MTH 234 Solutions to Exam 1 Oct 12 2015 Name Section Recitation Instructor READ THE FOLLOWING INSTRUCTIONS Do not open your exam until told to do so No calculators cell phones or any other electronic devices can be used on this exam Clear your desk of everything excepts pens pencils and erasers If you need scratch paper use the back of the previous page Without fully opening the exam check that you have pages 1 through 10 Fill in your name etc on this rst page Show all your work Write your answers clearly Include enough steps for the grader to be able to follow your work Don t skip limits or equal signs etc Include words to clarify your reasoning Do rst all of the problems you know how to do immediately Do not spend too much time on any particular problem Return to di cult problems later If you have any questions please raise your hand and a proctor will come to you There is no talking allowed during the exam You will be given exactly 90 minutes for this exam I have read and understand the above instructions SIGNATURE Page 1 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 Multiple Choice Circle the best answer No work needed No partial credit available 1 8 points Consider the surface given by equation below and answer the questions that follow a Horizontal traces or cross sections i e planes parallel to z k 1 of the surface 3z x2 2 y2 9 1 b Vertical traces or cross sections i e planes parallel to either x k or y k of the surface 2 4 points The contour plot level curves to the right could be from which function de ned by 1 are A Circles B Parabolas C Ellipses D Hyperbolas E None of the above de ned by 1 are A Circles B Parabolas C Ellipses D Hyperbolas E None of the above A f x y x sin y B f x y x sin y C f x y y sin x D f x y x cos y E f x y y sin x F None of the above Extra Work Space Probs 1 and 2 4 points for each part Page 2 of 10 xy MTH 234 Solutions to Exam 1 Oct 12 2015 Fill in the Blanks No work needed No partial credit available 3 10 points Find the velocity and position functions of a particle that satis es the following conditions a t sin t i 12e 2t j 8 k v 0 i 6 j r 0 3 j k a v t cos t i 6e 2t j 8t k b r t sin t i 3e 2t j 1 4t2 k 5 points for part a npc 5 points for part b except award 3 points for 4 8 points Let u 3 i 2 j k and v i 5 j 2 k Find each of the following a 2u 3v 9 i 11 j 4 k sin t i 3e 2t j 4t2 k b Let be the angle between u and v Then cos 9 14 30 4 points each except deduct 1 point only if missing the negative sign 5 6 points The line segment from 3 7 5 to 2 4 1 is given by the parametric equations x t 3 1 t 2t y t 7 1 t 4t z t 5 1 t t t 0 1 2 points each npc Extra Work Space Page 3 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 Standard Response Questions Show all work to receive credit Please BOX your nal answer 6 12 points Find the length of the curve below r t i 4t j t2 k 0 t 1 8t3 2 3 Solution so that It follows that v t 4 t i 4 j 2t k v t 2 16t 16 4t2 4 t 2 2 Arc length 2 t 2 dt cid 90 1 0 5 7 Let f x y 1 cid 112 2x2 1 y a 8 points Sketch the domain of f b 4 points Give the range of f Express your answer using interval notation Solution Ran f 0 y x y 2x2 1 Page 4 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 8 14 points Find the following limits or show that they do not exist We claim that the limit is zero For x y cid 54 0 0 x2 2x2 y2 It follows that a lim x y 0 0 x2 sin2 y 2x2 y2 0 Solution so that since sin2 y 0 Now b lim x y 0 0 xy2 8x2 3y4 Solution x2 0 2x2 y2 1 0 x2 sin2 y 2x2 y2 sin2 y lim x y 0 0 0 lim x y 0 0 sin2 y 0 The result follows by the Squeeze Law We claim the limit does not exist Let x my2 for m R xy2 lim x y 0 0 8x2 3y4 lim x y 0 0 my2 y2 8 my2 2 3y4 m 8m2 3 y4 y4 lim x y 0 0 m 8m2 3 The result now follows since m was arbitrary Page 5 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 9 Let f x y x2 ey 6y x a 12 points Find the linearization of f x y at the point 4 0 Solution Notice that fx 2xey and fy x2ey 6 x Thus 3y x L x y f 4 0 fx 4 0 x 4 fy 4 0 y 0 16 8 x 4 4 y 0 b 6 points Use the linearization from part a to approximate f 3 9 0 2 Solution f 3 9 0 2 L 3 9 0 2 16 8 0 1 4 0 2 16 0 8 0 8 Page 6 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 10 12 points Find the equation of the plane containing the points A B and C if A 3 0 1 B 1 2 2 C 6 1 3 Solution Let Then AB 2 i 2 j k AC 3 i j 2 k u v u v 5 i 7 j 4 k It follows that the equation of the plane is 5 x 3 7y 4 z 1 0 Many correct alternatives 11 10 points Let w xy2 yex 2 where x x s t and y y s t are di erentiable Suppose also that x 1 3 2 xt 1 3 4 xs 1 3 5 y 1 3 3 yt 1 3 6 ys 1 3 0 Find wt 1 3 6 Solution So Thus y2 yex 2 w x w x s t 1 3 6 2xy ex 2 w y w y s t 1 3 11 w t w x x t w y y t 6 4 11 6 Page 7 of 10 MTH 234 Solutions to Exam 1 Oct 12 2015 12 10 points Let g …