Cal3SampleFinal - CALCULUS III SAMPLE FINAL Problem 1(20 points Find parametric equations for the line of intersection of the planes z = x y 2x 5y z = 1

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

• Notes
• 7

This preview shows page 1 - 3 out of 7 pages.

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.