥ 0, z ≥ 0.   a Step 1. Write the Lagrange Equations. The gradient vectors are ∇ f = ax a−1 y b z c , by b−1 x z c , cz c−1 x a y b and ∇g = 2x, 2y, 2z, hence the Lagrange Condition ∇ f = λ ∇g gives the following equations: ax a−1 y b z c = λ (2x) by b−1 x a z c = λ (2y) cz c−1 x a y b =

(1)

λ (2z)

Step 2. Solve for λ in terms of x, y, and z. If x = 0, y = 0, or z = 0, f attains the minimum value 0, therefore we may assume that x  = 0, y  = 0, and z  = 0. The Lagrange equations (1) give

λ =

ax a−2 y b z c , 2

λ=

by b−2 x a z c , 2

λ=

cz c−2 x a y b 2

Step 3. Solve for x, y, and z using the constraint. Equating the expressions for λ , we obtain the following equations: ax a−2 y b z c = by b−2 x a z c ax a−2 y b z c = cz c−2 x a y b

(2)

We solve for x and y in terms of z. We first divide the first equation by the second equation to obtain 1=

b by b−2 x a z c = c cz c−2 x a y b

b y2 = z2 c

⇒

y=

z2 y2 

(3) b z c

Both equations (2) imply that a

by b−2 x z c = ax a−2 y b z c a

by b−2 x z c = cz c−2 x a y b Dividing the first equation by the second equation gives ax a−2 y b z c a z2 = c x2 cz c−2 x a y b  a a 2 2 z x = z ⇒ x= c c 1=

(4)

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We now substitute x and y from (3) and (4) in the constraint x 2 + y 2 + z 2 = 1 and solve for z. This gives 

a z c



2 + 

b z c

2 + z2 = 1

a b + + 1 z2 = 1 c c a+b+c 2 z =1 c

 ⇒

z=

c a+b+c

We find x and y using (4) and (3):     a c ac a = = c a+b+c c(a + b + c) a+b+c     b c bc b = = y= c a+b+c c(a + b + c) a+b+c

x=

We obtain the critical point:

 P=

a , a+b+c



b , a+b+c



c a+b+c



We examine the point where ∇g = 2x, 2y, 2z = 0, 0, 0, that is, (0, 0, 0): This point does not lie on the constraint, hence it is not a critical point. Step 4. Conclusions. We compute f (x, y, z) = x a y b z c at the critical point: b   a  c  a b c a a bb cc = f (P) = a+b+c a+b+c a+b+c (a + b + c)a+b+c Now, f (x, y, z) = x a y b z c is continuous on the set x 2 + y 2 + z 2 = 1, x ≥ 0, y ≥ 0, z ≥ 0, which is closed and bounded in R 3 . The minimum value is 0 (obtained at the points with at least one zero coordinate), therefore the critical point that we found corresponds to the maximum value. We conclude that the maximum value of x a y b z c subject to the constraint x 2 + y 2 + z 2 = 1, x ≥ 0, y ≥ 0, z ≥ 0 is  a a bb cc (a + b + c)a+b+c 25. Let f (x, y, z) = y + z − x 2 . Show that the Lagrange equations for f (x, y, z) = x 2 y + zy 2 subject to the constraint g(x, y) = x + yz = 4 (a) Find the solutions to the Lagrange equations for f subject to the constraint g(x, y, z) = x 2 − y 2 + z 3 = 0. Hint: have no solution. What can you conclude about the minimum and maximum values of f subject to g = 0? Show that at a critical point, λ , y, z must be nonzero and x = 0. (b) Show that f has no minimum or maximum subject to the constraint. Hint: Consider the values of f at the points (0, y 3 , y 2 ), which satisfy the constraint. (c) Does (b) contradict Theorem 1? SOLUTION

  (a) The gradient vectors are ∇ f = −2x, 1, 1 and ∇g = 2x, −2y, 3z 2 , hence the Lagrange Condition ∇ f = λ ∇g yields the following equations: −2x = λ (2x) 1 = λ (−2y)   1 = λ 3z 2

⇒

−2x = 2λ x

(1)

1 = −2λ y

(2)

1 = 3λ z 2

(3)

We solve the Lagrange equations. Equations (2) and (3) imply that y  = 0, z  = 0, λ  = 0, and that 1 2y

and

1 1 = 2 2y 3z

⇒

λ =−

λ=

1 3z 2

Equating and solving for y in terms of z gives −

3 y = − z2 2

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Equation (1) gives (2λ + 2)x = 0

λ = −1 or

→

x =0

By equation (3), λ must be positive, hence the solution λ = −1 is not appropriate. We check the solution x = 0. Substituting x = 0 and y = − 32 z 2 in the equation of the constraint x 2 − y 2 + z 3 = 0 gives  2 3 02 − − z 2 + z 3 = 0 2 9 − z4 + z3 = 0 4 Since z  = 0, we can divide by z 3 to obtain 9 − z+1=0 4

⇒

z=

4 9

We find y by the relation y = − 32 z 2 : 3 y=− · 2

 2 4 8 3 · 16 =− =− 9 2 · 81 27

We obtain the following solution of the Lagrange equations:  8 4 0, − , 27 9   Extreme values may also occur at points on the constraint where ∇g = 2x, −2y, 3z 2 = 0, 0, 0. 2

3

(b) The points (0, y 3 , y 2 ) satisfy the constraint since 02 − (y 3 ) + (y 2 ) = −y 6 + y 6 = 0. We examine the values of f (x, y, z) = y + z − x 2 at these points: f (0, y 3 , y 2 ) = y 3 + y 2 − 02 = y 3 + y 2 Since

lim (y 3 + y 2 ) = −∞ and lim (y 3 + y 2 ) = ∞, f has no absolute minimum or absolute maximum subject to

y→−∞

y→∞

the constraint. (c) Theorem 1 states that the solutions of the Lagrange equations are the only points where local extrema may occur subject to the constraint. As shown above, our function has no global extrema, but it does have local extrema. It’s not particularly hard to show that (0, 0, 0) is a local minimum and that (0, −8/27, 4/9) is a local minimum. There is no contradiction. 27. Let Q be multipliers the point on closest the+ellipse. It = was known to the the origin. Greek , y0 , zpoint theoutside plane ax by + cz d closest Use Lagrange to an findellipse the point P =to(xa 0given 0 ) on P Then calculate the distance P toBCE O.) that P Q is perpendicular to the tangent at Q (Figure 13). Explain in words mathematician Apollonius (thirdfrom century why this conclusion is a consequence of the method of Lagrange multipliers. Hint: The circles centered at P are level curves of the function to be minimized. P

Q

FIGURE 13 SOLUTION Let P = (x 0 , y0 ). The distance d between the point P and a point Q = (x, y) on the ellipse is minimum where the square d 2 is minimum (since the square function u 2 is increasing for u ≥ 0). Therefore, we want to minimize the function

f (x, y, z) = (x − x 0 )2 + (y − y0 )2 + (z − z 0 )2 subject to the constraint y2 x2 g(x, y) = 2 + 2 = 1 a b The method of Lagrange indicates that the solution Q is the point on the ellipse where ∇ f = λ ∇g, that is, the point on the ellipse where the gradients ∇ f and ∇g are parallel. Since the gradient is orthogonal to the level curves of the

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425

function, ∇g is orthogonal to the ellipse g(x, y) = 1, and ∇ f is orthogonal to the level curve of f passing through Q. But this level curve is a circle through Q centered at P, hence the parallel vectors ∇g and ∇ f are orthogonal to the ellipse and to the circle centered at P respectively. We conclude that the point Q is the point at which the tangent to the ellipse is also the tangent to the circle through Q centered at P. That is, the tangent to the ellipse at Q is perpendicular to the radius P Q of the circle. 29. Find the maximum value of f (x, y, z) = x y + x z + yz − 4x yz subject to the constraints x + y + z = 1 and Antonio has $5.00 to spend on a lunch consisting of hamburgers ($1.50 each) and French fries ($1.00 per orx, y, z ≥ 0. der). Antonio’s satisfaction from eating x 1 hamburgers and x 2 orders of French fries is measured by a function √ SOLUTION U (x 1 , x 2 ) = x 1 x 2 . How much of each type of food should he purchase to maximize his satisfaction (assume that each food can be purchased)? Stepfractional 1. Write amounts out the of Lagrange Equations. We have ∇ f = y + z − 4yz, x + z − 4x z, x + y − 4x y and ∇g = 1, 1, 1, hence the Lagrange Condition ∇ f = λ ∇g yields the following equations: y + z − 4yz = λ x + z − 4x z = λ x + y − 4x y = λ Step 2. Solve for x, y, and z using the constraint. The Lagrange equations imply that x + z − 4x z = y + z − 4yz x + y − 4x y = y + z − 4yz

x − 4x z = y − 4yz

⇒

x − 4x y = z − 4yz

(1)

We solve for x and y in terms of z. The first equation gives x − y + 4yz − 4x z = 0 x − y − 4z(x − y) = 0 (x − y)(1 − 4z) = 0

(2) ⇒

x=y

1 z= 4

or

The second equation in (1) gives: x − z + 4yz − 4x y = 0 x − z − 4y(x − z) = 0 (x − z)(1 − 4y) = 0

