Attempt QUESTIONS 1 and 2, and THREE other questions. penalised if you attempt additional questions.

Transcription

1 UNIVERSITY OF EAST ANGLIA School of Mathematics Main Series UG Examination CALCULUS AND MULTIVARIABLE CALCULUS MTHA4005Y Time allowed: 3 Hours Attempt QUESTIONS 1 and 2, and THREE other questions. penalised if you attempt additional questions. You will not be Notes are not permitted in this examination. Do not turn over until you are told to do so by the Invigilator. MTHA4005Y Module Contact: Dr Mark Cooker, MTH Copyright of the University of East Anglia Version: 1

2 (i) The complex numbers u and v are defined as follows: u = 3 + 4i ; v = 1 + 6i. Calculate the real part and the imaginary part of the following: u v ; u v ; u v. [6 marks] (ii) Now consider the complex number z = 1 2 i 3 2. First, confirm that z = e iπ/3. Hence show that z 6 = 1. (iii) Relative to an origin O of cartesian coordinates x, y, z, and their associated unit vectors i, j, k, the position vector of a general point is r = xi + yj + zk. Two planes, P 1 and P 2, are described by the following equations: P 1 : (r 3i) (i + j + 2k) = 0 (1) P 2 : r (4i j k) = 0. (2) The two planes intersect in a line L. Show that the cartesian coordinates of every point of L must satisfy the pair of equations: x + y + 2z = 3 (3) and 4x y z = 0. (4) [5 marks] (iv) Hence show that L can be described in terms of the real-valued parameter x = λ, as the vector equation r = 3j + 3k + λ(i + 9j 5k). [5 marks] MTHA4005Y Version: 1

3 (i) Evaluate each of the integrals: 1 x e x dx, 6 (ii) Find the general solution of the differential equation 3 1 dx. [6 marks] (x 1) 2 (x 2) d 2 y dx 2 5dy dx + 6y = 2e4x. [6 marks] (iii) Consider the integral I n = (1 x 3 ) n dx. By writing the integrand as 1 (1 x 3 ) n, or otherwise, derive the reduction formula: I n = x (1 x3 ) n 3n n 3n + 1 I n 1. Hence evaluate the indefinite integral I 3. [8 marks] MTHA4005Y PLEASE TURN OVER Version: 1

4 (i) Establish the first four terms of the following Maclaurin series for the function f(x) = (1 + x) 1/2 (1 + x) 1/2 = x x x3 1 < x < 1. [5 marks] Also, by making the substitution x = t 2, or otherwise, show that (1 t 2 ) 1/2 = t t t6 1 < t < 1. [1 mark] By carrying out a suitable integration of (1 t 2 ) 1/2, deduce that arcsin(θ) = θ θ θ θ7 1 < θ < 1. (ii) (a) Find the general solution of the differential equation (x + 1) dy dx = y x. (b) Use an integrating factor to find the general solution to the differential equation dy dx + y sin x = ecos x. (c) By choosing a suitable substitution, find the general solution to the following differential equation dy dx = y x + x y. [10 marks] MTHA4005Y Version: 1

5 (i) Consider the function f(x, y) = x 3 + y 3 3x 3y. (a) Calculate f x and f y. (b) Find the (x, y) coordinates of all the stationary points of f. (c) Classify each of the stationary points. [2 marks] (ii) (a) Consider the contour integral I = P (x, y)dx + Q(x, y)dy, C where P (x, y) = y 2, and Q(x, y) = x 2. The closed contour C is given by the straight line y = 0 from the origin to the point (1, 0), the circular arc x 2 + y 2 = 1 lying in the first quadrant of the (x, y) plane from (1, 0) to (0, 1), and the straight line x = 0 from (0, 1) to (0, 0). Hence, evaluate I. (b) Green s theorem in the plane states that C P (x, y)dx + Q(x, y)dy = S ( Q x P ) dxdy, y [5 marks] where S is the interior of the region bounded by the closed contour C. Verify Green s theorem in the plane for the functions and the region bounded by the contour C given in part (ii)(a). [5 marks] MTHA4005Y PLEASE TURN OVER Version: 1

