Calculus 222. Problems. Wednesday August 28, 2002

Transcription

1 Calculus Problems Wednesday August 8,

2 Integration is Antidifferentiation Evaluate the following integrals.. (6x 5 x 4 7x + 3/x 5 + 4e x + 7 x ). (x. a + a ) x + xa + a x + ax. ( x 3. 3 x ) x 6ex x. 5. (3x 5) e x. 7. (5x 4x + 3). 9. π/3 π/4 π π/ (y 9 y 5 + 3y) dy.. x 3/7. 3. t 6 t t 4 dt. 5. (x 3 ) ( x 3x ). (5y 4 6y + 4) dy. x. ( t ) t 4 dt. x +. x u( u + 3 u) du. 3 dt. 9. (x + /x). t4 x5 +.. (x )(3x + ). ( t / 8 ( ) t) dt r + 3 dr. r (x + ) e at dt. 7. x + x x ( 4 x5 + 5 ) x sin t dt. 33. sec x tan x π/ π/ π/3 x 4 x +. t dt. ( ) x + x. x 3. x (cos θ + sin θ) dθ. csc x cot x.

3 π/3 π/ π π csc θ dθ x (/x). 4. t dt. 43. x. 45. (x x ) 47. f(x) where f(x) = f(x) where f(x) = π π/4.5 ln 6 ln 3 e 3 e { x 4, if x <, x 5, if x. sec θ dθ. x. 8e x. x. x x. (x x ). { x, if π x, sin x, if < x π. 5. The change of variables formula for the definite integral is g(b) g(a) f(x) = b a f(g(u))g (u) du. Prove this formula and then use it to evaluate the following integrals. (a) u du + u. (b) x + x. (c) π/3 π/4 sin θ cos θ dθ. 5. Consider the set of functions of the form f(x) = e x (a cos x + b sin x) where a and b are constants. (a) Show that f (x) is a function of the same form, i.e., show that d [ e x (a cos x + b sin x) ] = e x[ A cos x + B sin x ] and find A and B in terms of a and b. (b) Use this to compute evaluate integrals: e x cos x, e x sin x, Similar ideas work for functions of the form g(x) = e x( a cos x + b sin x ). e x ( cos x 3 sin x). 3

4 5. Consider functions of the form f(x) = A sec x + B tan x where A and B are constants. (a) Show that f (x) = F (x) sec x where F (x) has the same form as f(x) (but with a different A and B). (b) Apply to the special case A = B = and show f (x) f(x) = sec x. (c) From this derive a formula for sec x. Remark. Another way to evaluate the integral in (c) is via the preposterous trick as follows: sec x(tan x + sec x) sec x + tan x sec x du sec x = = = tan x + sec x tan x + sec x u where u = tan x + sec x, du = (sec x + tan x sec x). Integration of Trig Functions 53. Use: cos(α + β) = cos α cos β sin α sin β sin(α + β) = sin α cos β + cos α sin β cos θ + sin θ = to prove: cos θ = + cos(θ), sin θ = cos(θ). Then evaluate these indefinite integrals: (a) cos x, (b) cos 4 x, (c) cos x sin x, (d) sin 3 x, (e) cos x sin x, (f) sin 6 x. 54. Compute the following integrals: + sin x (a). (b) + cos x sin x + cos x. 55. Express sin(α) cos(β) in terms of sin(α + β) and sin(α β). Then evaluate sin(3x) cos(5x). 4

5 56. Compute the following integrals when m and n are distinct integers. (a) (c) (e) π π π sin(mx) cos(nx). cos(mx) cos(nx). sin(mx) sin(nx). (b) (d) (f) π π π sin(nx) cos(nx). cos(mx) cos(nx). sin(mx) sin(nx). These facts are in many applications, especially in the study of wave motion (light, sound, economic cycles, clocks, oceans, etc.). They say that different frequency waves are independent. 57. Show that cos x + sin x = C cos(x + β) for suitable constants C and β and use this to evaluate the following integrals. (a) cos x + sin x, (b) (cos x + sin x), (c) A cos x + B sin x. where A and B are any constants. 3 Integration by Trig Substitution a 58. State and prove the Pythagorean Theorem. You may use the formulas for the areas of a square and a triangle. If a = cos θ and b = sin θ what is a + b? b c a b ( 59. Express in simplest form: (a) sin(cos (x)); (b) tan arccos ( log 6 ( 4 ))) ; (c) sin( arctan(a)); (d) arcsec(log (log 5 ( 6 5))). 6. The inverse sine function is the inverse function to the (restricted) sine function, i.e. when π/ θ π/ we have θ = sin (y) y = sin θ. Warning: sin (y) (sin y). The inverse sine function is sometimes called arcsine function and denoted θ = arcsin(y). (b) If y = sin θ, express sin θ, cos θ, and tan θ in terms of y when 5

6 (i) θ < π/; (ii) π/ < θ π; (iii) π/ < θ <. dy (c) Evaluate using the substitution y = sin θ, but give the final y answer in terms of y. 6. The inverse cosine function is the inverse function to the (restricted) cosine function, i.e. when θ π we have θ = cos (x) x = cos θ. Warning: cos (x) (cos x). The inverse cosine function is sometimes called arccosine function and denoted θ = arccos(x). (b) If x = cos θ, express sin θ, tan θ, and sec θ in terms of y when (i) θ < π/; (ii) π/ < θ π; (iii) π/ < θ <. (c) Evaluate using the substitution x = sin θ, but give the final x answer in terms of x. Then do it using the substitution x = cos θ. 6. Evaluate these indefinite integrals. (a), (b), x 4 x (c) x 4x 4. and these definite ones: (d) / /, (e) 4 x 3/, (f) 4 x 63. The change of variables formula for the definite integral is g(b) g(a) f(x) = b Prove this formula and then use it to evaluate a f(g(u))g (u) du. 3/ / x. x substitution x = sin u and then using the substitution x = cos u. first using the 64. Use integration to derive the formula for the area and length of a circle of radius r >. 65. The inverse tangent function is the inverse function to the (restricted) tangent function, i.e. for π/ < θ < π/ we have θ = tan (w) w = tan θ. Warning: tan (w) (tan w). The inverse tangent function is sometimes called arctangent function and denoted θ = arctan(w). 6

