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
²
and
2
=
±
2
,

5
,

1
²
respectively. The direction vector of the intersecting line is
±v
=
1
×
2
=
±
1
,
1
,

1
² × ±
2
,

5
,

1
²
=
±
6
,

1
,

7
²
To ±nd 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
²
=
±
1
,
0
,
1
²
+
t
±
6
,

1
,

7
²
=
±
1

6
t,

t,
1

7
t
²
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 ±rst line, and the point
Q
=(

1
,
0
,
0) is on the
second line. The direction vectors of these two lines are
±
1
,
1
,
1
²
and
±
1
,
2
,
3
²
respectively.
The vector
=
±
1
,
1
,
1
² × ±
1
,
2
,
3
²
=
±
1
,

4
,
3
²
is orthogonal to both lines. The distance
d
=
±
±
±
comp
±
PQ
±
±
±
=

±
·



=
±
1
,
0
,
0
² · ±
1
,

4
,
3
²
√
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
=
²
4
e
i
5
π/
6
³
5
=4
5
e
i
25
π
6
5
e
iπ/
6
(b) Cube roots of 1i: write

1

i
=
√
2
e
i
5
4
, then the three cube roots are
6
√
2
e
i
5
12
,
6
√
2
e
i
(5
12+2
3)
=
6
√
2
e
i
13
12
, and
6
√
2
e
i
(5
12+4
3)
=
6
√
2
e
i
21
12
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
CALCULUS III SAMPLE FINAL
Problem 4.
(20 points)
Solve the initialvalue problem
2
y
±
+5
y
±
+3
y
=0
,y
(0) = 3
±
(0) =

4
Solution:
The auxiliary equation is 2
r
2
r
+ 3 = 0 and it has too roots

1,

3
/
2. The general
solution to the diferential 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 o± the acceleration
vector:
±
r
(
t
)=
t
±
i
+ cos
2
t
±
j
+ sin
2
t
±
k
at time
t
=
π/
8.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Virdol
 Calculus, Equations, Derivative, Vectors, Parametric Equations, Partial differential equation, Parametric equation, Boundary, CALCULUS III SAMPLE

Click to edit the document details