Exam (elaborations) TEST BANK FOR Foundation Mathematics for the Physical Sciences By Riley K.F. and Hobson M.P.

Exam (elaborations) TEST BANK FOR Foundation Mathematics for the Physical Sciences By Riley K.F. and Hobson M.P. Foundation Mathematics for the Physical Sciences Student Solution Manual K. F. RILEY University of Cambridge M. P. HOBSON University of Cambridge Contents Preface page vii 1 Arithmetic and geometry 1 2 Preliminary algebra 14 3 Differential calculus 30 4 Integral calculus 43 5 Complex numbers and hyperbolic functions 54 6 Series and limits 67 7 Partial differentiation 82 8 Multiple integrals 99 9 Vector algebra 109 10 Matrices and vector spaces 122 11 Vector calculus 140 12 Line, surface and volume integrals 155 13 Laplace transforms 170 14 Ordinary differential equations 175 15 Elementary probability 198 A Physical constants 214 v 1 Arithmetic and geometry Powers and logarithms 1.1 Evaluate the following to 3 s.f.: (a) eπ, (b)πe, (c)log10(log2 32), (d) log2(log10 32). Parts (a) and (b) do no more than test the understanding of notation, and are found directly using a calculator. (a) eπ = 23.1, and (b) πe = 22.5. For the two other parts: (c) log10(log2 32) = log10(5) = 0.699. (d) log2(log10 32) = log2(1.505). We therefore need the value of x that satisfies 2x = 1.505. To find it, take logarithms and obtain x ln 2 = ln 1.505 ⇒ x = ln 1.505 ln 2 = 0.4088 0.6931 = 0.590. 1.3 Find the number for which the cube of its square root is equal to twice the square of its cube root. If a is the required number, then a3/2 = 2a2/3 ⇒ 2 = a(3/2)−(2/3) = a5/6. Now taking logarithms: 56 ln a = ln 2 ⇒ a = e(6 ln 2)/5 = e0.83177... = 2.297 . . . 1.5 By applying the rationalisation procedure twice, show that 131 3 − √ 5 + √ 7 = 9 − 11 √ 5 + 7 √ 7 + 6 √ 35. Initially treating √ 7 − √ 5 as one unit, we have 131 3 − √ 5 + √ 7 = 131[3 − ( √ 7 − √ 5)] 9 − ( √ 7 − √ 5)2 = 131[3 − ( √ 7 − √ 5)] 9 − 7 − 5 + 2 √ 35 . 1 2 Arithmetic and geometry Since the denominator is now −3 + 2 √ 35, as a second step we must multiply both numerator and denominator by 3 + 2 √ 35: 131 3 − √ 5 + √ 7 = 131[3 − ( √ 7 − √ 5)] (3 + 2 √ 35) −9 + (4 × 35) = 131(9 − 3 √ 7 + 3 √ 5 + 6 √ 35 − 14 √ 5 + 10 √ 7) 131 = 9 − 11 √ 5 + 7 √ 7 + 6 √ 35. 1.7 Solve the following for x: (a) x = 1 + ln x, (b) ln x = 2 + 4 ln 3, (c) ln(ln x) = 1. (a) By inspection of either the original equation or its exponentiated form, ex = e1eln x = ex, we conclude that x = 1. (b) By exponentiation, x = e2+4 ln 3 = e2e4 ln 3 = e234 = 81e2 = 598.5. (c) ln(ln x) = 1 ⇒ ln x = e ⇒ x = ee = 15.15. 1.9 Express (2n + 1)(2n + 3)(2n + 5) . . . (4n − 3)(4n − 1) in terms of factorials. Denoting the expression by f (n), f (n) = (4n)! (2n)! 1 (2n + 2)(2n + 4) . . . (4n − 2)(4n) = (4n)! (2n)! (n + 1)(n + 2) . . . (2n − 1)(2n) 2n = (4n)! n! (2n)! (2n)! 2n . 1.11 Measured quantities x and y are known to be connected by the formula y = ax x2 + b , where a and b are constants. Pairs of values obtained experimentally are x: 2.0 3.0 4.0 5.0 6.0 y: 0.32 0.29 0.25 0.21 0.18 Use these data to make best estimates of the values of y that would be obtained for (a) x = 7.0, and (b) x = −3.5. As measured by fractional error, which estimate is likely to be the more accurate? 3 Arithmetic and geometry In order to use this limited data to best advantage when estimating a and b graphically, the equation needs to be arranged in the linear form v = mu + c, since a straight-line graph is much the easiest form from which to extract parameters. The given equation can be arranged as x y = x2 a + b a , which is represented by a line with slope a −1 and intercept b/a when x2 is used as the independent variable and x/y as the dependent one. The required tabulation is: x 2.0 3.0 4.0 5.0 6.0 y 0.32 0.29 0.25 0.21 0.18 x2 4.0 9.0 16.0 25.0 36.0 x/y 6.25 10.34 16.00 23.81 33.33 Plotting these data as a graph for 0 ≤ x2 ≤ 40 produces a straight line (within normal plotting accuracy). The line has a slope 1 a = 28.1 − 2.7 30.0 − 0.0 = 0.847 ⇒ a = 1.18. The intercept