(3) ⇒

x=z

or

y=

1 4

We examine the possible solutions. 1. x = y, x = z. Substituting x = y = z in the equation of the constraint x + y + z = 1 gives 3z = 1 or z = 13 . We obtain the solution  1 1 1 , , 3 3 3 2. x = y, y = 14 . Substituting x = y = 14 in the constraint x + y + z = 1 gives 1 1 + +z =1 4 4 We obtain the solution



⇒

1 1 1 , , 4 4 2

z=

1 2

y=

1 2



3. z = 14 , x = z. Substituting z = 14 , x = 14 in the constraint gives 1 1 +y+ =1 4 4

⇒

We get the point 

1 1 1 , , 4 2 4



4. z = 14 , y = 14 . Substituting in the constraint gives x + 14 + 14 = 1 or x = 12 . We obtain the point  1 1 1 , , 2 4 4

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We conclude that the critical points are 

1 1 1 , , , 3 3 3  1 1 1 , , P3 = , 4 2 4 P1 =

 P2 =  P4 =

1 1 1 , , 4 4 2 1 1 1 , , 2 4 4

(4)

Step 3. Conclusions. The constraint x + y + z = 1, x ≥ 0, y ≥ 0, z ≥ 0 is the part of the plane x + y + z = 1 in the first octant. This is a closed and bounded set in R 3 , hence f (which is a continuous function) has minimum and maximum value subject to the constraint. The extreme values occur at points from (4). We evaluate f (x, y, z) = x y + x z + yz − 4x yz at these points: 1 1 1 3 4 5 1 1 1 1 1 1 · + · + · −4· · · = − = 3 3 3 3 3 3 3 3 3 9 27 27 1 1 1 3 1 1 1 1 1 1 f (P2 ) = f (P3 ) = f (P4 ) = · + · + · − 4 · · · = 4 4 4 2 4 2 4 4 2 16

f (P1 ) =

We conclude that the maximum value of f subject to the constraint is f (P2 ) = f (P3 ) = f (P4 ) =

3 . 16

31. With the same set-up as in the y z problem, find the plane that minimizes V if the plane is constrained to pass x previous + + = 1 (a, b, c > 0) together with the positive coordinate planes forms a plane Pwith equation throughAa point = (α a b c 1 tetrahedron of volume V = 6 abc (Figure 14). Find the plane that minimizes V if the plane is constrained to pass through the point P = (1, 1, 1).

Lagrange Multipliers: Optimizing with a Constraint

S E C T I O N 15.8

γ γ γ + + =1 c c c 3γ =1 c

(ET Section 14.8)

427

c = 3γ

⇒

We find a and b using (1): a=

α · 3γ = 3α , γ

b=

β · 3γ = 3β γ

We obtain the solution P = (3α , 3β , 3γ ) Step 3. Conclusions. Since V has a minimum value subject to the constraint, it occurs at the critical point. We substitute a = 3α , b = 3β , and c = 3γ in the equation of the plane ax + by + cz = 1 to obtain the following plane, which minimizes V: x y z x y z + + = 1 or + + =3 3α 3β 3γ α β γ 33. Let L be the minimum length of a ladder that can reach over a fence of height h to a wall located a distance b behind the wall. In a contest, a runner starting at A must touch a point P along a river and then run to B in the shortest time possible (Figure 15). The runner should choose the point P minimizing the total length of the path. (a) Use Lagrange multipliers to show that L = (h 2/3 + b2/3 )3/2 (Figure 16). Hint: Show that the problem amounts to 2+ (a) Define function (x, y) (y =+Ah) P 2+subject P B, where (x,ory). minimizing f (x,a y) = (x +f b) to y/bP==h/x x yRephrase = bh. the runner’s problem as a constrained optimization problem, assuming that the river is given by an equation g(x, y) = 0.−h) that is tangent to the graph of (b) Show that the value of L is also equal to the radius of the circle with center (−b, x y =(b) bh.Explain why the level curves of f (x, y) are ellipses. (c) Use Lagrange multipliers to justify the following statement: The ellipse through the point P minimizing the y length of the path is tangent to the river. xy = bh (d) Identify the point on the river in Figure 15 for which the length is minimal. y

L L

Wall h b

x

(−b, −h)

Ladder x Fence

FIGURE 16 SOLUTION

(a) We denote by x and y the lengths shown in the figure, and express the length l of the ladder in terms of x and y. A y D

E

h b o

x

C

B

Using the Pythagorean Theorem, we have   2 2 l = O A + O B = (y + h)2 + (x + b)2

(1)

Since the function u 2 is increasing for u ≥ 0, l and l 2 have their minimum values at the same point. Therefore, we may minimize the function f (x, y) = l 2 (x, y), which is f (x, y) = (x + b)2 + (y + h)2 We now identify the constraint on the variables x and y. (Notice that h, b are constants while x and y are free to change). Using proportional lengths in the similar triangles  AE D and DC B, we have AE DC

=

ED CB

That is, b y = h x

⇒

x y = bh

We thus must minimize f (x, y) = (x + b)2 + (y + h)2 subject to the constraint g(x, y) = x y = bh, x > 0, y > 0.

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Step 1. Write out the Lagrange Equations. We have ∇ f = 2(x + b), 2(y + h) and ∇g = y, x, hence the Lagrange Condition ∇ f = λ ∇g gives the following equations: 2(x + b) = λ y 2(y + h) = λ x Step 2. Solve for λ in terms of x and y. The equation of the constraint implies that y  = 0 and x  = 0. Therefore, the Lagrange equations yield

λ=

2(x + b) , y

2(y + h) x

λ=

Step 3. Solve for x and y using the constraint. Equating the two expressions for λ gives 2(y + h) 2(x + b) = y x We simplify: x(x + b) = y(y + h) x 2 + xb = y 2 + yh The equation of the constraint implies that y = bh x . We substitute and solve for x > 0. This gives  2 bh bh ·h + x 2 + xb = x x x 2 + xb =

b2 h 2 bh 2 + 2 x x

x 4 + x 3 b = b2 h 2 + bh 2 x x 4 + bx 3 − bh 2 x − b2 h 2 = 0 x 3 (x + b) − bh 2 (x + b) = 0   x 3 − bh 2 (x + b) = 0 Since x > 0 and b > 0, also x + b > 0 and the solution is x 3 − bh 2 = 0

x = (bh 2 )

⇒

1/3

We compute y. Using the relation y = bh x , y=

bh (bh 2 )

1/3

bh 1/3 = 1/3 2/3 = b2/3 h 1/3 = (b2 h) b h

We obtain the solution  1/3 , x = bh 2

 1/3 y = b2 h

(2)

Extreme values may also occur at the point on the constraint where ∇g = 0. However, ∇g = y, x = 0, 0 only at the point (0, 0), which is not on the constraint. bh Step 4. Conclusions. Notice that on the constraint y = bh x or x = y , as x → 0+ then y → ∞, and as x → ∞, then y → 0+. Also, as y → 0+, x → ∞ and as y → ∞, x → 0+. In either case, f (x, y) is increasing without bound. Using this property and the theorem on the existence of extreme values for a continuous function on a closed and bounded set (for a certain part of the constraint), one can show that f has a minimum value on the constraint. This minimum value occurs at the point (2). We substitute this point in (1) to obtain the following minimum length L:  2  2  1/3 1/3 L= + h + (bh 2 ) +b (b2 h) = = =

 2/3 1/3 2/3 1/3 (b2 h) + 2h(b2 h) + h 2 + (bh 2 ) + 2b(bh 2 ) + b2  4

4

4

4

b 3 h 2/3 + 2h 3 b2/3 + h 2 + b2/3 h 3 + 2b 3 h 2/3 + b2  4

4

3b 3 h 2/3 + 3h 3 b2/3 + h 2 + b2

Lagrange Multipliers: Optimizing with a Constraint

S E C T I O N 15.8

=



h 2/3

3

(ET Section 14.8)

429

2

2

3 + 3 h 2/3 b2/3 + 3h 2/3 b2/3 + b2/3

Using the identity (α + β )3 = α 3 + 3α 2 β + 3αβ 2 + β 3 , we conclude that  3/2

3  h 2/3 + b2/3 = h 2/3 + b2/3 . L= (b) The Lagrange Condition states that the gradient vectors ∇ f P and ∇g P are parallel (where P is the minimizing point). The gradient ∇ f P is orthogonal to the level curve of f passing through P, which is a circle through P centered at (−b, −h). ∇g P is orthogonal to the level curve of g passing through P, which is the curve of the constraint x y = bh. We conclude that the circle and the curve x y = bh, both being perpendicular to parallel vectors, are tangent at P. The radius of the circle is the minimum value L, of f (x, y). 35. Find the minimum and maximum of f (x, y, z) = x 2 + y 2 + z 2 subject to two constraints, x + 2y +2 z =2 3 and x − y =Find 4. the minimum and maximum of f (x, y, z) = y + 2z subject to two constraints, 2x + z = 4 and x + y = 1. SOLUTION

