Problem 1.1 For y = a + bx, y = 4 when x = 0, hence a = 4. When x increases by 4, y increases by 4b, hence b = 5 and y = 4 + 5x.

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1 Appendix B: Solutions to Problems Problem 1.1 For y a + bx, y 4 when x, hence a 4. When x increases by 4, y increases by 4b, hence b 5 and y 4 + 5x. Problem 1. The plus sign indicates that y increases as x increases. Problem 1.3 A quadratic equation with roots α and β is x α)x β), and so in this problem α + β) 6 and αβ 4. To obtain α 4 and β 4, use α + β) 4 α + αβ + β ) giving α 4 + β 4 + 4α 3 β + 6α β + 4αβ 3 α 4 + β 4 + 4αβα + β) α β, α 4 + β 4 α + β) 4 4αβα + β) + α β 75. Problem 1.4 a.b 4 5 cos cm. Problem 1.5 a b 4 5 sin cm. Problem. For f a + bx df fx + ) fx) a + bx + ) a bx) b b. Problem.3 For f a + bx + cx fx + ) fx) a + bx + ) + cx + ) a bx cx b + cx + c 1 b + cx.

2 Problem.4 For f x + lnx) df x + lnx) d x + lnx) x + lnx) x + 1 ) x 4x 3 + x + 4xlnx + lnx x. Problem.5 For f x + lnx) df 4x3 + x + 4xlnx + lnx x. d ) f x/x lnx 1x + + 4lnx x 1x lnx + x lnx x. Problem.6 The values of x where f x/3 + x ) has maxima or minima are values where the first differential of f is zero. df 3 + x ) 4x 3 + x ). Hence the numerator 6 x and x ± 3. There is no need in this case to examine the sign of the second differential at x ± 3 to see which corresponds to the maximum and which the minimum. The maximum value of the function is 3/ and the minimum 3/. Problem 3.1 where C is a constant. 3x 3x ) x 3 3 x x + 8x lnx + C, 4 1 3x 3x ) x 3 3 ) 4 x x + 8x lnx 1

3 ln4 ) 1 3 ) + 8 ln1, and integral one has the value ln x 3x ) ln ln4, x and integral two has the value 48- ln 8 + ln x 3x ) ln x 8 + ln1, and integral three has the value ln8 which equals the sum of the first two. Problem 3. For x e x, in the notation of equation 3.11 put fx) x and dgx) ex. Then x e x 1 x e x 1 e x x 1 x e x xe x. Now use equation 3.11 again for the last integral with to give fx) x and dgx) xe x 1 xex 1 ex e x 1 xex 1 4 ex and x e x 1 x e x 1 xex ex. Problem 3.3 Since the limits on the y integral depend on x do the y integral first. xx yx+1 x y 4xy + 3y )dy x xy + y 3) x+1 x x + x + 1)3x + 1) x 3 4 x x3 + 5 x + x. 3x 3 + 7x + 5x + 1) 3

4 Problem j)1 j) 4j + 3j j. Equation 4.3 gives the modulus of the complex number 8 j) to be Problem j)1 j) 3j) + j) 77 14j j ) 11 j) 7 4j) j j)7 + 4j) 7 4j)7 + 4j) j). 13 Problem 5.1 sin 1 sin ) cos 18 sin 3.5. tan 33 cos 1 cos9 + 3 sin 3.5. sin 33 cos 33 sin36 3 ) cos36 3 ) sin 3 cos Problem 5. tan θ sin θ cos θ sinθ/) cosθ/) cos θ/) sin θ/) sinθ/ + θ/) cosθ/ + θ/) tanθ/) 1 tan θ/). Problem 5.3 In Figure B.1, a d + f and d + f sin α b sin β. B.1) From the right-angled triangle formed by dropping a line from the point B to meet the line AC at right angles, d b cos γ, and c f cos β. Substitution in equation B.1 gives b cos γ + c cos β sin α b sin β 4

5 and sin α sin βb cos γ + c cos β) b sin β cos γ + c sin β cos β b ) sin γ sin β cos γ + sin β cos β sin β sin β cos γ + sin γ cos β. But β + γ π α and hence sinβ + γ) sin α and finally sinβ + γ) sin β cos γ + sin γ cos β. Problem 5.4 d dθ from equation.11, ) sin θ cos θ 1 + sin θ)cos θ sin θ) sin θ cos θ, 1 + sin θ 1 + sin θ) cos θ sin θ + sin θ cos θ sin θ sin θ cos θ 1 + sin θ) cos θ sin θ + sin θ cos θ sin θsin θ + cos θ) 1 + sin θ) cos θ sin θ + sin θcos θ 1) 1 + sin θ) cos θ sin θ sin 3 θ 1 + sin θ). Problem 5.5 sin θ cos 3 θdθ sin θ cos θdsin θ) sin θ1 sin θ)dsin θ) sin θ sin 4 θ)dsin θ) 1 3 sin3 θ 1 5 sin5 θ 1 3 sin3 θ 1 5 sin3 θ1 cos θ) 15 sin3 θ sin3 θ cos θ. 5

