SOLUCIONARIO

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13.6. DIRECTIONAL DERIVATIVE
107
(c) xfx + yfy = x(8xy 3 − 3y 4 + 5x4 ) + y(12x2 y 2 − 12xy 3 )
= 8x2 y 3 − 3xy 4 + 5x5 + 12x2 y 3 − 12xy 4 = 20x2 − 15xy 4 + 5x5
= 5(4x2 y 3 − 3xy 4 + x5 ) = 5f (x, y)
y
y
λy
= λ0 f
, we see that z = f
(d) By observing that f
=f
λx
x
x
of degree zero.
13.6
y
x
is homogeneous
Directional Derivative
1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j
2
2
2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j
3. ∇F =
y2
2xy
3xy 2
i
+
j
−
k
z3
z3
z4
4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk
5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j
x3 − 4y 3
27
3x2
5
6. ∇f = p
i+ p
j; ∇f (3, 2) = √ i − √ j
3
4
3
4
38
2
38
2 x y−y
2 x y−y
7. ∇F = 2xz 2 sin 4yi + 4x2 z 2 cos 4yj + 2x2 z sin 4yk
√
√
4π
4π
4π
i + 16 cos
j + 8 sin
k = 2 3i − 8j − 4 3k
∇F (−2, π/3, 1) = −4 sin
3
3
3
2x
2y
2z
4
3
1
i+ 2
j+ 2
k; ∇F (−4, 3, 5) = − i + j + k
x2 + y 2 + z 2
x + y2 + z2
x + y2 + z2
25
25
25
√
√
f (x + h 3/2, y + h/2) − f (x, y)
(x + h 3/2)2 + (y + h/2)2 )2x − y 2
9. Du f (x, y) = lim
= lim
h→0
h→0
h
h
√
2
2
√
√
h 3x + 3h /4 = hy + h /4
= lim
= lim ( 3x + 3h/4 + y + h/4) = 3x + y
h→0
h→0
h
√
√
f (x + h 2/2, y + h 2/2) − f (x, y)
10. Du f (x, y) = lim
h→0
h
√
√
3x + 34 2/2 − (y + h 2/2)2 − 3x + y 2
= lim
h→0
h
√
√
√
√
√
√
3h 2/2 − h 2yh2 /2
= lim
= lim (3 2/2 − 2y − h/2) = 3 2/2 − 2y
h→0
h→0
h
√
√
3 1
15 3
2 6
3 5
11. u =
i j; ∇f = 15x y i + 30x y j; ∇f (−1, 1) = 15i − 30j; Du f (−1, 1) =
− 15 =
2 2
2
√
15
( 3 − 2)
2
8. ∇F =
108
CHAPTER 13. PARTIAL DERIVATIVES
√
√
2 2
12. u =
i
j; ∇f = (4 + y 2 )i(2xy − 5)j; ∇f (3, −1) = 5i − 11j;
2 2
√
√
√
5 2 11 2
−
= −3 2
Du f (3, −1) =
2
2
√
√
x
1
3 10
−y
1
10
13. u =
i+ 2
j; ∇f (2, −2) = i + j;
i−
j; ∇f = 2
2
2
10
10
y
x +y
4
4
√
√ x +√
10 3 10
10
Du f (2, −2) =
−
=−
40
40
20
x2
6
8
y2
i
+
j; ∇f (2, −1) = i + 4j;
i + j; ∇f =
10
10
(x + y)2
(x + y)2
3 16
19
Du f (2, −1) = +
=
5
5
5
√
15. u = (2i + j)/ 5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j;
2(28)
42
98
Du f (3, 2) = √ + √ = √
5
5
5
√
√
16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) = 3i + j; Du f (1/2, π/3) = − 3
14. u =
1
1
17. u = √ j + √ k; ∇f = 2xy 2 (2z + 1)2 i2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k; ∇f (1, −1, 1) =
2
2
√
18
12
6
18i − 18j + 12k; Du f (1, −1, 1) = − √ + √ = − √ = −3 2
2
2
2
2
1
2x
1
2y
2y 2 − 2x2
18. u = √ i − √ j + √ k; ∇f = 2 i − 2 j +
k; ∇f (2, 4, −1) = 4i − 8j − 24k;
z
z
z3
6
6
6
√
4
16
24
Du f (2, 4, −1) = √ − √ − √ = −6 6
6
6
6
x2 + 4z
y2
i+ p
j+ p
k;
x2 y + 2y 2 z
2 x2 y + 2y 2 z
x2 y + 2y 2 z
∇f (−2, 2, 1) = −i + j + k; Du f (−2, 2, 1) = −1
19. u = −k; ∇f = p
xy
√
2
1
2
20. u = −(4i−4j+2k)/ 36 = − i+ j− k; ∇f = 2i−2yj+2zk; ∇f (4, −4, 2) = 2i+8j+4k;
3
3
3
4 16 4
8
Du f (4, −4, 2) = − +
− =
3
3
3
3
√
21. u = (−4i − j/ 17; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j;
16
4
12
Du f (4, 2) = − √ + √ = − √
17
17
17
√
22. u = (−2i + 5j/ 29; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j;
4
15
11
Du f (1, 1) = √ − √ = − √
29
29
29
√
√
2
2x
2x
23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2i +
j
2
p
√
√
√
√ 2
1/2
The maximum Du is ( 2) + ( 2/2)2
= 5/2 in the direction 2i + ( 2/2)j.
