STUDENT NUMBER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . JULY 2013 SEAT NUMBER . . . . . . . . . . . . . . . . . . . . . . DATE . . . . . . . . . . . . . . . . . . . . . . . Engineering Mathematics 2A SUBJECT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Engineering Math 238 FACULTY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CODE . . . . . . . . . . . . . . Durban CENTER (Westville/Durban) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . UNIVERSITY OF KWAZULU-NATAL DURATION: 3 hours TOTAL MARKS: INTERNAL EXAMINER: Dr. L. Scribani EXTERNAL EXAMINER: Dr. R. Narain 100 SPECIAL INSTRUCTIONS: QUESTION NUMBER 1 2 • Answer each question in the space provided, continuing overleaf if necessary. 3 • Pencils are allowed. 4 • Calculators are not allowed. They are not needed either. • This paper contains 19 pages. Make sure that no page is missing. Page 19 has been left blank; use it as you wish. 5 6 NOTE CAREFULLY (i) (ii) (iii) (iv) Candidates must use both sides of the paper for their answers. Any rough work must be clearly marked as such and ruled off. No part of this answer book is to be torn off. Under no circumstances may a candidate retain this book or remove it from the examination room. TOTAL: WARNING: A candidate will be disqualified: (a) If he/she takes into the examination room or has in his/her possession while in the room any books, memoranda or notes, or any papers whatsoever, except such answer books, or other books or papers as shall have been supplied by the Commissioner; (b) if he/she aids or attempts to aid another candidate, or obtains or attempts to obtain aid from another candidate, or communicates or attempts to communicate with another candidate or unauthorized person. Any candidate so disqualified may be further dealt with in such manner as the University Senate may determine. Typeset in TEX MARKS INT. EXT. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 2 Math 238 Question 1 (a) Find ṡ, T̂ , 19 points dT̂ dB̂ , κ, N̂ , B̂, and τ for the curve with parametric equations ds ds sin t2 x y = − cos t2 . z t2 (Note: cos t2 and sin t2 are standard abbreviations for cos(t2 ) and sin(t2 ), respectively.) Solution: 2t cos t2 ṙ = 2t sin t2 2t √ ṡ = 2 2 t cos t2 1 sin t2 T̂ = ṙ /ṡ = √ 2 1 −2t sin t2 dT̂ 1 ˙ = T̂ /ṡ = 2t cos t2 ds 4t 0 1 2 − sin t2 N̂ = cos t2 0 κ= cos t2 − cos t2 − sin t2 1 1 2 2 B̂ = T̂ × N̂ = √ sin t × cos t = √ − sin t2 2 2 1 1 0 2t sin t2 dB̂ 1 1 ˙ −2t cos t2 = − N̂ = B̂ /ṡ = ds 4t 2 0 τ= 1 2 (8) University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 3 (b) Describe the domain of the function f (x, y) = q 36 − x2 − 4y 2 . Hence, find an equation for the level curve of this function that passes through the point (0, 2). Sketch (or describe in words) this level curve. (5) Solution: 36 − x2 − 4y 2 ≥ 0 DOMAIN : 2 x 6 + 2 y 3 ≤1 The domain is the interior of the ellipse with semi-axes of length a = 6 and b = 3, and center at (0, 0). LEVEL CURVE: Substitute x = 0 and y = 2. It follows: Hence, setting and squaring both sides, it follows p 36 − 02 − 4 · 22 = q 36 − x2 − 4y 2 = √ √ 20. 20 x2 + 4y 2 = 16, 2 x 4 + 2 y 2 = 1, which is the ellipse with semi-axes of length a = 4 and b = 2, and center at (0, 0). University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 4 (c) Show that the following limits exist and find them. (i) lim (x,y)→(1,3) y6=3x 9x2 − y 2 3x − y (ii) x3 . (x,y)→(0,0) x2 + y 2 + y 4 lim (6) Solution: (i) This function is not defined along the line y = 3x, but in every neighborhood of (1, 3) there are points where it is defined, and at those points one may write 9x2 − y 2 (3x − y)(3x + y) = = 3x + y. 3x − y 3x − y The RHS clearly tends to 6 as (x, y) approaches (1, 3). (ii) Introducing polar coordinates, one gets r3 cos3 θ x3 = . x2 + y 2 + y 4 r2 + r4 sin4 θ Now, for (x, y) 6= (0, 0) the denominator of