Solution to Practice Problems for Test 2
1. Suppose f (x, y, z) is a differentiable function at the point (1, 1, 1) and
∂f
(1, 1, 1) = 2,
∂x
∂f
(1, 1, 1) = −1,
∂y
∂f
(1, 1, 1) = 2
∂z
(a) Find the directional derivative of f at the point (1, 1, 1) in the direction of a = 2i+2j+k.
∇f (1, 1, 1) = 2i − j + 2k.
1
1
4
a
= (2i + 2j + k) ⇒ directional derivative = (2i + 2j + k) · (2i − j + 2k) = .
|a|
3
3
3
4
directional derivative =
3
(b) Find a unit vector u in the direction in which f increases most rapidly at (1, 1, 1). What
is the rate of change in this direction?
|2i − j + 2k| = 3
1
u=
(2i − j + 2k)
3
rate =
3
2. Find all points (x, y) at which the tangent plane to the graph of
z = x2 + 2xy − 2y 2 − 5x − 2y
is parallel to the plane P : 3x − 6y + z = 8.
The tangent plane has normal ⟨ − fx , −fy , 1⟩. It is parallel to the plane P if and only if
⟨ − fx , −fy , 1⟩ = k⟨3, −6, 1⟩
⇔
k = 1 and
fx = −3k = −3 and fy = 6k = 6
{
{
⇒
{
⇒
fx = 2x + 2y − 5 = −3
fy = 2x − 4y − 2 = 6
2x + 2y = 2
2x − 4y = 8
x= 2
y = −1
1
3. Find and classify all critical points of f (x, y) = x2 + y 2 + x2 y + 4.

fx = 2x + 2xy = 0












⇒
⇒
2x(1 + y) = 0
⇒
x = 0 or y = −1
x2
2
√
√
x = 0, y = 0 or y = −1, x = 2, − 2
fy = 2y + x2 = 0
⇒
y=−
√
√
So the critical points are (0, 0), ( 2, −1) and (− 2, −1).
fxx (x, y) = 2(1 + y),
fyy (x, y) = 2,
fxy (x, y) = 2x
⇒
D(x, y) = 4(1 + y) − 4x2
D(0, 0) = 4 and fxx (0, 0) = 2 > 0 ⇒ f has a local minimum at (0, 0).
√
√
√
D(± 2, −1) = −8 < 0 ⇒ f has saddle points at ( 2, −1) and (− 2, −1).
(x, y)
D(x, y)
fxx
classification
(0, 0)
4
2
local
minimum
2
√
√
( 2, −1) (− 2, −1)
−8
−8
saddle
point
saddle
point
4. Find the maximum and minimum values of f (x, y) = 2x2 − y 2 subject to x3 + y 3 = 56.
∇f = λ∇g
⇒
4x = λ(3x2 )
(1)
−2y = λ(3y 2 )
(2)
If λ = 0, then x = 0 and y = 0 do not satisfy the constraint x3 + y 3 = 56. Therefore, we have
λ ̸= 0.
x = 0 ⇒ y = 561/3 ⇒ (x, y) = (0, 561/3 )
y = 0 ⇒ x = 561/3 ⇒ (x, y) = (561/3 , 0)
Suppose λ, x, y ̸= 0.
(1)
−2x
x2
⇒
= 2 ⇒ x = −2y.
(2)
y
y
Putting x = −2y into the constraint x3 + y 3 = 56, we have
−8y 3 + y 3 = 56
⇒
−7y 3 = 56
⇒
y = −2
⇒
x = −2(−2) = 4
( 1/3 ) (
)
1/3
(x, y)
56
,
0
0,
56
(4, −2)
( 2/3 )
2/3
f (x, y) 2 56
−56
28
)
(
Since 2 562/3 ≈ 29.27, we have
maximum =
)
(
2 562/3
minimum =
− 562/3
3
5. Find the volume of the solid in the first octant under the surface z = 2x + y 2 and bounded
by the co-ordinate planes and the plane x + 3y = 3.
volume =
=
=
∫ 1 ∫ 3(1−y)
0
∫1
0
∫1
0
0
2x + y 2 dx dy
3(1−y)
[x2 + y 2 x]0
dy
9(1 − y)2 + 3y 2 (1 − y) dy
[
3
= −3(1 − y) + y − y 4
4
3
=3+1−
4
3
=
]1
3
0
13
4
volume =
6. Reverse the order of integration in the integral
∫ 1 ∫ y3
0
0
f (x, y) dx dy =
∫1∫1
0
x1/3
∫ 1 ∫ y3
0
0
13
4
f (x, y) dx dy.
f (x, y) dy dx
7. Set up (but do not evaluate) an integral in cylindrical coordinates for the mass of the solid
between the paraboloids z = 2 − x2 − y 2 ad z = x2 + y 2 if the density function is δ(x, y, z) = z.
2 − x2 − y 2 = x2 + y 2 ⇒ 2(x2 + y 2 ) = 2 ⇒ x2 + y 2 = 1
mass =
4
∫ 2π ∫ 1 ∫ 2−r2
0
0
r2
zr dz dr dθ
8. Find the area of the part of the surface z = y 2 between y = x, y = 1, x = 0.
fx = 0, fy = 2y
√
⇒
=
√
fx2 + fy2 + 1 =
√
(2y)2 + 1
4y 2 + 1
⇒ Surface area
=
∫1∫y√
4y 2 + 1 dx dy
0 0
=
∫1 √
y 4y 2 + 1 dy
0
[
(4y 2 + 1)3/2
=
12
]1
0
53/2 − 1
=
12
9. Change the integral
z=
z=
∫ 2 ∫ √4−x2 ∫ √8−x2 −y2 2
√
(x + y 2 + z 2 )3/2 dz dy dx into spherical coordinates.
−2 0
x2 +y 2
√
√
8 − x2 − y 2 ⇔ x2 + y 2 + z 2 = 8 ⇔ ρ =
√
8
x2 + y 2 ⇔ ρ cos ϕ = ρ sin ϕ ⇔ cos ϕ = sin ϕ ⇔ ϕ =
∫ 2 ∫ √4−x2 ∫ √8−x2 −y2 2
√ 2 2 (x + y 2 + z 2 )3/2 dz dy dx
−2 0
x +y
=
∫ π/4 ∫ π ∫ √8
0
0
0
ρ3 ρ2 sin ϕ dρ dθ dϕ
5
π
4
∫∫
10. Evaluate
R
(3x − y)3 (x + y)4 dA, where R is the triangle with vertices (1, 0), (5, 0), (2, 3).
{
Use
u+v
−u + 3v
u = 3x − y
⇒x=
, y=
v = x+y
4
4
1 1 ( )( ) ( )( )
4 4 3
1
1
1
= 1
J(u, v) = −
−
=
4
4
4
4
4
−1 3 4
4
Boundary:
The line joining (1,0) and (2,3) has equation
3x − y = 3 ⇒ u = 3
The line joining (2,3) and (5,0) has equation
x+y =5 ⇒ v =5
The line joining (1,0) and (5,0) has equation
y=0⇒
∫∫
R
=
(3x − y)3 (x + y)4 dA
∫ 5 ∫ 3v
1
−u + 3v
= 0 ⇒ u = 3v
4
3
( )
1
du dv
uv
4
3 4
( )
[ ]3v
1 ∫ 5 4 u4
dv
=
v
4 1
4 3
34
= 2
4
∫
5
(
)
v 4 v 4 − 1 dv
1
[
]5
34 v 9 v 5
−
= 2
4 9
5 1
=
5477346
5
6