is at x/y = 2.7, and, as this is equal to b/a, it follows that b = 2.7 × 1.18 = 3.2. In fractional terms this is not likely to be very accurate, as b  x2 for all but two of the x-values used. (a) For x = 7.0, the estimated value of y is y = 1.18 × 7.0 49.0 + 3.2 = 0.158. (b) For x = −3.5, the estimated value of y is y = 1.18 × (−3.5) 12.25 + 3.2 = −0.267. Although as a graphical extrapolation estimate (b) is further removed from the measured values, it is likely to be the more accurate because, using the fact that y(−x) = −y(x), it is effectively obtained by (visual) interpolation amongst measured data rather than by extrapolation from it. 1.13 The variation with the absolute temperature T of the thermionic emission current i from a heated surface (in the absence of space charge effects) is said to be given by i = AT 2e −BT , where A and B are both independent of T . How would you plot experimental measurements of i as a function of T so as to check this relationship and then extract values for A and B? 4 Arithmetic and geometry The equation can be rearranged to read ln(i/T 2) = lnA − BT , and so y = ln(i/T 2) should be plotted against T , obtaining a straight-line graph if the relationship is valid. If so, the (negative) slope of the graph gives B and the intercept y0 on the y-axis gives A as A = ey0 . Dimensions 1.15 Three very different lengths that appear in quantum physics and cosmology are the Planck length p, the Compton wavelength λm, and the Schwarzchild radius rs . Given that p =  hG 2πcn , λm = h mc , rs = 2GM c2 , where m and M are masses, calculate the dimensions of the gravitational constant G and those of the Planck constant h. Deduce the value of n in the formula for the Planck length. Remembering that numerical constants do not contribute to ‘dimensional equations’, from the Schwarzschild radius formula we have [G] =  c2rs M  = (LT −1)2 L M = L3T −2M −1. From the Compton wavelength formula it follows that, [h] = [λmc] = L2T −1M. Finally, from the Planck length formula [cn] =  hG l2p  = L2T −1ML3T −2M −1 L2 = L3T −3, from which, since [c] = LT −1, it follows that n = 3. 1.17 According to Bohr’s theory of the hydrogen atom, the ionisation energy of hydrogen is mee4/82 0h2. Using Appendix A, show that this expression does have the dimensions of an energy and that its value when expressed in electron-volts is 13.8 eV. The dimensions of 0 seem difficult to determine, but one common equation containing it is that for the force F between two charges separated by a distance r, i.e. F = q1q2 4π0r2 ⇒ [F] = [q]2 [0]L2 (∗). The dimensions of the given expression for the ionization energy E contain [q]4/[0]2, and by squaring (∗) we deduce that they are ([F]L2)2. Using this, and the result from Problem 1.15 that [h] = ML2T −1, the dimensions of [E] are given by [E] = M (ML2T −1)2 × (MLT −2 × L2)2 = M3L6T −4 L4T −2M2 = ML2 T 2 = [energy]. 5 Arithmetic and geometry Substituting the numerical values given in Appendix A yields a value of 2.2 × 10−18 J, which is 2.2 × 10−18 ÷ 1.60 × 10−19 = 13.8 when expressed in electron-volts. 1.19 The electrical conductivity σ of a metal is measured in siemens per metre (Sm−1), where 1 S is the unit of conductance of an electrical component and is equivalent to 1AV−1. TheWiedemann–Franz law states that at absolute temperature T , and under certain conditions, σ is related to the thermal conductivity λ of the metal by the equation λ σT = π2 3  k e 2 . Verify that this equation is dimensionally acceptable and, using Appendix A, estimate the thermal conductivity of copper at room temperature, given that its electrical conductivity is 5.6 × 107 Sm−1. Dealing first with the LHS of the equation, the thermal conductivity is measured in joules per square metre per second for a unit temperature gradient (or watts per metre per kelvin). Its dimensions are therefore given by [λ] = ML2T −2 L −2 T −1 θL−1 = MLT −3θ −1. From the formula watts = volts × amps, we conclude that the dimensions of voltage are ML2T −2T −1 × I −1, and that