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MA 271 PRACTICE PROBLEMS
- If the line ℓ is parallel to the vector ~v = 2
i − 3
j + 7
k, and contains the point (2, 1 , −3), then its vector
equation is
A. ~r = (1 + 2t)
i − 3 t~j + (−2 + 7t)
k B. ~r = (2 + t)
i − 3
j + (7 − 2 t)
k
C. ~r = (2 + 2t)
i + (1 − 3 t)
j + (−3 + 7t)
k D. ~r = (2 + 2t)
i + (−3 + t)
j + (7 − 3 t)
k
E. ~r = (2 + t)
i +
j + (7 − 3 t)
k
- Find parametric equations of the line containing the points (1, − 1 , 0) and (− 2 , 3 , 5).
A. x = 1 − 3 t, y = −1 + 4t, z = 5t B. x = t, y = −t, z = 0
C. x = 1 − 2 t, y = −1 + 3t, z = 5t D. x = − 2 t, y = 3t, z = 5t
E. x = −1 + t, y = 2 − t, z = 5
- Find an equation of the plane that contains the point (1, − 1 , −1) and has normal vector
1
2
i + 2
j + 3
k.
A. x − y − z +
9
2
= 0 B. x + 4y + 6z + 9 = 0 C.
x− 1
1
2
y+
2
z+
3
D. x − y − z = 0 E.
1
2
x + 2y + 3z = 1
- Find an equation of the plane that contains the points (1, 0 , −1), (− 5 , 3 , 2), and (2, − 1 , 4).
A. 6 x − 11 y + z = 5 B. 6 x + 11y + z = 5 C. 11 x − 6 y + z = 0
D. ~r = 18
i − 33
j + 3
k E. x − 6 y − 11 z = 12
- Find parametric equations of the line tangent to the curve ~r(t) = t~i + t
2 ~ j + t
3 ~ k at the point (2, 4 , 8)
A. x = 2 + t, y = 4 + 4t, z = 8 + 12t B. x = 1 + 2t, y = 4 + 4t, z = 12 + 8t
C. x = 2t, y = 4t, z = 8t D. x = t, y = 4t, z = 12t E. x = 2 + t, y = 4 + 2t, z = 8 + 3t
- The position function of an object is
~r(t) = cos t~i + 3 sin t~j − t
2 ~ k
Find the velocity, acceleration, and speed of the object when t = π.
Velocity Acceleration Speed
A. −
i − π
2 ~ k − 3
j − 2 π
k
1 + π
4
B.
i − 3
j + 2π
k −
i − 2
k
10 + 4π
2
C. 3
j − 2 π
k −
i − 2
k
9 + 4π
2
D. − 3
j − 2 π
k
i − 2
k
9 + 4π
2
E.
i − 2
k − 3
j − 2 π
k
- A smooth parametrization of the semicircle which passes through the points (1, 0 , 5), (0, 1 , 5) and
(− 1 , 0 , 5) is
A. ~r(t) = sin t~i + cos t~j + 5
k, 0 ≤ t ≤ π B. ~r(t) = cos t~i + sin t~j + 5
k, 0 ≤ t ≤ π
C. ~r(t) = cos t~i + sin t~j + 5
k,
π
2
≤ t ≤
3 π
2
D. ~r(t) = cos t~i + sin t~j + 5
k, 0 ≤ t ≤
π
2
E. ~r(t) = sin t + cos t~j + 5
k,
π
2
≤ t ≤
3 π
2
- The length of the curve ~r(t) =
2
3
(1 + t)
3
2
i +
2
3
(1 − t)
3
2
j + t
k, − 1 ≤ t ≤ 1 is
A.
