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Solution manual calculus 8th edition varberg, purcell, rigdon ch12

CHAPTER

12

Derivatives for Functions of
Two or More Variables

12.1 Concepts Review

4. a.

6

1. real-valued function of two real variables

b. 12

2. level curve; contour map

c.


2

3. concentric circles

d.

(3cos 6)1/ 2 + 1.44 ≈ 3.1372

4. parallel lines

Problem Set 12.1
1. a.

5

b. 0
c.

6

d.

a6 + a 2

e.

2 x2 , x ≠ 0

f.

Undefined

5. F (t cos t , sec2 t ) = t 2 cos 2 t sec2 t = t 2 , cos t ≠ 0
6. F ( f (t ), g (t )) = F (ln t 2 , et / 2 )
= exp(ln t 2 ) + (et / 2 ) 2 = t 2 + et , t ≠ 0

7. z = 6 is a plane.

The natural domain is the set of all (x, y) such
that y is nonnegative.


2. a.

4

8. x + z = 6 is a plane.

b. 17
c.

17
16

d.

1 + a2 , a ≠ 0

e.

x 3 + x, x ≠ 0

f.

Undefined

The natural domain is the set of all (x, y) such
that x is nonzero.
3. a.

744

9. x + 2y + z = 6 is a plane.

sin(2π ) = 0

b.

⎛π⎞
4sin ⎜ ⎟ = 2
⎝6⎠

c.

⎛π⎞
16sin ⎜ ⎟ = 16
⎝2⎠

d.

π2 sin(π2 ) ≈ –4.2469

Section 12.1

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10. z = 6 – x 2 is a parabolic cylinder.

11. x 2 + y 2 + z 2 = 16 , z ≥ 0 is a hemisphere.

12.

x2 y 2 z 2
+
+
= 1 , z ≥ 0 is a hemi-ellipsoid.
4 16 16

13. z = 3 – x 2 – y 2 is a paraboloid.

14. z = 2 – x – y 2

Instructor’s Resource Manual

15. z = exp[–( x 2 + y 2 )]

16. z =

x2
,y>0
y

17. x 2 + y 2 = 2 z; x 2 + y 2 = 2k

18. x = zy, y ≠ 0 ; x = ky, y ≠ 0

19. x 2 = zy, y ≠ 0; x 2 = ky, y ≠ 0

Section 12.1

745

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20. x 2 = –( y – z ); x 2 = –( y – k )

21. z =

x2 + 1
x2 + y 2

24. ( x – 2)2 + ( y + 3)2 =

San Francisco and St. Louis had a
temperature between 70 and 80 degrees
Fahrenheit.

b.

Drive northwest to get to cooler
temperatures, and drive southeast to get
warmer temperatures.

c.

Since the level curve for 70 runs southwest
to northeast, you could drive southwest or
northeast and stay at about the same
temperature.

26. a.

The lowest barometric pressure, 1000
millibars and under, occurred in the region
of the Great Lakes, specifically near
Wisconsin. The highest barometric
pressure, 1025 millibars and over, occurred
on the east coast, from Massachusetts to
South Carolina.

b.

Driving northwest would take you to lower
barometric pressure, and driving southeast
would take you to higher barometric
pressure.

c.

Since near St. Louis the level curves run
southwest to northeast, you could drive
southwest or northeast and stay at about the
same barometric pressure.

k = 2: 2 x 2 + 2 y 2 = x 2 + 1
x2 y 2
+
= 1 ; ellipse
1
1
2

k = 4: 4 x 2 + 4 y 2 = x 2 + 1
1
3

+

y2
1
4

= 1; ellipse

V2

25. a.
, k = 1, 2, 4

k = 1: y 2 = 1 or y = ±1 ;
two parallel lines

x2

16

22. y = sin x + z; y = sin x + k

23. x = 0, if T = 0:
⎛1 ⎞
y 2 = ⎜ – 1⎟ x 2 , if y ≠ 0 .
⎝T


27. x 2 + y 2 + z 2 ≥ 16 ; the set of all points on and
outside the sphere of radius 4 that is centered at
the origin
28. The set of all points inside (the part containing
the z-axis) and on the hyperboloid of one sheet;
x2 y 2 z 2
+

= 1.
9
9
9
29.

746

Section 12.1

x2 y 2 z 2
+
+
≤ 1 ; points inside and on the
9 16 1
ellipsoid

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30. Points inside (the part containing the z-axis) or on
x2 y 2 z 2
the hyperboloid of one sheet,
+

= 1,
9
9 16
excluding points on the coordinate planes
31. Since the argument to the natural logarithm
function must be positive, we must have
x 2 + y 2 + z 2 > 0 . This is true for all ( x, y , z )

except ( x, y, z ) = ( 0, 0, 0 ) . The domain consists
3

all points in

except the origin.

32. Since the argument to the natural logarithm
function must be positive, we must have xy > 0 .

This occurs when the ordered pair ( x, y ) is in the
first quadrant or the third quadrant of the
xy-plane. There is no restriction on z. Thus, the
domain consists of all points ( x, y, z ) such that x
and y are both positive or both negative.
33. x 2 + y 2 + z 2 = k , k > 0; set of all spheres
centered at the origin

37. 4 x 2 – 9 y 2 = k , k in R;

k
100

+

y2
k
16

+

z2
k
25

= 1; set of all ellipsoids centered

k
4



y2
k
9

= 1, if k ≠ 0;

2x
(for k = 0) and all hyperbolic
3
cylinders parallel to the z-axis such that the ratio
⎛1⎞ ⎛1⎞
a:b is ⎜ ⎟ : ⎜ ⎟ or 3:2 (where a is associated
⎝ 2⎠ ⎝3⎠
with the x-term)

planes y = ±

38. e x

2 + y2 + z2

= k, k > 0

x 2 + y 2 + z 2 = ln k
concentric circles centered at the origin.

39. a.

All ( w, x, y, z ) except ( 0, 0, 0, 0 ) , which
would cause division by 0.

b.

All ( x1 , x2 ,… , xn ) in n-space.

c.

All ( x1 , x2 ,… , xn ) that satisfy
x12 + x22 +

+ xn2 ≤ 1 ; other values of

( x1 , x2 ,… , xn )

34. 100 x 2 + 16 y 2 + 25 z 2 = k , k > 0;
x2

x2

would lead to the square

root of a negative number.
40. If z = 0, then x = 0 or x = ± 3 y .

at origin such that their axes have ratio
⎛ 1 ⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 2:5:4.
⎝ 10 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠
35.

x2
1
16
2

+

y2
1
4
2



z2
= k ; the elliptic cone
1

x
y
z2
+
=
and all hyperboloids (one and two
9
9 16
sheets) with z-axis for axis such that a:b:c is
⎛1⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 3:3:4.
⎝ 4⎠ ⎝ 4⎠ ⎝ 3⎠

36.

x2
1
9
2



y2
1
4
2



z2
= k ; the elliptical cone
1

y
z
x2
+
=
and all hyperboloids (one and two
9 36 4
sheets) with x-axis for axis such that a:b:c is
⎛1⎞ ⎛1⎞
⎜ ⎟ : ⎜ ⎟ :1 or 2:3:6
⎝3⎠ ⎝ 2⎠

41. a.

b.

AC is the least steep path and BC is the most
steep path between A and C since the level
curves are farthest apart along AC and
closest together along BC.
AC ≈ (5750) 2 + (3000)2 ≈ 6490 ft
BC ≈ (580) 2 + (3000) 2 ≈ 3060 ft

42. Completing the squares on x and y yields the
equivalent equation
f ( x, y ) + 25.25 = ( x – 0.5) 2 + 3( y + 2)2 , an
elliptic paraboloid.

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Section 12.1

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(2 x – y 2 ) exp(– x 2 – y 2 )

43.

sin 2x 2 + y 2

46.

44.

sin x sin y
(1 + x 2 + y 2 )

sin( x 2 + y 2 )
x2 + y2

12.2 Concepts Review
[( f ( x0 + h, y0 ) – f ( x0 , y0 )]
; partial
h
h →0
derivative of f with respect to x

1. lim

2. 5; 1
3.
45.

∂2f
∂ y∂ x

4. 0

Problem Set 12.2
1.

f x ( x, y ) = 8(2 x – y )3 ; f y ( x, y ) = –4(2 x – y )3

2.

f x ( x, y ) = 6(4 x – y 2 )1/ 2 ;
f y ( x, y ) = –3 y (4 x – y 2 )1/ 2

748

Section 12.2

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3.

f x ( x, y ) =

f y ( x, y ) =

( xy )(2 x) – ( x 2 – y 2 )( y )
( xy )2

x2 + y2
x2 y

14.

( xy )2
( x2 + y2 )
xy 2

5.

f x ( x, y ) = e y cos x; f y ( x, y ) = e y sin x

6.

⎛ 1⎞
f x ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (6 x)
⎝ 3⎠

f x ( x, y ) = x ( x − y )

f x ( x, y ) = 4 xy 3 – 3x 2 y5 ;
f xy ( x, y ) = 12 xy 2 – 15 x 2 y 4
f y ( x, y ) = 6 x 2 y 2 – 5 x 3 y 4 ;
f yx ( x, y ) = 12 xy 2 – 15 x 2 y 4

18.

