Cal3SampleFinal - CALCULUS III SAMPLE FINAL Problem 1(20...

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 -1-i: 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

2 CALCULUS III SAMPLE FINAL
Problem 4. (20 points)
Solve the initial-value 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.