7 (b) If w = tan θ, express sin θ, cos θ, and sec θ in terms of w when (i) θ < π/; (ii) π/ < θ π; (iii) π/ < θ <. dw (c) Evaluate using the substitution w = tan θ, but give the final + w answer in terms of w. 66. Evaluate these indefinite integrals: (a) x +, (b) x + a, and these two definite ones: (d) 3 x +, (e) a 3 a x + a. (c) 7 + 3x, 67. The inverse secant function is the inverse function to the (restricted) secant function, i.e. for θ < π/ we have θ = sec (x) x = sec θ. The inverse secant function is sometimes called arc secant function and denoted θ = arcsec(x). Warning: sec (x) (sec x). (a) When we use the trig substitution x = sec θ, what are the possible values for θ? (I.e. What is the range of the function θ = sec (x)?) (b) If x = sec θ, express sin θ, cos θ, and tan θ in terms of x when (i) θ < π/; (ii) π/ < θ π; (iii) π/ < θ. (c) Evaluate x using x = sec θ, but give the final answer in terms x of x. 68. The hyperbolic sine and hyperbolic cosine are defined by sinh t = et e t, cosh t = et + e t. These hyperbolic functions are similar to the trigonometric functions cos θ and sin θ. For example, we have the following table of analogies: sin θ = cos θ cos θ = sin θ sinh t = cosh t cosh t = sinh t (cos θ) + (sin θ) = (cosh t) (sinh t) =. Prove the identities in the second column. 7

8 69. Show that the derivative of the hyperbolic sine function y = sinh t (see problem 68) is positive and that lim sinh t =, lim t sinh t =. t Conclude that there is a function t = sinh (y) with domain and range the set of all real numbers such that y = sinh t t = sinh y. The function t = sinh y is called the inverse hyperbolic sine function. Draw the graph of y = sinh t and of t = sinh y. Warning: sinh (y) (sinh y). 7. The inverse hyperbolic sine function can be computed explicitly. Show that y = u u u = y ± y +. (Hint: Multiply the equation on the left by u and use the quadratic formula.) Conclude that y = sinh t t = ln(y + y + ), i.e. sinh (y) = ln(y + y + ). dy 7. Evaluate the integral using the substitution y = sinh t (but y + express the answer as a function of y). Then evaluate it using the substitution y = tan θ (again express the answer as a function of y). Hint: To integrate the secant function see problem Evaluate the integral using the substitution x, x > x = sec θ, x = (tan θ), = (sec θ)(tan θ) dθ. (In the substitution we assume < θ < π/ since < x <.) Answer: Z x = = Z (sec θ)(tan θ) dθ tan θ Z sec θ dθ = ln(sec θ + tan θ) + C = ln(x + p x ) + C 8

9 73. Evaluate this same integral x, x > using the substitution x = cosh u, x = (sinh u) = sinh u du. (In the substitution we assume < u < since < x <.) Answer: Z x = = Z sinh u du sinh u Z du = u + C = cosh x + C. 74. Since the method of problem 7 and the method of problem 73 are both correct, the two answers must differ by a constant: for x >. Prove () by a direct argument. cosh x = ln(x + x ) + C () Answer: First cosh = so cosh = so both sides vanish when x = so equation () can hold only if C =. Therefore we show Now by the definition of inverse function for x >. Therefore we show The definition of cosh u is and e ln(x+ x ) = x + p x, cosh x = ln(x + p x ). () u = cosh x x = cosh u x = cosh ln(x + p x ). (3) cosh u = eu + e u e ln(x+ x ) = x + x, 9

10 so cosh ln(x + p x ) = x + p x + = (x + x ) + x + x x + x = (x + x x + x ) + x + x = x + x x x + x = x(x + x ) x + x = x as required. 4 Integration of Rational Functions 75. Express each of the following rational functions as a polynomial plus a rational function whose numerator s degree is lower than its denominator s degree. (a) x 3 x 3 4, (b) x3 + x x 3 4, (c) x3 x x 5 x Express each of the following rational functions f(x) as the product of linear and irreducible quadratic factors. (Hint: f() = f() = f() =.) (a) f(x) = 6x 5 6x 4 56x 3 + 4x + 6x (b) f(x) = 6x 9 48x 8 + 3x 7 5x x 4 5x Write ax + bx + c in the form a(x + p) + q, i.e. find p and q in terms of a, b, and c. Then evaluate the integrals (a) x + 6x + 8, (b) x + 6x +, (c) 5x + x (a) Use the method of equating coefficients to find numbers A, B, C such that x + 3 x(x + )(x ) = A x + B x + + C x x + 3 and then evaluate the integral x(x + )(x ).

11 Answer: We add A x + B x + + C x = A(x + )(x ) + Bx(x ) + Cx(x + ) x(x + )(x ) = (A + B + C)x + (C B)x A. x(x + )(x ) The numerators must be equal, i.e. x + 3 = (A + B + C)x + (C B)x A for all x, so equating coefficients gives a system of three linear equations in three unknowns A, B, C: so A = 3 and B = C =, i.e. and hence Z A + B + C = C B = = A = 3 x + 3 x(x + )(x ) = 3 x + x + + x x + 3 = 3 ln x + ln x ln(x ) + constant. x(x + )(x ) (b) Do this problem using the Heaviside trick. Answer: To solve multiply by x: x + 3 x(x + )(x ) = A x + B x + + C x, x + 3 (x + )(x ) = A + Bx x + + Cx x and plug in x = to get A = 3; then multiply by x + : x + 3 A(x + ) C(x + ) = + B + x(x ) x x and plug in x = to get B = ; finally multiply by x : and plug in x = to get C =. x + 3 A(x ) B(x ) = + + C, x(x + ) x x Find the integral x + 3 x (x ).

12 Answer: Apply the method of equating coefficients to the form x + 3 x (x ) = A x + B x + C x. In this problem, the Heaviside trick can still be used to find C and B; we get B = 3 and C = 4. Then A x 3 x + 4 Ax(x ) + 3(x ) + 4x = x x (x ) so A = 3. Hence Z x + 3 x (x ) = 3 ln x ln(x ) + constant. x Evaluate the following integrals: x 3 x 5 8. x x. 8. e 3x 83. e 4x. 84. e x e x 86. x(x + ). 87. x(x + ) (x )(x )(x 3) x (x )(x )(x 3). 9. Find the area of the region bounded by the curves x =, x =, y = x 5 x 4. e x e x + e x +. x (x ). x + (x )(x )(x 3). x 4x + 5, y = x 8x + 7 x 8x (a) Compute x(x h) where h is a small positive number. (b) What happens to your answer to (a) when h +? (c) Compute 94. (a) Evaluate x. of partial fractions. x by factoring the denominator and using the method (b) Evaluate this same integral using the substitution x = sec θ. (c) Show directly that the answer in (a) is the same as the answer in (b); i.e. that they differ by a constant.

