UNIVERSITY OF SOUTHAMPTON. A foreign language dictionary (paper version) is permitted provided it contains no notes, additions or annotations.

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1 UNIVERSITY OF SOUTHAMPTON MATH055W SEMESTER EXAMINATION 03/4 MATHEMATICS FOR ELECTRONIC & ELECTRICAL ENGINEERING Duration: 0 min Solutions Only University approved calculators may be used. A foreign language dictionary (paper version) is permitted provided it contains no notes, additions or annotations. Copyright 04 c University of Southampton Page of 7

2 MATH055W All questions similar to examples seen in weekly work, except: A3, A7, A4, which require a bit more thought. Solutions SECTION A A. ( marks) If y = sin x x 3, which of the following is dy (a) x3 sin x 3x cos x (x 3 ), (b) cos x 3x, (c) cos x x 3 dx? 3 sin x x 4, (d) 3x cos x x 3 sin x x 6, Quotient rule: f = sin x f = cos x, g = x 3 g = 3x so y = x3 cos x 3x sin x x 6 = cos x x 3 3 sin x x 4, so (c). A. ( marks) If f and g are functions which can be differentiated twice, and h(x) = f(x)g(x)), then h (x) = (a) f (x)g(x) + f(x)g (x), (b) f (x)g(x) + f (x)g (x) + f(x)g (x), (c) f (x)g (x) + f (x)g (x), (d) f (x)g (x) + f(x)g (x) + f (x)g(x), Product rule. h = f g + fg. Again: f = (f g) + (fg ) = (f g + f g ) + (f g + fg ), so (b). Copyright 04 c University of Southampton Page of 7

3 3 MATH055W A3. ( marks) If the graph of g(x) is as above, which of the following could be the graph of g (x)? (a) (b) (c) (d) Several ways to think about this. Shape cubic, expect derivative to have quadratic shape. Need g positive for x < a, then negative on ( a, a) then positive for x > a for some positive constant a. Only possible solution is (d). A4. ( marks) (a) (c) dx + x = tan ( x) + C, (b) tan (x) + C, tan( x) + C, (d) 4x + x + C, Substitute u = x, then u = x, du = dx and the integral becomes du + u = tan u + C, so (a). TURN OVER Copyright 04 c University of Southampton Page 3 of 7

4 4 MATH055W Questions A5 - A7 concern the function f(x) = x + 3 x +. A5. ( marks) The stationary point(s) of f(x) is(are) (a) (, 3) and (3, ), (b) (, 0), (c) (, ), (d) (, ) and ( 3, 6), Quotient rule: f = x(x + ) (x + 3) (x + ) = x + x 3 (x + ) = 0 when x + x 3 = (x + 3)(x ) = 0, hence x = 3,, and f( 3) = 6, f() =, so (d). A6. ( marks) The nature of the stationary point(s) of f(x) found in question A5 is as follows: (a) (, 3) is a maximum and (3, ) is a minimum (b) (, 0) is a maximum, (c) (, ) is a point of inflection, (d) (, ) is a minimum and ( 3, 6) is a maximum, Differentiate again, f = (x + )(x + ) (x + )(x + x 3) (x + ) 4 = (x + )(x + ) (x + x 3) (x + ) 3. So f 8 = (x + ) 3 and f () = > 0 so (, ) is a minimum; f ( 3) = < 0 so ( 3, 6) is a maximum, (d). Copyright 04 c University of Southampton Page 4 of 7

5 5 MATH055W A7. ( marks) Which of the following could be the graph of f(x)? (a) (b) (c) (d) Given the denominator x + in f, the only possible solution is (d). Check that the stationary points match: they do. Check that the asymptotes work: as x, f(x) x which can be seen in the picture, so (d) works. A8. ( marks) Given the matrix A = 0 4 0, the determinant of A is (a) 0, (b) 4, (c) 8, (d) 4, Expand along first row, A = ( 4)(4 ) + ( 0) = 0, so (e). TURN OVER Copyright 04 c University of Southampton Page 5 of 7