those of a siemens are therefore I 2T 3M −1L −2. This gives the dimensions of σ as [σ] = M −1L −3T 3I 2, from which it follows that [λ/σT ] = M2L4T −6I −2θ −2; recall that T in the given equation is a temperature, not a time. On the RHS, since the Boltzmann constant k has units of joules per kelvin, and charge= current × time, [k/e] = (ML2T −2θ −1)/(IT ) leading to [(k/e)2] = M2L4T −6I −2θ −2. Comparison with the result for the LHS shows that the stated formula is dimensionally acceptable. Taking room temperature as 293 K, and substituting other values from Appendix A, gives an estimate for λ of 402Wm−1 K−1. 1.21 The following is a student’s proposed formula for the energy flux S (the magnitude of the so-called Poynting vector) associated with an electromagnetic wave in a vacuum, the electric field strength of the wave being E and the associated magnetic flux density being B: S = 1 2  0 μ0 1/2 E2 +  μ0 0 1/2 B2  . The dimensions of 0, the permittivity of free space, areM −1L −3T 4I 2, and those of its permeability μ0 are MLT −2I −2. Given, further, that the force acting on a rod of length  that carries a current I at right angles to a field of magnetic flux density B is BI, determine whether the student’s formula could be correct and, if not, locate the error as closely as possible. From the force on a current-carrying rod, [B] = [force] [current] × L = MLT −2 IL = MT −2I −1, 6 Arithmetic and geometry whilst energy flux, measured in joules per square metre per second, has dimensions (ML2T −2)L −2T −1 = MT −3. Since the electric field E has dimensions [V ]L −1, and from the solution to Problem 1.19, [V ] = ML2T −3I −1, the dimensions of E are MLT −3I −1. Those of the ‘E2’ term are therefore  M −1L −3T 4I 2 MLT −2I−2 1/2 (MLT −3I −1)2 = MT −3. The corresponding calculation for the ‘B2’ term is  MLT −2I −2 M−1L−3T 4I 2 1/2 (MT −2I −1)2 = L2M3T −7I −4. Thus the electric term is compatible with an energy flux, but the magnetic one is not, and the error is almost certainly in the latter. Binomial expansion 1.23 Evaluate those of the following that are defined: (a) 5C3, (b) 3C5, (c) −5C3, (d) −3C5. (a) 5C3 = 5! 3! 2! = 10. (b) 3C5. This is not defined as 5 3 0. For (c) and (d) we will need to use the identity −mCk = (−1)km(m + 1) · · · (m + k − 1) k! = (−1)k m+k−1Ck. (c) −5C3 = (−1)3 5+3−1C3 = − 7! 3! 4! = −35. (d) −3C5 = (−1)5 5+3−1C5 = − 7! 5! 2! = −21. 1.25 By applying the binomial expansion directly to the identity (x + y)p(x + y)q ≡ (x + y)p+q , prove the result r t=0 pCr−t qCt = p+qCr = r t=0 pCt qCr−t which gives a formula for combining terms from two sets of binomial coefficients in a particular way (a kind of ‘convolution’, for readers who are already familiar with this term). 7 Arithmetic and geometry First, we write each term in the form (x + y)m = m s=0 mCsxsym−s , where m represents, in turn, p, q and p + q. Next we consider all the terms in the product of sums on the LHS that lead to terms containing xr . If the first sum contributes a term containing xr−t , with 0 ≤ t ≤ r, then the second sum must contribute one containing xt . The power of y that is in the same product term will be yp−(r−t ) × yq−t = yp+q−r . The full form of the term, including the relevant binomial coefficients, is therefore pCr−txr−typ−r+t × qCtxtyq−t . The sum of all these terms over t = 0, 1, . . . , r must also give the coefficient of xr in the expansion of (x + y)p+q, i.e. p+qCr ; this establishes the left-hand equality. The right-hand equality follows, either by symmetry or by interchanging the roles of p and q. Trigonometric identities 1.27 Prove that cos π 12 = √ 3 + 1 2 √ 2 by considering (a) the sum of the sines of π/3 and π/6, (b) the sine of the sum of π/3 and π/4. (a) Using sinA + sinB = 2 sin  A + B 2  cos  A − B 2  , we have sin π 3 + sin π 6 = 2 sin π 4 cos π 12 , √ 3 2 + 1 2 = 2 1 √ 2 cos π 12 , cos π 12 = √ 3 + 1 2 √ 2 . (b) Using, successively, the identities sin(A + B) = sinAcosB + cosAsin B, sin(π − θ) = sin θ, cos(12 π − θ) = sin θ, 8 Arithmetic and geometry