Math 113 Test IV Practice Problems (6.6, 6.7, 9.1 - 9.3)
fall 2013
Find all solutions of the equation.
1) 2 cos x + 2 = 0
Write a vector v in terms of i and j whose magnitude v
and direction angle θ are given.
17) v = 7, θ = 225°
2) 8 sin x + 6 2 = 6 sin + 5 2
18) v = 11, θ = 270°
Solve the equation on the interval [0, 2π).
3
3) sin 4x =
2
Solve the problem.
19) A child throws a ball with a speed of 6 feet per
second at an angle of 25° with the horizontal.
Express the vector described in terms of i and j.
If exact values are not possible, round
components to 3 decimals.
4) 2 sin 2 x = sin x
Solve the equation on the interval [0, 2π).
5) sin x - 2 sin x cos x = 0
20) Two forces, F1 and F2 , of magnitude 60 and 70
pounds, respectively, act on an object. The
direction of F1 is N40°E and the direction of F2
Find the magnitude v of the vector.
6) v = 9i + 12j
is N40°W. Find the magnitude and the
direction angle of the resultant force. Express
the direction angle to the nearest tenth of a
degree.
7) v = -4i + 4j
8) v = -i - j
Perform the indicated operation.
21) u = 9i + j, v = -4i - 2j, w = i - 8j; Find v - (u w).
A vector v has initial point P1 and terminal point P2 . Write
v in terms of ai + bj.
9) P1 = (-2, 3); P2 = (-5, -6)
Find the magnitude and direction angle (to the nearest
tenth) for the vector. Give the measure of the direction
angle as an angle in [0,360°].
22) -4, -3
10) P1 = (6, 6); P2 = (-5, -1)
Find the specified vector or scalar.
11) u = -2i - 6j, v = -6i + 8j; Find u + v.
Use the given vectors to find the specified scalar.
23) u = 13i - 7j and v = -6i + 7j; Find u · v.
12) u = -4i - 2j, v = 6i + 7j; Find u - v.
24) u = 9i - 6j, v = 3i - 3j, w = 6i - 8j; Find u · (v +
w).
13) v = 3i + 4j; Find 2v.
25) u = -5i + 3j and v = 2i - 6j, and w = -3i + 12j;
Find u · w + v · w.
14) u = 5i - 2j and v = 8i + 7j; Find v - u .
Find the angle between the given vectors.
26) u = 4i + 9j, v = 2i + 8j
Find the unit vector having the same direction as v.
15) v = -3i - 4j
27) u = -i + 2j, v = 5i - 6j
16) v = 3i + j
1
Find the angle between the given vectors and use to
determine whether the vectors are parallel, orthogonal, or
neither.
28) v = 3i + j, w = i - 3j
29) v = i +
35) Major axis horizontal with length 20; length of
minor axis = 12; center (0, 0)
Graph the ellipse.
(x - 1)2 (y - 2)2
36)
+
=1
9
4
3j, w = i - 2j
y
10
30) v = 2i + 4j, w = 4i + 8j
5
Graph the ellipse and locate the foci.
x2 y2
31)
+
=1
49
9
-10
-5
5
10
x
5
10
x
y
-5
10
-10
5
-10
-5
5
10
37) 16(x - 2)2 + 4(y - 1)2 = 64
x
y
-5
10
-10
5
Find the standard form of the equation of the ellipse and
give the location of its foci.
32)
-10
-5
-5
y
10
-10
5
-10
-5
5
10
Find the foci of the ellipse whose equation is given.
(x - 1)2 (y + 3)2
38)
+
=1
36
16
x
-5
Convert the equation to the standard form for an ellipse
by completing the square on x and y.
39) 25x2 + 36y2 + 150x - 72y - 639 = 0
-10
Find the vertices and locate the foci for the hyperbola
whose equation is given.
y2 x2
40)
=1
64 16
Find the standard form of the equation of the ellipse
satisfying the given conditions.
33) Foci: (-3, 0), (3, 0); vertices: (-7, 0), (7, 0)
34) Foci: (0, -3), (0, 3); y-intercepts: -4 and 4
Find the standard form of the equation of the hyperbola
satisfying the given conditions.
41) Foci: (-9, 0), (9, 0); vertices: (-6, 0), (6, 0)
2
42) Endpoints of transverse axis: (0, -8), (0, 8);
4
asymptote: y = x
7
Find the focus and directrix of the parabola with the given
equation.