6 6 MATH055W Questions A9 - A0 concern the matrix A = ( 3 α ). A9. ( marks) Given the matrix A as above, which of the following conditions on α ensures that the matrix equation Ax = b has a unique solution x? (a) α > 3, (b) α 3, (c) α < 3, (d) α = 3, For a unique solution, require the matrix to be invertible, so A = 3α 0, so (b). A0. ( marks) In the matrix A as above, let α = /3, and let b = ( β ). Which of the following conditions on β ensures that the matrix equation Ax = b is consistent? (a) β 3 4, (b) β 4 3, (c) β = 3 4, (d) β = 4 3, From A9 we now know that the matrix is not invertible! Expect a row of zeroes in the echelon form of A, and would like this to extend into the augmented form (A : b). Form the augmented matrix and use a single row operation (Gaussian elimination): (A : b) = ( 3 /3 β ) ( β 4/3 ). This is echelon form and the bottom-right entry is a pivot if it s non-zero. A pivot in the rightmost column of the echelon form shows inconsistent equations, so we require β = 4/3, so (d). Copyright 04 c University of Southampton Page 6 of 7

7 7 MATH055W A. ( marks) Which of the following is a particular integral for the differential equation d y dx + dy dx y = e x? (a) ex, (b) Ae x + Be x, (c) xe x, (d) e x, (e) none of the above? The auxiliary equation is m + m = 0 so the complementary function is Ae x + Be x, someone careless might choose (b)! As e x isn t in the complementary function, we can try y p = Ae x, so y p = Ae x and y p = Ae x. Substitute into the ODE to get so A = / and the answer is (d). Ae x Ae x Ae x = e x A. ( marks) Which of the following is a particular integral for d x dt x = t? (a) at + bt + c, (b) Ae t + Be t, (c) t + t, (d) t, Note that (b) is the complementary function. Again! Try y p = At + Bt + C so y p = A and substitute into the ODE: and comparing coefficients we get A (At + Bt + C) = t t : A =, t : B = 0 and constant: A C = 0 C = and the answer is (d). TURN OVER Copyright 04 c University of Southampton Page 7 of 7

8 8 MATH055W A3. ( marks) The integral t t t 3 3t dt, is equal to 6t 3 (a) 3 ln t3 3t 6t 3 + C, (c) (b) 3 t3 t t 4 t4 t 3 3t 3t + C, t 3t 6t 6 + C, (d) 3 ln t3 3t 6t 3 + C, Note that the denominator has derivative 3t 6t 6 = 3(t t ), so the substitution u = t 3 3t 6t 3 is required, giving so (a). 3 u du = ln u + C, 3 A4. ( marks) Which of the following functions f(t) has a Fourier series of the form f(t) = b n sin(nπt)? n= (a) f(t) = t in < t < and f(t + π) = f(t), (b) f(t) = t in π < t < π and f(t + ) = f(t), (c) f(t) = t sin t in < t < and f(t + ) = f(t), (d) f(t) = t in π < t < π and f(t + π) = f(t), The function needs to be odd for a sine-series, hence only (c) or (d) are possible. For period L, the sine terms would be of the form b n sin(nπx/l) and as the period of (c) is, the terms would be b n sin(nπx), matching the question. This is not the case for (d), hence (c) is correct. Copyright 04 c University of Southampton Page 8 of 7

9 9 MATH055W A5. ( marks) Given the functions f(t) = t 3 cos t and g(t) = sin ( t ), which of the following is true? (a) g is odd and f is even, (b) f is odd and g is even, (c) f is even and g is neither odd nor even, (d) both are odd, g( t) = sin[( t) ] = sin(t ) = g(t) so g is even. Similarly, f( t) = ( t) 3 cos( t) = t 3 cos(t) = f(t), as cosine is even, so f is odd, hence (b). A6. ( marks) A Fourier series is obtained for the function f(t) = { t if 0 t < 3, t 3 if 3 t < 6, Which of the following is the value of the series at t = 3? (a), (b).5, (c) 0, (d) 3, and f(t + 6) = f(t) for all t. At the discontinuity, the Fourier Series converges to the average value of the function on both sides: hence (e). (f(t) + f(t) + ) = ( t + t 3) = 3, A7. ( marks) Given that f(x, y) = x sin(xy), the partial derivative f x y (a) x sin(xy) cos(y)x sin(xy), (b) xy cos(xy) y sin(xy), (c) y cos(x), (d) x cos(xy) yx sin(xy), is equal to TURN OVER Copyright 04 c University of Southampton Page 9 of 7