3 B.
2 C.
1
2
3 D. 2
3 E.
- The level curves of the function f (x, y) =
1 − x
2 − 2 y
2 are
A. circles B. lines C. parabolas D. hyperbolas E. ellipses
- The level surface of the function f (x, y, z) = z −x
2
−y
2
that passes through the point (1, 2 , −3) intersects
the (x, z)-plane (y = 0) along the curve
A. z = x
2
2 − 8 C. z = x
2
2 − 8
E. does not intersect the (x, z)-plane
- Match the graphs of the equations with their names:
(1) x
2
2
2
= 4 (a) paraboloid
(2) x
2
2
= 4 (b) sphere
(3) x
2
2 = z
2 (c) cylinder
(4) x
2
2 = z (d) double cone
(5) x
2
2
2
= 1 (e) ellipsoid
A. 1b, 2c, 3d, 4a, 5e B. 1b, 2c, 3a, 4d, 5e C. 1e, 2c, 3d, 4a, 5b
D. 1b, 2d, 3a, 4c, 5e E. 1d, 2a, 3b, 4e, 5c
- Suppose that w = u
2
/v where u = g 1 (t) and v = g 2 (t) are differentiable functions of t. If g 1
g 2 (1) = 2, g
′
1
(1) = 5 and g
′
2
(1) = −4, find
dw
dt
when t = 1.
A. 6 B. 33 / 2 C. − 24 D. 33 E. 24
- If w = e
uv
and u = r + s, v = rs, find
∂w
∂r
A. e
(r+s)rs
(2rs + r
2
) B. e
(r+s)rs
(2rs + s
2
) C. e
(r+s)rs
(2rs + r
2
)
D. e
(r+s)rs (1 + s) E. e
(r+s)rs (r + s
2 ).
- Let f (x, y, z) =
x
2 y
4
x
4
8
when x 6 = 0 and f (0, 0) = 0. Which of the following are (is) true?
i) ∂f /∂x(0, 0) and ∂f /∂y(0, 0) exist at (0, 0).
ii) f (x, y) is continuous at (0, 0).
iii) The graph of z = f (x, y) has a tangent plane at (0, 0 , 0).
A. i) only B. i) and ii) only C. i) and iii) only D. i) ii), and iii) E. None
- The function f (x, y) = 2x
3
− 6 xy − 3 y
2
has
A. a relative minimum and a saddle point B. a relative maximum and a saddle point
C. a relative minimum and a relative maximum D. two saddle points
E. two relative minima.
- Consider the problem of finding the minimum value of the function f (x, y) = 4x
2
2
on the curve
xy = 1. In using the method of Lagrange multipliers, the value of λ (even though it is not needed) will
be
A. 2 B. − 2 C.
2 D.
1 √
2
E. 4.
- Evaluate the iterated integral
3
1
x
0
1
x
dydx.
A. −
8
9
B. 2 C. ln 3 D. 0 E. ln 2.
- Consider the double integral,
R
f (x, y)dA, where R is the portion of the disk x
2
2
≤ 1, in the upper
half-plane, y ≥ 0. Express the integral as an iterated integral.
A.
1
− 1
√
1 −x
2
−
√
1 −x
2 f (x, y)dydx B.
0
− 1
√
1 −x
2
0
f (x, y)dydx
C.
1
− 1
√
1 −x
2
0
f (x, y)dydx D.
1
0
√
1 −x
2
−
√
1 −x
2
f (x, y)dydx
E.
1
0
√
1 −x
2
0
f (x, y)dydx.
- Find a and b for the correct interchange of order of integration:
2
0
2 x
x
2 f (x, y)dydx =
4
0
b
a
f (x, y)dxdy.
A. a = y
2
, b = 2y B. a =
y
2
, b =
y C. a =
y
2
, b = y
D. a =
y, b =
y
2
E. cannot be done without explicit knowledge of f (x, y).
- Evaluate the double integral
R
ydA, where R is the region of the (x, y)-plane inside the triangle with
vertices (0, 0), (2, 0) and (2, 1).
A. 2 B.
8
3
C.
2
3
D. 1 E.
1
3
- The volume of the solid region in the first octant bounded above by the parabolic sheet z = 1 − x
2
, below
by the xy plane, and on the sides by the planes y = 0 and y = x is given by the double integral
A.
1
0
x
0
(1 − x
2 )dydx B.
1
0
1 −x
2
0
x dydx C.
1
− 1
x
−x
(1 − x
2 )dydx
D.
1
0
0
x
(1 − x
2
)dydx E.
1
0
1 −x
2
x
dydx.
- The area of one leaf of the three-leaved rose bounded by the graph of r = 5 sin 3θ is
A.
