THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK. Summer Examination 2009.

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1 OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK Summer Examination 2009 First Engineering MA008 Calculus and Linear Algebra for Engineers Prof. J.K. Langley Prof. J. Berndt Prof. M. Stynes Dr. S.J. Wills Answer Question and any other two questions from Section A. Answer Question 5 and any other two questions from Section B. Section A and Section B carry equal marks. Use separate answer books for each section. Time allowed: Three hours. Marks may be lost if not all your work is clearly shown or if you do not indicate where a calculator has been used. Page of 2

2 MA008 Summer 2009 Section A. (a) Solve the initial-value problem This question is compulsory xy + ( + x)y = e x for x >, y() = 0. Express your answer in the form y = f(x), simplifying it as much as possible. (b) (i) Determine whether the sequence a n = ( ) n n n + 2 for n =, 2,... converges; justify your answer. (ii) Use the result of part (i) and a test from the notes to determine whether the series ( ) n n n + 2 converges or diverges. n= (c) Find the interval of convergence of the series 2 n x n. n! n=0 2. (a) Evaluate the following limit without using l Hôpital s rule: (Hint: rationalize the numerator.) 3 x + 5 lim. x 4 x 4 (b) Apply the intermediate value theorem to find an interval in which the equation x 2 + x = has a root. (c) Find the derivative of y = 4 sinx 2x + cosx. Simplify your answer as much as possible. (d) Given that z = cos 2 (θ 3 ), find dz/dθ. Evaluate this derivative at θ = 3 π. (e) Use differentiation to find the absolute maximum and absolute minimum values of g(t) = 2t 3 + 3t 2 2t + 4 on the interval [ 3, 0], and state where these values are attained. Page 2 of 2

3 MA008 Summer (a) On the following graph of y = f(x) for.3 x.3, the five points marked with an asterisk have x-coordinates, 0.707, 0, 0.707,. Write down the intervals where f is (i) increasing (ii) decreasing (iii) concave up (iv) concave down. Then write down the x-coordinates of the point(s), if any, where f has a local maximum, local minimum and point of inflection (b) Use implicit differentiation to find dy/dx, given that ln(xy) + x + y = 2. (c) Draw a rough sketch of the region bounded by the curve y = e x, the coordinate axes, and the line x = 2, then find the area of this region. (d) Find the derivative of the function g(x) = x sinh x x 2 +, simplifying your answer as much as possible. (Hint: the log tables give the derivative of sinh x.) (e) Evaluate the integral 3 dx (x + 2) x 2 + 4x + 3 by completing the square, making a substitution, then consulting the log tables. Page 3 of 2

4 MA008 Summer (a) Evaluate the integral x sec 2 xdx using integration by parts and the log tables. (b) Compute the partial fraction expansion of the function but do not evaluate any integrals. 4x 2 (x )(x 2) 2 (c) The trapezoidal rule is, in the usual notation, b a f(x) dx T n = x 2 [f(x 0) + 2f(x ) + 2f(x 2 ) + + 2f(x n ) + f(x n )]. The error between this formula and the exact value of the integral satisfies the inequality b (b a)3 f(x) dx T n max 2n f (x). 2 x [a,b] a If the trapezoidal rule is used to approximate 2 (/x) dx, then use the above inequality to determine how large n should be to guarantee that the approximation is accurate to within dx (d) Evaluate the improper integral (4 x). 2 Page 4 of 2

5 MA008 Summer 2009 Section B This question is compulsory 5. (a) Find the angle between the vectors a = (, 4, 5) and b = (0, 3, 2). (b) Find the point of intersection that the line through the points P = (,, 2) and P 2 = (3, 0, ) has with the plane x 2y + z = 7. 3 (c) Compute the rank of the matrix (d) Compute the inverse of the matrix (e) Find the equation of the tangent plane to z = f(x, y) = xy 2 + 3xy at (x, y) = (, 3). (f) Use the chain rule to compute g v where g(u, v) = f(x, y) for f(x, y) = 4x2 y 3, and x = u 3 v sin u, y = 5u 2. (g) Let u(x, y) and v(x, y) be the real-valued functions defined by u(x, y) = Re(z 2 + 2iz), v(x, y) = Im(z 2 + 2iz), where z = x + iy. Determine u and v explicitly, and verify that they satisfy the Cauchy-Riemann equations everywhere in the complex plane. (h) Compute ( ) /4 and ( + i) 5. (i) Solve the boundary value problem d 2 x dt + 8dx + 6x = 0, x(0) =, x() = 3. 2 dt (j) Find a particular solution to the second order ODE d 2 y dx + 4dy 2y = 5. 2 dx Page 5 of 2

6 MA008 Summer Find the values of p for which the following system has (i) no solution, (ii) a unique solution, (iii) infinitely many solutions (i.e. involving parameter(s)). Give the solution(s) in cases (ii) and (iii). x + 2x 2 + 2x 3 = x 2 + px 3 = x + x 2 + px 3 = Find the eigenvalues and eigenvectors of the matrix Locate and classify the critical points of the function f(x, y) = 2x 2 y 3 2xy. 9. (a) Determine the radius of convergence of the following power series: (i) (z + i) n 2 n n=0 n! (ii) (z 2) n (n + 3) 2 n=0 n + 4i (b) Solve the equation e z = e z. 0. Find the general solution to the following second order ODEs: (i) d2 x dt 2 + x = cost + 2 (ii) x 3 d2 y dy + 3x2 dx2 dx = x4 x by writing v = dy and solving a first order equation dx for v. Page 6 of 2