48) y2 = -28x
43) Center: (6, 5); Focus: (0, 5); Vertex: (5, 5)
Graph the parabola.
49) x2 = -16y
Convert the equation to the standard form for a hyperbola
by completing the square on x and y.
44) y2 - 9x2 - 2y - 36x - 44 = 0
10
y
5
Use vertices and asymptotes to graph the hyperbola. Find
the equations of the asymptotes.
y2 x2
45)
=1
9
36
-10
-5
5
10 x
-5
y
10
-10
5
-10
-5
5
10
Find the standard form of the equation of the parabola
using the information given.
50) Vertex: (8, -1); Focus: (8, -3)
x
-5
Convert the equation to the standard form for a parabola
by completing the square on x or y as appropriate.
51) y2 + 2y + 4x - 3 = 0
-10
Find the vertex, focus, and directrix of the parabola with
the given equation.
52) (y - 3)2 = -4(x - 2)
Find the location of the center, vertices, and foci for the
hyperbola described by the equation.
(y + 4)2 (x - 3)2
46)
=1
9
4
53) (x + 2)2 = -20(y + 3)
Use the center, vertices, and asymptotes to graph the
hyperbola.
(y - 2)2 (x + 1)2
47)
=1
9
4
y
10
5
-10
-5
5
10
x
-5
-10
3
Answer Key
Testname: MATH 113 TEST IV PRACTICE (6.6, 6.7, 9.1-9.3) FALL 2013
1) x =
3π
5π
+ 2nπ or x =
+ 2nπ
4
4
2) x =
5π
7π
+ 2nπ or x =
+ 2nπ
4
4
3)
31) foci at (2 10, 0) and (-2 10, 0)
y
10
π π 2π 7π 7π 13π 5π 19π
, ,
,
,
,
,
,
12 6 3 12 6 12 3
12
4) 0, π,
5
π 5π
,
6 6
-10
-5
5) First factor out sin x
π
5π
0, , π,
3
3
17) v = -
3
i+
10
x
5
10
x
5
10
x
-5
-10
6) 15
7) 4 2
8) 2
9) v = -3i - 9j
10) v = -11i - 7j
11) -8i + 2j
12) -10i - 9j
13) 6i + 8j
14) 3 10
3
4
15) u = - i - j
5
5
16) u =
10
5
32)
x2 y2
+
=1
16 81
foci at ( 0, - 65) and (0,
x2 y2
33)
+
=1
49 40
1
j
10
34)
x2 y2
+
=1
7
16
35)
x2
y2
+
=1
100 36
65)
36)
y
10
7 2
7 2
ij
2
2
5
18) v = -11j
19) 5.438i + 2.536j
20) F = 99.37 pounds; θ = 93.7° or N3.7oW
21) -12i - 11j
22) 5; 216.9°
23) -127
24) 147
25) -27
26) 9.9°
27) 166.8°
28) θ = 0 o; orthogonal
-10
-5
-5
-10
37)
y
10
29) θ = 123.4o; neither
30) θ = 0 o; parallel
5
-10
-5
-5
-10
4
Answer Key
Testname: MATH 113 TEST IV PRACTICE (6.6, 6.7, 9.1-9.3) FALL 2013
49)
38) foci at (1 + 2 5, -3) and (1 - 2 5, -3)
(x + 3)2 (y - 1)2
39)
+
=1
36
25
10
40) vertices: (0, -8), (0, 8)
foci: (0, - 4 5), (0, 4 5)
x2 y2
41)
=1
36 45
42)
5
-10
y2
x2
=1
64 196
50) (x - 8)2 = -8(y + 1)
51) (y + 1)2 = -4(x - 1)
52) vertex: (2, 3)
focus: (1, 3)
directrix: x = 3
53) vertex: (-2, -3)
focus: (-2, -8)
directrix: y = 2
1
45) Asymptotes: y = ± x
2
y
10
5
-5
5
10
x
-5
-10
46) Center: (3, -4); Vertices: (3, -7) and (3, -1); Foci: (3, -4
- 13) and (3, -4 + 13)
47)
y
10
5
-10
5
-10
(y - 1)2
- (x + 2)2 = 1
9
-10
-5
-5
(y - 5)2
43) (x - 6)2 =1
35
44)
y
-5
5
10
x
-5
-10
48) focus: (-7, 0)
directrix: x = 7
5
10 x