{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This 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 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
Image of page 1

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

View Full Document Right 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 , 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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern