MA 261 EXAM I Spring 1998 Page 1/6NAMESTUDENT ID #INSTRUCTORINSTRUCTIONS1. There are 6 different test pages (including this cover page). Make sure you have acomplete test.2. Fill in the above items in print. I.D.# is your 9 digit ID (probably your social securitynumber). Also write your name at the top of pages 2–6.3. Do any necessary work for each problem on the space provided or on the back of thepages of this test booklet. Circle your answers in this test booklet. No partial creditwill be given but the work on the test booklet may be used in borderline cases.4. No books, notes or calculators may be used on this exam.5. Each problem is worth 10 points. The maximum possible score is 100 points.6. Usinga#2pencil, fill in each of the following items on your answer sheet:(a) On the top left side, write your name (last name, first name), and fill in the littlecircles.(b) On the bottom left side, under SECTION, write in your division and sectionnumber and fill in the little circles. (For example, for division 9 section 1, write0901. For example, for division 38 section 2, write 3802).(c) On the bottom, under STUDENT IDENTIFICATION NUMBER, write in yourstudent ID number, and fill in the little circles.(d) Using a #2 pencil, put your answers to questions 1–10 on your answer sheet byfilling in the circle of the letter of your response. Double check that you have filledin the circles you intended. If more than one circle is filled in for any question,your response will be considered incorrect. Use a #2 pencil.(e) Sign your answer sheet.7. After you have finished the exam, hand in your answer sheet andyour test booklet toyour instructor.MA 261 EXAM I Spring 1998 Name: Page 2/61. The angle between the vectors~i +~j and (√3 − 1)~i +(√3+1)~j is:A. 0B.π6C.π4D.π3E.π22. The symmetric equations for a line containing the point (2, −1, 4) and parallel to theline x =1− 3t, y =2+12t, z =5are:A.x − 2−3=y +112= z − 4B.x − 2−3=y +112, z =4C.x +32=y − 12−1=z − 54D.x +32=y − 12−1=z − 44E.x − 1−3=y − 212, z =5MA 261 EXAM I Spring 1998 Name: Page 3/63. The equation for the plane containing the linesx − 13=y − 25=z +12andx −12= y − 2=z +1−1is:A. −5x+7y−10z =19B. x + y + z =2C. 2x + y − z =5D. 3x +5y +2z =11E. −x + y − z =24. Which point is notin the domain of f where f(x, y)=(x2− y2)3/2ln |2x −4|?A. (1, 1)B. (1, 0)C. (0, 1)D. (−1, 1)E. (−1, −1)MA 261 EXAM I Spring 1998 Name: Page 4/65. Find the speed kv(2)k where ~a(t)=−10~k and v(1) =~i −~j +~k.A.√84B.√83C.√82D. 9E.√806. Find the length of the curve ~r(t)=t2~i +2t~j +lnt~k,1≤ t ≤ e.A. e2B. 2e +1C. 2e − 2D. 2e +2E. e2− 1MA 261 EXAM I Spring 1998 Name: Page 5/67. The intersection of the surface −10x2+ x +2y2+ z2= 4 with a plane parallel to theyz-plane isA. a circleB. a parabolaC. a hyperbolaD. a ellipseE. two lines8. Let z =px2+ y2, x = uv, y = u2− v2.Findzuzv.A.xv +2yuxu +2yvB.u2v +2v3px2+ y2C.2v3+ u2v2u3+ uv2D.uvu2− v2E.2u3− uv22v3− u2vMA 261 EXAM I Spring 1998 Name: Page 6/69. Suppose thatpx3y + x2y2= 10 implicitly defines y as a function of x.Finddydx.A.3x2y +2xy22px3y + x2y2B. −3x2y +2xy2x3+2x2yC.px3y + x2y2− 10x3− 3xy2D.x3y +2x2y3x2y +2xy2E.x2y +2xy2− 10px3y + x2y210. The directional derivative at (−1, 1) of f (x, y)=e−x22−y33in the direction of~a = −2~i +~j isA.15e−56B. −1√5~i −3√5~jC. −3√5e−56D. 2e76E. −25~i