1MA1 Introduction to the Maths Course

Transcription

1 1MA1/-1 1MA1 Introduction to the Maths Course Preamble Throughout your time as an engineering student at Oxford you will receive lectures and tuition in the range of applied mathematical tools that today s professional engineer needs at her or his fingertips. The MA series of lectures starts in the first term with courses in Calculus, Series, Complex Algebra, and Differential Equations. Many of the topics will be familiar, others less so, but inevitably the style and pace of teaching, involving lectures and tutorials, will be wholly new to you. To ease your transition, this introductory sheet provides a number of revision exercises, together with some questions that will require reading and that will introduce some of those new topics. The questions relate especially to the courses mentioned above. This sheet has not been designed to be completed in an evening, nor are all the questions entirely trivial so don t think for a moment as you sweat over some question that all your future colleagues have finished. Believe us, they have not! We hope that, with revision, the sheet represents about a week s work, but we suggest that you start the sheet around three weeks before you come up so that the revision has time to sink in, and so that the questions and answers are still fresh in your mind by 1st week of term, when your college tutors are likely to review your work. So, remember to bring your solutions to Oxford with you!

2 1MA1/0 Reading The ability to learn new material yourself is an important skill which you must acquire. But, like all books, mathematics for engineering texts are personal things. Some like the bald equations, others like to be given plenty of physical insight. However, two texts that are commonly recommended by lecturers because they cover the 1st year course quite concisely are: Title: Mathematical methods for science students Author: G. Stephenson Publisher: Longman, Harlow, 197 Edition: 2nd ed ISBN/ISSN: (paperback} Price: Author: Heading, John Title: Mathematical methods in science and engineering Publisher: Edward Arnold, London, 1970 Edition: 2nd ed. ISBN/ISSN: (paperback) You may wish to buy one or the other although all the material in this sheet may be more readily available to you in standard A-level texts.

3 1MA1/1 1MA1 Introduction to the Maths Course Prof G T Houlsby Q1-0 Dr D W Murray Q1-65 Revised /Dept/Teaching/Lectures/1MA1/exmath1.txt /Dept/Teaching/Lectures/1MA1/qsheet.tex 1 Differentiation

4 1MA1/2 2 Integration

5 1MA1/ Series 4 Functions You should be familiar with the properties of standard functions, such polynomials, rational functions (where both numerator and denominator are polynomials), exponential functions, logarithmic functions, and trigonometric functions and their identities. 1. (i) For what value(s) of x is the function f(x) = x/(x 2 1) undefined? Describe the behaviour of f as x approaches these values from above and below. (ii) Find the limits of f(x) and df/dx as x + and x. (iii) Does the function have stationary values? If so, find the values of x and f(x) at them. (iv) Now make a sketch of the function. 2. Repeat all the above for the function f(x) = x 2 /(x 1).. A quantity y(ω) varies as y = 1 2 log 10(1 + ω 2 ). What is the value of y when ω 0, when ω = 1 and when ω? Make a sketch of y versus log 10 ω. (You might like to evaluate y at ω = 0.01, 0.1, 0.2, 0.5, 1, 2,5, 10,100.) 4. Derive an approximation for the function y when ω. Does this tie up with your sketch? 5. Sketch y = e t and y = e t versus time t for 0 < t <. When a quantity varies as e t/τ, τ is called the time constant. What are the time constants of your two plots? Add to your sketch two curves showing the variation of a quantity with (i) a very short time constant, (ii) a very long time constant.

