MATH 201 Exam, Spring 2008

1 FINAL EXAMINATION (A) MATH 201 January 22, 2007; 8:00-10:00 P.M. Name: Signature: Student number: Section number: 17 18 19 20 Instructions: • No calculators are allowed. • There are two types of questions: PART I consists of six work-out problems. Give a detailed solution. PART II consists of eight multiple-choice questions each with exactly one correct answer. Circle the appropriate answer. Grading policy: • 10 points for each problem of PART I. • 5 points for each problem of PART II. • 0 point for no, wrong, or more than one answer of PART II. GRADE OF PART I /60 GRADE OF PART II /40 TOTAL GRADE /100 2 Part I (1). Use Lagrange Multipliers to find the point on the plane 2x + 3y + 4z = 12 at which f(x, y, z) = 4x2 + y2 + 5z2 has its least value. 3 Part I (2). Sketch the solid bounded by the cone z = √x2 + y2, the cylinder x2 + y2 = 4, and below by the xy−plane, and use triple integrals to find its volume. 4 Part I (3). Find the absolute maximum and minimum values of the function f(x, y) = 5 + 4x − 2x2 + 3y − y2 over the triangular region R bounded by the lines y = 2, y = x and y = −x. 5 Part I (4). Evaluate the integral � � R ex−y x + y dA, where R is the rectangle bounded by the lines y = x, y = x + 5, y = 2 − x and y = 4 − x by using the substitution x − y = u and x + y = v. 6 Part I (5). Use Green’s theorem to evaluate the line integral � C(7y − esin x)dx + [15x − sin(y3 + 8y)]dy, where C is the positively-directed cardioid r = 1 + cos θ. 7 Part I (6). Sketch the region of integration and evaluate the integral � 1 0 � 1 y ex2 dx dy. 8 Part II (1). The value of the double integral � � R cos(x2 + y2) dx dy, where R is the region bounded by the circle x2 + y2 = 9 is (a) 2π sin 3. (b) π sin 3. (c) 2π sin 9. (d) π sin 9. (e) None of the above. Part II (2). The mass of the solid bounded by the paraboloids z = x2 + y2 and z = 2 − x2 − y2 and whose density function is δ(x, y, z) = (x2 + y2)3/2 is given by the triple integral (a) � 2 −2 � √ 4−y2 −√ 4−y2 � 2−x2−y2 x2+y2 (x2 + y2)3/2 dz dx dy. (b) � 2 −2 � √ 4−x2 − √ 4−x2 � 2−x2−y2 x2+y2 (x2 + y2)3/2 dz dy dx. (c) � 1 −1 � √ 1−x2 − √ 1−x2 � 2−x2−y2 x2+y2 (x2 + y2)3/2 dz dy dx. (d) � 1 −1 � √ 1−x2 − √ 1−x2 � x2+y2 2−x2−y2(x2 + y2)3/2 dz dy dx. (e) None of the above. 9 Part II (3). The volume of the solid bounded above by the sphere x2 + y2 + z2 = 4, below by the xy−plane, and lies inside the cylinder x2 + y2 = 1 is given by the triple integral (a) � 2π 0 � π/3 0 � 2 0 ρ2 sin φ dρ dφ dθ + � 2π 0 � π/2 π/3 � 1/ cos φ 0 ρ2 sin φ dρ dφ dθ. (b) � 2π 0 � π/6 0 � 2 0 ρ sin φ dρ dφ dθ + � 2π 0 � π/2 π/6 � 1/ sin φ 0 ρ sin φ dρ dφ dθ. (c) � 2π 0 � π/6 0 � 2 0 ρ2 sin φ dρ dφ dθ + � 2π 0 � π/2 π/6 � sin φ 0 ρ2 sin φ dρ dφ dθ. (d) � 2π 0 � π/6 0 � 2 0 ρ2 sin φ dρ dφ dθ + � 2π 0 � π/2 π/6 � 1/ sin φ 0 ρ2 sin φ dρ dφ dθ. (e) None of the above. Part II (4). The approximate value of the integral � 0.1 0 sin x x dx with an error of magnitude less than 10−4 is (a) 0.1. (b) 0.2. (c) 0.3. (d) 0.4. (e) None of the above. 10 Part II (5). The power series ∞ � k=2 (−1)k (x − 1)k k ln k has interval of convergence (a) [0, 2]. (b) [0, 2[. (c) ]0, 2]. (d) ]0, 2[. (e) ] − ∞, ∞[. Part II (6). The direction of maximum change for the function f(x, y, z) = � x2 + y2 + z2 at the point (1, 2, −2) is (a) u = (2/3)i + (2/3)j + (−1/3)k. (b) u = (1/3)i + (−2/3)j + (2/3)k. (c) u = (1/3)i + (2/3)j + (−2/3)k. (d) u = (−1/3)i + (2/3)j + (2/3)k. (e) u = (−1/3)i + (−2/3)j + (−2/3)k. 11 Part II (7) The function f(x, y) = 2y2 − x3 − 2xy (a) has local minimum at (−1/3, −1/6) and local maximum at (0, 0). (b) has local minimum at (−1/3, −1/6) and saddle point at (0, 0). (c) has local maximum at (−1/3, −1/6) and saddle point at (0, 0). (d) has local maximum at (−1/3, −1/6) and local minimum at (0, 0). (e) has saddle point at (−1/3, −1/6) and local minimum at (0, 0). Part II (8) Equations of the tangent plane and normal line to the surface z = 6 − x2 − y2 at the point (1, 2, 1) are respectively (a) 2x + 4y + z = 11 and (x − 1)/2 = (y − 2)/4 = (z − 1). (b) x + 4y + 2z = 11 and (x − 1) = (y − 2)/4 = (z − 1)/2. (c) 4x + 2y + 3z = 11 and (x + 1)/2 = (3y − 2)/4 = (z + 1)/2. (d) 3x + 2y + 4z = 11 and (2x − 1)/2 = (y − 1)/2 = (2z − 1)/2. (e) 4x + 3y + z = 11 and (4x − 1)/3 = (y + 2)/4 = (4z − 1)/3.