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 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.