The constraint equations are g(x, y, z) = x + 2y + z − 3 = 0,

h(x, y) = x − y − 4 = 0

Step 1. Write out the Lagrange Equations. We have ∇ f = 2x, 2y, 2z, ∇g = 1, 2, 1, and ∇h = 1, −1, 0, hence the Lagrange Condition is ∇ f = λ ∇g + μ ∇h 2x, 2y, 2z = λ 1, 2, 1 + μ 1, −1, 0   = λ + μ , 2λ − μ , λ We obtain the following equations: 2x = λ + μ 2y = 2λ − μ 2z = λ Step 2. Solve for λ and μ . The first equation gives λ = 2x − μ . Combining with the third equation we get 2z = 2x − μ

(1)

The second equation gives μ = 2λ − 2y, combining with the third equation we get μ = 4z − 2y. Substituting in (1) we obtain 2z = 2x − (4z − 2y) = 2x − 4z + 2y 6z = 2x + 2y

⇒

z=

x+y 3

(2)

Step 3. Solve for x, y, and z using the constraints. The constraints give x and y as functions of z: x−y=4 ⇒ x + 2y + z = 3 ⇒

y = x −4 y=

3−x −z 2

Combining the two equations we get 3−x −z 2 2x − 8 = 3 − x − z x −4=

3x = 11 − z

⇒

x=

11 − z 3

We find y using y = x − 4: y=

−1 − z 11 − z −4= 3 3

We substitute x and y in (2) and solve for z: z=

11−z + −1−z 11 − z − 1 − z 10 − 2z 3 3 = =

3

9

9

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9z = 10 − 2z 11z = 10

⇒

z=

10 11

We find x and y: y=

−1 − 10 −1 − z 11 = − 21 = − 7 = 3 3 33 11

x=

11 − 10 11 − z 11 = 111 = 37 = 3 3 33 11

We obtain the solution

 P=

7 10 37 ,− , 11 11 11



Step 4. Calculate the critical values. We compute f (x, y, z) = z 2 + y 2 + z 2 at the critical point:  f (P) =

  2 37 2 10 138 7 2 1518 = ≈ 12.545 + − + = 11 11 11 121 11

As x tends to infinity, so also does f (x, y, z) tend to ∞. Therefore f has no maximum value and the given critical point P must produce a minimum. We conclude that the minimum value of f subject to the two constraints is f (P) = 138 11 ≈ 12.545.

Further Insights and Challenges 37. Assumptions Matter Consider the problem of minimizing f (x, y) = x subject to g(x, y) = (x − 1)3 − Suppose that both f (x, y) and the constraint function g(x, y) are linear. Use contour maps to explain why y 2 =f 0. (x, y) does not have a maximum subject to g(x, y) = 0 unless g = a f + b for some constants a, b. (a) Show, without using calculus, that the minimum occurs at P = (1, 0). (b) Show that the Lagrange condition ∇ f P = λ ∇g P is not satisfied for any value of λ . (c) Does this contradict Theorem 1? SOLUTION

(a) The equation of the constraint can be rewritten as (x − 1)3 = y 2

or

x = y 2/3 + 1

Therefore, at the points under the constraint, x ≥ 1, hence f (x, y) ≥ 1. Also at the point P = (1, 0) we have f (1, 0) = 1, hence f (1, 0) = 1 is the minimum value of f under the constraint.   (b) We have ∇ f = 1, 0 and ∇g = 3(x − 1)2 , −2y , hence the Lagrange Condition ∇ f = λ ∇g yields the following equations: 1 = λ · 3(x − 1)2 0 = −2λ y The first equation implies that λ  = 0 and x − 1 = ± √1 . The second equation gives y = 0. Substituting in the equation 3λ of the constraint gives  (x − 1)3 − y 2 =

±1 3 ±1 − 02 = = 0 √ 3λ (3λ )3/2

We conclude that the Lagrange Condition is not satisfied by any point under the constraint. (c) Theorem 1 is not violated since at the point P = (1, 0), ∇g = 0, whereas the Theorem is valid for points where ∇g P  = 0. Consider the utility function U (x 1 , x 2 ) = x 1 x 2 with budget constraint p1 x 1 + p2 x 2 = c. Marginal Utility Goods 1 and 2 are available at prices (in dollars) of p per unit of good 1 and p2 per unit of Show that the maximum of U (x 1 , x 2 ) subject to the budget constraint is equal1 to c2 /(4 p1 p2 ). good 2. A utility function U (x 1 , x 2 ) is a function representing the utility or benefit of consuming x j units of good Calculate the value of theofLagrange multiplier λ occurring in of (a).increase in utility per unit increase in the jth good. j. The marginal utility the jth good is ∂U/∂ x j , the rate Prove the following interpretation: λ is the rate of increase in utility per unit is increase of total c. following law of economics: Given a budget of L dollars, utility maximized at budget the consumption level (a, b) where the ratio of marginal utility is equal to the ratio of prices: SOLUTION

39. (a) (b) (c)

Ux1 (a, b) p Marginal utility of good 1 = = 1 Marginal utility of good 2 Ux2 (a, b) p2

S E C T I O N 15.8

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431

(a) By the earlier exercise, the utility is maximized at a point where the following equality holds: Ux1 p = 1 Ux2 p2 Since Ux1 = x 2 and Ux2 = x 1 , we get x2 p = 1 x1 p2

⇒

x2 =

p1 x p2 1

We now substitute x 2 in terms of x 1 in the constraint p1 x 1 + p2 x 2 = c and solve for x 1 . This gives p p1 x 1 + p2 · 1 x 1 = c p2 2 p1 x 1 = c

⇒

x1 =

c 2 p1

The corresponding value of x 2 is computed by x 2 = pp12 x 1 : x2 =

p1 c c · = p2 2 p1 2 p2

That is, U (x 1 , x 2 ) is maximized at the consumption level x 1 = 2 cp , x 2 = 2 cp . The maximum value is 1 2  c c c c2 c , · = U = 2 p1 2 p2 2 p1 2 p2 4 p1 p2 (b) The Lagrange Condition ∇U = λ ∇g for U (x 1 , x 2 ) = x 1 x 2 and g(x 1 , x 2 ) = p1 x 1 + p2 x 2 − c = 0 is x 2 , x 1  = λ  p1 , p2 

(1)

or x 2 = λ p1 x 1 = λ p2

⇒

x x λ= 2 = 1 p1 p2

In part (a) we showed that at the maximizing point x 1 = 2 cp , therefore the value of λ is 1 x c λ= 1 = p2 2 p1 p2 (c) We compute dU dc using the Chain Rule:   ∂U  ∂U  dU = x (c) + x (c) = x 2 x 1 (c) + x 1 x 2 (c) = x 2 , x 1  · x 1 (c), x 2 (c) dc ∂ x1 1 ∂ x2 2 Substituting in (1) we get   

dU = λ  p1 , p2  · x 1 (c), x 2 (c) = λ p1 x 1 (c) + p2 x 2 (c) dc

(2)

We now use the Chain Rule to differentiate the equation of the constraint p1 x 1 + p2 x 2 = c with respect to c: p1 x 1 (c) + p2 x 2 (c) = 1 Substituting in (2), we get dU =λ ·1=λ dc Using the approximation U ≈ dU dc c, we conclude that λ is the rate of increase in utility per unit increase of total budget L. 41. Let B > 0. Show that the maximum of f (x 1 , . . . , x n ) = x 1 x 2 · · · x n subject to the constraints x 1 + · · · + x n = B This exercise thefor multiplier beB/n. interpreted rate of change and x j ≥ 0 for j = 1,shows . . . , n that occurs x 1 = · · ·λ=may xn = Use thisastoa conclude that in general. Assume that the maximum value of f (x, y) subject to g(x, y) = c occurs at a point P. Then P depends on the value of c, so we may write P = (x(c), y(c)) and we have g(x(c), y(c)) =1/n c. a1 + · · · + an ≤ (a1 a2 · · · an ) (a) Show that n   for all positive numbers a1 , . . . , an . ∇g(x(c), y(c)) · x  (c), y  (c) = 1 Hint: Use the Chain Rule to differentiate the equation g(x(c), y(c)) = c with respect to c. (b) Show that

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SOLUTION We first notice that the constraints x 1 + · · · + x n = B and x j ≥ 0 for j = 1, . . . , n define a closed and bounded set in the nth dimensional space, hence f (continuous, as a polynomial) has extreme values on this set. The minimum value zero occurs where one of the coordinates is zero (for example, for n = 2 the constraint x 1 + x 2 = B, x 1 ≥ 0, x 2 ≥ 0 is a triangle in the first quadrant). We need to maximize the function f (x 1 , . . . , x n ) = x 1 x 2 · · · x n subject to the constraints g (x 1 , . . . , x n ) = x 1 + · · · + x n − B = 0, x j ≥ 0, j = 1, . . . , n. Step 1. Write out the Lagrange Equations. The gradient vectors are   ∇ f = x 2 x 3 · · · x n , x 1 x 3 · · · x n , . . . , x 1 x 2 · · · x n−1