13.6. DIRECTIONAL DERIVATIVE
109
24. ∇f = (xyex−y + yex−y i + (−xyex−y + xex−y j; ∇f (5, 5) = 30i − 20j
√
1/2
The maximum Du is 302 + (−20)2
= 10 13 in the direction 30i − 20j.
25. ∇f = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇f (1, 2, −1) = −2i + 2j − 4k
√
1/2
The maximum Du is (−2)2 + (2)2 + (−4)2
= 2 6 in the direction −2i + 2j − 4k.
26. ∇f = yzi + xzj + xyk; ∇f (3, 1, −5) = −5i − 15j + 3k
1/2 √
The maximum Du is (−5)2 + (−15)2 + (3)2
= 259 in the direction −5i − 15j + 3k.
2 2
2
2 2
2
27. ∇f =
p2x secp(x + y )ip+ 2y sec2 (x + y )j;
p
∇f ( π/6, π/6) = 2 p
π/6 sec (π/3)(i + j) = 8p π/6(i + j)
The minimum Du is −8 π/6(12 + 12 )1/2 = −8 π/3 in the direction −(i + j).
28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j)
√
1/2
The minimum Du is −12 12 + (−1)2
= −12 2 in the direction −(i − j) = −i + j.
√ y
√
√
ze
x
3
2
y
√
29. ∇f =
i + xze j + √ k; ∇f (16, 0, 9) = i + 12j + k. The minimum Du is
8
3
2 x
2 z
√
3
2
2
2
2 1/2
− (3/8) + 12 + (2/3)
= − 83281/24 in the direction − i − 12j − k.
8
3
1
1
1
i + j − k; ∇f (1/2, 1/6, 1/3) = 2i + 6j − 3k
x
y
z
1/2
The minimum Du is − 22 + 62 (−3)2
= −7 in the direction −2i − 6j + 3k.
30. ∇f =
31. Using implicit differentiation on 2x2 + y 2 = 9 we find y 0 =√−2x/y. At (2, 1) the slope of
the tangent line is −2(2)/1 = −4. √Thus, u√= ±(i − 4j)/
√ 17. Now, ∇f = i + 2yj and
∇f (3, 4) = i + 8j. Thus, Du = ±(1/ 17 − 32 17) = ±31/ 17.
2x + y − 1
x + 2y
3x + 3y − 1
√
√
32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) =
+ √
=
Solving
2
2
2
√
(3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1.
33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ±
3
4
i− j .
5
5
√
4
3
(b) u = (4i + 3j)/ 16 + 9 = i + j
5
5
4
3
(c) u = − i − j
5
5
34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6
35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j
3(3x2 − 6xy 2 ) −6x2 y + 3y 2
9x2 − 18xy 2 − 6x2 y + 3y 2
√
√
√
Du f (x, y) =
+
=
10
10
10
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110
CHAPTER 13. PARTIAL DERIVATIVES
3
3
(b) F (x, y) = √ (3x2 − 3xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy −
10
10
2x2 + 2y)j] 3
3
3
1
2
√
√
(6x − 6y − 4xy) + √
(−12xy − 2x2 + 2y)
Du F (x, y) = √
10
10
10
10
9
3
1
= (3x − 3y 2 − 2xy) + (−6xy − x2 + y) = (27x − 27y 2 − 36xy − 3x2 + 3y)
5
5
5
12
5
α − β = 7 and Dv f (a, b) =
36. Let ∇f (a, b) = αi + βj. Then Du f (a, b) = ∇f (a, b) · u =
13
13
5
12
∇f (a, b) · v =
α − β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus,
13
13
∇f (a, b) = 13i − (13/6)j.