10 0 MATH055W f y = x cos(xy) and using chain and product rules, hence (d) f x y = x cos(xy) yx sin(xy), A8. ( marks) If f(u, v) = sin u sinh v, where u = u(r) and v = v(r), which of the following is equal to df dr? (a) cos u sinh v du dr (c) sin u cosh v dv dr The chain rule needed is df f v + sin u cosh vdv dr, cos u sinh vdu dr, = sin u cosh v, the solution is (a). (b) sin u cosh vdu dr (d) cos v cosh udu dr dr = f du u dr + f dv f, and with v dr u + cos u sinh vdv dr, cos u sinh vdv dr, = cos u sinh v and A9. ( marks) dx x x is equal to (a) (x ) x x + C, (b) cosh (x + ) + C, (c) sinh (x ) + C, (d) x (x x) 3 + C, Complete the square hence (e). dx x x = dx (x ) = cosh (x ) + C Copyright 04 c University of Southampton Page 0 of 7

11 MATH055W A0. ( marks) x dx is equal to x (a) x ln x + C, (b) ln x + x + C, (c) x x + C, (d) x ln x + + C, x x dx = hence (b). x + dx = x + dx = x + ln x + C, x SECTION B B. (a) (5 marks) Find the following indefinite integrals (i) x + 3 dx, (ii) x + x + x + dx, (iii) x + x + dx. (i) ln x C. (ii) let u = x + x +, then the integral becomes u du = ln u + C = ln x + x + + C. (iii) partial fractions no good, as denominator doesn t factorize over R. Let u = x + then we have (x + ) + dx = u + du = tan u + C = tan (x + ) + C. (b) (7 marks) Using partial fractions and the results in part (a), if appropriate, find x x 5 (x + 3)(x + x + ) dx. (Question B. continued on next page) TURN OVER Copyright 04 c University of Southampton Page of 7

12 MATH055W Partial fractions gives x x 5 (x + 3)(x + x + ) = A x Bx + C x + x + = A(x + x + ) + (x + 3)(Bx + C) (x + 3)(x + x + ) so equating numerators we obtain x x 5 = A(x + x + ) + (x + 3)(Bx + C). Letting x = 3 we have 0 = A(5) + 0(Bx + C) A =. Letting x = 0 we obtain 5 = A() + 3C C = 3. Any other value of x will do. Choose x =, then 6 = A(5) + 4(B + C) 6 = 0 + 4B B =. The integral is now x + 3 x + 3 x + x + dx = x + 3 dx and using the results from (a) we have x + x + x + dx x + x + dx = ln x + 3 ln x + x + tan (x + ) + C. B. (a) (6 marks) Given the matrix A = 0 0, confirm that λ = is one of the eigenvalues of A, and find the other eigenvalues. Find the eigenvector corresponding to λ =. To find the eigenvalues, solve det(a λi) = 0 so we have λ 0 λ = 0, so expand by first column to get 0 λ ( λ)[( λ) 4] = 0 which immediately shows that λ = is an eigenvalue. The other eigenvalues are given by ( λ) 4 = 0 λ = ± λ =, 3. Copyright 04 c University of Southampton Page of 7

13 3 MATH055W To find the eigenvector, solve (A I)x = 0 so we solve 0 0 x y = z 0 This needs to be handled carefully: we have three equations, none involving x: y z = 0, y + z = 0 and y z = 0, with obvious solution y = z = 0. This leaves x as a free variable and the eigenvector is x = any real number. (b) (6 marks) Find the general solution of the differential equation C 0 0 where C is (xe t + cos t + sin x) dx dt + x e t x sin t + t = 0. Is the ODE exact? Let p = xe t + cos t + sin x and q = x e t x sin t + t. Then p t = xet sin t = q so the equation is exact. x To solve, integrate: H = p dx = xe t + cos t + sin x dx = x e t + x cos t cos x + f(t) and H = q dt = x e t x sin t + t dt = x e t + x cos t + t + g(x) where f(t) and g(x) are arbitrary functions. Equating both expressions, we see that f(t) = t and g(x) = cos x and the solution is H = x e t + x cos t + t cos x = constant. B3. (a) (6 marks) Find the solution of the differential equation which satisfies x = 0 when t = 0. x dx dt = ( + x )te 3t, (Question B3. continued on next page) TURN OVER Copyright 04 c University of Southampton Page 3 of 7

