Instructions and information

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1 Instructions and information. Check that this paper has a total of 5 pages including the cover page.. This is a closed book exam. Calculators and electronic devices are not allowed. Notes and dictionaries of any kind are not allowed. 3. Information on the computer card is to be entered with a soft lead pencil. Any erasing must be done cleanly. The computer will also accept black or blue ball point pens, but this is not advised as you will not be able to erase these to make corrections. 4. This exam paper is Version. Make sure that the Version column of your computer card has this same number filled in. If not, contact an invigilator. 5. Print your name and student number on the examination booklet provided. You should answer the written questions in the examination booklet. You may also use the examination booklet to do rough calculations for the multiple choice questions, but such work will not be evaluated. 6. Enter the requested ID information on the computer card. It is your responsibility to do this correctly. Carefully check that your student number is entered correctly. Sign the computer card in the space indicated. 7. Examination booklets and computer cards may not be removed from the exam room and must be handed in. 8. This exam paper has a total of 8 points and consists of two parts: Part is worth 64 points and consists of 6 multiple choice questions worth 4 points each. Only the answers entered on the computer card will count for Part. For each question there is only one correct answer. There is no penalty for incorrect answers. Part is worth 64 points and consists of 4 written answer questions worth 6 points each. Show and justify each step in the solution, and simplify answers. 9. The Examination Security Monitor Program detects pairs of students with unusually similar answer patterns on multiple-choice exams. Data generated by this program can be used as admissible evidence, either to initiate or corroborate an investigation or a charge of cheating under Section 6 of the Code of Student Conduct and Disciplinary Procedures. Part. Multiple Choice - Answer on the computer card. (4 points) Let f(x) = cos ( x ). The value of d8 f (0) is dx8 (A) 680, (B) 960, (C) 630, (D) 890, (E) /exam continued

2 . (4 points) Let sin(x) cos(x) = n=0 ( )n a n x n+ be the Maclaurin expansion of sin(x) cos(x). Then for n we have a n = (A) 4 n 4 (n+)!, (B) n (n)!, (C) n+ n+ (n+)!, (D) (n)!, n (E) (n)!. 3. (4 points) The series (n + ) n converges to n=0 (A) 4, (B) ln(4), (C) 3, (D), (E) ln(6). 4. (4 points) Let z(x, y) be defined implicitly by z 4 + z 3 x + zy = 5. Then z x point (x, y, z) = (,, ) takes the value at the (A) 6, (B) 6, (C) 3, (D) 3, (E) (4 points) The Maclaurin expansion of ( x) x is (A) + x + x x4 +, (B) + x + x + x 3 + 4x 4 +, (C) + x + 3 x x4 +, (D) + x + x + 3 x x4 +, (E) + x + x x (4 points) When the function f(x) = ln(3 + x) is expanded in powers of x, the radius of convergence of the resulting power series will be (A) 5, (B), (C) ln(), (D), (E). 7. (4 points) At a saddle point, the value of the Hessian matrix might be ( ) ( ) ( ) 3 4 (A), (B), (C), (D) ( ) (E). 3 3 ( 3 ), 3.../exam continued

3 8. (4 points) The length of the curve given by (x, y) = (t, 3 t 3 ) from the point given by t = 3 to the point given by t = 8 is (A) 38 3, (B) , (C), (D), (E) (4 points) The tangent plane to the surface z + x 3 y = at the point (x, y, z) = (,, ) contains which of the following lines: (A) x = + t, y = t, z = t, (B) x =, y = + t, z = + t, (C) x = + t, y = t, z = t, (D) x = + t, y = 3t, z = + t, (E) x = + t, y = 3t, z = t. 0. (4 points) Consider the function f(x, y, z) = (x + y + z)e xz The rate of greatest increase of f at the point (x, y, z) = (0,, 0) is (A) 3, (B), (C), (D) 3, (E).. (4 points) The equations u = x 3 + y, v = 3x y x have solutions for x and y in terms of u and v near (x, y) = (, ), (u, v) = (, ). At this point, the value of y (, ) is u (A) 4, (B) 4 3, (C) 4, (D), (E) 3.. (4 points) The osculating plane (i.e. the plane containing the tangent and normal directions) to the helix (x, y, z) = (cos(t), sin(t), t) at the point corresponding to t = 0 passes through the point (A) (,, ), (B) (,, ), (C) (,, ), (D) (0,, ), (E) (0,, ). 3. (4 points) At the point on the curve (x, y) = (sin(t), t + cos(t)) corresponding to t = π, the curvature is (A), (B) 4, (C), (D) 8, (E) (4 points) The value of x y ln(x)dxdy over the square 0 x, 0 y is (A), (B) e, (C), (D) e, (E) ln(). 5. (4 points) The area inside the polar loop r = cos(θ) is (A) 59π, (B) 49π, (C) 67π 6π 57π, (D), (E). 4.../exam continued