37.
38. ∇f = h2x, −5yi, |∇f | =
p
10x2 + 25y 2 = 10, 4x2 + 25y 2 = 100,
x2
y2
+
=1
25
4
y
x
39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the
maximum decrease in temperature) is in the direction −∇T (4, 3) = −16i − 4j.
40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of
a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be −∇T (x, y), we
have x0 (t)i + y 0 (t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtain
dx/x = 4dt, ln x = 4t + c1 , and x = C1 e4t . Separating variables in dy/dt = 2y, we obtain
dy/y = 2dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we have x = 4e4t
and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter,
x = y 2 , y ≥ 0.
41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of
a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be ∇T (x, y), we
have x0 (t)i + y 0 (t)j = ∇T (x, y) = −4xi − 2yj. Separating variables in dx/dt = −4x, we obtain
dx/x = −4dt, ln x = −4t + c1 , and x = C1 e−4t . Separating variables in dy/dt = −2y, we
obtain dy/y = −2dt, ln y = −2t + c2 , and y = C2 e−2t . Since x(0) = 3 and y(0) = 4, we have
x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the
parameter, 16x = 3y 2 , y ≥ 0.
13.6. DIRECTIONAL DERIVATIVE
111
42. Substituing x = 0, y = 0, z = 1, and T = 500 into t =
and T (x, y, z) =
k
we see that k = 500
x2 + y 2 + z 2
500
.
x2 + y 2 + z 2
1
1
2
2
h1, −2, −2i = i − j − k
3
3
3
3
1000y
1000z
1000x
i− 2
j− 2
k
∇T = − 2
2
2
2
2
2
2
(x + y + z )
(x + y + z )
(x + y 2 + z 2 )2
500
750
750
∇T (2, 3, 3) = −
i−
j−
k
121
121
121
1
500
2
750
2
750
2500
Du T (2, 3, 3) =
−
−
−
−
−
=
3
121
3
121
3
121
363
(a) u =
(b) The direction of maximum increase is
∇T (2, 3, 3) = −
500
750
750
252
i−
j−
k=
(−2i − 3j − 3k).
121
121
121
121
(c) The maximum rate of change of T is |∇T (2, 3, 3)| =
√
250 √
250 22
4+9+9=
.
121
121
Gmx
Gmy
Gm
i+ 2
j= 2
(xi + yj)
2
3/2
2
3/2
+y )
(x + y )
(x + y 2 )3/2
The maximum and minimum values of Du U (x, y) are obtained when u is in the directions
∇U and
−∇U , respectively. Thus, at a point (x,y), not (0,0), the directions of maximum and minimum
increase in U are xi+yj and −xi−yj, respectively. A vector at (x, y) in the direction ±(xi+yj)
lies on a line through the origin.
43. ∇U =
(x2
44. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain
f (x, y) = x3 + xy 3 + exy + g(y). Then fy = 3xy 2 + xexy + g 0 (y) = −2y 2 + 3xy 2 + xexy . Thus,
g 0 (y) = −2y 2 , g(y) = − 32 y 3 + c, and f (x, y) = x3 + xy 3 + exy − 23 + C.
45. ∇(cf ) =
∂
∂
(cf )i +
j = cfx i + cfy j = c(fx i + fy j) = c∇f
∂x
∂y
46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g
47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f
48. ∇(f /g) = (gfx − f gx )/g 2 i + (gfy − f gy )/g 2 j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2
= g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2
p
∂r
x
x
∂r
y
y
x2 + y 2 so
=p
= and
=p
=
2
2
2
2
∂x
r
∂y
r
x +y
x +y
Dx yE 1
r
This gives ∇r =
,
= hx, yi =
r r
r
r
49. r(x, y) =
112
50.