= 120 x 2 y ( x3 + y 2 )3
f y ( x, y ) = 5( x3 + y 2 ) 4 (2 y );
f yx ( x, y ) = 40 y ( x3 + y 2 )3 (3x 2 )
= 120 x 2 y ( x3 + y 2 )3

;

fu (u , v) = ve ; f v (u , v) = ue
uv

9. g x ( x, y ) = – ye
10.

– xy

19.

; g y ( x, y ) = – xe

20.

– xy

f x ( x, y ) = x(1 + x 2 y 2 ) –1;
f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2

f x ( x, y ) = 4[1 + (4 x – 7 y ) 2 ]–1;

21. Fx ( x, y ) =

2 –1

f y ( x, y ) = –7[1 + (4 x – 7 y ) ]

1
1–

w
z

=
1–

( wz )

2

( )

w 2
z

1–

13.

⎛1⎞
–1 ⎛ w ⎞
⎜ ⎟ + sin ⎜ ⎟
z
⎝ ⎠
⎝z⎠

1

( )

w 2
z

Fy ( x, y ) =

2

⎛ w⎞
⎜– 2 ⎟ =
⎝ z ⎠

( wz )
2
1 – ( wz )


2

2

2

2

f y ( x, y ) = –2 y sin( x + y ) + cos( x + y )

Instructor’s Resource Manual

( xy )

2

y2

=

2 2

=

x y

1
x2

;

1
9

( xy )(–1) – (2 x – y )( x)
( xy )

Fy (3, − 2) = −

f x ( x, y ) = –2 xy sin( x 2 + y 2 );
2

( xy )(2) – (2 x – y )( y )

Fx (3, − 2) =

⎛ w⎞
+ sin –1 ⎜ ⎟ ;
⎝z⎠

Fz = ( w, z ) = w

f x ( x, y ) = y (1 + x 2 y 2 ) –1;
f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2

f s ( s, t ) = 2 s ( s 2 – t 2 ) –1 ;

12. Fw ( w, z ) = w

f x ( x, y ) = 6e2 x cos y; f xy ( x, y ) = –6e2 x sin y
f y ( x, y ) = –3e2 x sin y; f yx ( x, y ) = –6e2 x sin y

uv

ft ( s, t ) = −2t ( s 2 − t 2 )−1

11.

f x ( x, y ) = 5( x3 + y 2 ) 4 (3 x 2 );
f xy ( x, y ) = 60 x 2 ( x3 + y 2 )3 (2 y )

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

8.

2 – s2

17.

⎛ 1⎞
f y ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (2 y )
⎝ 3⎠
⎛ 2y ⎞ 2
2 –4 / 3
= ⎜–
⎟ (3 x + y )
⎝ 3 ⎠

7.

; f s ( s, t ) = 2tet

f r (r , θ ) = 9r 2 cos 2θ ; fθ (r , θ ) = –6r 3 sin 2θ

= –2 x(3 x 2 + y 2 ) –4 / 3 ;

2 −1/ 2

2 – s2

16.

f x ( x, y ) = e x cos y; f y ( x, y ) = – e x sin y

2

f s ( s, t ) = –2set

15. Fx ( x, y ) = 2 cos x cos y; Fy ( x, y ) = –2sin x sin y

( xy )(−2 y ) − ( x 2 − y 2 )( x)

=−

4.

=

2

=

–2 x 2
2 2

x y

=–

2
x2

;

1
2

22. Fx ( x, y ) = (2 x + y )( x 2 + xy + y 2 ) –1;
Fx (–1, 4) =

2
≈ 0.1538
13

Fy ( x, y ) = ( x + 2 y )( x 2 + xy + y 2 ) –1 ;

Fy (–1, 4) =

7
≈ 0.5385
13

Section 12.2

749

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reproduced, in any form or by any means, without permission in writing from the publisher.


23.

f x ( x, y ) = – y 2 ( x 2 + y 4 ) –1;
fx

(

)

5, – 2 = –

kT
V
k
PT (V , T ) = ;
V

31. P (V , T ) =

4
≈ –0.1905
21

f y ( x, y ) = 2 xy ( x 2 + y 4 ) –1;
fy

24.

(

PT (100, 300) =

)

4 5
≈ –0.4259
5, – 2 = –
21

f x ( x, y ) = e y sinh x;

32. V [ PV (V , T )] + T [ PT (V , T )]
= V (– kTV –2 ) + T (kV –1 ) = 0

f x (–1, 1) = e sinh(–1) ≈ –3.1945

⎛ kT ⎞ ⎛ k ⎞ ⎛ V
PV VT TP = ⎜ –
⎟⎜ ⎟⎜
⎝ V 2 ⎠⎝ P ⎠⎝ k

f y ( x, y ) = e y cosh x;
f y (–1, 1) = e cosh(–1) ≈ 4.1945
x2 y 2
25. Let z = f ( x, y ) =
.
+
9
4
y
f y ( x, y ) =
2
The slope is f y (3, 2) = 1.

33.

⎛1⎞
27. z = f ( x, y ) = ⎜ ⎟ (9 x 2 + 9 y 2 − 36)1/ 2
⎝2⎠
9x
f x ( x, y ) =
2
2(9 x + 9 y 2 – 36)1/ 2
f x (2, 1) = 3
⎛5⎞
28. z = f ( x, y ) = ⎜ ⎟ (16 – x 2 )1/ 2 .
⎝4⎠
⎛ 5⎞
f x ( x, y ) = ⎜ – ⎟ x(16 – x 2 ) –1/ 2
⎝ 4⎠
5
f x (2, 3) = –
≈ –0.7217
4 3

f x ( x, y ) = 3 x 2 y – y 3 ; f xx ( x, y ) = 6 xy;

Therefore, f xx ( x, y ) + f yy ( x, y ) = 0.
34.

f x ( x, y ) = 2 x( x 2 + y 2 ) –1 ;
f xx ( x, y ) = –2( x 2 – y 2 )( x 2 + y 2 ) –1
f y ( x, y ) = 2 y ( x 2 + y 2 ) –1 ;
f yy ( x, y ) = 2( x 2 − y 2 )( x 2 + y 2 ) −1

35. Fy ( x, y ) = 15 x 4 y 4 – 6 x 2 y 2 ;
Fyy ( x, y ) = 60 x 4 y 3 − 12 x 2 y;
Fyyy ( x, y ) = 180 x 4 y 2 – 12 x 2

36.

f x ( x, y ) = [– sin(2 x 2 – y 2 )](4 x)
= –4 x sin(2 x 2 – y 2 )

f xx ( x, y ) = (–4 x)[cos(2 x 2 – y 2 )](4 x)
+ [sin(2 x 2 – y 2 )](–4)
f xxy ( x, y ) = –16 x 2 [– sin(2 x 2 – y 2 )](–2 y )
– 4[cos(2 x 2 – y 2 )](–2 y )
= –32 x 2 y sin(2 x 2 – y 2 ) + 8 y cos(2 x 2 – y 2 )

29. Vr (r , h) = 2πrh;
Vr (6, 10) = 120π ≈ 376.99 in.2

37. a.

2

30. Ty ( x, y ) = 3 y ; Ty (3, 2) = 12 degrees per ft
b.

c.

750

Section 12.2

kT
PV

=–
= –1
⎟=–
PV
PV


f y ( x, y ) = x3 – 3xy 2 ; f yy ( x, y ) = –6 xy

26. Let z = f ( x, y ) = (1/ 3)(36 – 9 x 2 – 4 y 2 )1/ 2 .
⎛ 4⎞
f y ( x, y ) = ⎜ − ⎟ y (36 − 9 x 2 − y 2 ) −1/ 2
⎝ 3⎠
8
≈ 0.8040.
The slope is f y (1, – 2) =
3 11

k
lb/in.2 per degree
100

∂3f
∂ y3
∂ 3y
∂ y∂ x 2
∂ 4y
∂ y 3∂ x

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38. a.

f yxx

b.

f yyxx

c.

f yyxxx

39. a.
b.

f x ( x, y, z ) = 6 xy – yz
f y ( x, y, z ) = 3 x 2 – xz + 2 yz 2 ;
f y (0, 1, 2) = 8

c.
40. a.
b.

45. Domain: (Case x < y)
The lengths of the sides are then x, y – x, and
1 – y. The sum of the lengths of any two sides
must be greater than the length of the remaining
side, leading to three inequalities:
1
x + (y – x) > 1 – y ⇒ y >
2
1
(y – x) + (1 – y) > x ⇒ x <
2
1
x + (1 – y) > y – x ⇒ y < x +
2

Using the result in a, f xy ( x, y, z ) = 6 x – z.
12 x 2 ( x3 + y 2 + z )3
f y ( x , y , z ) = 8 y ( x 3 + y 2 + z )3 ;
f y (0, 1, 1) = 64

c.

f z ( x, y, z ) = 4( x3 + y 2 + z )3 ;
2

2

f zz ( x, y, z ) = 12( x + y + z )

2

41.

f x ( x, y, x) = – yze – xyz – y ( xy – z 2 ) –1

42.

