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MTH 234 Solutions to Exam 1 Oct 12, 2015Name: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 first page.• Show all your work. Write your answers clearly! Include enough steps for thegrader to be able to follow your work. Don’t skip limits or equal signs, etc. Includewords to clarify your reasoning.• Do first all of the problems you know how to do immediately. Do not spend toomuch time on any particular problem. Return to difficult 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:. SIGNATUREPage 1 of 10MTH 234 Solutions to Exam 1 Oct 12, 2015Multiple 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.3z = x2+y29+ 2 (1)(a) Horizontal traces or cross-sections (i.e., planes parallel to z = k > 1) of the surfacedefined by (1) areA. CirclesB. ParabolasC. EllipsesD. HyperbolasE. None of the above(b) Vertical traces or cross-sections (i.e., planes parallel to either x = k or y = k) of the surfacedefined by (1) areA. CirclesB. ParabolasC. EllipsesD. HyperbolasE. None of the above2. (4 points) The contour plot (level curves) to the right could be from which function?xyA. f(x, y) =xsin yB. f(x, y) = x sin yC. f(x, y) = y sin xD. f(x, y) = x cos yE. f(x, y) =ysin xF. None of the aboveExtra Work Space.Probs 1 and 2: 4 points for each part.Page 2 of 10MTH 234 Solutions to Exam 1 Oct 12, 2015Fill in the Blanks. No work needed. No partial credit available.3. (10 points) Find the velocity and position functions of a particle that satisfies the following conditions.a(t) = −sin t i + 12e−2tj − 8 kv(0) = i − 6 jr(0) = 3 j + k(a) v(t) = cos t i − 6e−2tj − 8t k(b) r(t) = sin t i + 3e−2tj + (1 − 4t2) k5 points for part (a), npc. 5 points for part (b), except – award 3 points forsin t i + 3e−2tj − 4t2k4. (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(b) Let θ be the angle between u and v. Then cos θ =−9√14√304 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 equationsx(t) = 3(1 − t) − 2ty(t) = 7(1 − t) + 4tz(t) = 5(1 − t) + t , t ∈ [0, 1]2 points each, npcExtra Work Space.Page 3 of 10MTH 234 Solutions to Exam 1 Oct 12, 2015Standard Response Questions. Show all work to receive credit. Please BOX your final answer.6. (12 points) Find the length of the curve below.r(t) =8t3/23i − 4t j + t2k, 0 ≤ t ≤ 1Solution:v(t) = 4√t i − 4 j + 2t kso that|v(t)|2= 16t + 16 + 4t2= 4(t + 2)2It follows thatArc length = 2Z10(t + 2) dt=...= 57. Let f(x, y) =1p2x2− 1 − y.(a) (8 points) Sketch the domain of f .xyy < 2x2− 1(b) (4 points) Give the range of f. Express your answer using interval notation.Solution:Ran(f) = (0, ∞)Page 4 of 10MTH 234 Solutions to Exam 1 Oct 12, 20158. (14 points) Find the following limits or show that they do not exist.(a) lim(x,y)→(0,0)x2sin2y2x2+ y2= 0Solution:We claim that the limit is zero. For x, y 6= 0, 0 < x2≤ 2x2+ y2. It follows that0 <x22x2+ y2≤ 1so that0 <x2sin2y2x2+ y2≤ sin2ysince sin2y > 0. Nowlim(x,y)→(0,0)0 = lim(x,y)→(0,0)sin2y = 0The result follows by the Squeeze Law.(b) lim(x, y)→(0,0)xy28x2+ 3y4Solution:We claim the limit does not exist. Let x = my2for m ∈ R.lim(x,y)→(0,0)xy28x2+ 3y4= lim(x,y)→(0,0)(my2)y28(my2)2+ 3y4= lim(x,y)→(0,0)m8m2+ 3y4y4=m8m2+ 3The result now follows since m was arbitrary.Page 5 of 10MTH 234 Solutions to Exam 1 Oct 12, 20159. Let f(x, y) = x2ey− 6y√x.(a) (12 points) Find the linearization of f(x, y) at the point (4, 0).Solution:Notice that fx= 2xey−3y√xand fy= x2ey− 6√x. ThusL(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 + 0.8Page 6 of 10MTH 234 Solutions to Exam 1 Oct 12, 201510. (12 points) Find the equation of the plane containing the points A, B, and C ifA = (3, 0, 1), B = (1, 2, 2), C = (6, −1, 3)Solution:Letu =−→AB = −2 i + 2 j + kv =−→AC = 3 i − j + 2 kThenu × v = 5 i + 7 j − 4 kIt follows that the equation of the plane is5(x − 3) + 7y − 4(z − 1) = 0Many correct alternatives.11. (10 points) Let w = xy2+ yex−2where x = x(s, t) and y = y(s, t) are differentiable. Suppose also thatx(1, 3) = 2, xt(1, 3) = −4, xs(1, 3) = 5y(1, 3) = −3, yt(1, 3) = 6, ys(1, 3) = 0Find wt(1, 3) = 6 .Solution:So∂w∂x= y2+ yex−2,∂w∂y= 2xy + ex−2∂w∂x(s,t)=(1,3)= 6,∂w∂y(s,t)=(1,3)= −11Thus∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t= (6)(−4) + (−11)(6)Page 7 of 10MTH 234 Solutions to Exam 1 Oct 12, 201512. (10 points) Let g(x, y) = 5x2y3−ln 2yx. Find gxyand gyy.gxy= 30xy2+1x2ygyy= 30yx2+1y2xSolution:And here are the first partials.gx= 10xy3+ln(2y)x2gy= 15x2y2−1xy13. (12 points) Find the parametric equations for the line tangent to the curver(t) = h2 − ln t, t2+ 3, 1 − 4t3i at the point (2, 4, −3).x(t) = 2 − ty(t) = 4 + 2tz(t) = −3 − 12tSolution:Let P = (2, 4, −3). Notice that the curve passes throughP when t = 1. Nowv(t) =−1ti + 2t j − 12t2kThusv(1) = −i + 2 j − 12 kIt follows that parametric equations of the line tangent tothe curve at P arex(t) = 2 − t, y(t) = 4 + 2t, z(t) = −3 − 12tOf course, other parameterizations are possible.Page 8 of 10MTH 234 Solutions to Exam 1 Oct 12, 201514. (14 points) Let f(x, y) = 3xy2− 2y and answer the questions below.(a) Find the equation of the plane tangent to surface z = f (x, y) at the point P (2, 1, 4).Solution:Notice that fx= 3y2and fy= 6xy − 2 so that fx(2, 1) = 3 and fy(2, 1) = 10. Itfollows that the equation of the tangent plane at P is given byz − …

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