Department of Mathematics, IIT Bombay End-Semester Examination, MA 105 Autumn-2008

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1 Department of Mathematics, IIT Bombay End-Semester Examination, MA 105 Autumn-2008 Code: C-031 Date and time: 17 Nov, 2008, 9:30 A.M. - 12:30 P.M. Maximum Marks: 45 Important Instructions: 1. The question paper has two parts: I and II. Part I of the paper contains 25 multiple choice objective questions. For each question three choices A, B, C are given. For the questions marked with *, multiple correct/incorrect choices are possible. For all other questions, only one of the given choices is correct. Answers for all questions of part-i must be entered in the separate answer-sheet provided for this part. You should not only mark the correct choice by ( ) but also mark other choices by ( ). For example, if you think for a question, A is correct and B and C are wrong, then write Question Number A B C If a blank is left unmarked for a question, that question will not be checked. Check that all the pages of your question paper have the same code. 2. Maximum time allotted for Part I is 75 minutes, after which the answer sheets for Part I will be collected back. If you finish attempting Part I early, you can start answering Part II. 3. Part II of the paper is of descriptive type for 20 marks. Attempt all the questions. 4. Make an index for the part II of the answerbook. 1

2 Part I (Maximum time: 75 minutes) (1) Let D = R 3 \ {(0, 0,z) : z > 0}. (A) There exists a gradient field on D whose curl is non-zero. (B) Every vector field on D whose curl is zero must be a gradient field. (C) There exists a vector field on D whose curl is zero but which is not a gradient field. (2) Let D be a connected open subset of R 3. (A) Every vector field on D whose divergence is zero must be a curl field. (B) If D is simply connected, then every vector field on D whose divergence is zero must be a curl field. (C) Even if D is simply connected, there may exist a vector field on D whose divergence is zero but which is not a curl field. (3) If W is the solid ball x 2 + y 2 + z 2 1, then the value of the surface integral (x 2 + y + z)ds W with W oriented by the outward normal is (A) 4 3 π.. (B) π. (C) 4π. 3 (4) If the velocity field of a fluid over a region is given by F(x,,z) = (x, 2y,z), then (A) the point (1, 1, 1) is a source. (B) the point (1, 1, 1) is a sink. (C) the point (1, 1, 1) is neither a source nor a sink. (5) Let F be a non-constant vector field in R 3. (A) If div(f) is zero everywhere, then curl(f) must be non-zero somewhere. (B) If curl(f) is zero everywhere, then div(f) must be non-zero somewhere. (C) It is possible for both div(f) and curl(f) to be zero everywhere. (6) Let f and g be real-valued functions on R. (A) If lim x f(x) = a and lim x g(x) = b, then lim x [f(x) + g(x)] = a + b. (B) If lim x f(x) = + and lim x g(x) = +, then lim x [f(x) g(x)] = 0. (C) If lim x f(x) = + and lim x g(x) = +, then lim x [f(x) + g(x)] = +. 2

3 (7) Let f : [a,b] [c,d] be a continuous function. Then there exists x 0 [a,b] such that f(x 0 ) = x 0 (A) only when a = c,b = d. (B) if a c < d b. (C) if a c,d b. (8) Let f, g, h : R R be such that f is continuous at x = 0 and both g and h are differentiable at x = 0. If f(x) = g(x) for x < 0 and f(x) = h(x) for x > 0, then (A) h(0) = g(0) and f is differentiable at x = 0. (B) f is differentiable at x = 0 only if g(x) = h(x) for all x in a neighborhood of 0. (C) f +(0) = h (0), f (0) = g (0). (9) Let f(x) = (x 2) , x R. Then (A) x = 2 is a point of local maximum for f. (B) x = 2 is a point of inflection for f, and f has a horizontal tangent at x = 2. (C) x = 2 is a point of inflection for f, and f has a vertical tangent at x = 2. (10) Let f : [a,b] R be a bounded function such that f 2 is integrable on [a,b]. (A) f must be integrable on [a,b]. (B) f must be integrable on a closed non-degenerate subinterval of [a,b]. (C) f may not be integrable on any closed non-degenerate subinterval [a, b]. (11) Let f be a real-valued function on R 2. (A) If lim (x,y) (a,b) f(x,y) = L, then lim x a f(x,b) = L. (B) If lim x a f(x,b) = L, then lim (x,y) (a,b) f(x,y) = L. (C) If lim x a f(x,b) = L = lim y b f(a,y), then lim (x,y) (a,b) f(x,y) = L. (12) Let f(0, 0) = 0, and let f(x,y) = xy2 for (x,y) (0, 0). Then x 2 + y2 (A)both f x (0, 0) and f y (0, 0) do not exist. (B) both f x (0, 0) and f y (0, 0) exist but f is not continuous at (0, 0). (C)both f x (0, 0), f y (0, 0) exist and f is continuous at (0, 0). (13) The direction of maximum increase of the function f(x,y) = x 2 y 3 at (2, 1) and of maximum decrease of f(x,y) = x 2 y 3 at ( 4, 2) are (A) parallel to each other. (B) perpendicular to each other. (C) neither parallel nor perpendicular to each other. 3