∇g = 1, 1, . . . , 1 The Lagrange Condition ∇ f = λ ∇g yields the following equations: x2 x3 · · · xn = λ x1 x3 · · · xn = λ x 1 x 2 · · · x n−1 = λ Step 2. Solving for x 1 , x 2 , . . . , x n using the constraint. The Lagrange equations imply the following equations: x 2 x 3 · · · x n = x 1 x 2 · · · x n−1 x 1 x 3 · · · x n = x 1 x 2 · · · x n−1 x 1 x 2 x 4 · · · x n = x 1 x 2 · · · x n−1 .. . x 1 x 2 · · · x n−2 x n = x 1 x 2 · · · x n−1 We may assume that x j  = 0 for j = 1, . . . , n, since if one of the coordinates is zero, f has the minimum value zero. We divide each equation by its right-hand side to obtain xn =1 x1 xn =1 x2 xn =1 x3 .. . xn x n−1

x1 = xn x2 = xn ⇒

x3 = xn .. . x n−1 = x n

=1

Substituting in the constraint x 1 + · · · + x n = B and solving for x n gives x + xn + · · · + xn = B n ! " n

nx n = B

⇒

xn =

B n

Hence x 1 = · · · = x n = Bn . Step 3. Conclusions. The maximum value of f (x 1 , . . . , x n ) = x 1 x 2 · · · x n on the constraint x 1 + · · · + x n = B, x j ≥ 0, j = 1, . . . , n occurs at the point at which all coordinates are equal to Bn . The value of f at this point is  f

B B B , ,... , n n n



 =

B n n

It follows that for any point (x 1 , . . . , x n ) on the constraint, that is, for any point satisfying x 1 + · · · + x n = B with x j positive, the following holds:  n B f (x 1 , . . . , x n ) ≤ n That is,  x1 · · · xn ≤

x1 + · · · + xn n n

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433

or x + · · · + xn . (x 1 · · · x n )1/n ≤ 1 n 43. Given constants E, E 1 , E 2 , E 3 , consider the maximum of √ Let B > 0. Show that the maximum of f (x 1 , . . . , x n ) = x 1 + · · · + x n subject to x 12 + · · · + x n2 = B 2 is n B. S(x 1 , x 2 , x 3 ) = x 1 ln x 1 + x 2 ln x 2 + x 3 ln x 3 Conclude that √ subject to two constraints: |a1 | + · · · + |an | ≤ n(a12 + · · · + an2 )1/2 for all numbers a1 , . . . , an . x 1 + x 2 + x 3 = N , E 1 x 1 + E 2 x 2 + E 3 x 3 = E Show that there is a constant μ such that xi = A−1 eμ Ei for i = 1, 2, 3, where A = N −1 (eμ E 1 + eμ E 2 + eμ E 3 ). SOLUTION

The constraints equations are g (x 1 , x 2 , x 3 ) = x 1 + x 2 + x 3 − N = 0 h (x 1 , x 2 , x 3 ) = E 1 x 1 + E 2 x 2 + E 3 x 3 − E = 0

We first find the Lagrange equations. The gradient vectors are   1 1 1 = 1 + ln x 1 , 1 + ln x 2 , 1 + ln x 3  ∇ S = ln x 1 + x 1 · , ln x 2 + x 2 · , ln x 3 + x 3 · x1 x2 x3 ∇g = 1, 1, 1 ,

∇h = E 1 , E 2 , E 3 

The Lagrange Condition ∇ f = λ ∇g + μ ∇h gives the following equation:

  1 + ln x 1 , 1 + ln x 2 , 1 + ln x 3  = λ 1, 1, 1 + μ E 1 , E 2 , E 3  = λ + μ E 1 , λ + μ E 2 , λ + μ E 3

We obtain the Lagrange equations: 1 + ln x 1 = λ + μ E 1 1 + ln x 2 = λ + μ E 2 1 + ln x 3 = λ + μ E 3 We subtract the third equation from the other equations to obtain ln x 1 − ln x 3 = μ (E 1 − E 3 ) ln x 2 − ln x 3 = μ (E 2 − E 3 ) or x ln 1 = μ (E 1 − E 3 ) x3 x ln 2 = μ (E 2 − E 3 ) x3

⇒

x 1 = x 3 eμ (E 1 −E 3 ) x 2 = x 3 eμ (E 2 −E 3 )

(1)

Substituting x 1 and x 2 in the equation of the constraint g(x 1 , x 2 , x 3 ) = 0 and solving for x 3 gives x 3 eμ (E 1 −E 3 ) + x 3 eμ (E 2 −E 3 ) + x 3 = N We multiply by eμ E 3 :

 x 3 eμ E 1 + eμ E 2 + eμ E 3 = N eμ E 3 N eμ E 3 x3 = μ E e 1 + eμ E 2 + eμ E 3 Substituting in (1) we get N eμ E 3 N eμ E 1 x1 = μ E · eμ (E 1 −E 3 ) = μ E μ E μ E 1 2 3 1 e +e +e e + eμ E 2 + eμ E 3 N eμ E 3 N eμ E 2 x2 = μ E · eμ (E 2 −E 3 ) = μ E μ E μ E e 1 +e 2 +e 3 e 1 + eμ E 2 + eμ E 3 μ E3 μE μE Letting A = e 1 +e N 2 +e , we obtain

x 1 = A−1 eμ E 1 ,

x 2 = A−1 eμ E 2 ,

x 3 = A−1 eμ E 3

The value of μ is determined by the second constraint h(x 1 , x 2 , x 3 ) = 0. Boltzmann Distribution Generalize Exercise 43 to n variables: Show that there is a constant μ such that the maximum of

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CHAPTER REVIEW EXERCISES 

x 2 − y2 , x +3 (a) Sketch the domain of f . (b) Calculate f (3, 1) and f (−5, −3). (c) Find a point satisfying f (x, y) = 1. 1. Given f (x, y) =

SOLUTION

(a) f is defined where x 2 − y 2 ≥ 0 and x + 3  = 0. We solve these two inequalities: x 2 − y2 ≥ 0

⇒

x 2 ≥ y2

x + 3 = 0

⇒

x  = −3

⇒

|x| ≥ |y|

Therefore, the domain of f is the following set: D = {(x, y) : |x| ≥ |y|, x  = −3} y

x

−3

(b) To find f (3, 1) we substitute x = 3, y = 1 in f (x, y). We get  √ √ 32 − 12 8 2 = = f (3, 1) = 3+3 6 3 Similarly, setting x = −5, y = −3, we get f (−5, −3) =

 (−5)2 − (−3)2

√ =

−5 + 3

(c) We must find a point (x, y) such that

 f (x, y) =

16 = −2. −2

x 2 − y2 =1 x +3

We choose, for instance, y = 1, substitute and solve for x. This gives  x 2 − 12 =1 x +3  x2 − 1 = x + 3 x 2 − 1 = (x + 3)2 = x 2 + 6x + 9 6x = −10

⇒

x=−

    Thus, the point − 53 , 1 satisfies f − 53 , 1 = 1.

5 3

3. Sketch f (x, = x 2of: − y + 1 and describe its vertical and horizontal traces. Findthe thegraph domain andy)range √ √ SOLUTION (a) f (x, The y, z)graph = xis−shown y + iny the − zfollowing figure. 2 (b) f (x, y) = ln(4x − y) z

y x

Chapter Review Exercises

435

The trace obtained by setting x = c is the line z = c2 − y + 1 or z = (c2 + 1) − y in the plane x = c. The trace obtained by setting y = c is the parabola z = x 2 − c + 1 in the plane y = c. The trace obtained by setting z = c is the parabola y = x 2 + 1 − c in the plane z = c. Use a graphing utility to draw the graph of the function cos(x 2 + y 2 )e1−x y

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13.

(ET CHAPTER 14)

(2x + y)e−x+y

lim

(x,y)→(1,−3)

SOLUTION

The function f (x, y) = (2x + y)e−x+y is continuous, hence we evaluate the limit using substitution: (2x + y)e−x+y = (2 · 1 − 3)e−1−3 = −e−4

lim

(x,y)→(1,−3)

15. Let

(e x − 1)(e y − 1) x (x,y)→(0,2)

⎧ p ⎨ (x y) 4 f (x, y) = x + y 4 ⎩ 0

lim

(x, y)  = (0, 0) (x, y) = (0, 0)

Use polar coordinates to show that f (x, y) is continuous at all (x, y) if p > 2, but discontinuous at (0, 0) if p ≤ 2. SOLUTION We show using the polar coordinates x = r cos θ , y = r sin θ , that the limit of f (x, y) as (x, y) → (0, 0) is zero for p > 2. This will prove that f is continuous at the origin. Since f is a rational function with nonzero denominator for (x, y)  = (0, 0), f is continuous there. We have

lim

(x,y)→(0,0)

(r cos θ ) p (r sin θ ) p

r 2 p (cos θ sin θ ) p   = lim r →0+ (r cos θ )4 + (r sin θ )4 r →0+ r 4 cos4 θ + sin4 θ

f (x, y) = lim

= lim

(1)

r 2( p−2) (cos θ sin θ ) p

r →0+

cos4 θ + sin4 θ

We use the following inequalities: 4 cos θ sin4 θ ≤ 1 cos4 θ + sin4 θ =



cos2 θ + sin2 θ

= 1−

2

− 2 cos2 θ sin2 θ = 1 −

1 · (2 cos θ sin θ )2 2

1 1 2 1 sin 2θ ≥ 1 − = 2 2 2

Therefore, r 2( p−2) (cos θ sin θ ) p r 2( p−2) · 1 0≤ = 2r 2( p−2) ≤ 1 cos4 θ + sin4 θ 2