CHAPTER 13. PARTIAL DERIVATIVES
∂ (f (r))
df ∂r
∂ (f (r))
df ∂r
∂ (f (r)) ∂ (f (r))
=
and
=
so that ∇f (r) = h
,
i
∂x
dr ∂x
∂y
dr ∂y
∂x
∂y
df ∂r df ∂r
df ∂r ∂r
=h
,
i=
h ,
i = f 0 (r)∇r = f 0 (r)r/r
dr ∂x dr ∂y
dr ∂x ∂y
51. Let u = u1 i + u2 j and v = v1 i + v2 j.
Dv f = (fx i+ fy j) · v = v1 fx + v2 fy
∂
∂
Du Dv f =
(v1 fx + v2 fy )i +
(v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyz )i + (v1 fxy + v2 fyy )j] · u
∂x
∂y
= u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy
D − uf = (fx i + fy j) · u = u1 fx + u2 fy
∂
∂
(u1 fx + u2 fy )i +
(u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v
Dv Du f =
∂x
∂y
= u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy
Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f. [Note
that this result is a generalization fxy = fyx since Di Dj f = fyx and Dj Di f = fxy ]
i
∂
∂x
f1
k
∂
52. ∇ × F =
∂z
f3
∂f3
∂f2
∂f3
∂f1
∂f2
∂f1
=
−
i−
−
j+
−
k
∂y
∂z
∂x
∂z
∂x
∂y
13.7
j
∂
∂y
f2
Tangent Planes and Normal Lines
1. Since f (6, 1) = 4, the level curve is x − 2y = 4. ∇f = i − 2j;
∇f (6, 1) = i − 2j
y
x
2. Since f (1, 3) = 5, the level curve is y+2x = 5x or y = 3x, x 6= 0.
y
1
∇f = − 2 i + j; ∇f (1, 3) = −3i + j
x
x
y
x
www.elsolucionario.org
13.7. TANGENT PLANES AND NORMAL LINES
113
3. Since f (2, 5) = 1, the level curve is y = x2 + 1. ∇f = −2xi + j;
∇f (2, 5) = −4i + j
y
x
4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10.
∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j
y
x
5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /0 = 2
x
2y
x2 /8 + y 2 /18 = 1. ∇f = i + j; ∇f (−2, −3) = −i −
2
9
y
or
2
j
3
x
6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x 6= 0.
2y
y2
∇f = − 2 i + j; ∇f (2, 2) = −i + 2j
x
x
y
x
114
CHAPTER 13. PARTIAL DERIVATIVES
7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or
y 2 − (x − 1)2 = 1. ∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j
y
x
8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or
−(y − 1) cos x
1
y = 1 + sin x, sin x 6= 0. ∇f =
i+
j;
2
sin x
sin x
√
∇f (π/6, 3/2) = − 3i + 2j
y
x
9. Since f (3, 1, 1) = 2, the level curve is y + z = 2 ∇f = j + k;
∇f (3, 1, 1) = j + k
z
2
2
y
x
10. Since f (1, 1, 3) = −1, the level curve is x2 + y 2 − z = −1 or
z = 1 + x2 + y 2 . ∇f = 2xi + 2yj − k; ∇f (1, 1, 3) = 2i + 2j − k
z
y
x
11. Since F (3, 4, 0) = 5, the level curve is x2 + y 2 + z 2 = 25.
x
y
z
∇F = p
i+ p
j+ p
k;
2
2
2
2
2
2
2
x +y +z
x +y +z
x + y2 + z2
3
4
∇F (3, 4, 0) = i + j
4
5
z
5
5
y
x
13.7. TANGENT PLANES AND NORMAL LINES
115
12. Since F (0, −1, 1) = 0, the level curve is x2 − y 2 + z = 0 or z = y 2 − x2 .
∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2i + k
z
y
x
13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c 4i + j + 21 k or 2x =
4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Since
z = x2 + y 2 = 16 + 1 = 17, the point on the surface is (−4, −1, −17).
14. F (x, y, z) = x3 + y 3 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 =
27c, 2y = 8c, 1 = c. From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 =
15 − (±3)3 − 16 = −1 ∓ 27, the points on the surface are (3, 4, −28) and (−3, 4, 26).
15. F (x, y, z0 = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation
of the tangent plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9.
16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k.The equation
of the tangent plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4.
17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation
of the tangent plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5.
18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k.
The equation of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7.
19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of
the tangent plane is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50.
20. F (x, y, z) = xz; ∇F = zi + xk; ∇F (2, 0, 3) = 3i + 2k. The equation of the tangent plane is
3(x − 2) + 2(z − 3) = 0 or 3x + 2z = 12.
