Cal3SampleFinal

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online