⎛ 1 ⎞⎛ xy ⎞
f x ( x, y, z ) = ⎜ ⎟⎜ ⎟
⎝ 2 ⎠⎝ z ⎠

–1/ 2

⎛ 1 ⎞⎛ 1 ⎞
f x (–2, – 1, 8) = ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠⎝ 4 ⎠

⎛ y⎞
⎜ ⎟;
⎝z⎠

–1/ 2

1
⎛ 1⎞
⎜– ⎟ = –
8
⎝ 8⎠

43. If f ( x, y ) = x 4 + xy3 + 12, f y ( x, y ) = 3 xy 2 ;
f y (1, – 2) = 12. Therefore, along the tangent line

Δy = 1 ⇒ Δz = 12, so 0, 1, 12 is a tangent

vector (since Δx = 0). Then parametric equations
⎧ x =1 ⎫


of the tangent line are ⎨ y = –2 + t ⎬ . Then the
⎪ z = 5 + 12t ⎪


point of xy-plane at which the bee hits is
(1, 0, 29) [since y = 0 ⇒ t = 2 ⇒ x = 1, z = 29].
44. The largest rectangle that can be contained in the
circle is a square of diameter length 20. The edge
of such a square has length 10 2, so its area is
200. Therefore, the domain of A is
2

2

{( x, y ) : 0 ≤ x + y < 400}, and the range is
(0, 200].

The case for y < x yields similar inequalities
(x and y interchanged). The graph of DA , the
domain of A is given above. In set notation it is
1
1
1⎫

D A = ⎨ ( x, y ) : x < , y > , y < x + ⎬
2
2
2⎭

1
1
1⎫

∪ ⎨ ( x, y ) : x > , y < , x < y + ⎬ .
2
2
2⎭

Range: The area is greater than zero but can be
arbitrarily close to zero since one side can be
arbitrarily small and the other two sides are
bounded above. It seems that the area would be
largest when the triangle is equilateral. An
1
has
equilateral triangle with sides equal to
3

3
3⎤
. Hence the range of A is ⎜⎜ 0,
area
⎥ . (In
36
⎝ 36 ⎦
Sections 8 and 9 of this chapter methods will be
presented which will make it easy to prove that
the largest value of A will occur when the triangle
is equilateral.)
46. a.

u = cos (x) cos (ct):
u x = – sin( x ) cos(ct ) ; ut = – c cos( x) sin(ct )
u xx = – cos( x) cos(ct )
utt = – c 2 cos( x) cos(ct )

Therefore, c 2 u xx = utt .
u = e x cosh(ct ) :
u x = e x cosh(ct ), ut = ce x sinh(ct )
u xx = e x cosh(ct ), utt = c 2 e x cosh(ct )

Therefore, c 2u xx = utt .
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reproduced, in any form or by any means, without permission in writing from the publisher.


b.

u = e – ct sin( x) :

48. a.

sin( x + y 2 )

u x = e – ct cos x
u xx = – e – ct sin x
ut = – cect sin x

Therefore, cu xx = ut .
2
u = t –1/ 2 e – x / 4ct :

u x = t –1/ 2 e – x

u xx =
ut =

2 / 4ct

⎛ x ⎞
⎜–

⎝ 2ct ⎠

b.

Dx (sin( x + y 2 ))

c.

D y (sin( x + y 2 ))

d.

Dx ( D y (sin( x + y ) 2 ))

( x 2 – 2ct )
(4c 2 t 5 / 2 e x

2 / 4ct

)

2

( x – 2ct )
2

(4ct 5 / 2 e x / 4ct )
Therefore, cu xx = ut .

47. a.

Moving parallel to the y-axis from the point
(1, 1) to the nearest level curve and
Δz
, we obtain
approximating
Δy
f y (1, 1) =

4–5
= –4.
1.25 – 1

b. Moving parallel to the x-axis from the point
(–4, 2) to the nearest level curve and
Δz
approximating
, we obtain
Δx
1– 0
2
f x (–4, 2) ≈
= .
–2.5 – (–4) 3
c.

Moving parallel to the x-axis from the point
(–5, –2) to the nearest level curve and
Δz
approximately
, we obtain
Δx
1– 0
2
f x (–4, – 5) ≈
= .
–2.5 – (–5) 5

d. Moving parallel to the y-axis from the point
(0, –2) to the nearest level curve and
Δz
, we obtain
approximating
Δy
f y (0, 2) ≈

f y ( x, y , z )
= lim

Δy →0

b.

Δz →0

c.

f ( x , y + Δy , z ) − f ( x , y , z )
Δy

f z ( x, y , z )
= lim

0 –1
–19
8

8
= .
– (–2) 3

49. a.

f ( x , y , z + Δz ) − f ( x , y , z )
Δz

Gx ( w, x, y, z )
G ( w, x + Δx, y , z ) − G ( w, x, y, z )
Δx
Δx →0

= lim

752

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d.

e.


λ ( x, y , z , t )
∂z
λ ( x , y , z + Δz , t ) − λ ( x , y , z , t )
= lim
Δz
Δz →0

4. The limit does not exist because of Theorem A.
The function is a rational function, but the limit
of the denominator is 0, while the limit of the
numerator is -1.


S (b0 , b1 , b2 ,… , bn ) =
∂b2

5. –

⎛ S (b0 , b1 , b2 + Δb2 ,… , bn )





S
(
b
,
b
,
b
,
,
b
)
0 1 2
n ⎟
= lim ⎜

Δb2
Δb2 →0 ⎜





50. a.

b.

8.

tan( x 2 + y 2 )

lim

= (1)(1) = 1


wyz
+ 1 ⋅ ln ( wxyz )
⎡⎣ x ln ( wxyz ) ⎤⎦ = x ⋅
∂x
wxyz

λt ( x, y, z , t )

(1 + xyzt ) cos x − t ( cos x ) xyz
(1 + xyzt )2
cos x

(1 + xyzt )2

12.3 Concepts Review
1. 3; (x, y) approaches (1, 2).
lim

( x, y )→(1, 2)

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

3. contained in S
4. an interior point of S; boundary points

Problem Set 12.3
1. –18

3
⎛ 2π ⎞
= 2 cos 2 2π – sin ⎜ ⎟ = 2 –
≈ 1.1340
2
⎝ 3 ⎠

Instructor’s Resource Manual

1

2

2

( x + y ) cos( x + y 2 )

9. The limit does not exist since the function is not
defined anywhere along the line y = x. That is,
there is no neighborhood of the origin in which
the function is defined everywhere except
possibly at the origin.
10.

( x 2 + y 2 )( x 2 – y 2 )

lim

x2 + y 2

( x, y )→(0, 0)

=

lim

( x, y )→(0, 0)

( x2 – y 2 ) = 0

11. Changing to polar coordinates,
xy
r cos θ ⋅ r sin θ
lim
= lim
r
r →0
( x, y )→(0,0) x 2 + y 2
= lim r cos θ ⋅ sin θ = 0
r →0

12. If ( x, y ) approaches (0, 0) along the line y = x ,
x2

lim

2

=

2 2

lim

1

( x, x )→(0,0) 4 x 2
+x )
Thus, the limit does not exist.
( x, x )→(0,0) ( x

= +∞

13. Use polar coordinates.
x7 / 3

r


⎛ xy ⎞ ⎤
x cos 2 xy – sin ⎜ ⎟ ⎥

( x, y )→(2, π) ⎣
⎝ 3 ⎠⎦

2

( x, y )→(0, 0)

x +y

lim

sin( x 2 + y 2 )

lim

2

2. 3

( x2 + y2 )

( x, y )→(0, 0)

= cos w sin x cos y cos z

=

3.

7. 1

=

=

2.

6. −1


( sin w sin x cos y cos z )
∂w

= 1 + ln ( wxyz )

c.

5
2

1/ 3

2

=

r 7 / 3 ( cos θ )

( cos θ )

r
7/3

2

7/3

= r1/ 3 ( cos θ )

7/3

→ 0 as r → 0 , so the limit is 0.

14. Changing to polar coordinates,
r 2 cos 2 θ − r 2 sin 2 θ
lim r 2 cos θ sin θ ⋅
r →0
r2
= lim r 2 cos θ sin θ cos 2θ = 0
r →0

Section 12.3

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15.

f ( x, y ) =

r 4 cos 2 θ sin 2 θ

r 2 cos 2 θ + r 4 sin 4 θ
⎛ cos 2 θ sin 2 θ ⎞
= r2 ⎜

⎜ cos 2 θ + r 2 sin 4 θ ⎟


If cos θ = 0 , then f ( x, y ) = 0 . If cos θ ≠ 0 ,

hen this converges to 0 as r → 0 . Thus the
limit is 0.

26. Require 4 − x 2 − y 2 − z 2 > 0;
x 2 + y 2 + z 2 < 4. S is the space in the interior
of the sphere centered at the origin with radius 2.

27. The boundary consists of the points that form the
outer edge of the rectangle. The set is closed.

16. As ( x, y ) approaches (0,0) along x = y2,
y4

1
= . Along the x-axis,
2
y +y
0
however,
lim
= 0. Thus, the limit does
( x,0)→(0,0) x 2
not exist.
lim

( x, x )→(0,0)

17.