13 5 Integration by Parts 95. Evaluate x n ln x where n. Answer: Z x n ln x = xn+ ln x n + xn+ (n + ) + C. 96. Evaluate e ax sin bx where a + b. Hint: Integrate by parts twice. Answer: Z e ax sin bx = eax (a sin bx + b cos bx) + C. a + b 97. Evaluate e ax cos bx where a + b. Answer: Z e ax cos bx = eax (a cos bx b sin bx) + C. a + b 98. Prove the formula and use it to evaluate x n e x = x n e x n x e x. x n e x 99. Prove the formula sin n x = cos x sinn x + n n n sin n x, n and use it to evaluate sin x. Show that the answer is the same as the answer you get using the half angle formula.. Prove the formula cos n x = sin x cosn x + n n n cos n x, n and use it to evaluate cos 4 x. Show that the answer is the same as the answer you get using the half angle formula. 3

14 . (i) Prove the formula x m (ln x) n = xm+ (ln x) n and use it to evaluate (a) ln x, m + n m + (b) (ln x), (ii) Evaluate x ln x by another method. x m (ln x) n, m. (c) x 3 (ln x).. For an integer n > derive the formula tan n x = n tann x tan n x Using this, find tan 5 x by doing just one explicit integration. 3. Recall that the hyperbolic tangent is the analog of the trigonometric tangent: where tan θ = sin θ cos θ, sinh x tanh x = cosh x, sinh x = ex e x, cosh x = ex + e x. Verify that tanh n x = n tanhn x + tanh n x. 4. Evaluate arcsec x. Hint: Try u = arcsec x, v = x. 5. Evaluate x arcsec x. 6 Miscellaneous Problems 6. Show that the Weierstrass substitution leads to the formulas u = tan θ cos θ = u u du, sin θ =, dθ = + u + u + u. dθ Then evaluate sec θ dθ = using partial fractions. cos θ 4

15 Answer: See [TF] page Compute x by two different methods. (Perhaps partial fractions and trigonometric substitution.) 8. Let a be a positive constant and (a) Find F (x) if a. F (x) = x sin(aθ) cos(θ) dθ. (b) Find F (x) if a =. (Don t divide by zero.) 9. Find. Find Evaluate: 4 a x(x )(x )(x 3) and x sin x and a x cos x. (x 3 + ) x(x )(x )(x 3) x /3. 3 x.. x. /4 x x x. 4. x x + x + 7. x 4 x (x 36) / x 36. x 4 x x (36 x ) 3/ (x + ) (x + 3) 9. x 4 x.. x 4 x... e x (x + cos(x)) (x 3) / 3. (e x + ln(x)). 4. (x + 5) x + 5x. 3x + x x 4 5. x x 4 6. x (x ) 3 (x ) 3 (x + ) x x x + x sec x 3. sec 3 x 33. π/4 sin(3x) cos(x) π/3 sin 3 (x) cos (x).

16 π/ e e sin (x) cos (x). 36. x ln x. 38. x(ln x) π/3 e 3 e ln sin(x) cos (x). x ln x. (cosh x) 3. 6

17 7 Improper Integrals If f(x) is continuous or positive for x a then the limit a f(x) = lim b b a f(x) either exists and is finite or else it is infinite; the quantity b f(x) can t a bounce back and forth between two numbers as b. When the limit is finite we say that the improper integral f(x) converges, otherwise we say it a diverges.. Assume that f(x) and g(x) are continuous functions and that f(x) g(x) for x a. True or False? (a) If (b) If (c) If (d) If a a a a g(x) <, then f(x) <, then g(x) =, then f(x) =, then a a a a f(x) <. g(x) <. f(x) =. g(x) =.. Let f(x) = x x(x. Show that the integral ) f(x) diverges. Answer: For x we have f(x) /x. In fact, for x we have 3 < x so so Since x x = x x x = x > 3, 3x < x Z x = ln x + C, we have Z = 3 Z x x x = so = f(x). Z x f(x). 3. Let f(x) = x + x (x. Show that the integral + ) true. f(x) converges. In the context of a true-false question, true means always true and false means not always 7

18 Answer: For x we have f(x) /x. In fact, for x we have x + so Since Z x x + x + = x + x + + x + = + x + 3, = x + C, we have Z f(x) = x x + x + Z 3 x. x = x = so f(x) 3 Z x = 3. For each of the following improper integrals, say whether it converges or diverges. (Some of these integrals can be evaluated explicitly; for others you will need to prove inequalities and compare the given integral with another integral.) x 3 5. x 3 8. x 4/3. x 3 + sin x x /3 + sin x x + sin x e x 3. x / 6. x x x. x 3 sin x x /3 sin x x sin x 8 Sequences and Series e x 4. x e x 7. x /3 x /3 x 3/4 x 3 + sin x x /3 + sin x x + sin x e x x e x. 8. Evaluate 5 k=3 k, 5 n=3 n, 5 5 k, and k. k= k=3 8

19 An infinite series k= a k is said to converge iff the limit a k = lim k= n a k n k= exists and is finite; the series is said to diverge iff either the limit does not exist or is finite. Warning: Be sure to distinguish between the convergence of a sequence a k and the convergence of the corresponding series k= a k. The limit lim k a k is not the same as the limit k= a k. Most people say that the sequence a k converges when the limit lim k a k exists, but I will try to use the word converge only with series and not with sequences. 9. Compute the limit of each of the following sequences or show that the limit does not exist. (a) lim k k. (b) lim k k. (c) lim k k. 3. Which of the following infinite series converge? (a) k. (b) k. (c) k. (d) k= k= k= k= 3. True or False? (a) If the series a k converges, then lim a k =. k (b) If lim k a k =, then the series (c) If lim k a k =, then the series (d) If the series k= a k converges. k= a k diverges. k= a k diverges, then lim k a k =. (e) If lim k a k does not exist, then the series (f) If lim k a k = 3, then the series (g) If the series exist. true. n= a k diverges. k= a k diverges. k= (d) lim k ( )k. ( ) k. k= a n diverges, then lim n a n is either nonzero or else doesn t In the context of a true-false question, true means always true and false means not always 9

20 3. What is absolute convergence and what does it have to do with convergence? P P Answer: A series n= a n is said to converge absolutely if the series n= a n converges. P It is a theorem that a series which converges P absolutely converges, i.e. if n= a n < then there is a (finite) number L with n= a n = L. Moreover X X a n a n, n= n= i.e. X n= a n X n= a n X n= a n. 33. Does k= cos(k) k converge? Answer: The inequality X k= cos(k) k X k= k = shows that the series converges absolutely. Hance it converges. 34. True or false? (a) The series k= ( )n x n is absolutely convergent for < x <. (b) The series k= xk converges for x. (c) The series k= xk converges for x =. (d) The series k= xk converges for x =. 35. Give an example of a series which converges but does not converge absolutely. Answer: Let a n = ( ) n+ /n. Then by the integral test but X n= a n = X k= X n= (a k + a k ) = converges (again by the integral test). a n = X k= X n= n = k = k X k= (k )(k)

21 9 Convergence Tests 36. State the Comparison test for sums and use it to prove that the series 7 + sin(k) 5 k converges. k=3 Answer: If a k b k for k n then X k=n a k Hence the P P P series a k converges if the series b k does and the series b k diverges if the series a k does. For the series in the question we have X X X k=3 7 + sin(k) 5 k k=3 8 5 k = X k=n n= b k. 5 = 8 n The terms a k = 7 + sin(k) 5 k are positive so the series converges. <. 37. State the Integral Test and use it to prove that the series converges. (k + ) 4 k=8 Answer: If f (x) and f(x) for x N and a k = f(k) then so For example a k+ = f(k + ) f(x) f(k) = a k for k x k + X k=n X k=8 a k+ Z N (k + ) 4 f(x) Z so the series converges as the terms are positive. 8 X k=n a k. x 4 = State the ratio test. Answer: If the limit a k+ lim = ρ k a k P exists, then the series n= converges absolutely if ρ < and diverges if ρ >.