14 4 MATH055W The ODE is separable: x + x dx = te 3t dt. The l.h.s. integrates easily using u = + x : x + x dx = u du = ln u = ln( + x ). For the r.h.s. use integration by parts: te 3t dt = 3 te3t 3 e 3t dt = 3 te3t 9 e3t Putting the two together, we have ln( + x ) = 3 te3t 9 e3t + C. Now substitute x = 0 when t = 0 to obtain C = 9 so the solution is ln( + x ) = 3 te3t 9 e3t + ( [exp 9 x = ± 3 te3t 9 e3t + ) ] 9 (b) (6 marks) Express the complex number z = 6 in complex exponential form and hence find all the values of z 4, writing your answers in the form x + jy. Display these values on an Argand diagram. Easily, z = 6 and arg(z) = π. Thus, in complex exponential form, z = 6e πj. Hence z 4 = 6 4 e 4 (π+nπ)j where distinct solutions are given by n = 0,,, 3. The solutions are thus ( z 0 = e π 4 j = cos π 4 + j sin π ) = + j, ( 4 z = e 3π 4 j = cos 3π 4 + j sin 3π ) = + j, 4 ( z = e 5π 4 j = cos 5π 4 + j sin 5π ) = j, 4 and Copyright 04 c University of Southampton Page 4 of 7

15 5 MATH055W ( z 3 = e 7π 4 j = cos 7π 4 + j sin 7π 4 ) = j, And the solutions on the Argand Diagram are: B4. (a) (3 marks) If y = cosh x use the exponential definition of cosh y to show that y satisfies the equation e y xe y + = 0. If y = cosh x then x = cosh y = ey + e y. So ( e e y xe y + = e y y + e y ) e y + = e y ( e y + ) + = 0, as required. (b) (5 marks) Rewrite the above equation as a quadratic and hence deduce that cosh x = ln(x + x ). Let z = e y, then we can rewrite the equation as z xz + = 0, treating x as a coefficient. Solve this: e y = z = x ± 4x 4 = x ± x. ( Hence y = ln x ± ) x and we choose the positive root as cosh x 0, giving the desired result. (Question B4. continued on next page) TURN OVER Copyright 04 c University of Southampton Page 5 of 7

16 6 MATH055W (c) (4 marks) By differentiating the result in (b) verify that d dx (cosh x) = x. d [ ln(x + ] ( ) x dx ) = x + x x + x ( ) = x + x x + x = x required. x, as B5. (a) (6 marks) Using partial fractions and the table of Laplace transforms find { } L (s ). (s ) Solution: Equating numerators (s ) (s ) = A s + B (s ) + C s = A(s )(s )+B(s )+C(s ) (s ) (s ) () () = A(s )(s ) + B(s ) + C(s ) (3) s = : = 0 + B( ) + 0, B = (4) s = : = C(), C = (5) coeff of s : 0 = A + C, A = C = (6) Hence { } L (s ) (s ) = L { } { s L } (s ) + L { } s = e t t e t + e t (7) using the first shift theorem for the middle term. Copyright 04 c University of Southampton Page 6 of 7

17 7 MATH055W (b) (6 marks) Use Laplace transforms and part (a) to solve the differential equation d x dt 3 dx dt + x = et which satisfies the conditions x = 0 and dx dt = 0 when t = 0. Solution and solve for X(s): d x dt 3 dx dt + x = et. Take Laplace transforms, term by term, { d x L dt 3 dx } { d } { } dt + x x dx = L dt 3 L + L {x} = L {e t } dt [ using the formula sheet: s X sx(0) ẋ(0) ] 3 [sx x(0)] + X = ( ) initial conditions are zero, so collect like terms and simplify s X 3sX + X = (s 3s + )X = (s )(s )X = s X = (s )(s ) s Hence inverting { } x(t) = L {X(s)} = L (s ) (s ) = ( e t te t + e t), using (a) END OF PAPER Copyright 04 c University of Southampton Page 7 of 7