4

4

f ( x, y ) is continuous for all ( x, y ) since
for all ( x, y ), x 2 + y 2 + 1 ≠ 0.

18.

28. The boundary consists of the points of the circle
shown. The set is open.

f ( x, y ) is continuous for all ( x, y ) since
for all ( x, y ), x 2 + y 2 + 1 > 0.

19. Require 1 – x 2 – y 2 > 0; x 2 + y 2 < 1. S is the
interior of the unit circle centered at the origin.
20. Require 1 + x + y > 0; y > − x − 1. S is the set
of all ( x, y ) above the line y = − x − 1.
2

21. Require y – x ≠ 0. S is the entire plane except

the parabola y = x 2 .

29. The boundary consists of the circle and the
origin. The set is neither open (since, for
example, (1, 0) is not an interior point), nor
closed (since (0, 0) is not in the set).

22. The only points at which f might be
discontinuous occur when xy = 0.
sin( xy )
lim
= 1 = f (a, 0) for all nonzero
( x, y )→( a , 0) xy
a in , and then
sin( xy )
lim
= 1 = f (0, b) for all b in .
( x, y )→(0, b ) xy
Therefore, f is continuous on the entire plane.
23. Require x – y + 1 ≥ 0; y ≤ x + 1. S is the region
below and on the line y = x + 1.
24. Require 4 – x 2 – y 2 > 0; x 2 + y 2 < 4. S is the
interior of the circle of radius 2 centered at the
origin.
25.

f ( x, y, z ) is continuous for all ( x, y, z ) ≠ ( 0, 0, 0 )

since for all ( x, y, z ) ≠ ( 0, 0, 0 ) , x 2 + y 2 + z 2 > 0.

754

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30. The boundary consists of the points on the line
x = 1 along with the points on the line x = 4. The
set is neither closed nor open.

35. Along the x-axis (y = 0):

0

lim

( x, y )→(0, 0) x 2

+0

=0.

Along y = x:
x2

1 1
= .
( x, y )→(0, 0) 2 x
( x, y )→(0, 0) 2 2
Hence, the limit does not exist because for some
points near the origin f(x, y) is getting closer to 0,
1
but for others it is getting closer to .
2
lim

2

=

lim

0

36. Along y = 0: lim
31. The boundary consists of the graph of
⎛1⎞
y = sin ⎜ ⎟ along with the part of the y-axis for
⎝ x⎠
which y ≤ 1. The set is open.

x→ 0 x 2

lim

x 2 + x3

x→ 0 x 2

37. a.

+x

2

x→ 0 x 4

1+ x 1
= .
2
x→ 0 2

+ (mx)2
mx

= lim

x→ 0 x 2

b.

32. The boundary is the set itself along with the
origin. The set is neither open (since none of its
points are interior points) nor closed (since the
origin is not in the set).

c.

lim

+ m2

x2 ( x2 )

x→ 0 x 4

lim

= 0. Along y = x:

= lim

x 2 (mx)

lim

+0

2 2

+ (x )

x→ 0 x 4

+ m2 x 2

=0

= lim

x4

x→ 0 2 x

x2 y

( x, y )→( 0, 0) x 4

mx3

= lim

+y

4

1 1
=
x→ 0 2 2

= lim

does not exist.

38. f is discontinuous at each overhang. More
interesting, f is discontinuous along the
Continental Divide.
39. a.

{( x, y, z ) : x 2 + y 2 = 1, z in [1, 2]} [For
x 2 + y 2 < 1, the particle hits the hemisphere
and then slides to the origin (or bounds

toward the origin); for x 2 + y 2 = 1, it
2

33.

bounces up; for x 2 + y 2 > 1, it falls straight
down.]

2

x – 4y
( x + 2 y )( x – 2 y )
=
= x + 2 y (if x ≠ 2y)
x – 2y
x – 2y

If x = 2 y , x + 2 y = 2 x . Take g ( x ) = 2 x .

b.

{( x, y, z ) : x 2 + y 2 = 1, z = 1} (As one moves
at a level of z = 1 from the rim of the bowl
toward any position away from the bowl
there is a change from seeing all of the
interior of the bowl to seeing none of it.)

c.

{(x, y, z): z = 1} [f(x, y, z) is undefined
(infinite) at (x, y, 1).]

d.

φ (Small changes in points of the domain

34. Let L and M be the latter two limits.
[ f ( x, y ) + g ( x, y )] – [ L + M ]
≤ f ( x, y ) – L + f ( x , y ) – M ≤

ε

+

ε

2 2
for (x, y) in some δ-neighborhood of (a, b).
Therefore,
lim
[ f ( x, y ) + g ( x, y )] = L + M .
( x , y ) →( a , b )

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result in small changes in the shortest path
from the points to the origin.)

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40. f is continuous on an open set D and P0 is in D implies that there is neighborhood of P0 with radius r on which f
is continuous. f is continuous at P0 ⇒ lim f ( P) = f ( P0 ). Now let ε = f ( P0 ) which is positive. Then there is a δ
P → P0

such that 0 < δ < r and f ( p ) – f ( P0 ) < f ( P0 ) if P is in the δ-neighborhood of P0 . Therefore,
– f ( P0 ) < f ( p ) – f ( P0 ) < f ( P0 ), so 0 < f(p) (using the left-hand inequality) in that δ-neighborhood of P0 .

41. a.

⎧⎪( x 2 + y 2 )1/ 2 + 1 if y ≠ 0 ⎫⎪
f ( x, y ) = ⎨
⎬ . Check discontinuities where y = 0.
if y = 0 ⎭⎪
⎩⎪ x – 1

As y = 0, ( x 2 + y 2 )1/ 2 + 1 = x + 1, so f is continuous if x + 1 = x – 1 . Squaring each side and simplifying
yields x = – x, so f is continuous for x ≤ 0. That is, f is discontinuous along the positive x-axis.
b. Let P = (u, v) and Q = (x, y).
⎪⎧ OP + OQ if P and Q are not on same ray from the origin and neither is the origin ⎫⎪
f (u, v, x, y ) = ⎨
⎬.
otherwise
⎩⎪ PQ
⎭⎪

This means that in the first case one travels from P to the origin and then to Q; in the second case one travels
directly from P to Q without passing through the origin, so f is discontinuous on the set
{(u, v, x, y ) : u, v = k x, y for some k > 0, u , v ≠ 0, x, y ≠ 0}.

42. a.

⎛ hy ( h2 – y 2 )

⎜ 2 2 –0⎟
2
2
+
h
y
⎟ = lim y (h − y ) = − y
f x (0, y ) = lim ⎜
⎟ h →0 h 2 + y 2
h
h →0 ⎜





b.



f y ( x, 0) = lim ⎜
h →0 ⎜



c.

f yx (0, 0) = lim

d.

h →0

xh ( x 2 – h 2 )
x 2 + h2

h


–0⎟
y( x2 – h2 )
⎟ = lim
=x
⎟ h →0 x 2 + y 2



f y (0 + h, y ) – f y (0, y )
h

h–0
=1
h →0 h

= lim

f x ( x, 0 + h) – f x ( x, 0)
–h – 0
= lim
= –1
h
h
h →0
h →0
Therefore, f xy (0, 0) ≠ f yx (0, 0).
f xy (0, 0) = lim

43.

b.

44. a.

756

45.

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46. A function f of three variables is continuous at a
point (a, b, c ) if f (a, b, c) is defined and equal to
the limit of f ( x, y, z ) as ( x, y, z ) approaches
(a, b, c) . In other words,
lim
f ( x, y, z ) = f (a, b, c) .
( x , y , z ) → ( a ,b , c )

A function of three variables is continuous on an
open set S if it is continuous at every point in the
interior of the set. The function is continuous at
a boundary point P of S if f (Q) approaches
f (P) as Q approaches P along any path through
points in S in the neighborhood of P.
47. If we approach the point ( 0, 0, 0 ) along a straight

path from the point ( x, x, x ) , we have
lim

x ( x )( x)
3

3

3

=

lim

x3
3

x +x +x
( x , x, x )→(0,0,0) 3 x
Since the limit does not equal to f (0, 0, 0) , the
function is not continuous at the point (0, 0, 0) .
( x, x, x )→(0,0,0)

=

1
3

48. If we approach the point (0, 0, 0) along the
x-axis, we get
( x 2 − 02 )
x2
lim
(0 + 1)
=
lim
=1
( x,0,0)→(0,0,0)
( x 2 + 02 ) ( x,0,0)→(0,0,0) x 2
Since the limit does not equal f (0, 0, 0) , the
function is not continuous at the point (0, 0, 0).

12.4 Concepts Review
1. gradient
2. locally linear
3.

∂f
∂f
(p)i +
(p) j; y 2 i + 2 xyj
∂x
∂y

4. tangent plane

Problem Set 12.4
1.

2 xy + 3 y, x 2 + 3 x

2.

3 x 2 y , x3 – 3 y 2

3. ∇f ( x, y ) = ( x)(e xy y ) + (e xy )(1), xe xy x = e xy xy + 1, x 2
4.