4 (14) Let R be the region bounded by the graphs of y = x, x = 0 and y = 3. Then the area of the region R is given by 3 ( 3 ) (A) dy dx. 0 x 3 ( y ) (B) dx dy ( 6 ) (C) dx. 0 x dy (15) Suppose 2 f x y exists at a point(a,b) in R2. Then (A) f x must exist at (a,b). (B) f x exists at (a,b) if 2 f is continuous at (a,b). x y (C) f may not exist at (a,b) even if 2 f is continuous at (a,b). x x y (16) The area of the plane x + y + z = 2 that lies above the part of the disk x 2 + y 2 1 in the first quadrant is given by 3 3 (A) π. (B) 2 4 π. (C) 3π. (17) The directional derivative of f(x,y,z) = axy 2 + byz + cz 2 x 3 at the point (1, 2, 1) has a maximum value of 64 in the direction of the unit normal vector k (along the positive z-axis). Then we must have (A) (a,b,c) = (6, 24, 8) (B) (a,b,c) = ( 6, 24, 8) (C) (a,b,c) = (6, 24, 8). (18) All the critical points of f(x, y) = x sin y are (A) local maxima. (B) local minima. (C) saddle points. (19) If E is a compact subset of R 2 not intersecting the line x + y = 0, and D is the image of E under the mapping u = y x,v = y + x, then (A) exp( y x E y + x ) dx dy = 1 exp( u ) du dv. 2 D v (B) 1 exp( y x ) dx dy = exp( u ) du dv. 2 E y + x D v (C) exp( y x ) dx dy = exp( u ) du dv. y + x v E D 4

5 (20) The equation of the tangent plane to the surface x 2 + 2y 2 + 3z 2 = 10 at the point (1, 3, 1) is (A) x + 2 3y + 3z = 10. (B) x + 3y + 6z = 10. (C) 2x + 3y + z = 6. (21) Let F : R 3 R 3. (A) ( F) = 0. (B) If F is conservative, then curl(f) is conservative. (C) F + curl(f) is conservative. (22) For F, G : R 3 R 3 and f : R 3 R, one has (A) (ff) = f( F) + f F. (B) (F G) = ( F) G + F ( F). (C) ( f + F) = ( F). (23) For F(x,y) = 2xy i + x 2 j and the curves C 1 : r 1 (t) = t i + t 2 j, C 2 : r 2 (t) = t i + t j, C 3 : r 3 (t) = t i + t 3 j, 0 t 1, we have (A) F dr = F dr = F dr. C 1 C 2 C 3 (B) F dr = F dr F dr. C 1 C 2 C 3 (C) F dr F dr = F dr. C 1 C 2 C 3 (24) Let U be an open neighborhood of an oriented surface S and its smooth boundary C, and let G = (x,y,z). The equality F n ds = 1 (F G) dr S 2 C holds (A) for any smooth vector field F defined on U. (B) only if F satisfies curlf = 0 on S. (C) if F is a constant vector field on U. (25) Let F(x,y) = (2y +e x ) i+(x+siny 2 ) j and let C be the unit circle oriented counterclockwise. The value of F dr is (A) π. (B) π.. (C) π/2. C 5

6 Part II (1) Let u and v be scalar fields with continuous partial derivatives in R 3. (a) Verify directly (without using (b) below) that ( u v)=0. [1] (b) Verify that u v = (u v). [1] (c) If u(x,y,z) = x 3 y 3 +z 2 and v(x,y,z) = x+y +z, evaluate the surface integral ( u v) nds, S using Stokes Theorem, where S is the hemisphere x 2 + y 2 + z 2 = 1, z 0, and n is the unit normal to S with a non-negative z-component. [3] (2) (a) Suppose a scalar field φ that does not vanish anywhere has the properties φ 2 = 4φ and (φ φ) = 10φ. φ Evaluate ds where S is the surface of the unit sphere oriented by S n the outward normal. [2] (b) The ball x 2 + y 2 + z 2 25 is intersected by the plane z = 3. The smaller portion forms a solid W bounded by a closed surface S made up of two parts, a spherical part S 1 and a planar part S 2. Using the Divergence Theorem of Gauss, compute the surface integral xz dy dz + yz dz dx + dx dy S where S is oriented by the outward normal. [3] (3) (a) Let H(x,y,z) = x 2 y i+y 2 z j+z 2 x k. Find a vector field F and a scalar field g, such that H = curl(f) + (g). [3] (b) Use Green s Theorem to find the area of one loop of the four-leafed rose defined in polar coordinates by r = 3 sin(2θ). [2] (4) Define f(0, 0) = 0, and let (a) Prove that f(x,y) = x 2 + y 2 2x 2 y 4x6 y 2 for (x,y) (0, 0). (x 4 + y 2 ) 2 4x 4 y 2 (x 4 + y 2 ) 2 for all (x,y) R 2, and conclude that f is continuous. [2] (b) Show that the restriction of f to each line through (0, 0) has a strict local minimum at (0, 0), but that (0, 0) is not a local minimum for f. (Hint:For 0 θ 2π fixed, consider g θ (t) := f(t cos θ,tsin θ),t R.) [3] 6