5 π
6
B.
25 π
12
C.
25 π
6
D.
5 π
3
E.
25 π
3
- Find the area of the portion of the plane x + 3y + 2z = 6 that lies in the first octant.
A. 3
11 B. 6
7 C. 6
14 D. 3
14 E. 6
- A solid region in the first octant is bounded by the surfaces z = y
2 , y = x, y = 0, z = 0 and x = 4. The
volume of the region is
A. 64 B.
64
3
C.
32
3
D. 32 E.
16
3
- An object occupies the region bounded above by the sphere x
2
2
2 = 32 and below by the cone
z =
x
2
2
. The mass density at any point of the object is equal to its distance from the xy plane.
Set up a triple integral in rectangular coordinates for the total mass m of the object.
A.
4
− 4
√
16 −x
2
−
√
16 −x
2
32 −x
2 −y
2
−
x
2 +y
2
z dz dy dx B.
4
− 4
√
16 −x
2
−
√
16 −x
2
32 −x
2 −y
2
x
2 +y
2
z dz dy dx
C.
2
− 2
√
4 −x
2
−
√
4 −x
2
32 −x
2 −y
2
−
x
2 +y
2
z dz dy dx D.
4
0
√
16 −x
2
0
32 −x
2 −y
2
x
2 +y
2
z dz dy dx
E.
4
− 4
√
16 −x
2
−
√
16 −x
2
32 −x
2 −y
2
x
2 +y
2
xy dz dy dx.
- Do problem 34 in spherical coordinates.
A.
2 π
0
∫ π
4
0
√
32
0
ρ
3
cos ϕ sin ϕ dρ dϕ dθ B.
2 π
0
∫ π
4
0
√
32
0
ρ cos ϕ sin ϕ dρ dϕ dθ
C.
2 π
0
∫ π
4
0
√
32
0
ρ
3
sin
2
ϕ dρ dϕ dθ D.
2 π
0
∫ π
2
0
√
32
0
ρ
3
cos ϕ sin ϕ dρ dϕ dθ
E.
2 π
0
∫ π
4
0
√
32
0
ρ cos ϕ dρ dϕ dθ.
- The double integral
1
0
√
1 −x
2
0
y
2 (x
2
2 )
3 dydx when converted to polar coordinates becomes
A.
π
0
1
0
r
9
sin
2
θ dr dθ B.
∫ π
2
0
1
0
r
8
sin
2
θ dr dθ C.
π
0
1
0
r
8
sin θ dr dθ
D.
∫ π
2
0
1
0
r
8 sin θ dr dθ E.
∫ π
2
0
1
0
r
9 sin
2
θ dr dθ.
- Which of the triple integrals converts
2
− 2
√
4 −x
2
−
√
4 −x
2
2 √
x
2 +y
2
dz dy dx
from rectangular to cylindrical coordinates?
A.
π
0
2
0
2
r
r dz dr dθ B.
2 π
0
2
0
2
r
r dz dr dθ C.
2 π
0
2
− 2
2
r
r dz dr dθ
D.
π
0
2
0
2
r
r dz dr dθ E.
2 π
2
0
2
− 2
2
r
r dz dr dθ.
- If D is the solid region above the xy-plane that is between z =
4 − x
2 − y
2 and
z =
1 − x
2 − y
2 , then
D
x
2
2
2 dV =
A.
14 π
3
B.
16 π
3
C.
15 π
2
D. 8 π E. 15 π.
- If C goes along the x-axis from (0, 0) to (1, 0), then along y =
1 − x
2 to (0, 1), and then back to (0, 0)
along the y-axis, then
C
xy dy =
A. −
1
0
√
1 −x
2
0
y dy dx B.
1
0
√
1 −x
2
0
y dy dx C. −
1
0
√
1 −x
2
0
x dy dx
D.
1
0
√
1 −x
2
0
x dy dx E. 0
- Evaluate
C
F · d~r, if
F (x, y) = (xy
2 − 1)
i + (x
2 y − x)
j and C is the circle of radius 1 centered at (1, 2)
and oriented counterclockwise.