2 xy cos y, x 2 (cos y – y sin y )

5. x( x + y ) –2 y ( x + 2), x 2
6. ∇f ( x, y ) = 3[sin 2 ( x 2 y )][cos( x 2 y )](2 xy ), 3[sin 2 ( x 2 y )][cos( x 2 y )]( x 2 ) = 3x sin 2 ( x 2 y ) cos( x 2 y ) 2 y, x

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7. ( x 2 + y 2 + z 2 ) –1/ 2 x, y, z
8.

2 xy + z 2 , x 2 + 2 yz , y 2 + 2 xz

9. ∇f ( x, y ) = ( x 2 y )(e x – z ) + (e x – z )(2 xy ), x 2 e x – z , x 2 ye x – z (–1) = xe x – z y ( x + 2), x, – xy
10.

xz ( x + y + z ) –1 + z ln( x + y + z ), xz ( x + y + z ) –1 , xz ( x + y + z ) –1 + x ln( x + y + z )

11. ∇f ( x, y ) = 2 xy – y 2 , x 2 – 2 xy ; ∇f (–2, 3) = –21, 16
z = f (–2, 3) + –21, 16 ⋅ x + 2, y – 3 = 30 + (–21x – 42 + 16 y – 48)

z = –21x + 16y – 60
12. ∇f ( x, y ) = 3x 2 y + 3 y 2 , x3 + 6 xy , so ∇f (2, – 2) = (–12, – 16).

Tangent plane:
z = f (2, – 2) + ∇ (2, – 2) ⋅ x – 2, y + 2

= 8 + –12, – 16 ⋅ x – 2, y + 2 = 8 + (–12x + 24 – 16y – 32)

z = –12x – 16y
13. ∇f ( x, y ) = – π sin(πx) sin(πy ), π cos(πx) cos(πy ) + 2π cos(2πy )
1⎞

∇f ⎜ –1, ⎟ = 0, – 2π
2⎠

1⎞
1

z = f ⎜ –1, ⎟ + 0, – 2π ⋅ x + 1, y –
= –1 + (0 – 2πy + π);
2⎠
2

z = –2 π y + ( π – 1)

14. ∇f ( x, y ) =

2x
x2
,−
; ∇f (2, − 1) = −4, − 4
y
y2

z = f (2, – 1) + −4, – 4 ⋅ x – 2, y + 1

= –4 + (–4x + 8 –4y – 4)
z = –4x – 4y
15. ∇f ( x, y, z ) = 6 x + z 2 , – 4 y, 2 xz , so ∇f (1, 2, – 1) = 7, – 8, – 2

Tangent hyperplane:
w = f (1, 2, – 1) + ∇f (1, 2, – 1) ⋅ x – 1, y – 2, z + 1 = –4 + 7, – 8, – 2 ⋅ x – 1, y – 2, z + 1
= –4 + (7x – 7 – 8y + 16 – 2z – 2)
w = 7x – 8y – 2z + 3
16. ∇f ( x, y, z ) = yz + 2 x, xz , xy ; ∇f (2, 0, – 3) = 4, – 6, 0
w = f (2, 0, – 3) + 4, – 6, 0 ⋅ x – 2, y, z + 3 = 4 + (4x – 8 – 6y + 0)

w = 4x – 6y – 4
gf x − fg x , gf y − fg y , gf z − fg z
g fx , f y , fz – f gx , g y , gz
⎛ f ⎞
=
17. ∇ ⎜ ⎟ =
g2
g2
⎝g⎠

=

g ∇f – f ∇g
g2

18. ∇( f r ) = rf r –1 f x , rf r –1 f y , rf r –1 f z = rf r –1 f x , f y , f z = rf r –1∇f

758

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19. Let F ( x, y, z ) = x 2 − 6 x + 2 y 2 − 10 y + 2 xy − z = 0
∇F ( x, y, z ) = 2 x − 6 + 2 y, 4 y − 10 + 2 x, −1

The tangent plane will be horizontal if
∇F ( x, y, z ) = 0, 0, k , where k ≠ 0 . Therefore,

we have the following system of equations:
2x + 2 y − 6 = 0
2 x + 4 y − 10 = 0
Solving this system yields x = 1 and y = 2 .
Thus, there is a horizontal tangent plane at
( x, y ) = (1, 2 ) .
20. Let F ( x, y, z ) = x3 − z = 0
∇F ( x, y , z ) = 3 x 2 , 0, −1

The tangent plane will be horizontal if
∇F ( x, y, z ) = 0, 0, k , where k ≠ 0 . Therefore,
we need only solve the equation 3x 2 = 0 . There
is a horizontal tangent plane at ( x, y ) = (0, y ).
(Note: there are infinitely many points since y
can take on any value).
21. a. The point (2,1,9) projects to (2,1, 0) on the xy
plane. The equation of a plane containing this
point and parallel to the x-axis is given by
y = 1 . The tangent plane to the surface at the

point (2,1,9) is given by
z = f (2,1) + ∇f (2,1) ⋅ x − 2, y − 1
= 9 + 12,10 x − 2, y − 1
= 12 x + 10 y − 25
The line of intersection of the two planes is
the tangent line to the surface, passing
through the point (2,1,9) , whose projection in
the xy plane is parallel to the x-axis. This line
of intersection is parallel to the cross product
of the normal vectors for the planes. The
normal vectors are 12,10, −1 and 0,1, 0 for

the tangent plane and vertical plane
respectively. The cross product is given by
12,10, −1 × 0,1, 0 = 1, 0,12
Thus, parametric equations for the desired
tangent line are x = 2 + t
y =1
z = 9 + 12t
b. Using the equation for the tangent plane from
the previous part, we now want the vertical
plane to be parallel to the y-axis, but still pass
through the projected point (2,1, 0) . The
vertical plane now has equation x = 2 . The
normal equations are given by 12,10, −1 and

respectively. Again we find the cross product
of the normal vectors:
12,10, −1 × 1, 0, 0 = 0,10,10
Thus, parametric equations for the desired
tangent line are x = 2
y = 1 + 10t
z = 9 + 10t
c. Using the equation for the tangent plane from
the first part, we now want the vertical plane
to be parallel to the line y = x , but still pass

through the projected point (2,1, 0) . The
vertical plane now has equation y − x + 1 = 0 .
The normal equations are given by
12,10, −1 and 1, −1, 0 for the tangent and
vertical planes respectively. Again we find
the cross product of the normal vectors:
12,10, −1 × 1, −1, 0 = −1, −1, −22
Thus, parametric equations for the desired
tangent line are x = 2 − t
y = 1− t
z = 9 − 22t
22. a. The point (3, 2, 72) on the surface is the point
(3, 2, 0) when projected into the xy plane. The
equation of a plane containing this point and
parallel to the x-axis is given by y = 2 . The
tangent plane to the surface at the point
(3, 2, 72) is given by
z = f (3, 2) + ∇f (3, 2) ⋅ x − 3, y − 2
= 72 + 48,108 x − 3, y − 2
= 48 x + 108 y − 288
The line of intersection of the two planes is
the tangent line to the surface, passing
through the point (3, 2, 72) , whose projection
in the xy plane is parallel to the x-axis. This
line of intersection is parallel to the cross
product of the normal vectors for the planes.
The normal vectors are
48,108, −1 and 0, 2, 0 for the tangent plane

and vertical plane respectively. The cross
product is given by
48,108, −1 × 0, 2, 0 = 2, 0,96
Thus, parametric equations for the desired
tangent line are
x = 3 + 2t
y=2
z = 72 + 96t

1, 0, 0 for the tangent and vertical planes

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b. Using the equation for the tangent plane from
the previous part, we now want the vertical
plane to be parallel to the y-axis, but still pass
through the projected point (3, 2, 72) . The
vertical plane now has equation x = 3 . The
normal equations are given by
48,108, −1 and 3, 0, 0 for the tangent and

c. Using the equation for the tangent plane from
the first part, we now want the vertical plane
to be parallel to the line x = − y , but still pass

through the projected point (3, 2, 72) . The
vertical plane now has equation y + x − 5 = 0 .
The normal equations are given by
48,108, −1 and 1,1, 0 for the tangent and

vertical planes respectively. Again we find
the cross product of the normal vectors:
48,108, −1 × 3, 0, 0 = 0, −3, −324

vertical planes respectively. Again we find
the cross product of the normal vectors:
48,108, −1 × 1,1, 0 = 1, −1, −60

Thus, parametric equations for the desired
tangent line are
x=3
y = 2 − 3t
z = 72 − 324t

⎛ 1
⎛ 1
xy ⎞
xy ⎞
23. ∇f ( x, y ) = –10 ⎜
y ⎟ , – 10 ⎜
x⎟
⎜ 2 xy xy ⎟
⎜ 2 xy xy ⎟





Thus, parametric equations for the desired
tangent line are
x = 3+t
y = 2−t
z = 72 − 60t

=

–5 xy
xy

3/ 2

y, x


a
a ⎤
= .⎥
⎢ Note that
a
a ⎥⎦
⎣⎢

∇f (1, – 1) = –5, 5

Tangent plane:
z = f (1, – 1) + ∇f (1, – 1) ⋅ x – 1, y + 1 = –10 + –5, 5 ⋅ x – 1, y + 1 = –10 + (–5 x + 5 + 5 y + 5)

z = –5x + 5y
24. Let a be any point of S and let b be any other
point of S. Then for some c on the line segment
between a and b:
f (b) − f (a) = ∇f (c) ⋅ (b − a) = 0 ⋅ (b − a) = 0, so
f(b) = f(a) (for all b in S).