7 MA008 Summer 2009 Solutions to Section A In the solutions that follow, here and there you ll find a Note. This is not part of the answer expected of the student; it is a comment added here to give advice or to warn you about the most common mistakes that students made on the examination.. (a) Solve the initial-value problem xy + ( + x)y = e x for x >, y() = 0. Express your answer in the form y = f(x), simplifying it as much as possible. Answer: First, divide by x to get the ODE into the correct form: y + + x x y = e x x. () In the notation of the class notes, here P(x) = ( + x)/x. Thus ( ) + x P(x) dx = x dx = x + dx = ln x +x = ln x+x because x >, so e P(x) dx = e lnx+x = e ln x e x = xe x. Multiply () by the integrating factor xe x : ( xe x y + + x ) x y = xe x e x x, i.e., We can now integrate: yxe x = d dx (yxex ) =. dx = x + C for some constant C. In this equation use the fact that y() = 0: we get 0 = + C, i.e., C =. Thus yxe x = x, and consequently y = x ( = e x ). xe x x (b) (i) Determine whether the sequence a n = ( ) n n n + 2 for n =, 2,... converges; justify your answer. n Answer: As n, n + 2 = + 2 =. Hence the sequence + 0 n {a n } approaches the sequence,,,,,,... which is divergent because it does not approach a single value. Consequently the sequence {a n } is divergent. Page 7 of 2

8 Qu. continued... MA008 Summer 2009 (ii) Use the result of part (i) and a test from the notes to determine whether the series ( ) n n n + 2 n= converges or diverges. Answer: The series is a n. By part (i) we do not have lim n a n = 0, so the test for divergence tells us that this series is divergent. (Note: the alternating series test is inconclusive for this series recall that in any case this test can never be used to prove divergence and the comparison test cannot be used because some terms of the series are negative; the ratio test is also inconclusive.) (c) Find the interval of convergence of the series 2 n x n. n! n=0 Answer: As always for a power series, use the ratio test. lim a n+ n a n = lim 2 n+ x n+ n! n (n + )! 2 n x n = lim 2x n n + = 0 for all x. As the outcome of the ratio test is less than for all x, the series is (absolutely) convergent for all x, so the interval of convergence is (, ). 2. (a) Evaluate the following limit without using l Hôpital s rule: (Hint: rationalize the numerator.) Answer: 3 x + 5 lim. x 4 x 4 3 x x + 5 lim = lim x 4 x 4 x 4 x 4 9 x 5 = lim x x x + 5 (x 4)(3 + x + 5) = lim x x + 5 = 6. (b) Apply the intermediate value theorem to find an interval in which the equation x 2 + x = has a root. Answer: set f(x) = x 2 + x (i.e., move everything to one side in the given equation). Then f(0) = < 0, f() = > 0. As f is continuous and changes sign, by the intermediate value theorem f(x) = 0 for some x (0, ). That is, x 2 + x = for some x (0, ). Page 8 of 2

9 Qu. 2 continued... MA008 Summer 2009 (c) Find the derivative of y = 4 sinx 2x + cosx. Simplify your answer as much as possible. Answer: dy dx (2x + cosx)(4 cosx) (4 sin x)(2 sin x) = (2x + cosx) 2 8x cosx 8 sin x + 4 = (2x + cosx) 2 after multiplying out the numerator then using sin 2 x + cos 2 x =. (Note: after using the quotient rule, always multiply out the numerator but don t waste your time multiplying out the denominator.) (d) Given that z = cos 2 (θ 3 ), find dz/dθ. Evaluate this derivative at θ = 3 π. Answer: By the chain rule, dz dθ = [ 2 cos(θ 3 ) ][ sin(θ 3 ) ] (3θ 2 ) = 6θ 2 (cosθ 3 )(sin θ 3 ). When θ = 3 π, this formula gives dz dθ = 6π2/3 (cosπ)(sin π) = 6π 2/3 ( )(0) = 0. (e) Use differentiation to find the absolute maximum and absolute minimum values of g(t) = 2t 3 + 3t 2 2t + 4 on the interval [ 3, 0], and state where these values are attained. Answer: To find the critical points, first set 0 = g (t) = 6t 2 + 6t 2 = 6(t 2 + t 2) = 6(t + 2)(t ), which gives t = 2 and t =. Discard t = since / [ 3, 0]. The endpoints of the interval [ 3, 0] are also critical points. Thus the critical points are t = 2, 3, 0. As g is continuous on a closed interval it must have an absolute max and absolute min. To find the absolute max and min, evaluate g at the critical points: g( 2) = 24, g( 3) = 3, g(0) = 4. Choosing the largest and smallest values obtained, we see that the absolute max is 24 at t = 2 and the absolute min is 4 at t = 0. (Note: examining the signs of g or g to see where the functions is increasing or decreasing or has a max/min will not tell you what is an absolute max or min; furthermore, the second derivative test cannot be used to test endpoints of intervals. In other words, the method described above is the only way to do this problem.) 3. (a) On the following graph of y = f(x) for.3 x.3, the five points marked with an asterisk have x-coordinates, 0.707, 0, 0.707,. Write down the intervals where f is (i) increasing (ii) decreasing (iii) concave up (iv) concave down. Then write down the x-coordinates of the point(s), if any, where f has a local maximum, local minimum and point of inflection. Page 9 of 2