A. 2 B. π C. 0 D. −π E. − 2
- Green’s theorem yields the following formula for the area of a simple region R in terms of a line integral
over the boundary C of R, oriented counterclockwise. Area of R =
R
dA =
A. −
C
y dx B.
C
y dx C.
C
x dx D.
1
2
C
y dx − x dy E. −
x dy
- Evaluate the surface integral
S
x dσ where S is the part of the plane 2x + y + z = 4 in the first octant.
A. 8
6 B.
8
3
6 C.
8
3
14 D.
√
14
3
E.
√
10
3
- If S is the part of the paraboloid z = x
2
2 with z ≤ 4, ~n is the unit normal vector on S directed
upward, and
F (x, y, z) = x~i + y~j + z
k, then
S
F · ~n dσ =
A. 0 B. 8 π C. 4 π D. − 4 π E. − 8 π
- If
F (x, y, z) = cos z~i + sin z~j + xy
k, S is the complete boundary of the rectangular solid region bounded
by the planes x = 0, x = 1, y = 0, y = 1, z = 0 and z =
π
2
, and ~n is the outward unit normal on S, then
S
F · ~n dσ =
A. 0 B.
1
2
C. 1 D.
π
2
E. 2
- If
F (x, y, z) = x~i + y~j + z
k, S is the unit sphere x
2
2
2 = 1 and ~n is the outward unit normal on
S, then
S
F · ~n dσ =
A. − 4 π B.
2 π
3
C. 0 D.
4 π
3
E. 4 π
- Evaluate the limit lim
n→∞
[
n
n
]
A. 0 B. 1 C. − 1 D. 2 E. limit does not exist
- Evaluate the limit lim
n→∞
n
n +
n!
A. 0 B. 1 C. e D. 1/e E. limit does not exist
- If s =
∞ ∑
n=
n+
n
, then s =
A. 3 B. 6 C. 9 D. 2 E. 4/ 3
- If L =
∞ ∑
n=
n
∞ ∑
n=
n
n
, then L =
A. 1/ 3 B. 2/ 3 C. 1 D. 4/ 3 E. 5/ 3
∞ ∑
n=
(n
2
p
converges when
A. p > 1 B. p ≤ 1 C. p ≥ 1 D. p >
1
2
E. p ≤
1
2
∞ ∑
n=
n
p
converges for
A. p ≤ 1 B. p > 1 C. p < 0 D. p > 0 E. no values of p
- Which of the following series converge conditionally?
(i)
∞ ∑
n=
n
n
2
(ii)
∞ ∑
n=
n
n
ln n
(iii)
∞ ∑
n=
n
n
e
n
A. only (ii) B. only (i) and (iii) C. only (i) and (ii) D. all three E. none of them
- Which of the following series converge?
(i)
∞ ∑
n=
n
n
1
4
(ii)
∞ ∑
n=
n!
1 · 3 · 5 · · · (2n − 1)
(iii)
∞ ∑
n=
n
A. only (ii) B. only (i) and (iii) C. only (i) and (ii) D. all three E. none of them
- The interval of convergence for the power series
∞ ∑
n=
n
x
n
n ln n
is
A. −
1
3
≤ x <
1
3
B. −
1
3
< x ≤
1
3
C. 0 ≤ x ≤
1
3
D. − 1 ≤ x ≤ 2 E. − 1 < x < 1
- Find the interval of convergence of the power series
∞ ∑
n=
nx
n
n
A. −
1
2
< x <
1
2
B. − 2 < x < 2 C. − 2 ≤ x ≤ 2 D. − 2 < x ≤ 2 E. −∞ < x < ∞
- The fourth term of the Maclaurin series for
x
2
x− 1
is
A. −x
3
B. 3x
3
C. − 3 x
3
D. − 4 x
3
E. 4x
3
- The first three nonzero terms of the Maclaurin series for f (x) = (1 − x
2 ) sin x are
A. x −
5
6
x
3
31
150
x
5
B. 1 −
3
2
x
2
13
24
x
4
C. x −
7
6
x
3
31
150
x
5
D. x
2
−
7
6
x
3
1
25
x
5
E. x −
7
6
x
3
21
120
x
5