26. f(b) – f(a) = f 2, 6 − f 0, 0 = 0 − 2 = −2

−2 =

The value c = cx , c y will be a solution to
−5 = −2cx , −2c y

{

−2 =

2,1

}

c ∈ cx , c y : 4cx + 2c y = 5

In order for c to be between a and b, c must lie
on the line y = 12 x . Consequently, c will be the
solution to the following system of equations:
4cx + 2c y = 5 and c y = 12 c x . The solution is
c = 1, 12 .

760

Section 12.4

4 − x2

, 0 ; b – a = 2, 6

The value c = cx , c y will be the solution to

25. f(a) – f(b) = f 2,1 − f 0, 0 = 4 − 9 = −5
∇f ( x, y ) = −2 x, −2 y ; b – a = 2,1

−x

∇f ( x , y ) =

−c x
4 − cx 2
−2cx
4 − cx 2

,0

2, 6

⇒ cx = 2

Since c must be between a and b, c must lie on
the line y = 3x. Since cx = 2, c y = 3 2.
Thus, c =

2,3 2 .

27. ∇f (p) = ∇g (p) ⇒ ∇[ f (p) – g (p)] = 0
⇒ f (p) – g (p) is a constant.

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28. ∇f (p) = p ⇒ ∇f ( x, y ) = x, y
⇒ f x ( x , y ) = x, f y ( x, y ) = y

31. a.

(i)
∂( f + g ) ∂( f + g )
∂( f + g )
i+
j+
k
∂x
∂y
∂z
∂f
∂g
∂f
∂g
∂f
∂g
=
i+
i+
j+
j+ k + k
∂x
∂x
∂y
∂y
∂z
∂z
∂f
∂f
∂f
∂g
∂g
∂g
=
i+
j+ k +
i+
j+ k
∂x
∂y
∂z
∂x
∂y
∂z
= ∇f + ∇g

∇[ f + g ] =

1 2
x + α ( y ) for any function of y,
2
1
and f ( x, y ) = y 2 + β ( x) for any function of x.
2
1
⇒ f ( x, y ) = ( x 2 + y 2 ) + C for any C in .
2
⇒ f ( x, y ) =

∂[α f ] ∂[α f ] ∂[α f ]
i+
j+
k
∂x
∂y
∂z
∂[ f ]
∂[ f ]
∂[ f ]

i +α
j+α
k
∂x
∂y
∂z
= α∇f

29.

(ii) ∇[α f ] =

− xy

(iii)
∇[ fg ] =

∂ ( fg ) ∂ ( fg )
∂ ( fg )
i+
j+
k
∂x
∂y
∂z

∂f ⎞ ⎛ ∂g
∂f ⎞
⎛ ∂g
=⎜ f
+ g ⎟i + ⎜ f
+ g ⎟j
x
x
y



∂y ⎠

⎠ ⎝
∂f ⎞
⎛ ∂g
+⎜ f
+ g ⎟k
∂z ⎠
⎝ ∂z
⎛ ∂g
∂g
∂g ⎞
= f ⎜ i+
j+ k ⎟
∂y
∂z ⎠
⎝ ∂x

a.

⎛ ∂f
∂f
∂f ⎞
+g ⎜ i +
j+ k ⎟
∂y
∂z ⎠
⎝ ∂x
= f ∇g + g ∇f

The gradient points in the direction of
greatest increase of the function.

b. No. If it were, 0 + h – 0 = 0 + h δ (h) where

δ (h) → 0 as h → 0, which is possible.
30.

b. (i)
∇[ f + g ] =
+
=
+

sin(x) + sin(y) – sin(x + y)

∂( f + g )
∂( f + g )
i1 +
i2
∂x1
∂x2
+

∂( f + g )
in
∂xn

∂f
∂g
∂f
∂g
i1 +
i1 +
i2 +
i2
∂x1
∂x1
∂x2
∂x2
+

∂f
∂g
in +
in
∂xn
∂xn

=

∂f
∂f
i1 +
i2 +
∂x1
∂x2

+

∂f
in
∂xn

+

∂g
∂g
i1 +
i2 +
∂x1
∂x2

+

∂g
in
∂xn

= ∇f + ∇ g

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761

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(ii)
∂[α f ]
∂[α f ]
∂[α f ]
∇[α f ] =
i1 +
i2 + " +
in
∂x1
∂x2
∂xn


∂[ f ]
∂[ f ]
∂[ f ]
i1 + α
i2 + " + α
in
∂x1
∂x2
∂xn

= α∇f

(iii)
∇[ fg ] =

∂ ( fg )
∂ ( fg )
∂ ( fg )
i1 +
i2 + " +
in
∂x1
∂x2
∂xn

⎡⎛ 1 ⎞

= 2 x – 3 y , – 3 x + 4 y ⋅ ⎢⎜
⎟ 2, – 1 ⎥ ;
⎣⎝ 5 ⎠

27
Du f (–1, 2) = –
≈ –12.0748
5

⎡⎛ 1 ⎞

5. Du f ( x, y ) = e x sin y, cos y ⋅ ⎢⎜ ⎟ 1, 3 ⎥ ;
2




⎛ π⎞
Du f ⎜ 0, ⎟ =
⎝ 4⎠

(

2+ 6

) ≈ 0.9659

⎛ ∂g
⎛ ∂g
∂f ⎞
∂f ⎞
=⎜ f
+g
+g
⎟ i1 + ⎜ f
⎟ i2
∂x1 ⎠
∂x2 ⎠
⎝ ∂x1
⎝ ∂x2
⎛ ∂g
∂f ⎞
+" + ⎜ f
+g
⎟ in


x
xn ⎠
n

⎛ ∂g
∂g
∂g ⎞
= f⎜
i1 +
i2 + " +
in ⎟



x
x
xn ⎠
2
⎝ 1

6. Du f ( x, y ) = – ye – xy – xe – xy ⋅

⎛ ∂f

∂f
∂f
+g ⎜
i1 +
i2 + " +
in ⎟
∂x2
∂xn ⎠
⎝ ∂x1
= f ∇ g + g ∇f

7. Du f ( x, y, z ) =

12.5 Concepts Review
1.

4. Du f ( x, y )

4

Du f (1, –1) = e, – e ⋅
≈ –3.7132

–1, 3
2

–1, 3
2
=

–e – e 3
2

⎡⎛ 1 ⎞

= 3 x 2 y, x3 – 2 yz 2 , – 2 y 2 z ⋅ ⎢⎜ ⎟ 1, – 2, 2 ⎥ ;
⎣⎝ 3 ⎠

52
Du f (–2, 1, 3) =
3
⎡⎛ 1 ⎞
8. Du f ( x, y, z ) = 2 x, 2 y, 2 z ⋅ ⎢⎜ ⎟
⎣⎝ 2 ⎠

[ f (p + hu) – f (p)]
h


2, –1, –1 ⎥ ;


Du f (1, – 1, 2) = 2 – 1 ≈ 0.4142

2. u1 f x ( x, y ) + u2 f y ( x, y )

9. f increases most rapidly in the direction of the
3. greatest increase

gradient. ∇f ( x, y ) = 3 x 2 , – 5 y 4 ;

4. level curve

∇f (2, – 1) = 12, – 5

12, – 5

Problem Set 12.5
1. Du f ( x, y ) = 2 xy, x 2 ⋅

3 4
8
, − ; Du f (1, 2) =
5 5
5

⎡⎛ 1 ⎞

2. Du f ( x, y ) = x –1 y 2 , 2 y ln x ⋅ ⎢⎜
⎟ 1, – 1 ⎥ ;
⎣⎝ 2 ⎠

Du f (1, 4) = 8 2 ≈ 11.3137

a⎞
3. Du f ( x, y ) = f ( x, y ) ⋅ u ⎜ where u = ⎟

a ⎟⎠

1, – 1
= 4 x + y, x – 2 y ⋅
;
2
1, – 1
3
Du f (3, – 2) = 10, 7 ⋅
=
≈ 2.1213
2
2

762

Section 12.5

is the unit vector in that direction. The
13
rate of change of f(x, y) in that direction at that
point is the magnitude of the gradient.
12, – 5 = 13

10. ∇f ( x, y ) = e y cos x, e y sin x ;
3 1
⎛ 5π ⎞
∇f ⎜ , 0 ⎟ = –
,
, which is a unit vector.
2 2
⎝ 6

The rate of change in that direction is 1.

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16. At (2, 1), x 2 + 4 y 2 = 8 is the level curve.

11. ∇f ( x, y, z ) = 2 xyz , x 2 z , x 2 y ;

∇f ( x , y ) = 2 x , 8 y

f (1, – 1, 2) = –4, 2, – 1

A unit vector in that direction is
⎛ 1 ⎞

⎟ −4, 2, − 1 . The rate of change in that
⎝ 21 ⎠
direction is

∇f (2, 1) = 4 1, 2 , which is perpendicular to the

level curve at (2, 1).