√
21. F (x, y,√z) = cos(2x+y)−z; ∇F = −2 sin(2x+y)i−sin(2x+y)j−k;
∇F (π/2, π/4, −1 2) =
√
√
√ 2
π
2
π
1
2i+
j−k. The equation of the tangent plane is 2 x −
+
y−
− z+√
=
2
2
2
4
2
√
π π √
1
5π
0, 2 x −
+ y−
− 2 z+√
= 0, or 2x + y − 2z =
+ 1.
2
4
4
2
22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation
of the tangent plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13.
√
√
√
2x
2y
23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F = 2
i+ 2
j − k; ∇F (1/ 2, 1/ 2, 0) = 2i +
2
2
x +y
x +y
√
√
√
1
1
√
√
2j − k. The equation of the tangent plane is 2 x −
+ 2 y−
− (z − 0) =
2
2
√
√
√
1
1
+2 y− √
− 2z = 0, or 2x + 2y − 2z = 2 2.
0, 2 x − √
2
2
www.elsolucionario.org
116
CHAPTER 13. PARTIAL DERIVATIVES
−2y
24. F (x,
sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) =
√ y, z) = 8e
16 3i − 8j − k. The equation of the tangent plane is
√
√
√
2 3π
− 4.
16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3x − 8y − z =
3
25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to
the surface at (x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1
is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find c
so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = √
2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on
2
2
2
2
2. Thus, the points on the surface
the surface,
so
c
+
(2c)
+
(3c)
=
14c
=
7
and
c
=
±1/
√
√ √
√
√
√
are ( 2/2, 2, 3 2/2) and − 2/2, − 2, −3 2/2).
26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector
to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i − 4y0 j − 6z0 k. A normal vector to the
plane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the given
plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, √z0 = −c.
2
2
2
Now (x0 , y0 , z0 ) is on the surface,
2(−c)2 − 3(−c)
√so (4c)
√ −√
√ =
√ 11c
√ = 33 and c = ± 3. Thus,
the points on the surface are (4 3, − 3, − 3) and (−4 3, 3, 3).
27. The gradient of F (x, y, z) = x2 +4x+y 2 +z 2 −2z is ∇F = (2x+4)i+2yj+(2z−2)k, so a normal
to the surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck
for c 6= 0. Thus, we want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1.
Since (x0 , y0 , z0 ) is on the surface, (−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and
c = ±4. The points on the surface are (−2, 0, 5) and (−2, 0, −3).
28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so
a normal to the surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k.
(a) A normal to the xz plane is cj for c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) =
c, 8z0 = 0 or x0 = y0 , 3y0 −x0 = c/2, z0 = 0. Solving the first two equations, we obtain
2
x0 = y0 = c/4. Since (x0 , y0√
, z0 ) is on the surface, (c/4)2 +3(c/4)2 +4(0)√
−2(c/4)(c/4)
=
√
2
2c /16
=
16
and
c
=
±16/
2.
Thus,
the
points
on
the
surface
are
(4/
2,
4/
2,
0)
and
√
√
(−4 2, −4 2, 0).
(b) A normal to the yz-plane is ci for c 6= 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) =
0, 8z0 = 0 or x0 − y0 = c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we
obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )√is on the surface, (3c/4)2 + 3(c/4)2 +
4(0)2√− 2(3c/4)(c/4)
= 6c2 /16
√
√ = 16 and
√ c = ±16 6. Thus, the points on the surface are
(12/ 6, 4/ 6, 0) and (−12/ 6, −4/ 6, 0).
(c) A normal to the xy-plane is ckfor c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) =
0, 8z0 = c or x0 = y0 , 3y0 −x0 = 0, z0 = c/8. Solving the first two equations, we obtain
x0 = y0 = 0. Since (x0 , y0 , z0 ) is on the surface, 02 +3(0)2 +4(c/8)2 −2(0)(0) = c2 /16 = 16
and c = ±16. Thus, the points on the surface are (0, 0, 2) and (0, 0, −2).
29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and x0 , y0 , z0 )
is on the plane xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
13.7. TANGENT PLANES AND NORMAL LINES
117
∇F (x0 , y0 , z0 ) = (2x − 0/a2 )i + (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i +
(y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane,
the normal vectors are parallel and the plane is tangent to the surface.
