C3-sampleFinal08

C3-sampleFinal08 - CALCULUS III SAMPLE FINAL Problem 1(15...

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Unformatted text preview: CALCULUS III SAMPLE FINAL Problem 1: (15 points) (a) Find a parametric equation for the line of intersection of the planes z = x + y, 2 x − 5 y − z = 1 . (b) Find the distance between skew lines x = y = z and − x − 1 = y/ 2 = z/ 3. Solution: (a) The normal vectors to the two planes are vectorn 1 = ( 1 , 1 , − 1 ) and vectorn 2 = ( 2 , − 5 , − 1 ) respec- tively. 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 , , 1) on the line of intersection. A parametric equation of the line is ( x, y, z ) = ( 1 , , 1 ) + t (− 6 , − 1 , − 7 ) = ( 1 − 6 t, − t, 1 − 7 t ) (b) The point P = (0 , , 0) is on the first line, and the point Q = ( − 1 , , 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 , , ) · ( 1 , − 4 , 3 )| √ 1 2 + 4 2 + 3 3 = 1 √ 26 Problem 2. (10 points) (a) Find ( − 2 √ 3 + 2 i ) 5 using De Moivre’s Theorem. (b) Solve the initial-value problem 2 y ′′ + 5 y ′ + 3 y = 0 , y (0) = 3 , y ′ (0) = − 4 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) 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 − + c 2 e − = 3 , − c 1 e − − 3 c 2 e − / 2 = − 4 ....
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C3-sampleFinal08 - CALCULUS III SAMPLE FINAL Problem 1(15...

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