21 ≈ 4.5826.

12. f increases most rapidly in the direction of the
gradient. ∇f ( x, y, z ) = e yz , xze yz , xye yz ;
∇f (2, 0, – 4) = 1, – 8, 0
1, – 8, 0

is a unit vector in that direction.

65

1, – 8, 0 = 65 ≈ 8.0623 is the rate of change

of f(x, y, z) in that direction at that point.

13. –∇f ( x, y ) = 2 x, y ; –∇f (–1, 2) = 2 –1, 2 is

14. – ∇f ( x, y ) = –3cos(3 x – y ), cos(3 x – y ) ;
⎛π π⎞ ⎛ 1 ⎞
– ∇f ⎜ , ⎟ = ⎜
⎟ –3, 1 is the direction of
⎝6 4⎠ ⎝ 2⎠
most rapid decrease. A unit vector in that
⎛ 1 ⎞
direction is ⎜
⎟ –3, 1 .
⎝ 10 ⎠
y
x

2

y
x2

Du f ( x, y, z ) = y, x, 2 z ⋅

=2

2
2 1
,– ,
3
3 3

2
3

⎛ π⎞
18. ⎜ 0, ⎟ is on the y-axis, so the unit vector toward
⎝ 3⎠
the origin is –j.
Du ( x, y ) = – e – x cos y , – e – x sin y ⋅ 0, – 1
= e – x sin y;

3
⎛ π⎞
Du ⎜ 0, ⎟ =
3
2



19. a.

= k . For p = (1, 2),

k = 2, so the level curve through (1, 2) is

2
2 1
,– ,
3
3 3

Du f (1, 1, 1) =

the direction of most rapid decrease. A unit
⎛ 1 ⎞
vector in that direction is u = ⎜
⎟ –1, 2 .
⎝ 5⎠

15. The level curves are

17. u =

b.

or y = 2 x (x ≠ 0).
2

Hottest if denominator is smallest; i.e., at the
origin.
∇T ( x , y , z ) =

–200 2 x, 2 y, 2 z
(5 + x 2 + y 2 + z 2 )2

;

∇f (1, 2) = –4, 1 , which is perpendicular to the

⎛ 25 ⎞
∇T (1, – 1, 1) = ⎜ – ⎟ 1, – 1, 1
⎝ 4 ⎠
−1,1, −1 is one vector in the direction of

parabola at (1, 2).

greatest increase.

∇f ( x, y ) = –2 yx –3 , x –2

c.

Yes

20. – ∇V ( x, y, z )
= –100e –( x
= 200e

2 + y2 + z 2 )

–( x 2 + y 2 + z 2 )

–2 x, – 2 y, – 2 z
x, y, z is the direction of

greatest decrease at (x, y, z), and it points away
from the origin.

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Section 12.5

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21.

∇ f ( x, y , z )

(

= x x2 + y 2 + z 2

)

1
2

(

)

(

1

z2 2

y x2 + y 2 + z 2
z x2 + y2 +



)



1
2

24. The unit vector from (2, 4) toward (5, 0) is
3
4
, – . Then
5
5
3
4
Du f (2, 4) = –3, 8 ⋅ , –
= –8.2.
5
5

cos x 2 + y 2 + z 2 ,

cos x 2 + y 2 + z 2 ,

25. The climber is moving in the direction of
⎛ 1 ⎞
u=⎜
⎟ −1, 1 . Let
⎝ 2⎠

cos x 2 + y 2 + z 2

f ( x, y ) = 3000e−( x

1



= ⎜ x 2 + y 2 + z 2 2 cos x 2 + y 2 + z 2 ⎟ x, y, z




which either points towards or away from the
origin.

(

)

2 + 2 y 2 ) /100

∇f ( x, y ) = 3000e –( x

.

2 + 2 y 2 ) /100



x
y
,–
;
50
25

f (10, 10) = –600e –3 1, 2

She will move at a slope of
⎛ 1 ⎞
Du (10, 10) = –600e –3 1, 2 ⋅ ⎜
⎟ –1, 1
⎝ 2⎠

22. Let D = x + y + z be the distance. Then we
have
dT ∂D dT ∂D dT ∂D
∂T ∂T ∂T
∇T =
=
,
,
,
,
∂x ∂y ∂z
dD ∂x dD ∂y dD ∂z
2

=

(

2

2

dT
x x2 + y 2 +
dD

1
2 −2
z

)

,

(

dT
y x2 + y2 +
dD

1
2 −2
z

)

1
dT
2
2
2 −2
z x +y +z
dD

(

)

⎛ dT

⎟ x, y , z
=⎜
x2 + y2 +
⎜ dD



which either points towards or away from the
origin.

(

1
2 −2
z

)

(

)

= –300 2 e –3 ≈ –21.1229 .

She will descend. Slope is about –21.
,

26.

dx
dt

2x

=

dy
dt

–2 y

;

dx dy
=
; ln x = – ln y + C
x –y

At t = 0: ln –2 = – ln 1 + C ⇒ C = ln 2.
2
2
; x = ; xy = 2
y
y
Since the particle starts at (–2, 1) and neither x
nor y can equal 0, the equation simplifies
to xy = −2. ∇T (–2, 1) = –4, – 2 , so the particle
ln x = – ln y + ln 2 = ln

moves downward along the curve.

23. He should move in the direction of
1
1
–∇f (p) = – f x (p), f y (p) = – – , –
2
4
⎛1⎞
= ⎜ ⎟ 2, 1 . Or use 2, 1 . The angle α formed
⎝4⎠
⎛1⎞
with the East is tan –1 ⎜ ⎟ ≈ 26.57° (N63.43°E).
⎝2⎠

27. ∇T ( x, y ) = –4 x, – 2 y
dx
dy
= –4 x,
= –2 y
dt
dt
dx
dt

dy

= dt has solution x = 2 y 2 . Since the
–4 x –2 y
particle starts at (–2, 1), this simplifies to
x = –2 y 2 .

764

Section 12.5

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28. f(1, –1) = 5 -1, 1

D u , u f (1, –1) = u1 , u2 ⋅ –5, 5 = –5u1 + 5u2
1 2
a.

⎛ 1 ⎞
-1, 1 (in the direction of the gradient); u = ⎜
⎟ –1, 1 .
⎝ 2⎠

⎛ 1 ⎞
b. ± 1, 1 (direction perpendicular to gradient); u = ⎜ ±
⎟ 1, 1
2⎠


c.

Want Du f (1, – 1) = 1 where u = 1. That is, want –5u1 + 5u2 = 1 and u12 + u22 = 1. Solutions are u =

3 4
,
5 5

4
3
and − , − .
5
5

29. a.

∇T ( x , y , z ) = −
=–

10(2 x)
(x + y + z )
2

20
( x + y 2 + z 2 )2
2

r (t ) =

2

2 2

,−

10(2 y )
(x + y + z )
2

2

2 2

,−

10(2 z )
( x + y 2 + z 2 )2
2

x, y , z

t cos π t , t sin π t, t , so r(1) = -1, 0, 1 . Therefore, when t = 1, the bee is at (–1, 0, 1), and

∇T (–1, 0, 1) = –5 –1, 0, 1 .
r ′(t ) = cos πt – πt sin πt , sin πt + πt cos πt , 1 , so r ′(1) = –1, – π, 1 .

–1, – π, 1
r ′(1)
=
is the unit tangent vector at (–1, 0, 1).
r ′(1)
2 + π2
DuT (–1, 0, 1) = u ⋅∇T (–1, 0, 1)

U=

=

–1, – π, 1 ⋅ 5, 0, – 5

=–

10

≈ –2.9026
2+π
2 + π2
Therefore, the temperature is decreasing at about 2.9°C per meter traveled when the bee is at (–1, 0, 1); i.e.,
when t = 1 s.
2

b. Method 1: (First express T in terms of t.)
10
10
10
5
=
=
=
T=
2
2
2
2
2
2
2
x +y +z
t2
(t cos πt ) + (t sin πt ) + (t )
2t
T (t ) = 5t −2 ; T ′(t ) = −10t −3 ; t ′(1) = −10
Method 2: (Use Chain Rule.)
dT ds
10 ⎛
2⎞
Dt T (t ) =
= ( DuT ) ( r '(t ) ) , so Dt T (t ) = [ DuT (−1, 0, 1)] ( r '(1) ) = −
⎜ 2 + π ⎟ = −10
2 ⎝

ds dt
2+π
Therefore, the temperature is decreasing at about 10°C per second when the bee is at (–1, 0, 1); i.e., when
t = 1 s.

30. a.

