McGill University December Intermediate Calculus. Tuesday December 17, 2014 Time: 14:00-17:00

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1 McGill University December 214 Faculty of Science Final Examination Intermediate Calculus Math 262 Tuesday December 17, 214 Time: 14: - 17: Examiner: Dmitry Jakobson Associate Examiner: Neville Sancho Student name (last, first) Student number (McGill ID) INSTRUCTIONS 1. Please write your answers clearly in the space provided. 2. This exam is a total of 1 marks.. This is a closed book exam. 4. Translation dictionary is permitted. 5. Non-programmable calculators are permitted. This exam comprises the cover page, and 1 page of questions.

2 Math 262 Final Exam Page 2 December 17, 214 Problem 1 (1 points) Let r(t) = e t, 2t, e t, t 2π. i. Find T, N, B at the point (1,, 1). ii. Compute the length of the curve. Solution: Note when t =, r() = (1,, 1). Compute r (t) = e t, 2, e t, then Thus evaluate at t =, r (t) = (e t ) ( e t ) 2 = e t + e t. T(t) = r (t) r (t) = 1 e t + e t et, 2, e t, T() = 1 2 1, 2, 1. Next, d dt T(t) = 1 e t + e t et,, e t et e t (e t + e t ) 2 et, 2, e t 1 = (e t + e t ) 2, 2(e t e t ), 2, 2 then evaluate at t =, N(t) = d T(t) dt d T(t) = 1 e t + e 2, (e t e t ), 2, t dt N() = 1 2 2,, 2. Finally, B() = T() N() = 1 i j k = 1 2 1, 2, 1. The length of the required curve is r (t) dt = e t + e t dt = e 2π e 2π. Problem 2 (1 points) Evaluate the iterated integral I = 4 dy e x y

3 Math 262 Final Exam Page December 17, 214 Solution: We can simplify the iterated integral if we interchange the order of integration. We first express I as a double integral I = e x2 dy, where The region can also be specified as Reiterating in the opposite direction we get D D = {(x, y) y 4 and 1 2 y x 2}. D = {(x, y) x 2 and y 2x}. I = = = x e x2 y x=2 = e x2 x= = e 4 1. y=2x y= 2xe x2 e x2 dy Problem (1 points) Let R be the region x 2 + y 2 + z 2 4a 2, x 2 + y 2 a 2, where a > is a constant. Find the volume of R. Solution: The volume of R can be evaluated as Vol(R) = 1 dydz. It is more conivent to use the cylindrical coordinates, R x = r cos θ, y = r sin θ, z = z. Then R = {(r, θ, z) a r 2a, θ 2π, 4a 2 r 2 y 4a 2 r 2 }.

4 Math 262 Final Exam Page 4 December 17, 214 Thus Vol(R) = = R a = 2π dr a a a a 1 dydz rz 4a 2 r 2 rdθ 1 dz 4a 2 r 2 z= 4a 2 r 2 z= dr 4a 2 r 2 = 4π r 4a 2 r 2 dr a = 4π 1 (4a2 r 2 ) 2 = 4 πa. x=2a x=a Problem 4 (1 points) Find the maximum value of the function f(x, y, z) = xyz on the restricted domain: D = {(x, y, z) x 2 + y 2 + z 2 = r 2 (r > ), x >, y >, z > }. Solution: Note that f is smooth, positive in D, vanish on the boundary of D, therefore f must attain its maximum in D. Consider he Lagrangian function then its critical points are given by L(x, y, z, λ) = xyz + λ(x 2 + y 2 + z 2 r 2 ), = L z = L z = L z = L = yz + 2λx, = xz + 2λy, = xy + 2λz, λ = x2 + y 2 + z 2 r 2. Multiply x, y, z to the first, second and third equation respectively, and take the sum we obtain xyz = 2λ(x 2 + y 2 + z 2 ) = 2λr 2. Combine with the first equation, we obtain 2λx 2 = xyz = 2 λr2. It is obvious that λ, thus x = 1 r. Similarly, we have y = 1 r, z = 1 r. Since L has only one interior critical point, with the observation at the beginning, it is must be the maximum point of f. Thus f max = f( r, r, r) = ( 1 r) = 9 r.

