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Cal3SampleFinal

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

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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 vectorn 1 = ( 1 , 1 , 1 ) and vectorn 2 = ( 2 , 5 , 1 ) respectively. The direction vector of the intersecting line is vectorv = vectorn 1 × vectorn 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 vectorn = ( 1 , 1 , 1 ) × (− 1 , 2 , 3 ) = ( 1 , 4 , 3 ) is orthogonal to both lines. The distance d = vextendsingle vextendsingle vextendsingle comp vectorn vector PQ vextendsingle vextendsingle vextendsingle = | vector PQ · vectorn | | vectorn | = |(− 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 = parenleftBig 4 e i 5 π/ 6 parenrightBig 5 = 4 5 e i 25 π 6 = 4 5 e iπ/ 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

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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 differential 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: vector r ( t ) = t vector i + cos 2 t vector j + sin 2 t vector k at time t = π/ 8.
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