6 (i) By first reversing the order of integration, evaluate the double integral ( π/2 ) π/2 sin y y dy dx. 0 x (ii) (a) If x and y are related to u and v by u = xy v = x y, evaluate, in terms of u and v, the Jacobian (x, y) (u, v). (b) S is the region in the first quadrant of the (x, y) plane bounded by the curves xy = 1, xy = 4, y = x, y = 2x. Sketch the region in the (x, y) plane and the corresponding region, S, in the first quadrant of the (u, v) plane. (c) Hence, evaluate the area of the region S. [5 marks] [3 marks] (iii) The volume of the region lying inside the sphere x 2 + y 2 + z 2 the paraboloid z = x 2 + y 2 cylindrical polar coordinates (r, θ, z) = 6 and above can be evaluated from the following triple integral in [ ( V = ) ] dz rdr dθ, where x = r cos θ and y = r sin θ. Determine the limits of integration in the above expression without changing the order of integration. [Hint: Do not evaluate the integral.] MTHA4005Y Version: 1

7 Let the position vector be r = xî + yĵ + zˆk, where (î, ĵ, ˆk) are the unit coordinate vectors in the x, y and z directions respectively. (i) (a) For a general vector field v = v 1 î + v 2 ĵ + v 3ˆk, prove that ( v) = 0. (b) State the divergence theorem, being careful to define the relationship between the surface and the volume, and also how the direction of the normal is chosen. [6 marks] (ii) Let F be the vector field defined by F = (x y)î + (y + x)ĵ + (z + 2)ˆk. Let V be the region of space inside a hemisphere, given by x 2 + y 2 + z 2 < 4 and z > 0. (a) Show that F is a constant. Hence deduce that F dv = 16π, V where V is the interior of the hemisphere, as defined above. (b) Find an expression for the unit normal ˆn to the curved surface x 2 +y 2 +z 2 = 4, z > 0 of V, where the direction of ˆn is chosen so that ˆn ˆk > 0. Hence show that on that curved surface F ˆn = 2 + z. (c) By evaluating some appropriate surface integrals, show that the divergence theorem holds for the vector field F over the volume V. [14 marks] [ Hint: For part (ii)(c), it may be convenient to describe the curved surface of the hemisphere using spherical polar coordinates (r, θ, ϕ), defined by x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ. You may assume that, in these coordinates, the surface element for a surface of constant r has area ds = r 2 sin θ dθ dϕ. ] END OF PAPER MTHA4005Y Version: 1

8 MTHA4005Y Calculus and Multi-variable Calculus Examination Feedback 2017/18 Question 1 Part (i) was well done. Common errors were not securely identifying π/3 as the argument because arctan( 3 1/2 ) is either π/3 or 2π/3 within the usual interval for argument angles, from π to π. Part (ii): Mostly this was successfully done, using the relations given. Some failed attempts lacked explanation as to WHY the vector-product (cross-product) of the normals to the two given planes is a vector that points along the line L of intersection. Question 2 Part (i): Generally done well except some people integrated by parts on the first one, or chose the wrong form of the partial fractions for the second. Also integral of e u du is not e u /u as far too many put. Part (ii): Generally done well, although a fair number guessed a far too complicated form of the RHS. Part (iii): Most spotted that it was an integration by parts problem and could do the first step, but not many spotted the algebraic manipulation to obtain the given result. There were *lots* of imaginative, and entirely non-mathematically correct, ways demonstrated of massaging the expressions into the given result... Finding I 3 was done fairly well although several people didn t find I 0 or I 1 to start the sequence. Far too many people forgot to add the arbitrary constant in the indefinite integrals in (i) and end of (iii), and some randomly added an arbitrary constant for the definite integral in (ii) Question 3 (i) The Maclaurin series was well constructed, but many lost marks for not putting in + to mark an INFINITE series. Too many did not know (or could not show) that the integral of (1 t 2 ) 1/2 is arcsin(t) + c, where the constant c was also left out, or not evaluated. Part (ii) (a) Done well by most. Just about everyone noticed it was separable and could do the partial fractions. A few approached it using a substitution y = xv(x), which I hadn t thought of doing, but this also worked well and avoided partial fractions! (b) Completed fine by most, although several got the sign wrong on integrating sin(x), some forgot to multiply both LHS and RHS with integrating factor, and many forgot to add the constant of integration. (c) Most candidates obtained the right substitution, but quite a few didn t then calculate dy/dx properly, when doing the conversion. Question 4 (i) This part was generally well done, although there were quite a number of students who did not solve the system of equations correctly in (b) and ended up finding only 2 of the 4 stationary points. 1