22 39. For what values of x does the series converges? f(x) = k= (x 5) k (k + )3 k Answer: The limit of ratios of successive terms is x 5 k+ /((k + )3 k+ ) (k + ) x 5 x 5 lim = lim = k x 5 k /((k + )3 k ) k (k + )3 3 so the series converges when x 5 /3 <, i.e. when < x < (a) Does the integral (b) Does the series n= x(x + )(x + ) converge or diverge? n(n + )(n + ) converge or diverge? For each of the following say whether the given infinite series converges or diverges. Justify your answer. If the series is a convergent geometric series, evaluate it. If you use the one or more of convergence tests (comparison test, nth term test ratio test, root test, integral test), say which one(s) and prove the inequalities which justify their application. 4. n= n. Answer: Since n < x for n x n we have X n= n < X n= Z n n x = Z x = x = < so the series converges by the integral test. 4. n= ( ) n n n +. Answer: For large n the nth term is roughly ( ) n. The nth term test says that a series converges only if the limit of the nth term is zero. Hence the series does not converge. 43. n= cos n n n.

23 Answer: Since cos n we have X n= X cos n n n n 3/. n= The series on the right converges by the integral test as above. Hence the given series converges absolutely n= k= k= k= k= k= n. 45. k k + 3. k. k= 48. k= /k k= ( ) k k + k (ln k) 3. k ln k. k= 57. k= 6. n= ln k k. 46. k= 49. n= k!. 5. k= k sin k + k (ln 3) k. ( ) n ln n ln n. 6. k= 58. k= n= k 3 k. n n. k + k 5. k k k. (ln ) k. cos n n n. 6. A repeating decimal is a finite decimal plus a geometric series with r a power of. For example, = k= Find integers p and q such that = p/q. 78 k. 63. A rational number is a number which is a ratio of two integers. Show that a number is rational if and only if it is a repeating decimal. 64. Find a number larger than proof of the Integral test. n=5 n 3 and justify your answer. Hint: Use the 3

24 65. The infinite series k converges; its sum is approximately the same as k= the finite sum k, i.e. and the error k= k= k = k + k= k= k= k k is small. Show that this error is less than., i.e. that k= k <. k= 66. The infinite series k sin k is absolutely convergent; its sum is approximately the same as the finite sum k sin k = k= k sin k, i.e. k sin k + k= k= k= k sin k and the error k sin k is small. Show that this error is less than, i.e. that k= k sin k <. k= ( ) k 67. Show that the series k k= converges. Hint: /k /(k + ) =? 68. Prove that k= cos n (n + 8) 6 < Taylor Series 69. Find a second order polynomial (i.e. a quadratic function) Q(x) such that Q(7) = 43, Q (7) = 9, Q (7) =. 4

25 Answer: Q(x) = Ax + Bx + C, Q (x) = Ax + B, Q (x) = A, so A = /, etc. but it is easy to check that Q(x) = (x 7) + (x 7) satisfies Q(7) = 43, Q (7) = 9, Q (7) =. 7. (a) Compute P (), P (), P (), P () if P (x) = 3 + 5(x ) + 7 (x ) + 6 (x )3. (b) Compute P (a), P (a), P (a), P (a) if P (x) = c + c (x a) + c (x a) + c 3 6 (x a)3. (c) Compute P (k) (a), the kth derivative of P (x) evaluated at x = a, if P (x) = 3 k= c k k! (x a)k. 7. (a) Find a polynomial P (x) of degree three such that P () = 5, P () = 3, P () = 5, P () =. (b) Find a polynomial P (x) of degree three such that P (a) = c, P (a) = c, P (a) = c, P (a) = c 3. (c) Suppose that f is three times differentiable. Find a polynomial P (x) of degree three such that P (a) = f(a), P (a) = f (a), P (a) = f (a), P (a) = f (a). 7. Find numbers c, c,..., c 7 such that x 7 = 7 k= c k(x ) k. 73. Let f(x) = x. Find the polynomial P (x) of degree three such that P (k) (5) = f (k) (5) for k =,,, Complete the following paragraph by writing in the correct formulas. The Taylor series of a function f(x) at the point a is the infinite series k= The Maclaurin series is the special case a =. The Taylor series at a may be used to compute f(x) when x is sufficiently close to a. The nth degree 5

26 Taylor polynomial of f(x) about the point a is the polynomial P n (x) of degree n which best approximates f(x) for x near a. It is P n (x) = n k=. The nth Taylor remainder (error) is defined by If f(x) is analytic, then f(x) = P n (x) + R n (x). R n (x) = k=. If f(x) is (n + ) times differentiable, then R n (x) = for some between. The last is called Lagrange s formula for the remainder or the Extended Mean Value Theorem. 75. Write Taylor s formula and the Lagrange formula for the remainder. 76. The degree n Taylor polynomial of f(x) at a is the polynomial of degree n which best approximates f(x) near x = a. Explain. 77. True or false? The Taylor series for a polynomial is finite. 78. Write the Maclaurin series for e t. 79. Write the Maclaurin series for e at. 8. Write the Maclaurin series for sinh t. 8. Write the Maclaurin series for sin t. 8. Write the Maclaurin series for cosh t. 83. Write the Maclaurin series for cos t. 6

27 84. The binomial theorem gives the coefficients in the Maclaurin series ( + x) p = c k x k for the function f(x) = ( + x) p. Give the general formula for c k and find c, c, c, c 3, c 4 for p = 3 and p = / Find the coefficient of x 8 in the Binomial expansion (Maclaurin series) of ( + x) π. Do not do the arithmetic. What is the coefficient of x 8 in ( + x) 6? Power Series k= 86. Use the geometric series and the formula dt = ln( + t) + C + t to find the Maclaurin series for ln( + x). Answer: Substitute t for x in the geometric series; the result is + t = X k= Then integrate from t = to t = x; the result is Z x But R ( + t) dt = ln( + t) + C so ( ) k t k ( < t < ). dt X Z x + t = ( ) k t k dt = k= ln( + x) = Z x dt X + t = k= X k= ( ) k x k+. k + ( ) k x k+. k + After the change j = k + of the dummy variable this becomes the log series ln( + x) = X j= ( ) j x j j ( < x < ). 87. Use the geometric series and the formula dt + t = arctan(t) + C to find the Maclaurin series for arctan(x). 7