Since p − 2 > 0, lim 2r 2( p−2) = 0, hence by the Squeeze Theorem the limit in (1) is also zero. We conclude that f r →0+

is continuous for p > 2. We now show that for p < 2 the limit of f (x, y) as (x, y) → (0, 0) does not exist. We compute the limit as (x, y) approaches the origin along the line y = x. lim

(x,y)→(0,0) along y=x

p (x 2 ) x2p x 2( p−2) =∞ = lim = lim 2 x→0 x 4 + x 4 x→0 2x 4 x→0

f (x, y) = lim

Therefore the limit of f (x, y) as (x, y) → (0, 0) does not exist for p < 2. We now show that the limit does not exist for p = 2 as well. We compute the limits along the line y = 0 and y = x:

x 2 y2 4 4 x (x,y)→(0,0) +y

lim

x 2 y2 x 2 · 02 0 = lim = lim 4 = 0 x→0 x 4 + 04 x→0 x (x,y)→(0,0) x 4 + y 4 lim

along y=0

lim

x 2 y2

(x,y)→(0,0) x 4 + y 4 along y=x

x2 · x2 x4 1 = lim = 4 4 2 x→0 x + x x→0 2x 4

= lim

Since the limits along two paths are different, f (x, y) does not approach one limit as (x, y) → (0, 0). We thus showed that if p ≤ 2, the limit lim f (x, y) does not exist, and f is not continuous at the origin for p ≤ 2. (x,y)→(0,0)

 In Exercises 17–20, compute f x and f y . Calculate f x (1, 3) and f y (1, 3) for f (x, y) = 7x + y 2 . 17. f (x, y) = 2x + y 2

Chapter Review Exercises SOLUTION

437

To find f x we treat y as a constant, and to find f y we treat x as a constant. We get ∂ ∂x ∂ fy = ∂y fx =

  ∂  2 ∂ (2x) + 2x + y 2 = y =2+0=2 ∂x ∂x   ∂  2 ∂ 2x + y 2 = (2x) + y = 0 + 2y = 2y ∂y ∂y

19. f (x, y) = sin(x y)e−x−y f (x, y) = 4x y 3 SOLUTION We compute f x , treating y as a constant and using the Product Rule and the Chain Rule. We get fx =

 ∂

∂ ∂ sin(x y)e−x−y = (sin(x y)) e−x−y + sin(x y) e−x−y ∂x ∂x ∂x

= cos(x y) · ye−x−y + sin(x y) · (−1)e−x−y = e−x−y (y cos(x y) − sin(x y)) We compute f y similarly, treating x as a constant. Notice that since f (y, x) = f (x, y), the partial derivative f y can be obtained from f x by interchanging x and y. That is, f y = e−x−y (x cos(yx) − sin(yx)) . 21. Calculate f x x yz for f (x, y, z) = y sin(x + z). f (x, y) = ln(x 2 + x y 2 ) SOLUTION We differentiate f twice with respect to x, once with respect to y, and finally with respect to z. This gives ∂ (y sin(x + z)) = y cos(x + z) ∂x ∂ fx x = (y cos(x + z)) = −y sin(x + z) ∂x ∂ fx x y = (−y sin(x + z)) = − sin(x + z) ∂y fx =

f x x yz =

∂ (− sin(x + z)) = − cos(x + z) ∂z

23. Find an equation of the tangent to the graph of f (x, y) = x y 2 − x y + 3x 3 y at P = (1, 3). Fix c > 0. Show that for any constants α , β , the function u(t, x) = sin(α ct + β ) sin(α x) satisfies the wave SOLUTION equation The tangent plane has the equation 2 u − 1)∂+ 2 f (1, 3)(y − 3) z = f (1, 3) + f x (1,∂3)(x 2 u y = c ∂t 2 ∂x2 We compute the partial derivatives of f (x, y) = x y 2 − x y + 3x 3 y:

f x (x, y) = y 2 − y + 9x 2 y f y (x, y) = 2x y − x + 3x 3

⇒

(1)

f x (1, 3) = 32 − 3 + 9 · 12 · 3 = 33 f y (1, 3) = 2 · 1 · 3 − 1 + 3 · 13 = 8

Also, f (1, 3) = 1 · 32 − 1 · 3 + 3 · 13 · 3 = 15. Substituting these values in (1), we obtain the following equation: z = 15 + 33(x − 1) + 8(y − 3) or z = 33x + 8y − 42  2 2 + 70.1 using the linear approximation. Compare with a calculator value. 25. Estimate 4.94) Suppose7.1 that +f (4, = 3 and f x (4, 4) = f y (4, 4) = −1. Use the linear approximation to estimate f (4.1, 4) and f (3.88, 4.03). SOLUTION The function whose value we want to approximate is  f (x, y, z) = x 2 + y 2 + z We use the linear approximation at the point (7, 5, 70), hence h = 7.1 − 7 = 0.1, k = 4.9 − 5 = −0.1, and l = 70.1 − 70 = 0.1. We get f (7.1, 4.9, 70.1) ≈ f (7, 5, 70) + 0.1 f x (7, 5, 70) − 0.1 f y (7, 5, 70) + 0.1 f z (7, 5, 70) We compute the partial derivatives of f : x 2x =  f x (x, y, z) =  2 2 2 2 x +y +z x + y2 + z

⇒

7 7 f x (7, 5, 70) =  = 12 72 + 52 + 70

(1)

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(ET CHAPTER 14)

y 2y =  f y (x, y, z) =  2 x 2 + y2 + z x 2 + y2 + z

⇒

5 5 f y (7, 5, 70) =  = 12 72 + 52 + 70

1 1 1 ⇒ f z (7, 5, 70) =  = f z (x, y, z) =  2 2 2 2 24 2 x +y +z 2 7 + 5 + 70  Also, f (7, 5, 70) = 72 + 52 + 70 = 12. Substituting the values in (1) we obtain the following approximation:  5 1 1 7 − 0.1 · + 0.1 · = 12 ≈ 12.020833 7.12 + 4.92 + 70.1 ≈ 12 + 0.1 · 12 12 24 48 That is,  7.12 + 4.92 + 70.1 ≈ 12.020833 The value obtained using a calculator is 12.021647.  c 1.5 27. Jason earns S(h, c) = 20h z 1=+2x − y − 3dollars per month at a used where of hours worked Suppose that the plane is tangent to the graph of zcar = lot, f (x, y) at hPis=the (2,number 4). 100 Determine f (2, 4),sold. f x (2,He 4),has andalready f y (2, 4). and c(a) is the number of cars worked 160 hours and sold 69 cars. Right now Jason wants to go home but wonders how muchf more he might earn if he stays another 10 minutes with a customer who is considering buying a (b) Approximate (2.2, 3.9). car. Use the linear approximation to estimate how much extra money Jason will earn if he sells his 70th car during these 10 minutes. We estimate the money earned in staying for 16 hour more and selling one more car, using the linear approximation SOLUTION

S ≈ Sh (a, b)h + Sc (a, b)c

(1)

By the given information, a = 160, b = 69, h = 16 , and c = 1. We compute the partial derivative of the function:  c 1.5 S(h, c) = 20h 1 + 100  c 1.5 Sh (h, c) = 20 1 + ⇒ Sh (160, 69) = 43.94 100   c 0.5 c 0.5 1 = 0.3h 1 + · ⇒ Sc (160, 69) = 62.4 Sc (h, c) = 20h · 1.5 1 + 100 100 100 Substituting the values in (1), we get the following approximation: S = Sh (160, 69) ·

1 1 + Sc (160, 69) · 1 = 43.94 · + 62.4 ≈ $69.72 6 6

We see that John will make approximately $69.72 more if he sells his 70th car during 10 min. In Exercises 28–31, compute

d f (c(t)) at the given value of t. dt

29. f (x, y, z) = x z − yy2 , c(t) = (t, t 3 , 1 −2 t) f (x, y) = x + e , c(t) = (3t − 1, t ) at t = 2 SOLUTION We use the Chain Rule for Paths: d f (c(t)) = ∇ f c(t) · c (t) dt We compute the gradient of f :  ∇f =

 ∂f ∂f ∂f , , = z, −2y, x ∂ x ∂y ∂z

On the path, x = t, y = t 3 , and z = 1 − t. Therefore,

  ∇ f c(t) = 1 − t, −2t 3 , t

  Also, c (t) = 1, 3t 2 , −1 , hence by (1) we obtain       d f (c(t)) = 1 − t, −2t 3 , t · 1, 3t 2 , −1 = 1 − t + 3t 2 −2t 3 − t = −6t 5 − 2t + 1 dt y 31. f (x, y) = tan−1 3y , c(t) = (cos t, sin t), t = π f (x, y) = xe x − ye3x , c(t) = (et , ln t) at3 t = 1

(1)

Chapter Review Exercises SOLUTION

439

We use the Chain Rule for Paths. We have       −y 1 ∂f ∂f −y x x2 x , , , ∇f = = =