this fraction is r2 + r4 sin4 θ ≥ r2 while the in the numerator one finds |r3 cos3 θ| ≤ |r3 |. Therefore, 0≤ r3 cos3 θ r3 ≤ ; 4 r2 r2 + r4 sin θ the RHS tends to zero, and so the limit is zero (by the “sandwich theorem”). University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 Question 2 5 13 points (a) The sphere A described by the equation x2 + y 2 − 2y + z 2 − z = 0, √ and the sphere B with center at the origin and radius equal to 3, intersect each other along a circle γ. Write the equation for B and find the equations of the tangent line to γ at the point P (1, 1, 1). (6) Solution: The equation for B is x2 + y 2 + z 2 = 3. One needs normal vectors (not necessarily unit) to A and B at P . Writing 2x 2 2 2 ∇ (x + y − 2y + z − z) = 2y − 2 2z − 1 and 2x ∇ (x2 + y 2 + z 2 ) = 2y , 2z and substituting x = 1, y = 1, z = 1, one finds NA 2 = 0, 1 2 NB = 2 . 2 These are the required vectors, normal to A and B respectively. Hence −1 1 2 T = 0 × 1 = 1 . 2 1 1 is a vector in the direction of the tangent line to γ at P . The equations for the tangent may be written x = 1 − t or y =1+t z = 1 + 2t, x−1 y−1 z−1 = = . −1 1 2 Either form of the solution is acceptable. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 6 (b) A function z = z(x, y) is given by the implicit equation y 2 − xz + z 3 = 9. Find the partial derivatives zx and zy at the point where x = 0, y = 1 and z = 2. (4) (c) Hence find the directional derivative of z(x, y) at P (0, 1) in the direction of O(0, 0). (3) Solution: (b) The independent variables are x and y. Differentiating with respect to x and y, respectively, treating the other independent variable as a constant, one gets: −z − x zx + 3z 2 zx = 0 2y − x zy + 3z 2 zy = 0 Substituting x = 0, y = 1 and z = 2 one gets immediately: zx = 1 2 = , 3·4 6 zy = (c) The gradient of z at P (0, 1) is ∇z = " PO = " = " 1 6 − 1 6 # −2 1 =− . 3·4 6 . The vector in the direction from P to O is # 0 , −1 which is also a unit vector. Therefore, h Dv z(x, y) i @P (0,1) 1 6 − 1 6 # " · # 0 1 =+ . −1 6 University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 7 Math 238 Question 3 18 points (a) Five variables x, y, z, u and v in R5 are related by the equations x − 2y + u − v = 3 y + z + u3 + v 3 = 2. Find, in the usual notation, and v = 1. ∂u ∂x and yz ∂u ∂z xy at the point where x = 1, y = −1, z = 1, u = 1 Solution: (i) Differentiate with respect to x, treating y and z as constant, and v as v(x, y, z): 1 + ux − vx = 0, 3u2 ux + 3v 2 vx = 0. Substituting x = 1, y = −1, z = 1, u = 1 and v = 1, it follows: 1 + ux − v x = 0 ux + vx = 0. Solving this system by any method, one gets 1 ux = − . 2 (ii) Differentiate with respect to z, treating x and y as constant, and v as v(x, y, z): uz − v z = 0 1 + 3u2 uz + 3v 2 vz = 0. Substituting x = 1, y = −1, z = 1, u = 1 and v = 1, it follows: uz − v z = 0 1 + 3uz + 3vz = 0. Solving this system by any method, one gets 1 uz = − . 6 (6) University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 8 (b) Find the Jacobian J= of the transformation ∂(x, y, z) ∂(u, v, w) x = u + v + w2 y = u2 + v + w z = u + v2 + w (5) Solution: Routine calculations yield 1 ∂(x, y, z) = 2u ∂(u, v, w) 1 1 1 2v 2w 1 = 1 = 1 · (1 − 2v) − 1 · (2u − 1) + 2w · (4uv − 1) = = 2 − 2(u + v + w) + 8 uvw. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 9 (c) Apply the method of Lagrange’s multipliers to find the minimum value of f (x, y, z) = x2 + y 2 + z 2 subject to the constraints 2x + z = 5 and 2y + z = 1. (7) x2 Solution: Consider the Lagrangian function L(x, y) = + Straight differentiation yields 2x − 2λ = 0 2y − 2µ = 0 2z − λ − µ = 0. y2 + z2 − λ(2x + z − 1) − µ(2y + z − 1). These equations, together with the constraints, form a system of 5 linear equations with 5 unknowns. However, the first couple of equations yield immediately x=λ and y = µ. The third equation then