28 88. Use the exponential series and the formulas sinh x = ex e x, cosh x = ex + e x to find the Maclaurin series for sinh x and cosh x. 89. Prove that the derivative of the hyperbolic sine is the hyperbolic cosine by differentiating the Maclaurin series of the hyperbolic sine. 9. Prove that cosh t sinh t =. 9. In trigonometry you (should have) learned the formula e iθ = cos θ + i sin θ where i =. (Some trigonometry books write cis θ instead of e iθ.) Prove the trigonometric addition formulas for sine and cosine by equating real and imaginary parts in the identity e i(α+β) = e iα e iβ. 9. Use the exponential series and the facts that cos θ is the real part of e iθ and that sin θ is the imaginary part of e iθ to find the Maclaurin series for sin x and for cos x. 93. Prove that the derivative of the sine is the cosine by differentiating the Maclaurin series of the sine. 94. Show that ( x) 4 = n= (n + )(n + )(n + 3) x n Find the Maclaurin series for sinh x = x dt + t. 96. Find the Maclaurin series for x sin(t ) dt. Answer: See [TF] page 8, but note that [TF] treats the indefinite integral, not the definite integral. 8

29 97. Show that n= x n x n = ln( t) dt. t by differentiating the Maclaurin series termwise. 98. Show that the function y = e x satisfies the differential equation dy = y by differentiating the Maclaurin series termwise. Answer: The series is y = X n= dy X = nx n = n! n= x n. Termwise differentiation gives n! X n= x n X (n )! = x k k! = y. k= 99. Show that the functions y = cos x and y = sin x satisfy the differential equation d y + y = by differentiating the Maclaurin series termwise.. Show that the functions y = cosh x and y = sinh x satisfy the differential equation d y y = by differentiating the Maclaurin series termwise.. Show that the Bessel function J (x) = satisfies the differential equation ( ) n x n n= n n! x d y + dy + xy =.. Find the radius of convergence for the series k= xk. Justify your answer. 9

30 Answer: The radius of convergence is r =. Any of the four following paragraphs constitutes justification. i) The formula x = n X k= x n + xn+ x holds by elementary algebra. Hence the series converges if lim n x n+ = and this is true for x <. If x the limit lim n x n is nonzero (or does not exist) so the series diverges by the nth term test. ii) The limit of ratios of successive terms is x k+ lim = x k x k so the series converges for x < and diverges for x > by the ratio test. iii) The limit of kth root of the kth term is lim k x k /k = x so the series converges for x < and diverges for x > by the root test. P iv) For x > the series has form k= f(k) where f(t) = x t = e t ln x The integral R b f(t) dt is Z b f(t) dt = so the improper integral is given by et ln x ln x t=b = xb t= ln x Z (ln x) if < x < f(t) dt = if x >. The series converges for < x < and diverges for < x by the integral test. 3. Find the radius of convergence r for the series n= nx n n. Answer: The limit of the ratio of absolute values of successive terms is (n + )x n+ / n+ x (n + ) lim = lim = x n nx n / n n n. By the ratio test the series converges if this limit is less than one (i.e. if x < ) and diverges if this limit is greater than one (i.e. if x > ). Hence r =. 4. Find the radius of convergence r for the series 3 n= ( ) n x n+. (n + )!

31 Answer: The limit of the ratio of absolute values of successive terms is x n+3 /(n + 3)! lim n x n+ /(n + )! = lim x n (n + 3)(n + ) =. Hence by the ratio test the series converges for all x, i.e. r =. For each of the following power series specify the largest interval on which converges, i.e. the interval of convergence. You need not concern yourself with convergence at the endpoints of the interval. JUSTIFY YOUR ANSWER. 5. x n. 6. nx n x n. 7. n n= (x 5) n n. 9. n= n= x n n 3. n= n=. (x + 3) n. 5. n= 7. Let f(x) = n= n= n n (x )n.. n= n= x n e n n +. x n n!. 3. x n e n (n )!. n!x n. 6. n= n= n x n. n= x n What is the radius of convergence? What is the (n!) power series of the function f () (x)? What is the power series (up to a constant) of f(x)? What is f (48) ()? Taylor s Formula - Estimating the Error 8. Find the fourth degree Taylor polynomial P 4 (x) about the point for the function f(x) = cos x and estimate the error cos x P 4 (x) for x <. Answer: f(x) = f (4) (x) = cos x, f (x) = f (5) (x) = sin x, f (x) = cos x, f (3) (x) = sin x, so f() = f (4) () =, f () = f (3) () =, f () =. and hence the fourth degree Taylor polynomial is The error is P 4 (x) = 4X k= f (k) ()x k k! R 4 (x) = f (5) (z 5 )x 5 5! 3 = x! + x4 4!. = (cos z 5)x 5 5!

32 for some unknown z 5 between and x. As cos z 5 we have cos x x! + x4 4! = R4(x) x5 < 5! 5! for x <. 9. Find the eighth degree Taylor polynomial P 8 (x) about the point for the function f(x) = cos x and estimate the error cos x P 8 (x) for x <.. Use Taylor s formula with f(x) = x, n =, a = 9 to calculate approximately. Show that the error is less than /6. Answer: The value of a is chosen so that the derivatives f (k) (a) can be calculated exactly: since f(x) = x /, f (x) = x / we have f(9) = 3, f (9) = /6. Thus for x near 9 with an error, f (x) = x 3/, 4 x P(x) = f(9) + f (9)(x 9) = 3 + R (x) = f (z )(x 9) In particular, = 3 + ( 9)/6 + R () where R () = f (z )( 9) (x 9) 6 = z 3/ 4 (x 9). = z 3/ ( 9) 4 for some unknown z with 9 < z <. Now P () = 3 + = and since 6 9 < z, R () = z 3/ 4 ( 9) < 9 3/ 4 = 6 = Since R () < we have P () R () < < P (), so 3.6 < < Note that we don t try to find z, but we used the inequality 9 < z < to prove that R () < /6, i.e. that = P () up to two decimal places. Compare the value f() = given by your calculator with the approximate value P () = Repeat problem using n =. Show that the error is less than.3. 3

33 Answer: The third derivative is f (x) = 3x 5/ 8 and f (9) = /8. Thus x P(x) = f(9) + f (9)(x 9) + f (9)(x 9) = 3 +! with an error R (x) = f (z 3)(x 9) 3 3! = z 5/ (x 9)3. (x 9) 6 (x 9) 6 Since R () > we have P () < < P () + R (). Now P () = = and since z > 9 we have 6 < R () < 9 5/ ( 9) 3 48 = 8 48 <.3 so 3.6 < < Find the 4th degree Taylor polynomial P 4 (x) about the point for the function f(x) = sin x. Estimate the error sin x P 4 (x) for x <. 3. Let f(x) = x /3. Find the polynomial P (x) of degree two which best approximates f(x) near x = 8. Evaluate P () and use the Extended Mean Value Theorem to prove that /3 P () Hint: The function g(x) = x 8/3 is decreasing so g() < g(8). 4. Find a polynomial P (x) of degree three such that sin(x) P (x) lim x x 3 =. Use the Extended Mean Value Theorem to show that sin(x) P (x) x Use Taylor s formula of degree (with an appropriate expansion point) to approximate the cube root of to within an error of 9. Justify your answer. 33