3
4
, – ⋅ f x ⋅ f y = –6, so
5
5
3 f x – 4 f y = –30.
Du f =

4 3
⋅ f x , f y = 17, so
,
5 5
4 f x + 3 f y = 85.
Dv f =

The simultaneous solution is
f x = 10, f y = 15, so ∇f = 10, 15 .
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b. Without loss of generality, let
u = i and v = j . If θ and φ are the angles

between u and ∇f , and between v and ∇f ,
then:
π
1. θ + φ = (if ∇f is in the 1st quadrant).
2
π
2. θ = + φ (if ∇f is in the 2nd quadrant).
2

3. φ + θ =
(if ∇f is in the 3rd quadrant).
2
π
4. φ = + θ (if ∇f is in the 4th quadrant).
2
In each case cos φ = sin θ or cos φ = –sin θ,

Therefore, f is not differentiable at the origin. But
Du f (0, 0) exists for all u since
f (0 + h, 0) – f (0, 0)
0–0
= lim
h
h →0
h →0 h
= lim (0) = 0, and
f x (0, 0) = lim
h →0

f (0, 0 + h) – f (0, 0)
0–0
= lim
h
h →0
h →0 h
= lim (0) = 0, so ∇f (0, 0) = 0, 0 = 0. Then
f y (0, 0) = lim
h →0

Du f (0, 0) = ∇f (0, 0) ⋅ u = 0 ⋅ u = 0.

33. Leave: (–0.1, –5)

so cos 2 φ = sin 2 θ . Thus,
( Du f ) 2 + ( Dv f )2 = (u ⋅∇f )2 + ( v ⋅∇f )2
= ∇f

2

cos 2 θ + ∇f

2

= ∇f

2

(cos 2 θ + cos 2 φ )

= ∇f

2

cos2 θ + sin 2 θ = ∇f

cos 2 φ

2

.

x2 – y 2

31.

a.

A′(100, 120)

b.

B ′(190, 25)

c.

20 – 30
1
= – ; f y ( D) = 0;
230 – 200
3
40 – 30 2
Du f ( E ) ≈
=
25
5
f x (C ) ≈

32. Graph of domain of f
⎧0, in shaded region ⎫
f ( x, y ) = ⎨

⎩1, elsewhere


lim

( x, y )→(0, 0)

f ( x, y ) does not exist since

( x, y ) → (0, 0) :

along the y-axis, f(x, y) = 0, but along the y = x 4
curve, f(x, y) = 1.
766

34. Leave (–2, –5)

Section 12.5

x – x3

9 – y2

35. Leave: (3, 5)
36. (4.2, 4.2)

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.


12.6 Concepts Review
1.
2.

∂ z dx ∂ z dy
+
∂ x dt ∂ y dt
y 2 cos t + 2 xy (– sin t )
= cos3 t – 2sin 2 t cos t

3.

∂z∂x ∂z ∂y
+
∂ x ∂t ∂ y ∂t

4. 12

Problem Set 12.6
1.

dw
= (2 xy 3 )(3t 2 ) + (3x 2 y 2 )(2t )
dt
= (2t 9 )(3t 2 ) + (3t10 )(2t ) = 12t11

2.

dw
= (2 xy – y 2 )(– sin t ) + ( x 2 – 2 xy )(cos t )
dt
= (sin t + cos t )(1 − 3sin t cos t )

3.

dw
= (e x sin y + e y cos x)(3) + (e x cos y + e y sin x)(2) = 3e3t sin 2t + 3e2t cos 3t + 2e3t cos 2t + 2e2t sin 3t
dt

4.

sec2 t
sec2 t – 2 tan 2 t 1 – tan 2 t
dw ⎛ 1 ⎞ 2 ⎛ 1 ⎞
– 2 tan t =
=
= ⎜ ⎟ sec t + ⎜ – ⎟ (2sec2 t tan t ) =
tan t
tan t
tan t
dt ⎝ x ⎠
⎝ y⎠

5.

dw
= [ yz 2 (cos( xyz 2 )](3t 2 ) + [ xz 2 cos( xyz 2 )](2t ) + [2 xyz cos( xyz 2 )](1)
dt
= (3 yz 2 t 2 + 2 xz 2 t + 2 xyz ) cos( xyz 2 ) = (3t 6 + 2t 6 + 2t 6 ) cos(t 7 ) = 7t 6 cos(t 7 )

6.

dw
= ( y + z )(2t ) + ( x + z )(−2t ) + ( y + x)(−1) = 2t (2 – t – t 2 ) – 2t (1 – t + t 2 ) – (1) = –4t 3 + 2t –1
dt

7.

∂w
= (2 xy )( s) + ( x 2 )(–1) = 2st ( s – t ) s – s 2t 2 = s 2 t (2s – 3t )
∂t

8.

∂w

⎛ s ⎞⎤
= (2 x – x –1 y )(– st –2 ) + (– ln x)( s 2 ) = s 2 ⎢1 – 2t –3 – ln ⎜ ⎟ ⎥
∂t
⎝ t ⎠⎦


9.

2 2
2 2
2 2
∂w
= e x + y (2 x)( s cos t ) + e x + y (2 y )(sin s) = 2e x + y ( xs cos t + y sin s )
∂t

= 2( s 2 sin t cos t + t sin 2 s) exp( s 2 sin 2 t + t 2 sin 2 s )

10.

∂w
2e s (t +1) ( st –1)
= [( x + y ) −1 − ( x − y )−1 ](e s ) + [( x + y )−1 + ( x − y )−1 ]( se st ) =
∂t
t 2 e2 s – e2 st

Instructor's Resource Manual

Section 12.6

767

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.


11.

∂w
x(– s sin st )
y ( s cos st )
z(s2 )
=
+
+
= s 4 t (1 + s 4 t 2 ) –1/ 2
∂ t ( x 2 + y 2 + z 2 )1/ 2 ( x 2 + y 2 + z 2 )1/ 2 ( x 2 + y 2 + z 2 )1/ 2

12.

2
∂w
= (e xy + z y )(1) + (e xy + z x)(−1) + (e xy + z )(2t ) = e xy + z ( y − x + 2t ) = e s (0) = 0
∂t

13.

∂z
⎛∂ z⎞
= (2 xy )(2) + ( x 2 )(–2 st ) = 4(2t + s )(1 − st 2 ) − 2 st (2t + s )2 ; ⎜
= 72

∂t
⎝ ∂ t ⎠ (1, − 2)

14.

∂z
⎛∂ z ⎞
= ( y + 1)(1) + ( x + 1)(rt ) = 1 + rt (1 + 2s + r + t ); ⎜
=5

∂s
⎝ ∂ s ⎠ (1, –1, 2)

15.

dw
= (2u – tan v)(1) + (– u sec2 v)(π) = 2 x – tan πx – πx sec2 πx
dx
dw
1+ π
⎛1⎞
⎛π⎞
= ⎜ ⎟ –1 – ⎜ ⎟ = –
≈ –2.0708
dx x = 1 ⎝ 2 ⎠
2
2
⎝ ⎠
4

16.

∂w
= (2 xy )(– ρ sin θ sin φ ) + ( x 2 )( ρ cosθ sin φ ) + (2 z )(0) = ρ 3 cos θ sin 3 φ (–2sin 2 θ + cos 2 θ );
∂θ
⎛∂w⎞
= –8


⎝ ∂θ ⎠ (2, π, π )
2

17. V (r , h) = πr 2 h,

dr
= 0.5 in./yr,
dt

19. The stream carries the boat along at 2 ft/s with
respect to the boy.

dh
= 8 in./yr
dt
dV
⎛ dr ⎞
⎛ dh ⎞
= (2πrh) ⎜ ⎟ + (πr 2 ) ⎜ ⎟ ;
dt
⎝ dt ⎠
⎝ dt ⎠
⎛ dV

⎝ dt
=


= 11200π in.3/yr

⎠ (20, 400)

11200π in.3 1 board ft
×
≈ 244.35 board ft/yr
1 yr
144 in.3

18. Let T = e – x –3 y .
dT
dx
dy
= e – x –3 y (–1) + e – x –3 y (–3)
dt
dt
dt
= e – x –3 y (–1)(2) + e – x –3 y (–3)(2) = –8e – x –3 y
dT
= –8, so the temperature is decreasing
dt (0, 0)
at 8°/min.

768

Section 12.6

dx
dy
= 2,
= 4, s 2 = x 2 + y 2
dt
dt
⎛ ds ⎞
⎛ dx ⎞
⎛ dy ⎞
2s ⎜ ⎟ = 2 x ⎜ ⎟ + 2 y ⎜ ⎟
⎝ dt ⎠
⎝ dt ⎠
⎝ dt ⎠
ds (2 x + 4 y )
=
dt
s

When t = 3, x = 6, y = 12, s = 6 5. Thus,
⎛ ds ⎞
= 20 ≈ 4.47 ft/s
⎜ ⎟
⎝ dt ⎠ t =3
dh
⎛1⎞
20. V (r , h) = ⎜ ⎟ πr 2 h,
= 3 in./min,
dt
⎝3⎠
dr
= 2 in./min
dt
dV ⎛ 2 ⎞
⎛ dr ⎞ ⎛ 1 ⎞
⎛ dh ⎞
= ⎜ ⎟ πrh ⎜ ⎟ + ⎜ ⎟ πr 2 ⎜ ⎟ ;
dt ⎝ 3 ⎠
dt
3
⎝ ⎠ ⎝ ⎠
⎝ dt ⎠
20,800π
⎛ dV ⎞
=
≈ 21, 782 in.3/min


3
⎝ dt ⎠ (40, 100)

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.


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