10 Qu. 3 continued... MA008 Summer Answer: increasing on [.3, ], [,.3] decreasing on [, ] concave up on [ 0.707, 0], [0.707,.3] concave down on [.3, 0.707], [0, 0.707] local max at,.3. local min at.3,. points of inflection at 0.707, 0, (b) Use implicit differentiation to find dy/dx, given that ln(xy) + x + y = 2. Answer: You can differentiate this as it stands by using the chain rule, but it s easier if you simplify it first using a rule of logs: lnx + ln y + x + y = 2. Then differentiating with respect to x yields x + y dy dx + + dy dx = 0. Solve for dy/dx: ( ) dy dx y + = x dy dx + y = + x y x dy + x) = y( dx x( + y). (c) Draw a rough sketch of the region bounded by the curve y = e x, the coordinate axes, and the line x = 2, then find the area of this region. Page 0 of 2

11 Qu. 3 continued... MA008 Summer 2009 Answer: Sketch the region yourself. (Note: x = 2 is a vertical line. The region is bounded on top by y = e x, on the right by x = 2, on the bottom by the x-axis, on the left by the y-axis.) Then the desired area is 2 (d) Find the derivative of the function 0 e x dx = e x 2 x=0 = e2 e 0 = e 2. g(x) = x sinh x x 2 +, simplifying your answer as much as possible. (Hint: the log tables give the derivative of sinh x.) Answer: Using the product rule (and the log tables) to differentiate the first term, and the chain rule for the second term, g (x) = x x2 + + sinh x 2 (x2 + ) /2 (2x) = sinh x because the first and last terms cancel. (e) Evaluate the integral 3 dx (x + 2) x 2 + 4x + 3 by completing the square, making a substitution, then consulting the log tables. Answer: Completing the square, x 2 + 4x + 3 = (x + 2) = (x + 2) 2. 3 dx Hence the integral is (x + 2). Make the substitution u = x+2, (x + 2) 2 so du/dx =, i.e., du = dx. The integral becomes 3 du u u 2 = 3 sec u + C = 3 sec (x + 2) + C, where we used the log tables to evaluate the integral. (Note: don t try to skip the step let u = x+2 and jump directly to the answer; this shortcut can often go wrong.) 4. (a) Evaluate the integral x sec 2 xdx using integration by parts and the log tables. Answer: Let u = x, dv/dx = sec 2 x. Then du/dx =, v = tan x. Hence x sec 2 xd = uv dx = uv vu dx = x tan x tan xdx = x tan x ln sec x + C. Page of 2

12 Qu. 4 continued... MA008 Summer 2009 (b) Compute the partial fraction expansion of the function but do not evaluate any integrals. Answer: 4x 2 (x )(x 2) 2 4x 2 (x )(x 2) = A 2 x + B x 2 + C (x 2) 2 so 4x 2 = A(x 2) 2 + B(x )(x 2) + C(x ). Put x = : this gives 4 = A( ) 2, i.e., A = 4. Put x = 2: this gives 6 = C(), i.e., C = 6. Put x = 0: this gives 0 = 4A + 2B C, i.e., B = (C 4A)/2 = 0. Thus the answer is 4 x + 6 (x 2). 2 (c) The trapezoidal rule is, in the usual notation, b a f(x) dx T n = x 2 [f(x 0) + 2f(x ) + 2f(x 2 ) + + 2f(x n ) + f(x n )]. The error between this formula and the exact value of the integral satisfies the inequality b (b a)3 f(x) dx T n max 2n f (x). 2 x [a,b] a If the trapezoidal rule is used to approximate 2 (/x) dx, then use the above inequality to determine how large n should be to guarantee that the approximation is accurate to within Answer: f(x) = /x = x, so f (x) = x 2 and f (x) = 2x 3. Thus max x [a,b] f (x) = max x [,2] 2x 3 = 2() 3 = 2 and We want n = 3. (b a) 3 2n 2 ( 0)3 max = (2) = x [a,b] 2n 2 6n 2. 6n , i.e., n2 000/6 = 66.7, i.e., n 2.9. The answer is (d) Evaluate the improper integral dx (4 x) 2. dx Answer: (4 x) = lim dx 2 a a (4 x) 2. Let u = 4 x so du/dx =. Change the limits of integration: when x =, u = 4 = 3 and when x = a, u = 4 a. Then lim a a dx (4 x) 2 = lim a = lim a 3 4 a du u 2 [ 3 4 a = lim ] = 3 0 = 3. a u 3 u=4 a Page 2 of 2 End of Exam