CALCULUS III SAMPLE FINAL
Problem 1:
(20 points) Find parametric equations for the line of intersection of the planes
z
=
x
+
y,
2
x
-
5
y
-
z
= 1
.
Solution:
The normal vectors to the two planes are
n
1
=
1
,
1
,
-
1
and
n
2
=
2
,
-
5
,
-
1
respectively. The direction vector of the intersecting line is
v
=
n
1
×
n
2
=
1
,
1
,
-
1
×
2
,
-
5
,
-
1
=
-
6
,
-
1
,
-
7
To find a point on the line of intersection, try to let
y
= 0, then we get a system
z
=
x
,
2
x
-
z
= 1, and by solving it we get
x
=
z
= 1, so we get a point (1
,
0
,
1) on the line of
intersection. The parametric equation of the line is
x, y, z
=
1
,
0
,
1 +
t
-
6
,
-
1
,
-
7
=
1
-
6
t,
-
t,
1
-
7
t
Problem 2:
(20 points) Find the distance between skew lines
x
=
y
=
z
and
-
x
-
1 =
y/
2 =
z/
3.
Solution:
The point
P
= (0
,
0
,
0) is on the first line, and the point
Q
= (
-
1
,
0
,
0) is on the
second line. The direction vectors of these two lines are
1
,
1
,
1
and
-
1
,
2
,
3
respectively.
The vector
n
=
1
,
1
,
1
× -
1
,
2
,
3
=
1
,
-
4
,
3
is orthogonal to both lines. The distance
d
=
comp
n
PQ
=
|
PQ
·
n
|
|
n
|
=
|
-
1
,
0
,
0
·
1
,
-
4
,
3
|
√
1
2
+ 4
2
+ 3
3
=
1
√
26
Problem 3.
(20 points)
(a) Find (
-
2
√
3 + 2
i
)
5
using De Moivre’s Theorem:
(b) Find the cube roots of
-
1
-
i
and sketch the roots in the complex plane:
Solution:
(a)
(
-
2
√
3 + 2
i
)
5
=
4
e
i
5
π
/
6
5
= 4
5
e
i
25
π
6
= 4
5
e
i
π
/
6
(b) Cube roots of -1-i: write
-
1
-
i
=
√
2
e
i
5
π
/
4
, then the three cube roots are
6
√
2
e
i
5
π
/
12
,
6
√
2
e
i
(5
π
/
12+2
π
/
3)
=
6
√
2
e
i
13
π
/
12
, and
6
√
2
e
i
(5
π
/
12+4
π
/
3)
=
6
√
2
e
i
21
π
/
12
1
2
CALCULUS III SAMPLE FINAL
Problem 4.
(20 points)
Solve the initial-value problem
2
y
+ 5
y
+ 3
y
= 0
,
y
(0) = 3
,
y
(0) =
-
4
Solution:
The auxiliary equation is 2
r
2
+ 5
r
+ 3 = 0 and it has too roots
-
1,
-
3
/
2. The general
solution to the di
ff
erential equation is
c
1
e
-
x
+
c
2
e
-
3
x/
2
. Plug in the initial conditions,
c
1
e
-
0
+
c
2
e
-
0
= 3
,
-
c
1
e
-
0
-
3
c
2
e
-
0
/
2 =
-
4
.
Solve this system we get
c
1
= 1,
c
2
= 2. So the solution to the IVP is
e
-
x
+ 2
e
-
3
x/
2
.
Problem 5.
(20 points) Find the tangential and normal components of the acceleration
vector:
r
(
t
) =
t i
+ cos
2
t j
+ sin
2
t k
at time
t
=
π
/
8.