6 1MA1/4 6. A quantity varies as y = 100e 10t + e t/10. Which part controls the behaviour of y at short time scales (ie when t is just above zero), and which at long times-scales? When does the switch over occur? 7. A quantity y 1 varies with time t as y 1 = 2 cos ωt. A second quantity y 2 varies as y 2 = cos(2ωt+ π ). Plot y 4 1 and y 2 versus ωt, for 2π < ωt < 2π. What are the amplitudes and frequencies of y 1 and y 2? 8. Show that tan 1 x + tan [ ] 1 1 x 1+x = π. (Note: 4 tan 1 x = arctanx. Hint: try writing A = tan 1 x; B = tan ( ) 1 1 x 1+x ; then evaluating tan(a + B).) You may not have come across hyperbolic functions before. The hyperbolic cosine is defined as cosh x = 1 2 (ex + e x ), and the hyperbolic sine is defined as sinh x = 1 2 (ex e x ). Other hyperbolic functions are defined by analogy with trigonometric functions: eg, the hyperbolic tangent is tanh x = sinh x/ cosh x. 9. Show that (i) cosh 2 x sinh 2 x = 1; (ii) (1 tanh 2 x) sinh 2x = 2 tanh x. 40. Find d d cosh x and sinh x. (Express your results as hyperbolic functions.) dx dx 5 Complex Algebra Complex Algebra is no longer in the Common Core at A-level, though some courses contain it. Whether you are familiar with it or not, you can read about it in Chapter 2 of James, or Chapter 7 of Stephenson or indeed most A-level texts, such as Bostock and Chandler. We will use the notation that a complex number z = (x + iy), where x is the Real part of z, y is the Imaginary part of z, and i is the Imaginary unit such that i 2 = 1. That is, x = Re(z) and y = Im(z). 41. Evaluate (i) (1+2i)+(2+i); (ii) (1+2i)(2+i); (iii) (1+2i) and plot the resulting complex numbers on an Argand diagram. 42. If z = (x + iy), its complex conjugate is z = (x iy). Show that zz = (x 2 + y 2 ). 4. By multiplying top and bottom of the complex fraction by the complex conjugate of (+4i), evaluate i + 4i 44. If we have a function of complex numbers z = f(z 1, z 2, z,...) then its complex conjugate z is obtained as f(z 1, z 2, z,...). Verify this by evaluating (i) (1 2i)(2 i) and (ii) (1 2i) and comparing with Q41. Prove that (z 1 /z 2 ) = z 1 /z Using the usual quadratic formula, find the two complex roots of z 2 + 2z + 2 = 0. (Hint: as i 2 = 1 we have that 1 = ±i.) Are complex solutions to a quadratic equation always conjugates?

7 1MA1/5 46. Show that z a = 1, z b = i and z c = i are all cube roots of unity. (Ie z a = 1, z b = 1 and z c = 1.) 47. From the Argand diagram shown, it is obvious that any complex number (x + iy) can be rewritten as A(cos θ + i sin θ). Express the three cube roots of unity in this form, and plot them on an Argand diagram. Imag y z=x+iy A θ Real x 48. (i) Show, using the familiar trigonometrical identities, that (cos θ + i sin θ) 2 = (cos 2θ + i sin 2θ). In fact, de Moivre s Theorem tells us that (cos θ + i sin θ) n = (cos nθ + i sin nθ). Verify this for the cube roots of unity. 49. The complex number z(t) = [cos(ωt) + i sin(ωt)] is a function of time t. Plot z(t) for ωt = 0, π/6, 2π/6, π/2...2π in the Argand plane. You will see that z(t) rotates round and around as t increases. What motion is the real part, Re(z(t)), performing? 50. A complex exponential is defined by e iθ = (cos θ + i sin θ). Rewrite the three cube roots of unity in the form e iθ. Confirm using de Moivre s Theorem, that [ e iθ ] n = e inθ.

8 1MA1/6 6 Vectors You should be familiar with the vector algebra of points lines and planes, and with the scalar product. In the following i, j and k are unit vectors in the x, y and z directions. 51. Find the unit vector ˆv in the direction i j + 2k 52. Find the coordinates of point P if OP = and OP is in the direction of (i) i + j + k, (ii) i 2j + k. (O is the origin.) 5. Write down the vector equation of the straight lines (i) parallel to i + j + k and through the origin, (ii) parallel to i 2j + k and through the point (1,1,1). 54. Find the point on the line i + j + k that is nearest to the point (, 4, 5). 55. Show that the line joining (, 4, 5) to this closest point is perpendicular to the original line i + j + k. 56. A sphere of radius 1 unit is centred on the point (1,1,1). Describe the locus r of points on the sphere in vector form. 57. What is the angle between the vectors (i + 2j + k) and (i + 2j + k)? 58. Find the vector position of a point 1/ of the way along the line between (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ), and nearer (x 1, y 1, z 1 ). 59. What are the vector equations of the planes which have vector (i+j+k) as normal and (i) which contains the origin amd (ii) which contains the point (1, 2, ). For (ii) what is the perpendicular distance from the plane to the origin? 60. At time t = 0 two forces f 1 = (i + j) and f 2 = (2i 2j) start act on a point body of unit mass lying stationary at point (1, 2) of the x, y plane. Describe the trajectory r(t) of the particle.