  2 2 ∂ x ∂y x 2 + y2 x 2 + y2 1 + xy 1 + xy

On the path, x = cos t and y = sin t. Therefore,   sin t cos t , = − sin t, cos t ∇ f c(t) = − cos2 t + sin2 t cos2 t + sin2 t c (t) = − sin t, cos t At the point t = π3 we have

 √   π π 3 1 ∇ f c π  = − sin , cos = − , 3 3 3 2 2

and

c

π  3

 √   π π 3 1 = − sin , cos = − , 3 3 2 2

Therefore,  √   √  π  3 1 3 1 3 1 d f (c(t)) = − , · − , = + =1 = ∇ f c π  · c π 3 dt 3 2 2 2 2 4 4 t= 3 In Exercises 32–35, compute the directional derivative at P in the direction v. 33. f (x, y, z) = zx3−4x y 2 , P = (1, 1, 1), v = 2, −1, 2 f (x, y) = x y , P = (3, −1), v = 2i + j SOLUTION We first normalize v to obtain a unit vector u in the direction of v:   2, −1, 2 2 1 2 u=  = ,− , 3 3 3 22 + (−1)2 + 22 We compute the directional derivative using the following equality: Du f (1, 1, 1) = ∇ f (1,1,1) · u The gradient vector at the point (1, 1, 1) is the following vector:     ∇ f = f x , f y , f z = z − y 2 , −2x y, x

⇒

∇ f (1,1,1) = 0, −2, 1

Hence,  Du f (1, 1, 1) = 0, −2, 1 ·

 2 1 2 4 2 2 ,− , =0+ + = 3 3 3 3 3 3

35. f (x, y, z) = sin(x y + z), SOLUTION We=normalize f (x, y) e x +y , 2

2

P =√(0, 0, √0), v = j + k 2 2 vPto=obtain a, vector, u vin=the3,direction of v: −4 2 2 1 1 · 0, 1, 1 = √ 0, 1, 1 u=  2 2 2 2 0 +1 +1

By the Theorem on Evaluating Directional Derivatives, Dv f (P) = ∇ f P · u

(1)

We compute the gradient vector:   ∂f ∂f ∂f , , ∇f = = y cos(x y + z), x cos(x y + z), cos(x y + z) ∂ x ∂y ∂z Hence, ∇ f P = 0, 0, 1 . By (1) we conclude that 1 1 Dv f (P) = ∇ f P · u = 0, 0, 1 · √ 0, 1, 1 = √ . 2 2 2 Find the unit vector e at P = (0, 0, 1) pointing in the direction along which f (x, y, z) = x z + e−x +y increases

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(ET CHAPTER 14)

37. Find an equation of the tangent plane at P = (0, 3, −1) to the surface with equation SOLUTION

ze x + e z+1 = x y + y − 3 The surface is defined implicitly by the equation F(x, y, z) = ze x + e z+1 − x y − y + 3

The tangent plane to the surface at the point (0, 3, −1) has the following equation: 0 = Fx (0, 3, −1)x + Fy (0, 3, −1)(y − 3) + Fz (0, 3, −1)(z + 1)

(1)

We compute the partial derivatives at the given point: Fx (x, y, z) = ze x − y

⇒

Fx (0, 3, −1) = −1e0 − 3 = −4

Fy (x, y, z) = −x − 1

⇒

Fy (0, 3, −1) = −0 − 1 = −1

Fz (x, y, z) = e x + e z+1

⇒

Fz (0, 3, −1) = e0 + e−1+1 = 2

Substituting in (1) we obtain the following equation: −4x − (y − 3) + 2(z + 1) = 0 −4x − y + 2z + 5 = 0 2z = 4x + y − 5

⇒

z = 2x + 0.5y − 2.5

∂f ∂f andthe tangent , whereplane x = uto−the v and y =xun + v. 39. LetLet f (x, y) 0=be (xan −integer y)e x . Use Ruleconstant. to calculate n = andthe r anChain arbitrary Show that surface yn + zn = r ∂u ∂v at P = (a, b, c) has equation SOLUTION First we calculate the Primary Derivatives: a n−1 x + bn−1 y + cn−1 z = r ∂f ∂f x = e (x − y) + e x = e x (x − y + 1), = −e x ∂x ∂y ∂y Since ∂∂ux = 1, ∂u = 1, ∂∂vx = −1, and ∂∂vy = 1, the Chain Rule gives

∂f ∂ ∂f = ∂u ∂x

Chapter Review Exercises

∂x = 0, ∂z

∂y = 0, ∂z

441

∂z =1 ∂z

Substituting these derivatives in the Chain Rule gives ∂f ∂f ∂f ∂f ∂f ∂f = cos θ + sin θ + ·0= cos θ + sin θ ∂r ∂x ∂y ∂z ∂x ∂y ∂f ∂f ∂f ∂f ∂f ∂f (−r sin θ ) + r cos θ + · 0 = − r sin θ + r cos θ = ∂θ ∂x ∂y ∂z ∂x ∂y 43. Let g(u, v) = f (u 3 − v 3 , v 3 − u 3 ). Prove that ∂f ∂f and Let f (x, y, z) = (x 2 + y 2 )e−x z and P = (1, 0, 2). Use the result of Exercise 41 to calculate at the ∂r ∂ θ ∂g ∂g 2 2 − u = 0 v point P. ∂u ∂v SOLUTION We are given the function f (x, y), where x = u 3 − v 3 and y = v 3 − u 3 . Using the Chain Rule we have the following derivatives:

∂g ∂ f ∂x ∂ f ∂y = + ∂u ∂ x ∂u ∂y ∂u ∂ f ∂x ∂ f ∂y ∂g = + ∂v ∂ x ∂v ∂y ∂v

(1)

We compute the following partial derivatives: ∂x ∂y = 3u 2 , = −3u 2 ∂u ∂u ∂x ∂y = −3v 2 , = 3v 2 ∂v ∂v Substituting in (1) we obtain   ∂f ∂f ∂f ∂g ∂f  = · 3u 2 + −3u 2 = 3u 2 − ∂u ∂x ∂y ∂x ∂y      ∂f ∂f ∂f ∂g ∂f = − −3v 2 + 3v 2 = −3v 2 ∂v ∂x ∂y ∂x ∂y Therefore, v2

∂g ∂g + u2 = 3u 2 v 2 ∂u ∂v



∂f ∂f − ∂x ∂y



− 3u 2 v 2

∂f ∂f − ∂x ∂y

=0

∂z , where xe z + ze y = x 2+ y. 2 45. Calculate Let f (x, ∂ xy) = g(u), where u = x + y and g(u) is a differentiable function in one variable. Prove that y  2 0 defines  z implicitly SOLUTION The function F(x, y, z) = xe z +∂ ze as a function of x and y. Using ∂g 2 f 2− x −∂ fy = + = 4u implicit differentiation, the partial derivative of z with respect to x is ∂x ∂y ∂u Fx ∂z =− (1) ∂x Fz We compute the partial derivatives Fx and Fz : Fx = e z − 1 Fz = xe z + e y Substituting in (1) gives ez − 1 ∂z =− z . ∂x xe + e y In Exercises 47–50, find4the critical points of the function and analyze them using the Second Derivative Test. Let f (x, y) = x − 2x 2 + y 2 − 6y. f and use the Second Derivative Test to determine whether they are local minima or 47. (a) f (x,Find y) =the x 2critical + 2y 2 points − 4x y of + 6x maxima. (b) Find the minimum value of f without calculus by completing the square.

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(ET CHAPTER 14)

SOLUTION To find the critical points of the function f (x, y) = x 2 + 2y 2 − 4x y + 6x, we set the partial derivatives equal to zero and solve. This gives

f x (x, y) = 2x − 4y + 6 = 0 f y (x, y) = 4y − 4x = 0 By the second equation y = x. Substituting in the first equation gives 2x − 4x + 6 = 0 −2x = −6

⇒

x =3

There is one critical point P = (3, 3). We now apply the Second Derivative Test to examine the critical point. We first find the second-order partials: f x x (x, y) = 2,

f yy (x, y) = 4,

f x y (x, y) = −4

Since D = f x x f yy − f x2y = 2 · 4 − 16 = −8 < 0, the point (3, 3) is a saddle point. 49. f (x, y) = e x+y3 − xe2y f (x, y) = x + 2y 3 − x y SOLUTION We find the critical point by setting the partial derivatives of f (x, y) = e x+y − xe2y equal to zero and solve. This gives f x (x, y) = e x+y − e2y = 0 f y (x, y) = e x+y − 2xe2y = 0 The first equation gives e x+y = e2y and the second equation gives e x+y = 2xe2y . Equating the two expressions, dividing by the nonzero function e2y , and solving for x, we obtain e2y = 2xe2y