34 3 Differential Equations Classify each of the following as homogeneous linear, inhomogeneous linear, or nonlinear and specify the order. For each linear equation say whether or not the coefficients are constant.. y + y =.. xy y =. 3. xy y = x. 4. xy + yy =. 5. xy + yy = x. 6. y + y = xe x. 7. (a) Show that y = x + 5 is a solution of xy y =. (b) Show that y = C x + C is a solution of xy y =. 8. (a) Show that y = (tan(c x+c ))/c is a solution of yy = (y ) y. Show that y = tan(x) and y = are solutions of this equation, but that y + y is not. Is the equation linear homogeneous? 9. (a) Show that y = 4e x + 7e x is a solution of y 3y + y =. (b) Show that y = C e x + C e x is a solution of y 3y + y =. (c) Find a solution of y 3y + y = such that y() = 7 and y () = 9.. The differential equation is called the Logistic Equation. dy dt = 4 y 4 (a) Find the solution y which satisfies y() =. (b) Find lim t y. (c) Draw the phase line to check your answer. (The phase line is a diagram indicating where dy/dt = and showing the direction the solution is moving between the zeros.). Find the function y of x which satisfies the conditions dy + x y =, y() = 5.. Consider the differential equation dy = y. (a) Find the solution y which satisfies y () = 5. (b) Find the solution y which satisfies y () =. (c) Find lim x y. (d) Find lim x y. (e) Find lim x y. 34

35 3. Find the function y of x which satisfies the conditions dy (cos x)y = esin x, y() = Find a nonconstant solution of y dy + x3 =. Answer: We rewrite in the form Integrate y dy + x 3 =. y x4 4 = c. Various values of c give different solutions. We take c = 7 and solve for y: 3x 3 /3 y = Find all solutions of dy + y =. Answer: By separation of variables dy y where y = e C. = so ln y = x + C so y = ye x 6. Find all solutions of dy + y = e x. Answer: Use the integrating factor ρ = e x so d (ex y) = e x dy + y = e x e x = e x. Integrate to get e x y = e x + C or y = e x + Ce x. 7. Find y if dy + y = e x and y = 7 when x =. Answer: y = e x + Ce x so 7 = + C so C = 6. Hence y = e x + 6e x. 35

36 4 Linear Homogeneous 8. Find all real solutions of d y dt 6dy + y =. dt Answer: We try y = e rt. This is a solution only if r 6r + = or Thus r = 6 ± 36 4 = 3 ± i. z = e (3+i)t = e 3t (cos t + i sin t), z = e (3 i)t = e 3t (cos t i sin t), are complex solutions so y = z + z = e 3t cos t, y = are real solutions. The most general real solution is z z i = e 3t sin t, where A and A are (real) constants. y = A e 3t cos t + A e 3t sin t 9. Find y if y 6y + y =, and in addition y satisfies the initial conditions y() = 7, and y () =. Answer: The general solution is and its derivative is y(t) = C e 3t cos t + C e 3t sin t y = dy dt = C(3e3t cos t e 3t sin t) + C (3e 3t sin t + e 3t cos t). Evaluate y and y at t = : Hence C = so the answer is 7 = C, = 3C + C. y = 7e 3t cos t e 3t sin t.. (a) Find the general solution y = y(x) of the differential equation (HINT: r 4 = (r )(r + ).) d 4 y 4 = y. 36

37 Answer: The roots of the equation r 4 = are r = i, r = i, r 3 =, r 4 =, so the general complex solution is y = c e ix + c e ix + c 3 e x + c 4 e x where c, c, c 3, c 4 are complex. Using the relations e ix = cos x+i sin x, e ix = cos x+i sin x, cos x = eix + e ix, sin x = eix e ix, i we can write the most general real solution in the form y = a cos x + a sin x + c 3e x + c 4e x where a = c + c and a = (c c )i are real. Using the relations e x = cosh x+sinh x, e x = cosh x+sinh x, cosh x = ex + e x, sinh x = ex e x, we can also write the most general real solution in the form where a 3 = c 3 + c 4 and a = c 3 c 4. y = a cos x + a sin x + a 3 cosh x + a 4 sinh x (b) Find the solution which satisfies the initial conditions y() =, y () =, y () =, y () = 8. Answer: y = a cos x + a sin x + a 3 cosh x + a 4 sinh x y = a sin x + a cos x + a 3 sinh x + a 4 cosh x y = a cos x a sin x + a 3 cosh x + a 4 sinh x y = a sin x a cos x + a 3 sinh x + a 4 cosh x so at x = we have = a + a 3 = a + a 4 = a + a 3 8 = a + a 4 From the third equation we get a = a 3 so a = a 3 = by the first equation. From the second equation we get a = a 4 so a 4 = 4, a = 4 by the last equation. The answer is y = cos x 4 sin x + cosh x + 4 sinh x. Find the general solution y = y(x) of the differential equation d 4 y 4 = d y.. Find a linear homogeneous differential equation with constant coefficients which has xe 3x cos(x) as a solution. 37

38 Answer: The function e 3x cos(x) is the real part of the function e (3+i)x which is a solution of the equation (D (3 + i))(d (3 i))y = (D 6D + 3)y = where D = d/. Hence xe 3x cos(x) solves (D 6D + 3) y = (D 4 D 3 + 6D 56D + 69)y =. In other words d 4 y d3 y d y dy y =. Solve each of the following initial value problems. Your final answer should not use complex numbers, but you may use complex numbers to find it. 3. y + 9y =, y() =, y () = y + 9y =, y() = 3, y () =. 5. y 5y + 6y =, y() =, y () =. 6. y 5y + 6y =, y() =, y () =. 7. y + 5y + 6y =, y() =, y () =. 8. y + 5y + 6y =, y() =, y () =. 9. y 6y + 5y =, y() =, y () =. 3. y 6y + 5y =, y() =, y () =. 3. y + 6y + 5y =, y() =, y () =. 3. y + 6y + 5y =, y() =, y () =. 33. y 4y + 5y =, y() =, y () =. 34. y 4y + 5y =, y() =, y () =. 35. y + 4y + 5y =, y() =, y () =. 36. y + 4y + 5y =, y() =, y () =. 37. Suppose that (b/a) (c/a) = ω and k = b/a so that the solutions of ar + br + c = are r = k ± ωi. 38

39 Show that the general solution of is where and c and c are constants. a d y dt + bdy dt + cy = y = e kt( c cos(ωt) + c sin(ωt) ) 5 Linear Inhomogeneous 38. (a) Find a solution of y 3y + y = e 3x. Answer: We use the method of undetermined coefficients. written in the form (D 3D + )y = e 3x. Factoring gives Apply (D 3) to get Therefore, every solution must have form (D )(D )y = e 3x. (D 3)(D )(D )y =. y = c e x + c e x + c 3 e 3x. The equation can be Subtracting c e x + c e x from a solution gives another solution (by the principle of superposition) so we know that there is some solution of form y = c 3 e 3x. We plug in and solve for c 3 : (D 3D + )c 3 e 3x = c 3 e 3x = e 3x = c 3 = /. A particular solution of () is y = e 3x /. (b) Find a particular solution of y 3y + y = e x. Answer: The equation can be written in the form Factoring gives Apply (D ) to get Therefore, every solution must have form (D 3D + )y = e x. (D )(D )y = e x. (D )(D ) y =. y = c e x + c e x + c 3xe x. Subtracting c e x + c e x from a solution gives another solution (by the principle of superposition) so we know that there is some solution of form y = c 3xe x. We plug in and solve for c 3 : (D 3D + )c 3xe x = c 3e x = e x = c 3 =. A particular solution of (b) is y = xe x. 39

