Test 3 Review
1. Prove that the direction of maximal rate increase of a function f (x, y) is in the direction of ∇f and that the maximal
rate of decrease is in the direction of -∇f . What are the rates of increase in these directions?
2. Find the normal and tangent lines to the curve x2 y + xy 3 = 6 at (2, 1).
Answer: The normal line is r(t) = (2 + t)i + (1 + 2t)j. The tangent line is r(t) = (2 + 2t)i + (1 − t)j.
3. Find the normal line and tangent plane to the surface 3x2 + 2y 2 + z 2 = 6 at the point (1, 1, 1).
Answer: The normal line is r(t) = (1 + 3t)i + (1 + 2t)j + (1 + t)k. The tangent plane is 3x + 2y + z = 6.
4. Let f (x, y) = −3x2 + 6x − 3 + 2xy − 2y − 2y 2 . Find and classify the critical points.
Answer: The only critical point is (x, y) = (1, 0) which is a local maximum.
5. Let f (x, y) = x4 − 2x2 + 4x2 y − 4y + 3y 2 + 1. Find all the critical points and classify them as local minima, local
maxima, or saddle points.
Answer: There is a local minimum at (0, 23 ) and saddle points at (1, 0) and (−1, 0).
∇f = (4x3 − 4x + 8xy, 4x2 − 4 + 6y), so
2
(x, y) = (0, ), (1, 0), (−1, 0)
3
2
∂ f
4
A=
= 12x2 − 4 + 8y = , 8, 8
∂x2
3
2
∂ f
B=
= 8x = 0, 8, 8
∂y∂x
C=
∂2f
=6
∂y 2
D = AC − B 2 = 12, −16, −16
so (0, 23 )is a local minimum, and the other two are saddles.
6. Maximize 2x + y subject to the condition that
√
Answer: 17.
x2
4
+ y 2 = 1.
7. Find the points on y 2 x = 9 which are closest to (0, 0) .
√ √ √
Answer: ( 3 92 , 6 2 3 9).
8. Maximize 3x − 2y + z on the sphere x2 + y 2 + z 2 = 14.
Answer: The maximum is f (3, −2, 1) = 14.
9. Find the absolute extreme values taken on by f (x, y) = 2x2 − 4x + y 2 − 4y + 3 on the closed triangular region bounded
by the lines x = 0, y = 2, and y = 2x.
Answer: The absolute maximum is f (0, 0) = 3. The absolute minimum is f (1, 2) = −3.
10. Find the absolute extreme values taken on by f (x, y) = x4 + 2y 3 on the region x2 + y 2 ≤ 1.
Answer: The absolute maximum is f (0, 1) = 2. The absolute minimum is f (0, −1) = −2.
11. Find the differential df where f (x, y) = 3x3 − 5x2 y 2 + 2x − y.
Answer: df = (9x2 − 10xy 2 + 2) dx + (−10x2 y − 1) dy.
12. Determine whether or not the vector function is a gradient. If so, final all the functions with that gradient.
(a) (2xy + 3 − y sin x)i + (x2 + 2y + 1 + cos x)j
(b) (xy 2 + 2y 2 )i + (2y 3 − x2 y + 2x)j
(c) (ey sin z + 2x)i + (xey sin z − y 2 )j + (xey cos z)k
Answer:
(a) Yes, f (x, y) = x2 y + y cos x + y 2 + 3x + y + C.
(b) No.
(c) Yes, f (x, y, z) = xey sin z + x2 − 13 y 3 + C.
13. Evaluate
4
3
Σ Σ (i2 − j)
j=1i=1
Answer: 26
14. Evaluate
∫
∫
1
√
y
xy 2 dx dy
0
Answer:
1
40
15. Find
∫
1
0
Answer:
y
1
3 (1
∫
( )
sin x3 dx dy
1
√
y
− cos 1)
∫ ∫
16. Evaluate
(x − y) dx dy
Ω
where Ω is the region between the curves y = 3x and y 2 = 4 − x.
2
Answer:
√
48 3
5
17. Use polar coordinates to evaluate
∫
1
√
1−x2
∫
arctan(y/x) dy dx.
−1
Answer:
0
π2
4
18. A region in the plane is inside
r = sin (θ) , θ ∈ [0, π]
where θ is the usual polar angle measured from the positive x axis in the counter clockwise direction. Find the area of
the region.
Answer:
π
4
(
)
19. Find the volume of the three dimensional region bounded by the surfaces z = 4 − x2 + 3y 2 and z = 3x2 + y 2 .
Answer: 2π
20. A homogeneous plate is in the shape of an isosceles triangle of base b and height h.
(a) Find the centroid of the plate.
(b) Find the moment of inertia about the base.
(c) Find the moment of inertia about the axis of symmetry of the triangle.
Answer:
(a) Along the axis of symmetry, two thirds the distance from the vertex to the base.
(b)
bh3
12
or
Ah2
6 .
(c)
hb3
48
or
Ab2
24 .
21. Evaluate
∫ ∫ ∫
x dx dy dz
T
where T is the solid bounded by the planes x = 0, y = 0, z = 0, y + z = 1 and x + z = 1.
∫ 1 ∫ 1−z ∫ 1−z
Answer: 0 0
x dy dx dz = 81
0
22. Find
∫
1
∫
0
Answer:
∫
4
0
∫
2x
∫
0
1
0
sin x
dz dy dx +
x
23. Rewrite the integral
10−2z
∫
5−z
sin x
dx dy dz.
x
y
2
0
∫
5
∫
4
∫
∫
2x
0
∫
1
5−x
0
1
∫
sin x
dz dy dx = 2 + 2 sin 4 − 2 sin 5.
x
1−y
dz dy dx
−1
x2
0
as an equivalent integral with differential order
(a) dz dx dy
(b) dy dz dx
(c) dy dx dz
(d) dx dy dz
(e) dx dz dy
Answer:
∫ 1 ∫ √y ∫ 1−y
(a) 0 −√y 0
dz dx dy
∫ 1 ∫ 1−x2 ∫ 1−z
dy dz dx
(b) −1 0
x2
∫ 1 ∫ √1−z ∫ 1−z
(c) 0 −√1−z x2 dy dx dz
∫ 1 ∫ 1−z ∫ √y
√ dx dy dz
(d) 0 0
− y
√
∫ 1 ∫ 1−y ∫ y
√ dx dz dy
(e) 0 0
− y
24. Find the x coordinate of the center of mass of the solid whose density is δ (x, y, z) = x which occupies the three
dimensional region below the plane z = y and above the triangular region in the xy-plane determined by x ∈ [0, 1] and
0 ≤ y ≤ x.
∫ 1∫ x∫ y
Answer: The mass is M =
x dz dy dx. The center of mass is
0
(
(x, y, z) =
1
M
∫
0
1∫
0
x∫
y
x2 dz dy dx,
0
0
0
1
M
∫
1
∫
x
∫
y
xy dz dy dx,
0
0
0
1
M
∫
1
∫
x
∫
)
y
xz dz dy dx
0
0
0
√
x2 + y 2 and below the paraboloid
25. Find the mass and center of mass of the solid that lies above the cone z =
2
2
z = 1 − x − y where density is the distance to the origin.
∫ 2π ∫ 1 ∫ 1−r2 √
Answer: The mass is M =
r r2 + z 2 dz dr dθ. The center of mass is
0
0
r
(
(x, y, z) =
1
0, 0,
M
∫
2π
∫
1
∫
1−r 2
zr
0
0
r
√
)
r2 + z 2 dz dr dθ