⇒

1 = 2x

⇒

x=

1 2

We now substitute x = 12 in the first equation and solve for y, to obtain 1

e 2 +y − e2y = 0

⇒

1

e 2 +y = e2y

⇒

1 + y = 2y 2

⇒

y=

1 2

  There is one critical point, 12 , 12 . We examine the critical point using the Second Derivative Test. We compute the second derivatives at this point:  1 1 1 1 , f x x (x, y) = e x+y ⇒ f x x = e2+2 = e 2 2  1 1 1 1 1 1 , f yy (x, y) = e x+y − 4xe2y ⇒ f yy = e 2 + 2 − 4 · e2· 2 = −e 2 2 2  1 1 1 1 1 , f x y (x, y) = e x+y − 2e2y ⇒ f x y = e 2 + 2 − 2e2· 2 = −e 2 2 Therefore the discriminant at the critical point is  1 1 , D = f x x f yy − f x2y = e · (−e) − (−e)2 = −2e2 < 0 2 2   We conclude that 12 , 12 is a saddle point. 51. Prove that f (x, y) = (x + 2y)e x y has no critical points. 1 (x + y 2by ) setting the partial derivatives of f (x, y) = (x + 2y)e x y equal to zero and f (x, y) sin(xthe + critical y) − points SOLUTION We=find 2 solving. We get   f x (x, y) = e x y + (x + 2y)ye x y = e x y 1 + x y + 2y 2 = 0   f y (x, y) = 2e x y + (x + 2y)xe x y = e x y 2 + x 2 + 2x y = 0 We divide the two equations by the nonzero expression e x y to obtain the following equations: 1 + x y + 2y 2 = 0

Chapter Review Exercises

443

2 + 2x y + x 2 = 0 The first equation implies that x y = −1 − 2y 2 . Substituting in the second equation gives   2 + 2 −1 − 2y 2 + x 2 = 0 2 − 2 − 4y 2 + x 2 = 0 x 2 = 4y 2

⇒

x = 2y

or

x = −2y

We substitute in the first equation and solve for y: x = 2y

x = −2y

1 + 2y 2 + 2y 2 = 0

1 − 2y 2 + 2y 2 = 0

1 + 4y 2 = 0

1=0

y 2 = − 14 In both cases there is no solution. We conclude that there are no solutions for f x = 0 and f y = 0, that is, there are no critical points. 53. Find the global extrema of f (x, y) = 2x y − x − y on the domain {y ≤ 4, y ≥ x 2 }. Find the global extrema of f (x, y) = x 3 − x y − y 2 + y on the square [0, 1] × [0, 1]. SOLUTION The region is shown in the figure. y y=4

A

B

y = x2 −2

x 0

2

Step 1. Finding the critical points. We find the critical points in the interior of the domain by setting the partial derivatives equal to zero and solving. We get f x = 2y − 1 = 0 f y = 2x − 1 = 0

⇒

x=

1 , 2

y=

1 2

  2  The critical point is 12 , 12 . (It lies in the interior of the domain since 12 < 4 and 12 > 12 ). Step 2. Finding the global extrema on the boundary. We consider the two parts of the boundary separately. The parabola y = x 2 , −2 ≤ x ≤ 2: y 8

−2

x 0

1

2

f (x, x ) = 2x − x 2 − x 2

3

−8

On this curve, f (x, x 2 ) = 2 · x · x 2 − x − x 2 = 2x 3 − x 2 − x. Using calculus in one variable or the graph of the function, we see that the minimum of f (x, x 2 ) on the interval occurs at x = −2 and the maximum at x = 2. The corresponding points are (−2, 4) and (2, 4). The segment AB: On this segment y = 4, −2 ≤ x ≤ 2, hence f (x, 4) = 2 · x · 4 − x − 4 = 7x − 4. The maximum value occurs at x = 2 and the minimum value at x = −2. The corresponding points on the segment AB are (−2, 4) and (2, 4) Step 3. Conclusions. Since the global extrema occur either at critical points in the interior of the domain or on the boundary of the domain, the candidates for global extrema are the following points:  1 1 , , (−2, 4), (2, 4) 2 2

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We compute the values of f = 2x y − x − y at these points:  1 1 1 1 1 1 1 , =2· · − − =− f 2 2 2 2 2 2 2 f (−2, 4) = 2 · (−2) · 4 + 2 − 4 = −18 f (2, 4) = 2 · 2 · 4 − 2 − 4 = 10 We conclude that the global maximum is f (2, 4) = 10 and the global minimum is f (−2, 4) = −18. 2 2 55. UseFind Lagrange multipliers findy,the and maximum value of f (x, the maximum of to f (x, z) minimum = x yz subject to the constraint g(x, y, y) z) = = 3x 2x − + 2y y +on4zthe =circle 1. x + y = 4. SOLUTION

Step 1. Write out the Lagrange Equations. The constraint curve is g(x, y) = x 2 + y 2 − 4 = 0, hence ∇g = 2x, 2y and ∇ f = 3, −2. The Lagrange Condition ∇ f = λ ∇g is thus 3, −2 = λ 2x, 2y. That is, 3 = λ · 2x −2 = λ · 2y Note that λ  = 0. Step 2. Solve for x and y using the constraint. The Lagrange equations gives 3 = λ · 2x − 2 = λ · 2y

3 2λ 1 y=− λ

x= ⇒

(1)

We substitute x and y in the equation of the constraint and solve for λ . We get 

 3 2 1 2 + − =4 2λ λ 9 1 + 2 =4 2 4λ λ

√

1 13 =4 · λ2 4

⇒

λ=

13 4

√

λ =−

or

13 4

Substituting in (1), we obtain the points 6 x= √ , 13 6 x = −√ , 13

4 y = −√ 13 4 y= √ 13

The critical points are thus  P1 =

6 4 √ ,−√ 13 13



 4 6 P2 = − √ , √ 13 13 Step 3. Calculate the value at the critical points. We find the value of f (x, y) = 3x − 2y at the critical points: −4 26 6 f (P1 ) = 3 · √ − 2 · √ = √ 13 13 13 −6 4 −26 f (P2 ) = 3 · √ − 2 · √ = √ 13 13 13 Thus, the maximum value of f on the circle is √26 , and the minimum is − √26 . 13

13

2 + 9y 2 = 36. 57. FindFind the the minimum and value maximum values of xf y(x, y) = xto2 ythe onconstraint the ellipse5x 4x− minimum of f (x, y) = subject y = 4 in two ways: using Lagrange multipliersWe and setting = 5x − 4 in fand (x,maximum y). SOLUTION must findythe minimum values of f (x, y) = x 2 y subject to the constraint g(x, y) = 4x 2 + 9y 2 − 36 = 0.

Chapter Review Exercises

445

  Step 1. Write out the Lagrange Equations. The gradient vectors are ∇ f = 2x y, x 2 and ∇g = 8x, 18y, hence the Lagrange Condition ∇ f = λ ∇g gives   2x y, x 2 = λ 8x, 18y = 8λ x, 18λ y We obtain the following Lagrange Equations: 2x y = 8λ x x 2 = 18λ y Step 2. Solve for λ in terms of x and y. If x = 0, the equation of the constraint implies that y = ±2. The points (0, 2) and (0, −2) satisfy the Lagrange Equations for λ = 0. If x  = 0, the second Lagrange Equation implies that y  = 0. Therefore the Lagrange Equations give 2x y = 8λ x

⇒

x 2 = 18λ y

λ=

y 4

λ =

⇒

x2 18y

Step 3. Solve for x and y using the constraint. We equate the two expressions for λ to obtain x2 y = 4 18y 18y 2 = 4x 2 We now substitute 4x 2 = 18y 2 in the equation of the constraint 4x 2 + 9y 2 = 36 and solve for y. This gives 18y 2 + 9y 2 = 36 27y 2 = 36

y2 =

⇒

36 27

⇒

2 y1 = √ , 3

2 y2 = − √ 3

2 We find the x-coordinates using x 2 = 9y2 :

9y 2 2 4 9 x2 = · = 6 2 3

x2 =

⇒

x1 =

√

6,

√ x2 = − 6

We obtain the following critical points:  √

2 6, √ P1 = (0, 2), P2 = (0, −2), P3 = 3    √ √ √ 2 2 2 6, − √ , P5 = − 6, √ , P6 = − 6, − √ P4 = 3 3 3 Step 4. Conclusions. We evaluate the function f (x, y) = x 2 y at the critical points: f (P1 ) = 02 · 2 = 0 f (P2 ) = 02 · (−2) = 0 12 2 f (P3 ) = f (P5 ) = 6 · √ = √ 3 3  2 12 f (P4 ) = f (P5 ) = 6 · − √ = −√ 3 3 Since the min and max of f occur on the ellipse, it must occur at critical points. Thus, we conclude that the maximum 12 and − √ 12 respectively. and minimum of f subject to the constraint are √ 3

3

59. Find the extreme values of f (x, y, z) = x + 2y + 3z subject−1 to the two constraints x + y + z = 1 and x 2 + y 2 + z 2 = 1. Find the point in the first quadrant on the curve y = x + x closest to the origin. We must find the extreme values of f (x, y, z) = x + 2y + 3z subject to the constraints g(x, y, z) = x + y + z − 1 = 0 and h(x, y, z) = x 2 + y 2 + z 2 − 1 = 0.