40 (c) Find the general solution of y 3y + y = 4e 3x + 5e x. Answer: By the principle of superposition and parts (a) and (b) (D 3D + )(4y + 5y ) = 4e 3x + 5e x. The general solution of the inhomogeneous equation is a particular solution of the inhomogeneous equation plus the general solution of the homogeneous equation. The general solution of the inhomogeneous equation is therefore y = c e x + c e x + 4y + 5y = c e x + c e x + e 3x 5xe x. 39. Find all solutions of the nonhomogeneous equation d y y =. Answer: Both the method of undetermined coefficients and the method of variation of parameters work for this problem. To use the method of undetermined coefficients we should try to find a particular solution of (N) among the solutions of Dy = (since y = satisfies Dy = ). The solutions of Dy = are the constants. The constant function y = is a solution of the nonhomogeneous equation (N) and the general solution of the homogeneous equation d u u = is u = C e x + C e x so the general solution of the nonhomogeneous equation (N) is y = + u = + C e x + C e x. To use the method of variation of constants we note that the general solution of the homogeneous equation (H) is u = C u + C u = C e x + C e x so we look for a solution of the inhomogeneous equation (N) in the form y = v u + v u = v e x + v e x where v u + v u =. Then = y y = v u + v u so (N) (H) so so so so as before. v u + v u = v e x + v e x = v u + v u = v e x v e x = e x v =, e x v =, v = e x, v = e x, v = e x + C, v = e x + C, y = ( e x + C )e x + ( e x + C )e x = + C e x + C e x 4

41 4. Find the general solution y(t) of the differential equation d y + 9y = cos 3t. dt Answer: We use the method of undetermined coefficients. Let D = d/dt represent the operation of differentiation with respect to the independent variable t. We are trying to find all solutions of (D + 9)y = cos 3t. ( ) Since (D + 9) cos 3t =, every solution of ( ) is a solution of (D + 9) y =, (#) but not conversely. The most general complex solution of (#) is y = c e 3it + c e 3it + c 3 te 3it + c 4 te 3it where c, c, c 3, c 4 are complex constants; the most general real solution is y = C cos 3t + C sin 3t + C 3t cos t + C 4t sin 3t ( ) where C, C, C 3, C 4 are real constants. If y is given by ( ), then (D + 9)y = 6C 3 sin 3t + 6C 4 cos 3t. To find a particular solution of (D +9)y = cos 3t we take 6C 3 = and 6C 4 =. The general solution is this particular solution plus the general solution of the homogeneous equation (D + 9)y =, i.e. y = t sin 3t 6 + C cos 3t + C sin 3t 4. Find the general solution y(t) of the differential equation d y + 9y = cos t. dt 4. Find the general solution y(t) of the differential equation d y + y = cos t. dt 43. Find the general solution y(t) of the differential equation d y + y = cos 3t. dt 44. Find y if 4

42 (a) d y + dy + y = and y() =, y () = 3. (b) d y + dy + y = e x and y() =, y () =. (c) d y + dy + y = xe x and y() =, y () =. (d) d y + dy + y = e x + xe x and y() =, y () = 3. Hint: Use the principle of superposition to save work. 45. Suppose that (b/a) (c/a) = ω and k = b/a, i.e. that r = k ± ωi are the solutions of ar + br + c =. (a) Show that the general solution of a d y dt + bdy dt + cy = t is y = t c b c + e kt (c cos(ωt) + c sin(ωt)) where and c and c are constants. (b) Find the general solution of 6 Applications a d y dt + bdy + cy =. dt 46. A population of bacteria grows at a rate proportional to its size. Write and solve a differential equation which expresses this. If there are bacteria after one hour and bacteria after two hours, how many bacteria are there after three hours? 47. Rabbits in Madison have a birth rate of 5% per year and a death rate (from old age) of % per year. Each year rabbits get run over and 7 rabbits move in from Sun Prairie. (a) Write a differential equation which describes Madison s rabbit population at time t. Answer: Let P = P (t) denote the rabbit population at time t. In a tiny time interval of duration dt,.5p dt rabbits are born,.p dt rabbits die from old age, dt rabbits are run over, and 7 dt rabbits move in from Sun Prairie for a net change of dp =.5P dt.p dt dt + 7 dt = (.3P 3) dt. The differential equation is dp dt =.3P 3. 4

43 (b) If there were, rabbits in Madison in 99, how many are there in 994? Answer: To solve the differential equation we separate variables: integrate: dp =.3 dt P, ln(p, ) =.3t + C and choose the constant C of integration so that that the equation is true when t = (which corresponds to 99). ln(p, ) =.3t + ln(,, ). The answer P (3) (which corresponds to 994) satisfies ln(p (3), ) =.9 + ln(, ). Exponentiate and add, to both sides to get P (3) =, +, e According to Newton s law of cooling the rate dt/dt at which an object cools is proportional to the difference T A between its temperature T and the ambient temperature A. The differential equation which expresses this is dt dt where k < and A are constants. = k(t A) (a) Solve this equation and show that every solution satisfies lim T = A. t (b) A cup of coffee at a temperature of 8 o F sits in a room whose temperature is 75 F. In five minutes its temperature has dropped to 5 o F. When will its temperature be 9 o F? What is the limit of the temperature as t? 49. Retaw is a mysterious living liquid; it grows at a rate of 5% of its volume per hour. A scientist has a tank initially holding y gallons of retaw and removes retaw from the tank continuously at the rate of 3 gallons per hour. (a) Find a differential equation for the number y(t) of gallons of retaw in the tank at time t. (b) Solve this equation for y as a function of t. appear in your answer.) (c) What is lim t y(t) if y =? (The initial volume y will (d) What should the value of y be so that y(t) remains constant? 43