9 In (c) some people mixed up the criteria that need to be satisfied for saddle points with the criteria for minima and maxima. (ii) This part of the question was the most poorly attempted. There were quite a range of different attempts on this question, and indeed there are different ways of answering the problem, depending on whether one chooses to use x, or y, as the integration variables for the different parts of the contour integral. For those that proceeded to make the substitutions, many people became stuck by the integral expressions they were faced with and couldn t proceed to evaluate the final result. However, a large number of students clearly did not understand the concept of the contour integral and made very elementary errors. (iii) Attempts on this part varied greatly from correct evaluation of the double integral, to setting the integral with the incorrect limits, where many used the interval (0, 1) for both x and y coordinates, which implies an integration over a square region rather than a quarter of a disc. Others mainly got stuck in performing the integration after setting up the integral correctly. Question 5 (i) This was generally well answered, although there were the inevitable cases of people literally swapping the inner and outer integrals without working out the new limits and so turning an expression for a number into an expression for a function, despite the many attempts to warn students of this type of error in the lectures. (ii) There were a range of answers here. Although this was reasonably well attempted, one of the main errors that students made was to re-arrange the given expression for u and v to write them for example as x = u/y or y = x/v and then evaluate partial derivatives of x and y with respect to u and v from such expressions. Again explicit examples were constructed in lectures to explain to students why this is incorrect namely you need to be careful of what is being held constant when evaluating partial derivatives. This error led to incorrect expressions for the Jacobian. (iii) This was poorly attempted. Only a few students managed to get this correctly. The majority appeared to struggle in visualising the volume that is contained within the two surfaces, and so were unable to work out the correct limits of integration. Question 6 This question was only attempted by a few candidates. There was a range of answers submitted, from partial attempts to complete solutions. Part (i)(a) was standard bookwork and was generally answered very well. A few students were unable to recall the correct definitions for divergence and curl. Some students used the correct definitions, but then incorrectly thought that e.g. the first component of the curl (namely v3 z v2 z ) had no x- dependence and so made no contribution to the divergence. In fact the terms all contribute in general, but pairs of terms cancel due to the commutativity of the mixed partial derivatives (e.g. 2 v 3 z x 2 v 3 x z ). Part (i)(b) was again bookwork and answered reasonably well. Some marks were lost for not including the relationship between the surface S and the volume V (S is the closed surface enclosing the volume V ) and/or not defining ˆn (it is the unit outward pointing normal to the surface S). A small number of students mixed up the divergence theorem with Stokes theorem, at least when it came to describing the normal. 2

10 In part (ii)(a) almost everyone was able to obtain the divergence as 3. Around half of those who attempted the question then failed to realise they could evaluate the integral simply by multiplying 3 by the volume of the hemisphere. (This was surprising as this trick was emphasised in the lectures and the revision session, and was used in the problem sheet questions.) Some of those who did spot this, failed to halve the formula 4 3 πr3 for the volume of a sphere, and/or quoted it incorrectly as 4 3 πr2. Some of those who did not spot the trick were able to successfully evaluate the integral by switching to spherical polar coordinates. A small number tried to use Cartesian coordinates, but failed to obtain the correct limits. (The integral can be done this way, but it is a little fiddly to get right.) In (ii)(b), most recalled the expression ± φ/ φ for a unit normal to a surface defined by φ(x, y, z) = const, and correctly defined φ = x 2 + y 2 + z 2. Some then messed up the algebra and so failed to obtain the final answer. The key point was spotting that x 2 + y 2 + z 2 could always be replaced by 4, since we are dealing with points on the surface. Part (ii)(c) was the least well done part of the question. Very few students realised that the surface bounding the volume comprised two parts: the curved surface of the hemisphere and the flat circular base. The integral of v ˆn needed to be evaluated over both and then summed. Most students only completed the integral over the curved surface. For this integral, most were able to follow the hint and do the change of variables to spherical polars, and correctly identified the limits as 0 r < 2, 0 θ < π/2 and 0 φ < 2π. A common sign error lead some to get 16π rather than the correct answer of 24π for the curved-surface integral. 3