SOLUTION

446

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D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S

(ET CHAPTER 14)

Step 1. Write out the Lagrange Equations. We have ∇ f =< 1, 2, 3 >, ∇g =< 1, 1, 1 >, ∇h =< 2x, 2y, 2z >, hence the Lagrange condition ∇ f = λ ∇g + μ ∇h gives < 1, 2, 3 > = λ < 1, 1, 1 > +μ < 2x, 2y, 2z >=< λ + 2μ x, λ + 2μ y, λ + 2μ z > or 1 = λ + 2μ x 2 = λ + 2μ y 3 = λ + 2μ z Step 2. Solve for λ and μ . The Lagrange Equations give 1 = λ + 2μ x

λ =1 − 2μ x

2 = λ + 2μ y

λ =2 − 2μ y

⇒

3 = λ + 2μ z

λ =3 − 2μ z

Equating the three expressions for λ , we get the following equations: 1 − 2μ x = 2 − 2μ y

⇒

1 − 2μ x = 3 − 2μ z

2μ (y − x) = 1

μ (z − x) = 2

1 , and the second implies that μ = 2 . Equating the two expressions for μ , The first equation implies that μ = 2(y−x) z−x we get

1 2 = 2(y − x) z−x z − x = 4y − 4x

⇒

z = 4y − 3x

Step 3. Solve for x, y, and z using the constraints. We substitute z = 4y − 3x in the equations of the constraints and solve to find x and y. This gives x + y + (4y − 3x) = 1 x 2 + y 2 + (4y − 3x)2 = 1

y=

⇒

1 + 2x 5

10x 2 + 17y 2 − 24x y = 1

Substituting in the second equation and solving for x, we get 1 + 2x 5

y=

 1 + 2x 2 1 + 2x =1 − 24x · 10x 2 + 17 5 5

250x 2 + 17(1 + 2x)2 − 120x (1 + 2x) = 25 39x 2 − 26x − 4 = 0 26 ±

x 1,2 =

√

1300

78 ⇒ x1 =

√ 1 5 13 + ≈ 0.8, 3 39

x2 =

√ 1 5 13 − ≈ −0.13 3 39

We find the y-coordinates using y = 1+2x 5 . y1 =

1 + 2 · 0.8 = 0.52, 5

y2 =

1 − 2 · 0.13 = 0.15 5

Finally, we find the z-coordinate using z = 4y − 3x: z 1 = 4 · 0.52 − 3 · 0.8 = −0.32,

z 2 = 4 · 0.15 + 3 · 0.13 = 0.99

We obtain the critical points: P1 = (0.8, 0.52, −0.32), .

P2 = (−0.13, 0.15, 0.99)

Chapter Review Exercises

447

Step 4. Conclusions. We evaluate the function f (x, y, z) = x + 2y + 3z at the critical points: f (P1 ) = 0.8 + 2 · 0.52 − 3 · 0.32 = 0.88 f (P2 ) = −0.13 + 2 · 0.15 + 3 · 0.99 = 3.14

(1)

The two constraints determine the common points of the unit sphere x 2 + y 2 + z 2 = 1 and the plane x + y + z = 1. This set is a circle that is a closed and bounded set in R 3 . Therefore, f has a minimum and maximum values on this set. These extrema are given in (1). 61. Find the dimensions of the box of maximum volume with its sides parallel to the coordinate planes that can be Use Lagrange multipliers to find the dimensions of a cylindrical can of fixed volume V with minimal surface area inscribed in the ellipsoid (Figure 4) (including the top and bottom of the can).  x 2  y 2  z 2 + + =1 a b c z

y x

FIGURE 4 SOLUTION

We denote the vertices of the box by (±x, ±y, ±z), where x ≥ 0, y ≥ 0, z ≥ 0. The volume of the box is V (x, y, z) = 8x yz

The vertices of the box must satisfy the equation of the ellipsoid, hence, x2 y2 z2 g(x, y, z) = 2 + 2 + 2 − 1 = 0, a b c

x ≥ 0,

y ≥ 0,

z ≥ 0.

We need to maximize V due to the constraint: g(x, y, z) = 0, x ≥ 0, y ≥ 0, z ≥ 0.

  Step 1. Write out the Lagrange Equations. We have ∇V = 8 yz, x z, x y and ∇g = 2x2 , 2y2 , 2z2 , hence the Lagrange a b c Condition ∇V = λ ∇g gives the following equations: 2x yz = λ 2 a 2y xz = λ 2 b 2z xy = λ 2 c Step 2. Solve for λ in terms of x, y, and z. If x = 0, y = 0, or z = 0, the volume of the box has the minimum value zero. We thus may assume that x  = 0, y  = 0, and z  = 0. The Lagrange equations give

λ=

a 2 yz , 2x

λ=

b2 x z , 2y

λ=

c2 x y 2z

Step 3. Solve for x, y, and z using the constraint. Equating the three expressions for λ yields the following equations: c2 x y a 2 yz = 2 x 2 z c2 x y b2 x z = 2 y 2 z

⇒

  y c2 x 2 − a 2 z 2 = 0   x c2 y 2 − b2 z 2 = 0

Since x > 0 and y > 0, these equations imply that c2 x 2 − a 2 z 2 = 0 c2 y 2 − b2 z 2 = 0

az c bz y= c

x= ⇒

(1)

448

C H A P T E R 15

D I F F E R E N T I AT I O N I N S E V E R A L VA R I A B L E S

(ET CHAPTER 14)

We now substitute x and y in the equation of the constraint and solve for z. This gives 

az 2 c

+

a2

 bz 2 c z2 + 2 =1 2 b c

z2 z2 z2 + 2 + 2 =1 2 c c c 3z 2 =1 c2

c z= √ 3

⇒

We find x and y using (1): x=

a a c √ = √ , c 3 3

y=

b c b √ = √ c 3 3

We obtain the critical point: 

a b c √ ,√ ,√ 3 3 3

P=



Step 4. Conclusions. The function V = 8x yz is a polynomial, hence it is continuous. The constraint defines a closed and compact set in R 3 , hence f has extreme values on the constraint. The maximum value is obtained at the critical point P. We find it: b c abc a V (P) = 8 √ · √ · √ = 8 √ 3 3 3 3 3 We conclude that the dimensions of the box of maximum volume with sides parallel to the coordinate planes, which can be inscribed in the ellipsoid, are b y= √ , 3

a x= √ , 3

c z= √ . 3

63. A bead hangs on a string of length  whose ends are fixed by thumbtacks located at points (0, 0) and (a, b) on a Given n nonzero numbers σ , . . . , σn , show that the minimum value of bulletin board (Figure 5). The bead1rests in the position that minimizes its height y. Use Lagrange multipliers to show 2 2 2 that θ1 = θ2 . that the two sides of the string make equal angles horizontal, that is, 2show f (x 1 , .with . . , xthe n ) = x σ + · · · + x n σn As an aside, note that the locus of the bead when pulled taut1is1an ellipse with foci at O and P, and the tangent line at ⎞−1another proof of the reflective property of the ellipse (a ⎛ provides the lowest point is horizontal. Therefore, this exercise n  lightsubject ray emanating from one focus and bouncing off the ellipse −2 to x 1 + · · · + x n = 1 is c, where c = ⎝ σ ⎠ is. reflected to the other focus). j

j=1 y

P = (a, b)

x

O 1

2

FIGURE 5 SOLUTION

We denote by l1 and l2 the lengths shown in the figure. y

E (a, b)

2

A

B

1

1

F

C

2

2

b

D

x −y

Chapter Review Exercises

449

Hence, l = l1 + l2

(1)

We find l1 and l2 in terms of y, θ1 , and θ2 . Since y < 0, we have −y = cos θ1 l1

⇒

l1 = −

b−y = cos θ2 l2

⇒

l2 =

y cos θ1

b−y cos θ2

Substituting in (1), we obtain l=

b−y −y + cos θ1 cos θ2

(2)

Rewriting the above equation, we obtain our first constraint equation: l = −y sec θ1 + (b − y) sec θ2

(3)

For our second constraint, we note that tan θ1 = AB/(−y) and tan θ2 = B D/(b − y). Since AB + B D = a, we have a = −y tan θ1 + (b − y) tan θ2

(4)

Our objective is to minimize f (y, θ1 , θ2 ) = −y with the two constraint equations g1 (y, θ1 , θ2 ) = −y sec θ1 + (b − y) sec θ2 g2 (y, θ1 , θ2 ) = −y tan θ1 + (b − y) tan θ2 We use the equation ∇ f (y, θ1 , θ2 ) = λ1 ∇g1 (y, θ1 , θ2 ) + λ2 ∇g2 (y, θ1 , θ2 ), which becomes 1, 0, 0 = λ1 − sec θ1 − sec θ2 , −y sec θ1 tan θ1 , (b − y) sec θ2 tan θ2  + λ2 − tan θ1 − tan θ2 , −y sec2 θ1 , (b − y) sec2 θ2  Note that λ1 and λ2 can’t both be zero. If λ1 = 0, then we get 0 = λ2 (−y sec2 θ2 ), which doesn’t happen. A similar argument holds for λ2 . Thus, we can assume that both λ1 and λ2 are nonzero. Looking at the second coordinate of the above equation, we get λ1 (y sec θ1 tan θ1 ) = λ2 (−y sec2 θ1 ), which means λ1 /λ2 = − sec θ1 / tan θ1 = −1/ sin θ1 . A similar analysis of the third coordinate gives λ1 /λ2 = − sec θ2 / tan θ2 = −1/ sin θ2 . We conclude that sin θ1 = sin θ2 , and hence θ1 = θ2 , as desired.