44 5. A gallon vat is full of 5% solution of acid. Starting at time t = a 4% solution of acid is pumped into the vat at gallons per minute. The solution is kept well mixed and drawn off at gallons per minute so as to maintain the total value of gallons. Derive an expression for the acid concentration at times t >. As t what percentage solution is approached? 5. The volume of a lake is V = 9 cubic feet. Pollution P runs into the lake at 3 cubic feet per minute, and clean water runs in at cubic feet per minute. The lake drains at a rate of 4 cubic feet per minute so its volume is constant. Let C be the concentration of pollution in the lake; i.e. C = P/V. (a) Give a differential equation for C. (b) Solve the differential equation. Use the initial condition C = C when t = to evaluate the constant of integration. (c) There is a critical value C with the property that for any solution C = C(t) we have lim t C = C. Find C. If C = C, what is C(t)? 5. A philanthropist endows a chair. This means that she donates an amount of money B to the university. The university invests the money (it earns interest) and pays the salary of a professor. Denote the interest rate on the investment by r (e.g. if r =.6, then the investment earns interest at a rate of 6% per year) the salary of the professor by a (e.g. a = $5, per year), and the balance in the investment account at time t by B. (a) Give a differential equation for B. (b) Solve the differential equation. Use the initial condition B = B when t = to evaluate the constant of integration. (c) There is a critical value B with the property that () if B < B, then there is a t > with B(t) = (i.e. the account runs out of money) while () if B > B, then lim t B =. Find B. (d) This problem is like the pollution problem except for the signs of r and a. Explain. 53. A 3 gallon tank is full of milk containing % butterfat. Milk containing % butterfat is pumped in a gallons per minute starting at : AM and the well mixed milk is drained off at 5 gallons per minute. What is the percent butterfat in the milk in the tank 5 minutes later at :5 AM? Hint: How much milk is in the tank at time t? How much butterfat is in the milk at time t =? 54. A sixteen pound weight is suspended from the lower end of a spring whose upper end is attached to a rigid support. The weight extends the spring by half a foot. It is struck by a sharp blow which gives it an initial downward velocity of eight feet per second. Find its position as a function of time. 44

45 Answer: To do this problem we need some elementary physics. There are two forces on the weight: gravity and the tension in the spring. Let F be the sum of these two forces, m be the mass of the weight, and a be the acceleration of the weight. Then Newton s law is F = ma. The force of gravity is mg where g = 3ft/sec. We say that the weight is at the equilibrium position when it is suspended motionless so that the force of gravity and the tension balance. We measure the height y so that y = when the spring is at equilibrium and positive values of y correspond to positions below equilibrium. We assume Hooke s law which says that the tension in the spring is a linear function k(s + y) where k is a constant (called the Hooke s constant) and s is the amount by which the weight stretches the spring in the equilibrium position; in this problem s = /. When y = these forces balance so mg = ks. The fact that the weight is sixteen pounds means that mg = 6. Since the force of gravity is down and the tension pulls the weight up, the net force is F = ky for any displacement y. In this problem mg = 6 and s = / so k = 3. Since g = 3 we have m = / and so the equation F = ma of motion is (/)y = 3y. The general solution is y = C sin 8t + C cos 8t. Since y() = we get C = and since y () = 8 we get C =. Hence y = sin t. See [TF] page 878 or [VPR] page A sixteen pound weight is suspended from the lower end of a spring whose upper end is attached to a rigid support. The weight extends the spring by half a foot. The weight is pulled down one feet and released. Find its position as a function of time. 56. The equation for the displacement y(t) from equilibrium of a spring subject to a forced vibration of frequency ω is d y + 4y = sin(ωt). (#) dt (a) Find the solution y = y(ω, t) of (#) for ω if y() = and y () =. Answer: We use the method of undetermined coefficients. If we apply D + to both sides, we get zero on the right so the solution must be a solution of (D +)(D +4)y = and hence must have the form Applying d /dt + 4 gives y = c cos(t) + c sin(t) + c 3 cos(ωt) + c 4 sin(ωt) c 3( ω + 4) cos(ωt) + c 4( ω + 4) sin(ωt) = sin(ωt) so c 3 = and c 4 = /(4 ω ). Hence the solution we seek has the form y = c cos(t) + c sin(t) + sin(ωt) 4 ω. Now y() = c =, y () = c + ω 4 ω =, so the solution we seek is y(ω, t) = ω sin(t) (4 ω ) + sin(ωt) 4 ω. 45

46 (b) What is lim ω y(ω, t)? Answer: By simple algebra y(ω, t) = Use l Hôpital (with ω as the variable) to get sin(ωt) ω sin(t) (4 ω ) t cos(ωt) sin(t) lim y(ω, t) = lim = ω ω 4ω (c) Find the solution y(t) of t cos(t) sin(t). 8 d y + 4y = sin(t) ( ) dt if y() = and y () =. (Hint: Compare with (#).) Answer: We could use the the method of undetermined coefficients as before; we know that the solution must have form y = c cos(t) + c sin(t) + c 3 t cos(t) + c 4 t sin(t) and we proceed as usual. However, taking ω = in (#) gives ( ) which suggests that the answer is the answer to (b): Sure enough, dy dt so d y dt so y () = and y () = and d y dt y = y(, t) = t cos(t) sin(t). 8 = 4t sin(t) + cos(t) cos(t) 8 t cos(t) + sin(t)) = = t sin(t) = t cos(t) + sin(t) sin(t) t cos(t) sin(t) + 4y = t cos(t) = sin(t) Suppose that an undamped spring is subjected to an external periodic force so that its position y at time t satisfies the differential equation d y dt + ω y = c sin(ωt). (a) Show that the general solution is c y = C cos ω t + C sin ω t + ω ω sin ωt. when ω ω. (b) Solve the equation when ω = ω. (c) Show that in part (a) the solution remains bounded as t but in part (b) this is not so. (This phenomenon is called resonance and accounts for many engineering disasters.) 46

47 7 Conic Sections. Find an equation for the ellipse whose foci are at F (, ) and F (, 3) an whose semi-major axis is 3. Do not simplify this equation. Answer: p (x ) + (y ) + p (x ) + (y 3) = 6.. Find an equation for the ellipse whose foci are at F (, ) and F (, ) and whose semi-major axis is 4. Do high school algebra to transform it into a polynomial equation, i.e. one without square root signs. Answer: An equation for the ellipse (with square root signs) is p (x ) + y + p (x + ) + y = 8 The following algebra transforms this to an equation without square root signs. p (x ) + y = 8 p (x + ) + y (x ) + y = (8 p (x + ) + y ) (x ) + y = 64 6 p (x + ) + y ) + (x + ) + y 8x 64 = 6 p (x + ) + y ) x + 8 = p (x + ) + y x + 6x + 64 = 4((x + ) + y ) 48 = 3x + y. 3. Find an equation for the ellipse whose foci are at F (c, ) and F ( c, ) and whose semi-major axis is a. Do high school algebra to transform it into a polynomial equation, i.e. one without square root signs. Answer: An equation for the ellipse (with square root signs) is p (x c) + y + p (x + c) + y = a The following algebra transforms this to an equation without square root signs. p (x c) + y = a p (x + c) + y (x c) + y = (a p (x + c) + y ) (x c) + y = 4a 4a p (x + c) + y ) + (x + c) + y 4cx 4a = 4a p (x + c) + y ) 6c x + 3ca x + 6a 4 = 6a ((x + c) + y ) 6a (a c ) = 6(a c )x + 6a y. The equation can be further simplified by defining b = a c and dividing by 6a b to get = x a + y b. 47