6 Problem 5.6 From equations 5.18 and 5.19 cos θ cos φ 1 4 ejθ + e jθ )e jφ + e jφ ) 1 4 ejθ+φ) + e jθ+φ) + e jθ φ) + e jθ φ) ). B.) sin θ sin φ 1 4 ejθ e jθ )e jφ e jφ ) 1 4 ejθ+φ) + e jθ+φ) e jθ φ) e jθ φ) ). B.3) Addition of equations B. and B.3 gives cos θ cos φ + sin θ sin φ cosθ φ). Problem 5.7 sin θ + φ) cos sin θ cos φ + sin φ cos θ θ φ) ) cos θ cos φ + sin θ sin φ ). Multiplying out the brackets and using equation 5.7 gives sin θ + φ) cos θ φ) sin θ cos θ + sin φ cos φ sin θ + sin φ. Problem 5.8 3j)1 + j) 4 + 3j 8 + j 4 + 3j Expressing z in polar form, the modulus r 8 + j)4 3j) /5) + 4/5) j 5 and the angle θ is given by tan θ 4/5 7/5 4/7. 6

7 Problem 6.1 Hence x6 1) x 1) and 3x6 1) 3x 1) x 1) 3 x 1) x 1). 3 x x 4 1). x 1) Problem 6. 1 j sin x 1 j ejx e jx) 1 + jx x! jx3 3! 1 j ) x4 4! + jx5 5!... ) 1 jx x! + jx3 + x4 3! 4! jx5 5!... x x3 3! + x5 5!... Problem 6.3 Expand 1 x) 1 using the binomial theorem with n 1. 1 x) ) x) + 1) ) x)! + 1) ) 3) 3! x) 3 + 1) ) 3) 4) x) ! 1 + x + x + x 3 + x 4... Problem 6.4 ln 1 + x ln1 + x) ln1 x). 1 x For x < 1 the result of Example 6. can be used to expand the logarithmic terms and subtracting them gives the answer. 7

8 Problem 6.5 Use Taylor s theorem to expand f and g close to zero at x δx. fδx) gδx) f) + δxdf/) +... g) + δxdg/) +... where the differentials are evaluated at x and higher differentials than the first can be ignored as we take the limit of δx when f) g) df/) dg/). Problem 7.1 When one half the sample has decayed, N N / and equation 7. gives the time t 1/ as the solution to N / N e λt 1/. Hence e λt 1/ and t1/ ln/λ. Problem 7. di λi hence di I λ. As for equation 7.1 the solution is I I e λx and thus when x d the number of gamma rays in the beam is Id) I e λd. Problem 7.3 Integrating both sides of dn/n λdt gives 1 N λt + C. At t, N N and the constant C 1/N. Hence N 1/λt + 1/N ) N N λt). Problem 7.4 The integrating factor for the equation L di dt + RI V is e Rt/L. Multiplying both sides by this factor gives di dt + R ) L I e Rt/L V L ert/l 8

9 or Integrating both sides and d dt IeRt/L ) V L ert/l. Ie Rt/L V R ert/l + C I V R + Ce Rt/L. But I for t and hence the constant C V/R and I V R 1 e Rt/L ). Problem 7.5 y α sinωx + β) α sin ωx cos β + α sin β cos ωx. Hence A α cos β and B α sin β. Squaring and adding results in α A + B and tan β B/A. Problem 7.6 y αe γx sinωx + β) dy/ αγe γx sinωx + β) + αωe γx cosωx + β) d / αγ e γx sinωx + β) α ωγe γx cosωx + β) αγωe γx cosωx + β) αω e γx sinωx + β) Substitution in equation 7.7 gives [ aαγ e γx aαω e γx bαγe γx + cαe γx] sinωx + β) + [ aαωγe γx aαωγe γx + bαωe γx] cosωx + β). For values of x such that ωx + β) nπ where n is an integer, the cosine term is ±1 and the sin term is zero. This gives the result aαωγ + ωb and γ b/a. For values of x such that ωx + β) n + 1/π) where n is an integer, the sine term is ±1 and the cos term is zero. This gives aγ aω bγ + c and ω c/a b /4a. 9

10 Problem 7.7 Try as solution d y + αy x. y a + bx + cx ex 4. Substitution of it and its derivatives into the equation gives c cx + αa + αbx + αcx + α 3 + αex 4 x. Equating the coefficients of the different powers of x on both sides of the above gives d e ; aα + c ; 6d + bα, and 1e + cα 1. From these relationships, b ; c 1/α and a /α, and the solution to the differential equation is y /α + x /α. 1