becomes 2z − x − y = 0. The problem has been reduced to the 3 x 3 linear system −x − y + 2z = 0 2y + z = 1 2x + z = 5 One could at this point use the standard methods of first-year algebra, but this system is so simple that one may as well substitute y = (1 − z)/2 from the second equation, and x = (5 − z)/2 from the third, into the first, and so get immediately z = 1. It then follows x = 2, y = 0. So, finally, fmin = 22 + 02 + 12 = 5. Note: The students are not required to prove that f = 5 is a minimum rather than a maximum. However, this is easy to see geometrically, because the two constraints determine a line in space, and obviously fmin is the minimum distance from the origin of this line. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 10 Math 238 Question 4 15 points (a) Find ZZ T 2y ex dx dy, x2 where T is the triangle in the xy plane with vertices at O(0, 0), A(1, 1) and B(1, 2). Choose the order of integration you deem appropriate. (5) Solution: There is only one way this problem can be done using y y=2x elementary calculus, and that is by integrating in dy first. Routine calculations give: Z = 1 0 Z ex x2 1 0 Z 2x 2y dy dx = x Z 1 0 i ex h 2 2 dx = 4x − x x2 h = 3 ex i1 0 Z ex h 2 i2x y dx = x x2 1 0 2 ex · 3x2 dx = x2 = 3(e − 1). y=x T 1 0 1 x University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A ZZ (b) Find S 11 Math 238 ex y dy dx where S is the region of the xy plane bounded above by the line y = 1 and sin2 x below by the graph of y = cos x, between the points x = 0 and x = 2π. (5) Solution: One gets immediately: ZZ S = Z 2π 0 ex y dy dx = sin2 x Z 2π 0 ex h 1 2 i 1 dx = y cos x sin2 x 2 = 1 2 Z = 2π ex sin2 x 1 2 Z 2π 0 Z e2π − 1 . 2 y dy dx = cos x ex (1 − cos2 x) dx = sin2 x ex dx = 0 1 y 1 S 0 2π x University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A (c) Find the area of the region inside one loop (pictured on the right) of the curve described in polar coordinates by the equation r = sin 6θ 12 Math 238 y (5) x Solution: The Jacobian for polar coordinates is J =r Observe that r = 0 when θ = 0. The values of θ for which r = 0 are obtained by solving the equation sin 6θ = 0, which gives θ = kπ/6, k = 1, 2, . . .. The first solution, θ = π/6, corresponds to the end of the first loop. It follows immediately: A= = 1 2 Z Z 0 π/6 0 π/6 Z sin 6θ r dr dθ = 0 sin2 6θ dθ = 1 4 Z π/6 0 Z 0 π/6 h isin 6θ 1 2 r 2 0 = (1 − cos 12θ) dθ = π . 24 University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 13 Math 238 Question 5 20 points (a) Calculate x2 y2 ZZZ z2 z 2 dx dy dz, S a2 . where S is the solid sphere + + ≤ You may do the calculation in spherical, cylindrical or cartesian coordinates (your choice). If you choose spherical or cylindrical, you may state without proof the appropriate Jacobian. (5) Solution: In spherical coordinates (the best choice) one gets immediately J = ρ2 sin φ, and hence ZZZ z 2 dx dy dz = S = 2π Z π Z 2π 0 cos2 φ sin φ dφ 0 = 2π Z π 0 Z 1 Z a ρ2 cos2 φ ρ2 sin φ dρ dφ dθ = 0 ρ4 dρ = 2π 0 h 1 + 1 i h a5 i 3 5 = h 1 3 cos3 φ 4πa5 . 15 iπ h 1 0 5 ρ5 ia 0 = University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 14 Math 238 (b) Let F = (2x + y) ı̂ + x ̂ + 2z k̂ and γ be the path parametrized by the equations y = t3 x=t z = t2 [0 ≤ t ≤ 1], leading from O(0, 0, 0) to B(1, 1, 1) . Calculate the value of (i) As a line integral; Z γ F · dr (5) (ii) Showing that F is conservative, and finding a scalar function φ(x, y, z) such that F = ∇φ. Solution: (i) Let t r = t3 t2 Therefore 2t + t3 F = t . 2t2 1 ṙ = 3t2 , 2t =⇒ (4) F · dr = (2t + t3 + 3t3 + 4t3 ) = (2t + 8t3 ) dt; It follows that Z C F · dr = Z 1 (2t + 8t3 ) dt = 3. 0 (ii) 0 2x + y ∂/∂x ∇ × F = ∂/∂y × x = 0 . 