5 Math 262 Final Exam Page 5 December 17, 214 Problem 5 (1 points) i. Write the Taylor series about y = of sin(y). You do not need to justify your answer. ii. Use part i. to write the Taylor series about x = of sin(x 2 ). iii. Write the Taylor series about x = of f(x) = x sin(t2 )dt. Solution: We have sin y = n= Next, we substitute y = x 2 to get sin x 2 = n= Integrating term by term, we get x sin(t 2 )dt = n= ( 1) n y 2n+1 (2n + 1)! ( 1) n x 4n+2 (2n + 1)! = y y /6 + y 5 /5! ±... = x 2 x 6 /6 + x 1 /5! ±... ( 1) n x 4n+ (4n + )(2n + 1)! = x / x 7 /42 + x 11 /(11 5!) + ±... Problem 6 (1 points) The system of equations x + yz =, xz + y = defines y and z implicitly as functions of x near the point (x, y, z) = (1, 1, 1). Find dz/ and dy/ at that point. Solution: We differentiate the two equations with respect to x, regrading y and z as functions of x: At the point (x, y, z) = (1, 1, 1), we have = d ( x + yz ) = x 2 + z dy + y dz, = d ( ) xz + y = z + x dz dy + y2. = + dy dz, = 1 + dz + dy. Adding these two equations, eliminating dz, we obtain plug in any of these two equations, we have dy = 1, dz = 2.

6 Math 262 Final Exam Page 6 December 17, 214 Problem 7 (1 points) Find all local maxima, local minima, and saddle points of the function f(x, y) = cos x + cos y in the region {(x, y) : π/2 < x < π/2, π/2 < y < π/2}. Solution: Note the function is smooth, the necessary condition for a interior point to be a local maxima, local minima, and saddle points is that the gradient vanish at there, From this, we obtain gradf = ( sin x, sin y) =. x = mπ, y = nπ, m, n =, ±1, ±2,. In the restricted region {(x, y) : π/2 < x < π/2, π/2 < y < π/2}, we have The Hessian matrix of f is Note that (x, y) {(, ), (π, ), (, π), (π, π)}. Hessf = Hessf(, ) = [ ] cos x. cos y [ ] 1 1 is negative definite, then (, ) is a local maximum; [ ] 1 Hessf(π, ) = 1 is indefinite, then (π, ) is a saddle point; Hessf(, π) = is indefinite, then (, π) is a saddle point; Hessf(π, π) = is positive definite, then (π, π) is a local minimum. Problem 8 (1 points) Evaluate the area of the ellipse x 2 /6 + y 2 /25 = 1. [ ] 1 1 [ ] 1 1

7 Math 262 Final Exam Page 7 December 17, 214 Solution: By definition, we have Area = Consider the change of variables, x 2 /6+y 2 /25 1 dy, x = 6r cos θ, y = 5r sin θ, r 1, θ 2π, then the corresponding Jacobian is defined according (x, y) (r, θ) = 6 cos θ 5 sin θ 6r sin θ 5r cos θ = r, thus Area = 1 = 6π = π. dr 1 rdr rdθ Problem 9 (1 points) Compute x dv where T denotes the tetrahedron bounded by the coordinate planes and T the plane x/2 + y/4 + z/5 = 1. Solution: The limits of integration for the tetrahedron can be chosen as follows (not that the choice is not unique; other choices are possible): x 2, y 4 2x and z 5 5x/2 5y/4. The integral is equal to 4 2x 5 5x/2 5y/4 The last expression is equal to Opening the brackets, we get xdzdy = 4 2x (5x 5x 2 5xy/4)dy. [(5x 5x 2 /2)(4 2x) (5x/8)(4 2x) 2 ] [2x 1x 2 1x 2 + 5x 1x + 1x 2 5x /2] = Integrating, we get (5x 4 /8 1x / + 5x 2 ) 2 x= = 1 8/ + 2 = 1/. (5x /2 1x 2 + 1x).

8 Math 262 Final Exam Page 8 December 17, 214 Problem 1 (1 points) Use spherical coordinates to find B (x 2 + y 2 )dv, where B is the ball {(x, y, z) : x 2 + y 2 + z 2 16}. Solution: The limits of integration are θ 2π, φ π and ρ 4. We have x 2 + y 2 = φ 2 sin 2 φ(cos 2 θ + sin 2 θ) = ρ 2 sin 2 φ. The volume element is ρ 2 sin φdρdφdθ The integral is thus equal to 4 π ρ 4 sin φdρdφdθ = 4 ρ 4 dρ dθ π sin φdφ = 2π 45 5 π sin φdφ. The last integral we can rewrite as π (1 cos2 φ) sin φdφ. Let u = cos φ, then du = sin φdφ. The integral becomes 1 1 (1 u 2 )du = 1 1 The answer is thus equal to (8192π)/15. (1 u 2 )du = (u u /) 1 u= 1 = 4/.