30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /b2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 )
is on the plane xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i −
(y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane,
the normal vectors are parallel, and the plane is tangent to the surface.
31. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric
equations of the line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t.
32. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric
equations of the line are x = 3 + 12t; y = −2 + 16t, z = 2 − t.
33. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric
y − 1/3
z−3
x − 1/2
=
=
.
equations of the line are
4
6
−1
34. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric
x−3
y−4
z−5
equations of the line are
=
=
.
6
8
−10
35. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface
at (x0 , y0 , z0 ) is x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x −
x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on
the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus, the equation of the tangent plane is
x0 x + y0 y − z0 z = 0, which passes through the origin.
1
1
1
√ √
x+ y + z. Then ∇F = √ i+ √ j+ √ k and a normal to the surface
2 x 2 y
2 z
1
1
1
at (x0 , y0 , z0 ) is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is
2 x0
2 y0
2 z0
1
1
1
1
1
1
√
√
√
√ (x−x0 )+ √ (y−y0 )+ √ (z−z0 ) = 0 or √ x+ √ y+ √ z = x0 + y0 + z0 =
2 x0
2 y0
2 z0
x0
y0
z0
√
√ √
√ √
√
√ √
√ √
√
√a. The
√ sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a =
a · a = a.
36. Let F (x, y, z) =
√
37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i + 2y0 j + 2z0 k. Parametric
equations of the normal line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting
t = −1/2, we see that the normal line passes through the origin.
38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i+Fy j+Fz k and Gx i+Gy j+Gz k,
respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, the
surfaces are orthogonal at P if and only if Fx Gx + Fy Gy + Fz Gz = 0.
39. We have F (x, y, z) = x2 + y 2 + z 2 and G(x, y, z) = x2 + y 2 − z 2 .
∇F = h2x, 2y, 2zi =
6 0 except at the origin
118
CHAPTER 13. PARTIAL DERIVATIVES
∇G = h2x, 2y, −2zi =
6 0 except at the origin
Therefore, the gradient vectors are nonzero at each of the intersection points. Now
Fx Gx + Fy Gy + Fx Gz = (2x)(2x) + (2y)(2y) + (2z)(−2z)
= 4x2 + 4y 2 − 4z 2
= 4(x2 + y 2 + z 2 ) = 4(0) = 0
The second to last equality follows from the fact that the intersection points lie on both
surfaces and hence satisfy the second equation x2 + y 2 − z 2 = 0.
40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then
Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1)
= −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z).
For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0)
and the surfaces are orthogonal at points of intersection.
13.8
Extrema of Multivariable Functions
1. fx = 2x; fxx = 2; fxy = 0; fy = 2y; fyy = 2; D = 4. Solving fx = 0 and fy = 0, we
obtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx (0, 0) = 2 > 0, f (0, 0) = 5 is a
relative minimum.
2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, we
obtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx (0, 0) = 8 > 0, f (0, 0) = 0 is
a relative minimum.
3. fx = −2x + 8; fxx = −2; f xy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 and
fy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 and
fxx (4, 3) = −2 < 0, f (4, 3, ) = 25 is a relative maximum.
4. fx = 6x − 6; f xx = 6; f xy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 and
fy = 0, we obtain the critical point (1, −2). Since D(1, −2) = 24 > 0 and
fxx (1, −2) = 6 > 0, f (1, −2) = −11 is a relative minimum.
5. fx = 10x + 20; fxx = 10; fxy = 0; fy = 10y − 10; fyy = 10; D = 100. Solving
fx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 and
fxx (−2, 1) = 10 > 0, f (−2, 1) = 15 is a relative minimum.
6. fx = −8x − 8; fxx = −8; fxy = 0; fy = −4y + 12; fyy = −4; D = 32. Solving
fx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 and
fxx (−1, 3) = −8 < 0, f (−1, 3) = 27 is a relative maximum.
7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y 2 − 3; fyy = 6y; D = 144xy. Solving
fx = 0 and fy = 0, we obtain the critical points (−1, −1), (−1, 1), (1, −1), and (1, 1).
Since D(−1, 1) = −144 < 0 and D(1, −1) = −144 < 0, these points do not give relative
extrema. Since D(−1, −1) = 144 > 0 and fxx (−1, −1) = −24 < 0, f (−1, −1) = 10 is a