4 6. (4 points) The volume of the region in 3-space given by the condition x + y z is (A) π 3, (B) π 3, (C) 4π 5, (D) 3π 4, (E) π 4. Part. Written - Write your solutions in the booklets provided. 7. (6 points) Use Lagrange multipliers to find the absolute maximum value of f(x, y, z) = xy on the locus {(x, y, z); x + y = 9, y 3 z = 8}. We note that the equation x + y = 9 forces (x, y) to be bounded. Then, since y is bounded, the second equation forces z to be bounded. Therefore the absolute maximum must be attained and it suffices to consider all possible critical points relative to the locus. Lagrange s equations are y = λx xy = λy + 3µy 0 = µz From the last of these, either z = 0 or µ = 0. Case z = 0. Then y = and it follows that x = ± 5. Also f(± 5,, 0) = ±4 5. Case µ = 0. Then from the second of the Lagrange equations either y = 0 or λ = x. If y = 0 then the corresponding value of f will be zero, so that this will not be the absolute max. (We already know a point where f takes the value 4 5.) Alternatively, if y = 0, then z = 8 which has no solution. We discard this case and substitute λ = x in the first of the Lagrange equations to get y = x. Substituting this in the first constraint, we find x = ± 3 and y = ± 6 (independent ±). Substituting in the second constraint gives z = (± 6) 3 8 which can be solved for z if y = 6 since 6 3 > 8 since 6 > 64, but not if y = 6. It comes that f(± 3, 6) = ±6 3. Finally, we need to figure that 6 3 > 4 5 since 08 > 80. Hence the absolute maximum value is 6 3. Strictly speaking one should also run a check to see if the locus is everywhere parametrizable. This can only fail if the jacobian matrix ( ) x y 0 J = 0 3y z has rank one on the locus. But if J has rank one, then two of x, y, z are zero and the coordinate axes do not intersect the locus. Hence the locus is everywhere parametrizable. Some students observed that the objective function xy and the first constraint x +y = 9 only involve x and y. So they threw away the second constraint. Manipulating objective functions and constraints in this (and other) ways is often appealing and often dangerous. Here, the second constraint forces y 3 8, i.e. y. So actually, an additional restriction is imposed. One should also check y =. In the problem, the absolute max is taken where y >, so it s not an issue. Nevertheless students who did this should have checked that the second constraint was solvable for z and lost marks if they didn t. 5.../exam continued

5 8. (6 points) Find and classify all the critical points of the function defined in the plane. At a critical point we will have Subtracting these equations gives f(x, y) = x 3 + y 3 3x 3y 4xy f x = 3x 6x 4y = 0 f y = 3y 6y 4x = 0 3(x y ) 6(x y) 4(y x) = 0 which factors as (3x + 3y )(x y) = 0. If x = y we get 3x 0x = 0 leading to (x, y) = (0, 0) and (x, y) = ( 0 3, 0 3 ). On the other hand, if 3x + 3y = 0, then y = x + 3 and we find that 9x 6x 8 = 0 which factors as (3x + )(3x 4) = 0 leading to (x, y) = ( 3, 4 3 ) and (x, y) = ( 4 3, 3 ). The general hessian is ( ) 6x y 6 and it has determinant 36(x )(y ) and trace 6(x + y ). Therefore At (x, y) = (0, 0), the determinant is positive and the trace is negative. This is a local maximum point. At (x, y) = ( 0 3, 0 3 ), the determinant is positive and the trace is positive. This is a local minimum point. At (x, y) = ( 3, 4 3 ), the determinant is negative. This is a saddle point. At (x, y) = ( 4 3, 3 ), the determinant is negative. This is a saddle point. 6.../exam continued

6 9. (6 points) Find the area of the region in the plane defined by the inequalities x y 4x and y x y. Hint: Use the substitution u = x y, v = y x. The Jacobian matrix is ( u x v x u y v y ) ( ) x = 3 y x y xy 3 with determinant 3x y. Therefore du dv = 3x y dx dy. We get dx dy = 3 x y du dv. This has to be expressed in terms of u and v. We see that uv = x y, and therefore dx dy = 3 u v du dv The desired area is therefore 3 u v du dv = ( 4 3 [,4] [,] ) ( ) u du v dv = ( [ u ] ) ( 4 [ v ] ) = (6 points) Find the surface area of the part of the surface z = y + x that lies above the triangle with corners at (x, y) = (0, 0), (0, ), (, ) in the xy-plane. We will call the triangle T. Then the desired area is ( ) ( ) z z + + da = x y T on substituting u = 5 + 4y, du = 8y dy = = = 8 T y=0 y= y da u=5 y x= y dx dy y 5 + 4y dy u du = [ ] 9 u 3 = u=5 (7 5 5). 7