Faculty of Engineering, Mathematics and Science School of Mathematics

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1 Faculty of Engineering, Mathematics and Science School of Mathematics SF Engineers SF MSISS SF MEMS MAE: Engineering Mathematics IV Trinity Term 18 May,??? Sports Centre??? ??? Prof. Sergey Frolov Instructions to Candidates: ATTEMPT QUESTION 1 and THREE OTHER QUESTIONS Materials Permitted for this Eamination: Formulae and Tables are available from the invigilators, if required. Non-programmable calculators are permitted for this eamination, please indicate the make and model of your calculator on each answer book used. You may not start this eamination until you are instructed to do so by the Invigilator. Page 1 of 14

2 Each question is worth 5 marks. 1. Consider the following initial value problem y + π y = 3π ut 4 4πδt 7, y = 3, y = π. a 16 marks. Solve the initial value problem by the Laplace transform. b 3 marks. Write the following function in a piece-wise form, and plot it 3 ut 3 ut 5 + 3ut 7. c 6 marks. Write the following function in a piece-wise form, and plot it yt = cos πt + ut 41 cos πt + ut 7 sin πt. Show the details of your work. Solution : a We denote Y s = Ly, and then using the formulae Ly = s Y s sy y, Lut a = e as s, Lδt a = e as, we get the algebraic equation s + π Y s = 3π e 4s s 4πe 7s + 3s π. Solving the equation for Y, we get 5 marks. Y s = 3s s + π Then we represent π s + π + 3π e 4s s s + π 4πe 7s s + π. marks. π ss + π = 1 s s s + π. Finally we use the formulae of the inverse Laplace transform 1 s ω L 1 = 1, L 1 = cos ωt, L 1 = sin ωt, s s + ω s + ω L 1 e as F s = ft aut a, Page of 14 1

3 to get 6 marks. yt = 3 cosπt sinπt + 3ut 41 cosπt 4 4ut 7 sinπt 7 = 3 cosπt sinπt + 3ut 41 cosπt + 4ut 7 sinπt, 3 cos πt sin πt if < t < 4 3 marks. yt = 3 sin πt if 4 < t < sin πt if t > 7 b 3 marks.. The plot of the function is shown below c 6 marks.. The plot of the solution is shown on the left, and the plot of the given function is shown on the right Page 3 of 14

4 . Consider the function f, y, z = y z 3 y 6 + siny 3z, and the point P = 1, 3,. a 1 mark. Calculate f1, 3,. b 7 marks. Calculate f 1, 3,. c 7 marks. Calculate f y 1, 3,. d 7 marks. Calculate f z 1, 3,. e marks. Find unit vectors in the directions in which f increases and decreases most rapidly at the point P. f 1 mark. Find the rate of change of f at the point P in these directions. Show the details of your work. Solution: a-d f increases most rapidly in the direction of its gradient, so we compute 1 mark. f1, 3, = sin 3 3 = 3, P 7 marks. f P = 3, 7 marks. f y P = 5, 7 marks. f z P = 1. Thus, the gradient and its magnitude are equal to fp = 3, 5, 1, fp = e marks. Therefore, the unit vector in the direction of the gradient is u = 3 35, 5, 1.57,.845,.169. f decreases most rapidly in the direction opposite to its gradient, so the unit vector is v = 3 35, 5, 1.57,.845,.169. Page 4 of 14

5 f 1 mark. The rate of change of f at P in the direction of u is equal to 35 fp =.958, and the rate of change of f at P in the direction of v is equal to 35 fp = a Consider the surface z = f, y = ln i. 1 mark. Calculate z f1, 3. ii. 8 marks. Calculate f 1, 3. iii. 8 marks. Calculate f y 1, y 3 3 y + 1 iv. 1 marks. Find an equation for the tangent plane to the surface at the point P = 1, 3, z where z = f1, 3.. b Consider the plane z = g, y = 6 3y i. marks. Find points of intersection of the plane with the -, y- and z-aes. ii. 1 mark. Sketch the plane, and show the point P = 1, 1, z on it where z = g1, 1. iii. 4 marks. Find parametric equations for the normal line to the plane at the point P = 1, 1, z. Show the details of your work. Solution: ai-aiv. We first find at the point p = 1, 3 and then simplify z = ln 1 mark. y 3 y + 1 p = 7, z =, 1 3 y 3 3 y + 1 Page 5 of 14 = 1 3 lny 3 y + 1 ln 3.