0 2z ∂/∂z Therefore F is conservative. To find the potential φ, integrate Z (2x + y) dx = x2 + yx + k(y, z). Differentiate the RHS with respect to y and compare the result with F2 : x+ ∂k =x ∂y =⇒ k(y, z) = ℓ(z) and so φ = x2 + yx + ℓ(z). Differentiate the RHS with respect to z and compare the result with F3 : ℓ′ (z) = 2z ℓ(z) = z 2 . =⇒ So finally, up to a constant φ(x, y, z) = x2 + yx + z 2 and every line integral from the origin to B(1, 1, 1) yields Z γ h F · dr = x2 + yx + z 2 iB(1,1,1) O(0,0,0) = 3. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 15 Math 238 z (c) Find the surface area of the portion of the helicoid parametrized by the equations x = u cos v y = u sin v z = v, z =2π pictured on the right, lying between the planes z = 0 and z = 2π, and inside the cylinder x2 + y 2 ≤ 1. (6) y Solution: One finds immediately: x cos v ru = sin v , 0 −u sin v rv = u cos v , 1 sin v ru × rv = − cos v . u The orientation of ru × rv is irrelevant in this context. It follows that p |ru × rv | = Therefore, A= Z 2π 0 Substituting u = sinh k one obtains A = 2π h Z Z 0 1p arsinh 1 u2 + 1 du dv = 2π cosh2 dk = π 0 = π k + 12 sinh 2k u2 + 1. iarsinh 1 Z h Z 0 1p u2 + 1 du. arsinh 1 (1 + cosh 2k) dk = 0 iarsinh 1 = = π k + sinh k cosh k 0 0 √ = π(arsinh 1 + 2 ). √ Note: The students may prefer to write ln(1 + 2) instead of arsinh 1. Both forms of the solution are acceptable. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 16 Math 238 Question 6 15 points (a) Apply Green’s theorem in the plane to find the area enclosed by one loop of the line represented parametrically by the equations x = sin t y = t sin t, where t ranges from 0 to π. y (5) Solution: By Green’s theorem, it may be shown that the area inside a loop is A= 1 2 I (x dy − y dx). In this problem, one finds: x x = sin t y = t sin t dx = cos t dt, dy = (sin t + t cos t) dt. It follows immediately that A= 1 2 Z π 0 [sin t(sin t + t cos t) − t sin t cos t] dt = = 1 2 Z π 0 sin2 t dt = π . 4 University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A 17 Math 238 (b) Let ZZ F · dS S z z =9 be the total outward flux of the field F = (6x + y) ı̂ − (x + z) ̂ − 5z k̂ S across the surface S of the region (pictured on the right) bounded above by the paraboloid z = 9 − (x2 + y 2 ) and below the plane z = 0. Without calculating this integral, apply the divergence theorem to transform it into an appropriate triple integral, and then evaluate such integral. (5) x y Solution: The divergence of F is ∂(6x + y) ∂(x + z) ∂(−5z) + + = 6 − 5 = 1. ∂x ∂y ∂z ∇·F = By the divergence theorem, it follows that ZZ S F · dS = ZZZ 1 dx dy dz, B where B is the portion of the paraboloid above the xy plane. (This is simply the volume of B.) In order to use p polar coordinates, one must find the base radius, i.e., solve the equation 9 = x2 + y 2 , which yields x2 + y 2 = 3. So, in polar coordinates r ranges from 0 to 3. The RHS becomes Z Z Z 2π 0 3 0 (9 − r2 ) r dr dθ = 2π 1 9 = 2π r2 − r4 2 4 = 3 0 3 0 (9r − r3 ) dr = 81 81 = = 2π − 2 4 81π . 2 University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A z (c) Applying Stokes’ theorem, transform the line integral I Q 18 Math 238 B F · dr, where C F = (sin x − z) ı̂ + cos y ̂ + (tan z + x) k̂ and Q is the square path ABCD pictured on the right, with vertices at A(0, 1, 0), B(0, 1, 1), C(1, 1, 1) and D = D(1, 1, 0), into an appropriate surface integral. Hence, evaluate such integral. (5) A x y D Solution: Calculate the curl of F : 0 sin x − z ∂/∂x = −2 . cos y ∇ × F = ∂/∂y × 0 tan z + x ∂/∂z Observe that Q is parallel to the zx plane, hence obviously any normal vector to Q is parallel to ̂. As Q is covered counterclockwise as seen from the right, one must take N̂ = +̂. To make use of Stokes’ theorem, one must calculate the flux of ∇ × F through the whole square. Observing that 0 0 ∇ × F · N̂ = −2 · 1 = −2, 0 0 one gets immediately: I Q F · dr = Z 0 1Z 1 0 (−2) dx dz = −2. University of KwaZulu-Natal Examinations: 27 June 2013 Engineering Mathematics 2A Math 238 19