9 1MA1/7 7 Some answers and hints

10 1MA1/8 1. (i) f(x) = x/(x 2 1) undefined at x = ±1. Asymtotic behaviour as in sketch. (ii) As x +, f(x) 0 from above. As x, f(x) 0 from below. Gradients both tend to zero. (iii) df/dx = (x 2 + 1)/(x 2 1) 2 is nowhere 0 hence no turning points. 2. f(x) = x 2 /(x 1) undefined at x = 1. Asymptotic behaviour as sketch. (ii) f(x) ± as x ±. df/dx = x(x 2)/(x 1) 2 +1 at both extremes. (iii) df/dx = 0 at x = 0, 2: max at (0, 0), min (2, 4).. y = 0 as ω = 0. y = 0.15 at ω = 1, y as ω. ω y Approximation is y = log 10 ω. 5. y = e t and y = e t : time constants 1 and 1/ respectively e 10t dominates at small t. Cross over when 100e 10t = e t/10 or e 9.9t = Hence t = Amplitude 2, frequency f = ω/2π; Amplitude 1, frequence f = ω/π. 8. Put A = tan 1 x; B = tan ( ) 1 1 x 1+x. tan(a + B) = (tan A + tan B) (1 tan A tan B) = x + 1 x 1+x 1 x 1 x 1+x = x x 2 = 1 Hence A + B = tan 1 1 = π/4. 9. (i) cosh 2 x = (e 2x + e 2x + 2)/4; sinh 2 x = (e 2x + e 2x 2)/4; cosh 2 sinh 2 = 4/4 = 1. (ii) 1 tanh 2 = 1/ cosh 2 ; sinh 2x = 2 cosh x sinh x; Hence (1 tanh 2 x) sinh 2x = 2 cosh x sinh x/ cosh 2 x = 2 tanh x. d 40. dx (ex + e x )/2 = (e x e x )/2. Hence d d cosh x = sinh x and similarly sinh x = cosh x. dx dx

11 1MA1/9 41. (i) ( + 5i); (ii) ( 4 + 7i); (iii) ( 11 2i); 42. Note (x 2 + y 2 ) is the modulus of z (and of z too for that matter). 4. (11/25) + i(2/25) 44. ( ) z1 z 2 ( ) z1 z 2 = z 2 z 2 But z 2 z 2 is real so = 1 z 2 z 2 (z 1 z 2 ). Now so that z 1 z 2 = (x 1 + iy 1 )(x 2 iy 2 ) = (x 1 x 2 + x 2 y 2 + i(y 1 x 2 y 2 x 1 )) (z 1 z 2 ) = (x 1 x 2 + x 2 y 2 i(y 1 x 2 y 2 x 1 )) = z 1 z 2 So ( ) z1 z 2 = 1 z 2 z 2 z 1 z 2 = z 1 z Solutions are ( 1±i). Yes: for a complex soln. The usual formula gives roots as ( b ± b 2 4ac)/2a. For complex roots, b 2 4ac < 0, giving the imaginary part and ± signs always gives conjugate pairs with the same real part. Note though if b 2 4ac > 0 the two real solutions are different A = 1 and θ = 0, 2π/ (120 ), 4π/ (240 ). 48. (i) Square to find (cos 2 θ sin 2 θ + 2i sin θ cos θ), hence result. 49. Complex number describes circle. Real part is cos ωt: simple harmonic motion. 50. θ = 0, 2π/, 4π/ again. 51. ˆv = 1 6 (i j + 2k). 52. (i) (,, ), 14 (1, 2, ). 5. (i) r = α (i + j + k), where parameter α is any real number. (NB: strictly no need for the, but using makes α measure distance). (ii) r = (1 + 6 α ) ( i + 1 2α ) j + (1 + α ) 6 k) 6 (Again no real need for 6.

12 1MA1/ Vector from point to a general point on line is ( α )i + ( α 4)j + ( α 5)k. We want α corresponding to minimum distance, or minimum squared-distance. Squared distance is d 2 = ( α ) 2 + ( α 4) 2 + ( α 5) 2.. Diff wrt α and set to zero, cancelling factor of 2/, gives ( α ) + ( α 4) + α 5) = 0, so that α = 4. Thus the closest point is (4, 4, 4). 55. Vector from (, 4, 5) to (4, 4, 4) is (i k). Scalar product with (i+j+k) is ( ) = 0. Hence perpendicular. 56. Locus r is such that r (i + j + k) = Take scalar product of UNIT vectors. cos 1 (10/14) = (x 1, y 1, z 1 ) + 1 [(x 2, y 2, z 2 ) (x 1, y 1, z 1 )]. Ie 1 [(2x 1 + x 2 ), (2y 1 + y 2 ), (2z 1 + z 2 )]. 59. Equation of plane with UNIT normal ˆn and perpendicular distance from origin is r ˆn = ±D, where sign chosen according to sign of normal. So (i) r (i + j + k) = 0. (ii) r (i + j + k) = d where d chosen to make (1, 2, ) lie on plane. d = 6. To get a unit normal, use 1 (i+j+k), so that D = 6/ is perpendicular distance from origin to plane. 60. Total force is (i j) so for unit mass, d 2 x dt 2 = ; d2 y dt 2 = 1. Thus dx dt = t + a; dy dt = t + b(where a = b = 0, as stationary at t = 0). and x = t 2 /2 + c; y = t 2 /2 + d(wherec = 1, d = 2from initial posn). So, r(t) = (t 2 /2 + 1)i + ( t 2 /2 + 2)j.