6 Then, we compute the partial derivatives at p 8 marks. 8 marks. The tangent plane equation is given by z p = 1 9. y z p = mark. z = 1 9 bi. marks. 3,,,,,,,, y 3 = y. bii. 1 mark. The tangent plane is the one through the points in b. biii. We have z =, z y = 3, and therefore the normal line to the plane is given by 4 marks. r = i + j + k + t i + 3j + k. 4. Consider a plain lamina in the y-plane bounded by the curves y = and y = with the density a marks. Sketch the lamina. δ, y = + y e y cos π, b 4 marks. Show that the density of the lamina is positive. c 6 marks. Find lim δ, 1 d 13 marks. Use double integration to find the mass of the lamina. Show the details of your work. Solution: a marks. Below is the lamina. Page 6 of 14

7 b 4 marks. For all points of the lamina and y. Since for nonnegative e cos π we get δ, y = + y e y cos π = + y e cos π, as required. c 6 marks. We have δ, 1 = + e cos π = + e cos π, Thus lim + e cos π = + lim e cos π = + lim 1 π = 3, where we used that e t 1 + t + t + t3 6 +, cos t 1 t + t Page 7 of 14

8 d 13 marks. The mass of the lamina is given by the following double integral + y e y cos π 6 marks. M = da R [ 1 ] + y e y cos π = dy d = y + e cos π y y= d = d + 1 = 3 3/ = y= e 1 + cos π d. The remaining integral is computed by using integration by parts e cos π1 d e 1 cos π1 d 3 marks. e 1 d = d e 1 + e 1 d = e e 1 d = 1 e 1 1 = + e, and 4 marks. cos π d = d 1 π sin π 1 sin π d π = 1 π sin π 1 1 π sin π d = + 1 π cos π 1 = π. Thus the mass is equal to M = e 1 π = e 1 π Page 8 of 14

9 5. a Let the path C between the points 3π/, π/3 and π/, π be a curve formed from two line segments C 1 and C, where C 1 is joining 3π/, π/3 and π/, π/3, and C is joining π/, π/3 and π/, π. i. marks. Plot the path C, and show its orientation on the plot. ii. 5 marks. Parameterise C 1, and evaluate I 1 = 4y 5π 3 sin3 d + + 3π 3 cos 3y C 1 iii. 5 marks. Parameterise C, and evaluate I = 4y 5π 3 sin3 d + + 3π 3 cos 3y C dy. dy. iv. 1 mark. Compute the sum I = I 1 + I. b 1 mark. Show that for any integration path C the integral above depends only on the initial and terminal points 3π/, π/3 and π/, π of the path C. c 9 marks. Find a potential function φ, y. d marks. Use the Fundamental Theorem of Line Integrals to find the value of the integral above for an integration path between the initial point 3π/, π/3 and the terminal point π/, π. Does it agree with the result in aiv? Show the details of your work. Solution: a i. marks. Above is the path C. a ii. 5 marks. The line segments C 1 is parametrised as = t, y = π/3, 3π/ t π/, and one gets C 1 4y 5π 3 sin3 d + + 3π 3 cos 3y dy = 8π 3 / a iii. 5 marks. The line segments C is parametrised as = π/, y = t, π/3 t π, and one gets C 4y 5π 3 sin3 d + + 3π 3 cos 3y Page 9 of 14 dy = 7π 3 /

10 a iv. 1 mark. The sum gives 3 π b 1 mark. We have f, y = 4y 5π 3 sin3, g, y = + 3π 3 cos 3y. y f, y = 4, g, y = 4 y f, y = g, y, and therefore the integral is independent of the path. c To compute the integral we find a potential function φ, y 4 marks. To find Cy we use that 4 marks. Thus, we get φ = 4y 5π3 sin3 φ, y = y+5/3π 3 cos3+cy. φ y = + dcy dy = + 3π 3 cos 3y Cy = π 3 sin 3y + C. dcy dy 1 mark. φ, y = y π3 cos 3 + π 3 sin 3y + C. d marks. By using the formula, we obtain = 3π 3 cos 3y π/,π 3π/, π/3 4y 5π 3 sin3 d + + 3π 3 cos 3y which agrees with the result in aiv. dy = φ, y π/,π 3π/, π/3 = 3 π3, Page 1 of 14

11 Useful Formulae 1. Let rt be a vector function with values in R 3 : rt = ft i + gt j + ht k. a Its derivative is dr = df, dg, dh dt dt dt dt. b The magnitude of this vector is dr dt = df dt c The unit tangent vector is T = dr dt dr dt. + dg dt + dh dt. d The vector equation of the line tangent to the graph of rt at the point P =, y, z corresponding to t = t on the curve is Rt = r + t t v, where r = rt and v = dr dt t. e The arc length of the graph of rt between t 1 and t is L = 1 dr dt. dt f The arc length parameter s having rt as its reference point is s = t t. Let σ be a surface in R 3 : z = f, y dr du. du a The slope k of the surface in the -direction at the point, y is k = z, y. b The slope k y of the surface in the y-direction at the point, y is k y = z y, y. c The equation for the tangent plane to the surface at the point P =, y, z is z = z + k + k y y y. d Parametric equations for the normal line to the surface at P =, y, z are rt = r + t k i k y j + k, r = i + y j + z k. e The volume under the surface and over a region R in the y-plane is V = R f, y da. f The area of the portion of the surface that is above a region R in the y-plane is S = ds = 1 + z σ R + z y da. g The mass of the lamina with the density δ, y, z that is the portion of the surface that is above a region R in the y-plane is M = δ, y, z ds = 1 δ, y, z + z σ R + z y da. Page 11 of 14

12 3. The local linear approimation of the function z = f, y at the point, y is L, y = f, y + f, y + f y, y y y. 4. Let f, y, z be a function of three variables a The gradient of f is f = f, f y, f z. b f increases most rapidly in the direction of its gradient, and the rate of change of f in this direction is equal to f. c If f is smooth then its critical points satisfy f = f y = f z =. 5. Let R be a region in the y-plane bounded by the curves y = g, y = h, = a, = b, and g h for a b. Then the double integral over the region is R f, y da = [ b ] h f, ydy d. a g 6. Let R be a region in the y-plane bounded by the curves in polar coordinates r = r 1 θ, r = r θ, θ = α, θ = β and r 1 r for α θ β. Then the double integral over the region is R fr, θ da = R fr, θ rdr dθ = β α 7. Let R be a plain lamina with density δ, y. a Its mass is equal to M = R δ, y da. [ r θ r 1 θ fr, θrdr ] dθ. b The -coordinate of its centre of gravity is equal to cg = 1 M c The y-coordinate of its centre of gravity is equal to y cg = 1 M R R δ, y da. y δ, y da. 8. Let G be a simple solid whose projection onto the y-plane is a region R. G is bounded by a surface z = g, y from below and by a surface z = h, y from above. a The triple integral over the solid is G f, y, z dv = R [ h,y g,y ] f, y, z dz da. b The volume of the solid is V = G dv = R [h, y g, y] da. 9. Let G be a solid enclosed between the two surfaces in spherical coordinates r = gθ, φ, r = hθ, φ. Page 1 of 14

13 a The triple integral over the solid is G fr, θ, φ dv = π [ π ] hθ,φ fr, θ, φ gθ,φ r dr b The volume of the solid is V = G dv = π π sin φ dφ [ hθ,φ gθ,φ dθ. ] r dr sin φ dφ dθ. c The mass of the solid with the density δr, θ, φ is M = δr, θ, φ dv. G 1. Let a region R y in the y-plane be mapped to a region R uv in the uv-plane under the change of variables u = u, y, v = v, y. a The magnitude of the Jacobian of the change is u,v,y = u v u v y y. b The integral over R y is R y f, y da y = R uv f u, v, yu, v u,v,y 11. The area of the surface that etends upward from the curve = t, y = yt, a t b in the y-plane to the surface z = f, y is given by the following line integral A = z ds = b ft, yt d C a dt + dy dt dt. 1. Consider a line integral C f, y d + g, y dy, and let P = P, y P and Q = Q, y Q be the endpoints of the curve C. a The line integral is independent of the path if y f, y = g, y. b Then there is a potential function φ, y satisfying φ = f, y, φ y c and the Fundamental Theorem of Line Integrals says that = g, y, 1 da uv. C f, y d + g, y dy = Q P φ P, y P. φ φ d + dy = φ, y y Q P = φ Q, y Q 13. Let a closed curve C be oriented counterclockwise, and be the boundary of a simply connected region R in the y-plane. By Green s Theorem we have da C f, y d + g, y dy = R g,y f,y y 14. Let F, y, z = M i + N j + P k be a vector field. a If σ is the surface z = f, y, oriented by upward unit normals n, and R is the projection of σ onto the y-plane then flu = F n ds = M f N f + P da. σ R y Page 13 of 14

14 b If σ is the surface z = f, y, oriented by downward unit normals n, and R is the projection of σ onto the y-plane then flu = F n ds = M f + N f P da. σ R y c According to the Divergence Theorem the flu of F across a closed surface σ with outward orientation is flu = σ F n ds = V div F dv, M div F = + N + P. y z d If σ is an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation then, according to Stokes Theorem C F dr = curl F n ds, curl F = P N i+ M P σ y z z j+ N M k. y 15. The Laplace transform of a function ft is the function F s defined by F s = Lft = e st ftdt, ft = L 1 F s. F unction T ransf orm F unction T ransf orm e at 1 s a e at t n n! s a n+1 e at sin ωt e at sinh ωt t sin ωt ut a ω s a +ω e at cos ωt s a s a +ω ω e at s a cosh ωt s a ω s a ω ωs s t cos ωt ω s +ω s +ω e as s δt a e as 16. Let F s = Lft, then Lft aut a = e as F s; L e at df s ft = F s a ; L tft = ; L fkt = 1F s ds k k. 17. Let Y s = Ly, then Ly = sy s y, Ly = s Y s sy y. 18. Convolution. Let ft gt = t fτgt τ dτ. Then Lft gt = F sgs Page 14 of 14